Lecture Notes - Chemistry 1110Dr. Luther Giddings

Last Updated: August 21, 2017

Table of Contents

Chapter 1: Matter and Measurements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1What is chemistry? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1Why is chemistry relevant? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2What is matter? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3How is matter classified? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4What kinds of properties does matter have? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6An introduction to the elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7The natural states of the elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9The Periodic Table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11Chemistry is an empirical science . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13Scientific notation and powers of 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13Measurements and significant figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15SI units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18Measurements and measured quantities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19Derived quantities-volume . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20Derived quantities-density . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20Energy and heat . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21Dimensional analysis and problem solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22Organic or inorganic - a postscript . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

Chapter 2: Atoms and the Periodic Table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28The structure of the atom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28The Nucleus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28Atoms and protons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29Atoms, neutrons, and isotopes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30Atoms and electrons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34The octet rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35Fundamentals of electrons in atoms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37Ground states and excited states . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42Electron configurations: full configurations and the periodic table . . . . . . . . . . . . . . . . 42Electron configurations: Noble Gas configurations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52Electron configurations: electron spin diagrams (arrow diagrams) . . . . . . . . . . . . . . . . . 53Inner shell, outer shell, and valence electrons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58The electron configurations of ions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64Atomic properties and periodic trends . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

Chapter 3: Ionic Compounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69Chemical bonds and chemical compounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69Lewis electron dot structures (Lewis structures) for atoms . . . . . . . . . . . . . . . . . . . . . . . 72Ionic compound nomenclature: naming ionic compounds . . . . . . . . . . . . . . . . . . . . . . . 73Ionic compounds: cation names . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74Ionic compounds: monatomic and polyatomic anion names . . . . . . . . . . . . . . . . . . . . . . 75A digression: old ionic compound nomenclature systems . . . . . . . . . . . . . . . . . . . . . . . . 78

The names and molecular formulas of ionic compounds . . . . . . . . . . . . . . . . . . . . . . . . 80Ionic compounds: going from names to molecular formulas . . . . . . . . . . . . . . . . . . . . . 83Ionic compounds: going from molecular formulas to names . . . . . . . . . . . . . . . . . . . . . 86The nomenclature of binary molecular compounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92The nomenclature of acids and bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94Common nomenclature problems - some troubleshooting help . . . . . . . . . . . . . . . . . . . 96

Chapter 4: Molecular Compounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97Covalent bonds and molecular compounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97Covalent bonds and the Periodic Table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98Multiple covalent bonds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98Coordinate covalent bonds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99Molecular formulas and structural formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100The Lewis structures of ionic compounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103The Lewis structures of covalent compounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103Exceptions to the octet rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113The shapes of molecules and VSEPR - Valence Shell Electron-Pair Repulsion . . . . . . 116Two Sets of Electrons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119Three Sets of Electrons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119Four Sets of Electrons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121Five Sets of Electrons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122Six Sets of Electrons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125Polar covalent bonds and polar molecules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126Bond polarity and molecular polarity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128Summary of determining molecular polarity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128

Chapter 5: Classification and Balancing of Chemical Reactions . . . . . . . . . . . . . . . . . . . . . . . 129Chemical equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129Balancing chemical equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132Strong electrolytes, weak electrolytes, and non-electrolytes . . . . . . . . . . . . . . . . . . . . . 138Molecular, ionic, and net ionic equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140Double displacement reactions: precipitation formation and solubility rules . . . . . . . . 143Double displacement reactions and water formation . . . . . . . . . . . . . . . . . . . . . . . . . . . 145Double displacement reactions and gas formation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148Electron transfer (redox) reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150Post-script: responses to student questions about Chapter 5 and 6 concepts . . . . . . . . 156

Chapter 6: Chemical Reactions - Mole and Mass Relationships . . . . . . . . . . . . . . . . . . . . . . . . 160Molecular formulas and empirical formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160Molecular formulas, molecular weights, and formula weights . . . . . . . . . . . . . . . . . . . 161The mole and Avogadro's number . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162Molecular weight and molar mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164Stoichiometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166Theoretical yield and percent yield . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169

Chapter 7: Chemical Reactions - Energy, Rates, and Equilibrium . . . . . . . . . . . . . . . . . . . . . . 171An introduction to thermodynamics and kinetics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171Energy, work, and other thermodynamic definitions . . . . . . . . . . . . . . . . . . . . . . . . . . 171Internal energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174Enthalpy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175Entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178Gibbs free energy and the spontaneity of reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180The rates of chemical reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183Reaction rates and molecular collisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184Rate Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186Activation energy and activated complexes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186The reversibility of chemical reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189Catalysts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190Chemical equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192The equilibrium constant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193Writing equilibrium constant expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197Equilibrium constants and Gibbs free energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199LeChatelier's principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200

Chapter 8: Gases, Liquids, and Solids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204General gas properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204Ideal gases and the Ideal Gas Law (Universal Gas Law) . . . . . . . . . . . . . . . . . . . . . . . 207The ideal gas law and molar mass calculations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209The ideal gas law and stoichiometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210The combined gas law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211Dalton's Law of Partial Pressures and mole fractions . . . . . . . . . . . . . . . . . . . . . . . . . . 216States of matter, phase transitions, and enthalpies of phase transitions . . . . . . . . . . . . 218Intermolecular forces: attractive interactions between molecules . . . . . . . . . . . . . . . . . 222London (dispersion) forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223The magnitude of London forces increases with molecular size and surface area . . . . 225Dipole-dipole interactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226Hydrogen bonds - a special case of dipole-dipole interactions . . . . . . . . . . . . . . . . . . . 227Ion-dipole interactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 228Predicting the intermolecular forces in compounds . . . . . . . . . . . . . . . . . . . . . . . . . . . 229Intermolecular forces and solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233Intermolecular forces and physical properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234

Chapter 9: Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 238General solution properties and definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 238Calculating concentration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 240Dilutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243Some properties of liquids and solutions: surface tension and capillary action . . . . . . 246Some properties of liquids and solutions: vaporization and vapor pressure . . . . . . . . . 247Some properties of liquids and solutions: diffusion . . . . . . . . . . . . . . . . . . . . . . . . . . . 249Some properties of liquids and solutions: osmosis . . . . . . . . . . . . . . . . . . . . . . . . . . . . 250

Chapter 10: Acids and Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252General acid-base information: a review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252The Arrhenius acid-base theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252The Brønsted-Lowry acid-base theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252The Lewis acid-base theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254Strong and weak acids and bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 257The behavior of acids and conjugate bases in water . . . . . . . . . . . . . . . . . . . . . . . . . . . 258The behavior of bases and conjugate acids in water . . . . . . . . . . . . . . . . . . . . . . . . . . . 262The acid-base behavior of water . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264The pH scale . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 265Calculating [H3O

+], [OH-], and pH . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 266Buffer solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 270

IUPAC 2015

Name Symbol     At No     Atomic Wt Name Symbol     At No     Atomic Wt

Actinium Ac 89 [227] Mendelevium Md 101 [258]

Aluminium Al 13 26.981 5385(7) Mercury Hg 80 200.592(3)

Americium Am 95 [243] Molybdenum Mo 42 95.95(1)

Antimony Sb 51 121.760(1) Moscovium Mc 115 [289]

Argon Ar 18 39.948(1) Neodymium Nd 60 144.242(3)

Arsenic As 33 74.921 595(6) Neon Ne 10 20.1797(6)

Astatine At 85 [210] Neptunium Np 93 [237]

Barium Ba 56 137.327(7) Nickel Ni 28 58.6934(4)

Berkelium Bk 97 [247] Nihonium Nh 113 [286]

Beryllium Be 4 9.012 1831(5) Niobium Nb 41 92.906 37(2)

Bismuth Bi 83 208.980 40(1) Nitrogen N 7 14.007

Bohrium Bh 107 [270] Nobelium No 102 [259]

Boron B 5 10.81 Oganesson Og 118 [294]

Bromine Br 35 79.904 Osmium Os 76 190.23(3)

Cadmium Cd 48 112.414(4) Oxygen O 8 15.999

Caesium Cs 55 132.905 451 96(6) Palladium Pd 46 106.42(1)

Calcium Ca 20 40.078(4) Phosphorus P 15 30.973 761 998(5)

Californium Cf 98 [251] Platinum Pt 78 195.084(9)

Carbon C 6 12.011 Plutonium Pu 94 [244]

Cerium Ce 58 140.116(1) Polonium Po 84 [209]

Chlorine Cl 17 35.45 Potassium K 19 39.0983(1)

Chromium Cr 24 51.9961(6) Praseodymium     Pr 59 140.907 66(2)

Cobalt Co 27 58.933 194(4) Promethium Pm 61 [145]

Copernicium Cn 112 [285] Protactinium Pa 91 231.035 88(2)

Copper Cu 29 63.546(3) Radium Ra 88 [226]

Curium Cm 96 [247] Radon Rn 86 [222]

Darmstadtium Ds 110 [281] Rhenium Re 75 186.207(1)

Dubnium Db 105 [270] Rhodium Rh 45 102.905 50(2)

Dysprosium Dy 66 162.500(1) Roentgenium Rg 111 [281]

Einsteinium Es 99 [252] Rubidium Rb 37 85.4678(3)

Erbium Er 68 167.259(3) Ruthenium Ru 44 101.07(2)

Europium Eu 63 151.964(1) Rutherfordium Rf 104 [267]

Fermium Fm 100 [257] Samarium Sm 62 150.36(2)

Flerovium Fl 114 [289] Scandium Sc 21 44.955 908(5)

Fluorine F 9 18.998 403 163(6) Seaborgium Sg 106 [269]

Francium Fr 87 [223] Selenium Se 34 78.971(8)

Gadolinium Gd 64 157.25(3) Silicon Si 14 28.085

Gallium Ga 31 69.723(1) Silver Ag 47 107.8682(2)

Germanium Ge 32 72.630(8) Sodium Na 11 22.989 769 28(2)    

Gold Au 79 196.966 569(5) Strontium Sr 38 87.62(1)

Hafnium Hf 72 178.49(2) Sulfur S 16 32.06

Hassium Hs 108 [270] Tantalum Ta 73 180.947 88(2)

Helium He 2 4.002 602(2) Technetium Tc 43 [97]

Holmium Ho 67 164.930 33(2) Tellurium Te 52 127.60(3)

Hydrogen H 1 1.008 Tennessine Ts 117 [293]

Indium In 49 114.818(1) Terbium Tb 65 158.925 35(2)

Iodine I 53 126.904 47(3) Thallium Tl 81 204.38

Iridium Ir 77 192.217(3) Thorium Th 90 232.0377(4)

Iron Fe 26 55.845(2) Thulium Tm 69 168.934 22(2)

Krypton Kr 36 83.798(2) Tin Sn 50 118.710(7)

Lanthanum La 57 138.905 47(7) Titanium Ti 22 47.867(1)

Lawrencium Lr 103 [262] Tungsten W 74 183.84(1)

Lead Pb 82 207.2(1) Uranium U 92 238.028 91(3)

Lithium Li 3 6.94 Vanadium V 23 50.9415(1)

Livermorium Lv 116 [293] Xenon Xe 54 131.293(6)

Lutetium Lu 71 174.9668(1) Ytterbium Yb 70 173.045(10)

Magnesium Mg 12 24.305 Yttrium Y 39 88.905 84(2)

Manganese Mn 25 54.938 044(3) Zinc Zn 30 65.38(2)

Meitnerium Mt 109 [278] Zirconium Zr 40 91.224(2)

Chapter 1: Matter and Measurements

What is chemistry?

If you refer to various chemistry text books you will find a variety of definitions ofchemistry. I prefer to define chemistry as the study of things made up of atoms and molecules. Itis the study of why things made up of atoms and molecules behave the way they do. It is also thestudy of how to make them behave more usefully. As everything in the world around us iscomposed from atoms, by gaining a fundamental understanding of chemistry we can begin tounderstand many important and interesting phenomena that affect us on a daily basis.

While one may read of a variety of different types of chemistry such as forensicchemistry, geochemistry, food chemistry, and etc., there are really just five fundamentalbranches of chemistry:

Analytical chemistry is the study of what is in a substance, and how much of a particularthing is in a substance. Most of the things in the world around us are not chemically pure butconsist of dozens or even hundreds or thousands of different chemically distinct components.Gasoline looks like it is simply a liquid, but in fact it is composed of several hundred differentchemicals. The smell of an orange is due to the presence of nearly two hundred differentcompounds. And so on. Analytical chemistry is used all around us, every day, and in a variety ofways. The study of pollutants in the environment, the analysis of the composition of anexpensive perfume, the testing of a patient's blood in a hospital or of an athlete's urine during acompetition, and a check of the composition of the various liquids used during the manufactureof gasoline, are a few examples of how analytical chemistry is used.

Inorganic chemistry is the study of all of the elements and their compounds exceptcarbon and its compounds.

Organic chemistry is the study of carbon and its compounds. Since there are 118 knownelements, it often seems odd that an entire branch of chemistry is devoted to a single element andits compounds while the other 117 elements and their compounds are all lumped together in aseparate discipline, but there is a very good reason for this. There are about 1.5 million knowninorganic compounds. This is a lot of compounds, and you will not be required to know them all forthis class, although it may seem like it before you're finished. The number of known organiccompounds varies, depending on the reference. These sources state that there are from 16 to 40million or more known organic compounds. Carbon is the basic element of life, and livingcreatures have developed an astonishing array of different organic compounds. A traditionaldefinition of organic compounds are those that contain carbon, i.e., compounds in which carbonis found in the molecular formula. But there are a small number of carbon-containing compoundsthat are classified as inorganic, such as the cyanides, carbonates, and bicarbonates (we’ll learnmore about these in Chapter 4). So a better working definition of organic compound is acompound that contains both carbon and hydrogen in its molecular formula, with the exception

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of the bicarbonates (hydrogen carbonates). Inorganic compounds are compounds that do notcontain both carbon and hydrogen. This is a general way of classifying organic compounds and thereare some notable exceptions in the “real world,” but it will be the way by which we make thedistinction between organic and inorganic compounds in this class. I’ll expand on this thought at the endof this chapter. It’s a little thing, but something I expect you to learn.

Biochemistry is the study of the chemistry of life and living things. This is a field of greatimportance to those who practice in the health sciences. The ways the atoms and moleculesbehave in living systems is predicated on the rules of organic chemistry.

Physical chemistry is the study of how the laws of physics affect things as small as atomsand molecules. During the early decades of the 20th Century researchers were astonished todiscover that very small particles, like atoms and molecules, do not obey the laws of physicsexpounded by Sir Isaac Newton. This spurred the development of the field known as quantummechanics, which explains the behavior of things that are very small.

Why is chemistry relevant?

We live in a world that grows more and more technical by the day. Even if you're notinterested in medicine or how the body works, how do you know if the claims made by someoneselling a particular health care product or beauty product are legitimate or bogus? Are youwilling to trust an industry that releases substances into the environment when it says they aresafe, or will you believe those who might be willing to distort or misrepresent the truth in orderto give weight to their argument that the chemicals released are dangerous? Do you trust peopleor government agencies that assure you that exposure to a particular chemical is not dangerous,particularly when their arguments may be based on economic or political considerations and noton pure science? When medicine is prescribed for a sick family member or friend, are you surethey will receive the appropriate medication and not something that may aggravate their illnessor even kill them?

Whether you're interested in science or not, whether you're interested in chemistry or not,you need to know some of the fundamentals of the science in order to be a good consumer, agood citizen, and a good family member and friend. You need to know a little bit aboutchemistry in order to protect yourself. You need to know a little bit about chemistry in order tobe able to think clearly and correctly about many important everyday issues.

To many people chemistry is a dirty word, but that attitude is terribly naive anduninformed. Everything around us is chemicals, from the air we breath, the water we drink, andthe food we eat, to our very bodies. All living things are made up entirely of chemicals. All deadthings, and all things that never have or never will live (such as rocks) are made up of chemicals.The earth on which we live is nothing more, at its most fundamental, than a great mass ofchemicals. The sun on which we rely for light and heat and life itself is a giant globe of reacting

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chemicals. And so on. It is true, the incorrect use of chemistry is responsible for many importanthealth and environmental issues. But chemistry also makes the world a better place.

Compare your life with that of someone living a year ago, ten years ago, fifty years ago,or a hundred or more years ago. What things do you have that they did not? How are medicineand health care different for you? personal hygiene? transportation? communication? the wayyou spend your leisure time? food and nutrition? the clothes you wear? Almost inevitably,anything you have that someone in the past did not have is attributable, either directly orindirectly, to advances made in chemistry.

What do chemists do to make the world a better place? In order to address this questionlet me tell you a little bit about the work of some of my friends who hold masters and doctoraldegrees in chemistry. They are involved in the study of:

• The analysis and identification of DNA and large biological molecules.• The chemistry of blood clotting mechanisms.• The chemistry of cardiovascular disease.• The chemical mechanisms of infectious diseases.• New chemical methods for breast cancer detection.• The development of new and improved antibiotics and other medicinal compounds.• The development of new and improved fertilizers and pesticides.• Food chemistry.• Personal hygiene products & cosmetics.• The chemistry of surfaces, which sounds terribly arcane until one realizes that many

important reactions will only occur on certain types of surfaces.• Composite materials, such as graphite fiber products.• The production of gasoline and other petroleum-based products.• Lubricants and fuel additives for automobiles.• The detection of drugs and poisons.This is just a small sample of the sorts of things that chemists do to make our world a betterplace.

What is matter?

As is the case with chemistry, there are a number of different definitions of matter. It iscommon to define matter as anything that has both mass and volume. I prefer to define matter asanything made up of atoms, which are the basic building blocks of the world around us. It is truethat there are types of matter smaller than atoms, such as protons, neutrons, electrons, andquarks, but we leave the study of particles smaller than atoms to courses in physics.

The study of matter is the study of chemistry. This means that the study of matter is thestudy of things made up of atoms and molecules. We will learn more about this in Chapter 2.

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How is matter classified?

Matter can be classified either by its state or by its composition.

There are three common states of matter, solids, liquids, and gases. We do not concernourselves with more exotic states of matter such as plasmas or Bose-Einstein condensates whenwe study elementary chemistry.

We will use the word "particle" as we discuss the three common states of matter, sincesubstances can be made up either of atoms, molecules, or ions. Ions are atoms or molecules with anelectric charge. We’ll discuss ions in the very near future. Particle is used to generically represent thevarious small units of matter we may find in a substance.

There are various ways to make distinctions between the three common states of matter.For our purposes, the most useful comparison at the particle level is as follows:

In the solid state the particles are relatively very close together, usually well-ordered(tidily arranged), and are held together by relatively strong attractive interactions between thesolid particles.

In the liquid state the particles are not quite as close together, not quite as well-ordered,and the attractive interactions are not quite as strong.

In the gaseous state the particles are relatively far apart, chaotically arranged (i.e., notordered), and attractive interactions between the particles are either negligible or nonexistent.

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There are specific names for the transitions from one state of matter to another. Thesetransitions are also called phase changes, or changes of state. These refer to state changes in anymaterial, and not just in water. The names of these changes of state are listed in the graphic at thetop of page 4. You do need to know all six of them, melting, vaporization, sublimation, condensation,freezing, and deposition.

Matter can also be classified based on its composition.

An element is a substance in which all of the atoms are the same. In other words, anelement is atomically pure. Elements may exist as single atoms or in molecular (i.e. more thansingle atom) form, such as Fe, O2, P4 and S8. Notice that the molecular formulas of elementsreflect their composition from one and only one type of atom. A molecular formula is ashorthand way of representing the types of atoms - and how many of each - are found in asubstance. The molecular formula “Fe” tells us our substance is made up exclusively of ironatoms. When we discuss O2, we are talking about two oxygen atoms that have chemicallycombined (i.e, they two atoms have chemically bonded to each other) to form a molecule. Eventhough O2 molecules contain two atoms, this is still representative of an element since both ofthe atoms are of the same element. This is also true of phosphorus, P4, and sulfur, S8.

Compounds are substances made from the atoms of two or more elements that are heldtogether by chemical bonds. There are millions of compounds in the everyday world around us,including water (H2O), carbon dioxide (CO2), and glucose (C6H12O6). Each compound has its ownmolecular formula. What does the molecular formula of water tell you about it’s composition? Howabout he chemical compositions of carbon dioxide and glucose? Notice that the molecular formulasof compounds reflect that they are made up of two or more different types of atoms. Asmentioned, it is essential to understand that the different types of atoms in compounds are heldtogether by chemical bonds, a topic we will examine in Chapter 3. One important thing toremember about compounds is that the atoms in a compound can only be separated from eachother by chemical reactions.

A mixture is a substance composed of two or more components which can be separatedby physical means, i.e., based on the different physical properties of the components. Thosecomponents may be elements, compounds, or other mixtures. As an example, table salt (NaCl,sodium chloride) dissolved in water can be recovered by evaporating the water. The ability toseparate the components of a mixture by taking advantage of differences in physical propertiesmeans that chemical bonds do not exist between the components. We will refine this concept morecorrectly in Chapter 8 but for the time being, this will suffice for our purposes.

There are two types of mixtures. They are classified by how evenly dispersed theparticles are at the molecular level.

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In homogeneous mixtures there is uniformity in composition, properties, and appearancethroughout the mixture down to the particle level. Examples of homogeneous mixtures includetable salt dissolved in water, red wine, gasoline, and alloys such as 18 karat gold.

Heterogeneous mixtures are characterized by discontinuities in composition, properties,and appearance at levels well before the molecular level. In other words, heterogeneous mixturesdo not have uniform composition, properties, and appearance throughout the mixture. Examplesof heterogeneous mixtures include pizza, granite, a nice fizzy soda in a glass filled with ice, anda peanut butter and jelly sandwich. What do you see when you look at a PBJ sandwich? Does it lookthe same throughout, or do you see layers? What does the presence of layers mean about itscomposition?The separation of the components of a mixture can be accomplished using physical changes, i.e.,changes based on the physical properties of the substances. The separation of the components ofa compound can only be achieved by chemical reactions. Elements are pure and cannot beseparated into atoms of any type other those atoms that are unique to the element.

What kinds of properties does matter have?

Matter has physical properties, which are properties that can be observed and measuredwithout changing the composition of the material. The physical properties of a substance includeits color, odor, state of matter, melting point, boiling point, heats of vaporization and fusion,density, solubility, metallic character, electrical and thermal conductivity, magnetic properties,crystal shape, malleability, ductility, and viscosity. Most common physical changes eitherinvolve changes of state (phase transitions) or are the consequence of mechanical processing(e.g. grinding, crushing, slicing, pulverizing, gluing pieces together, etc.)

The chemical properties of a material are the chemical reactivity of the substance, that is,how a substance changes its composition when it chemically reacts with other substances.

There is a difference between changes of state (physical changes) and chemical reactions.Physical changes are those involving physical properties, or, changes in which there is no changeto the chemical composition of the material(s) under observation. When chemical changes (or,chemical reactions) occur, the starting materials are consumed and new materials are formed dueto the breaking and making of chemical bonds. Chemical reactions may be indicated by colorchanges, the absorbing or release of energy (heat, light, sound, electric), or by the formation ofnew materials such as gases, pure liquids, or solids (precipitates). These are general indicationsthat a chemical reaction has occurred and they are not absolutely correct in all instances. Use them tohelp you identify chemical reactions in the world around you, but use them with care!

As an example, The molecular formula of water is H2O. This tells us that each watermolecule consists of two hydrogen atoms and one oxygen atom. If we take water out of ourkitchen faucet, it is in the liquid state. If we take liquid water and cool it by placing it in thefreezer it will eventually change from the liquid to the solid state. But while its state has

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changed, the water remains H2O, even though we call solid water “ice.” If we take solid water(ice) and put it in a pan on the stove, as it warms it will first change from the solid to the liquidstate and then from the liquid to the gaseous state. However, as these changes of state occur, thewater remains chemically unchanged. It remains H2O.

There is another way to think of the distinction between physical and chemical changes.If a substance maintains its same molecular formula throughout a process, it is a physical change.On the other hand, if a process results in changes to the molecular formulas of participatingsubstances, a chemical reaction has taken place. When water freezes, the process is H2O(l) =>H2O(s). The (l) indicates that our substance is in the liquid state, while (s) indicates our substance isnow in the solid state. Even though a change of state has occurred, the molecular formula of waterremains unchanged, indicating this is a physical process.

On the other hand, if we take a glass of liquid water and place a nine volt battery in it,something interesting occurs. This is an experiment you can safely do yourself. Should you elect totry it, I’d suggest you use a clear drinking glass or some other clear glass container. It will notharm the glass. Add about one teaspoon of table salt to one to two pints of tap water, and placethe battery in the water. Streams of bubbles begin to emanate from the battery terminals almostimmediately. As you observe the formation of bubbles, you can see that a far greater volume ofbubbles is being produced at one terminal than at the other. This is because the electrical energyof the battery is being used to break the chemical bonds that hold the hydrogen atoms andoxygen atoms in water molecules together. Hydrogen gas is being produced at the terminalproducing the greater volume of gas, while oxygen gas is being produced at the other terminal.Why would you expect to see more hydrogen gas produced than oxygen gas? Hint: think about themolecular formula of water.

2 H2O(l) => 2 H2 (g) + O2 (g)

This chemical equation described the process. It indicates that while we start with water, thebubbles that are formed consist of hydrogen gas and oxygen gas. Some water has been consumedduring this process and converted into different substances. This is typical of chemical reactions.One or more substances are consumed as one or more new substances are produced.

An introduction to the elements

As we stated earlier, elements are substances in which all of the atoms are the same.There are 118 known elements, of which about 90 are naturally occurring. This is the first ofseveral controversial statements you’re going to encounter in this class. You should know there is somedisagreement from one source to the next as to the number of naturally-occurring versus syntheticelements. Don’t worry about it. This is an early introduction to a reality of chemistry: there are manyareas in which chemists disagree with one another. There are sound reasons for these disagreements. I

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expect you to know what I tell you in these lecture notes and to understand that as you look beyondthem, you will at times find slightly contradictory information, in the text and elsewhere. It is not thatthese other sources are right and I am wrong, or visa versa. It is that we respectfully disagree with oneanother. While these disagreements are common, they are usually also minor. We as chemists aretypically united in the larger, more important points that we teach our students in class.

The twenty most abundant elements in the earth's crust by mass are: Note: I do not expect you tomemorize this table.

element percent composition

oxygen 45.5

silicon 27.2

aluminum 8.3

iron 6.2

calcium 4.7

magnesium 2.8

sodium 2.3

potassium 1.8

titanium 0.6

hydrogen 0.2

phosphorus 0.1

manganese 0.1

fluorine 0.05

barium 0.04

strontium 0.04

sulfur 0.03

carbon 0.02

zirconium 0.02

vanadium 0.01

Chlorine 0.01

all others 0.05

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A table of the fifteen most abundant elements in the human body is found below. It isinteresting to note the difference in the composition of the earth’s crust and that of the humanbody. Again, I do not expect you to memorize this information. It comes from a more detailed tablefound at "Elemental Composition of the Human Body”http://web2.iadfw.net/uthman/elements_of_body.html should you find yourself interested.

element mass of element ina

70 kg person (kg)

percentcomposition

(mass percent)

oxygen 43 61.4

carbon 16 22.9

hydrogen 7.0 10.0

nitrogen 1.8 2.6

calcium 1.0 1.4

phosphorus 0.8 1.1

potassium 0.1 0.2

sulfur 0.1 0.2

sodium 0.1 0.1

chlorine 0.09 0.1

magnesium 0.02 0.02

iron 0.004 0.006

fluorine 0.003 0.004

zinc 0.002 0.003

silicon 0.001 0.001

An important source of information about the elements is found in the periodic table. Wewill we will discuss it in a moment.

The natural states of the elements

How do the elements occur naturally? In other words, when we venture out into the realworld, in what state or condition might we find a sample of a particular element? Knowing thenatural states of the elements is often very important in understanding chemical reactions. You

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should know the following by heart, and soon. Questions regarding this information frequently find theirway into quizzes and exams.

In their elemental state, most elements occur as large numbers of individual atoms of thesubstance, chemically bonded one to another. An element is represented using its chemicalsymbol, which we can find in the periodic table. So, as an example, if we talk about copper wemay represent it as “Cu,” which is its atomic symbol. If we work with potassium in a chemicalreaction we’ll represent it with it’s symbol “K.” And so on. We’re going to talk a great dealabout the periodic table in this class. It is an absolutely essential tool to anyone studyingchemistry. I do not expect you to memorize it, and will always provide you with a copy onquizzes and exams. You can find a copy of the periodic table I will give you on exams at the followingaddress: http://www.slcc-science.org/chem/giddings/chem1110int/info/periodic-table-2016-revised.pdf. Ivery strongly suggest you print a copy of it now and keep it in front of you at all times as you work yourway through this course. Bear in mind that while I do not expect you to memorize the table, you arerequired to know what the table means and represents, how to use it, and to understand a number ofthings about it that we will discuss as we move through class. As an important word of warning: youshould never attempt any quiz, Mastering Chemistry exercise, or exam without a copy of the periodictable immediately in front of you. To do so is an exercise in self-abuse, as well an invitation to trouble.However, you’re the one paying to take this course. You can heed my advise, or you can learn the hardway, through your own mistakes, as you prefer.

Seven elements occur naturally in diatomic (i.e., molecules made up of two atoms) form.These include hydrogen, nitrogen, and oxygen (H2, N2, and O2 respectively) and the halogensfluorine, chlorine, bromine, and iodine (F2, Cl2, Br2, and I2 respectively). You must remember thatthese substances occur as diatomic molecules or you will come to grief in this class repeatedly! Amnemonic device shared with me by a student was: “Have No Fear Of Ice Cold Beer.” In this phrase,the first letter of each word corresponds to an element that exists as diatomic molecules, (H)ave =hydrogen, (N)o = nitrogen, (F)ear = fluorine, (O)f = oxygen, (I)ce = iodine, (C)old = chlorine, and (B)eer= bromine. There are several others. Learn one and remember it.

Phosphorus and sulfur occur naturally as P4 and S8 molecules respectively. However, wewon’t worry much, if at all, about this in this class. In other words, if we deal with elementalphosphorus or sulfur in this class, we’ll simply represent them as P or S respectively unlesswe’re instructed otherwise.

In the real world it is generally more common to find elements chemically combined withatoms of other elements, rather than in their pure form. In other words, in the real world most ofthe elements occur as compounds with other elements. Finding hydrogen and oxygen atomsbonded together in water (H2O), sodium and chlorine bonded together in sodium chloride(NaCl), and aluminum and oxygen bonded together in an aluminum ore called bauxite (Al2O3)

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are three common examples. Note that compounds are represented using molecular formulas,short-hand devices used to indicate the types of atoms - and how many of each - are found in asingle molecule of the compound. H2O, NaCl, and Al2O3 are examples of molecular formulas.H2O tells us that a single water molecule consists of two hydrogen atoms and one oxygen atom.NaCl informs us the there is one sodium atom and one chlorine atom found in a single moleculeof sodium chloride. In a single molecule of bauxite, also known as aluminum oxide, there aretwo aluminum atoms and three oxygen atoms chemically bonded to each other. We’ll discusscompounds at greater length in the near future.

The Periodic Table

The Periodic Table is a table of the elements arranged in rows and columns in sequenceof increasing atomic number (i.e., the number of protons in the nucleus). The rows are calledperiods and the columns are called groups. Periods and groups are vocabulary words you needto know.

The elements are represented by symbols in the periodic table.

In most cases the relationship between the name and the symbol is obvious. For example,O-oxygen, H-hydrogen, N-nitrogen. For a few elements the symbol is derived from anon-English language like Latin or German, e.g., K-potassium (kalium), Fe-iron (ferrum),Au-gold (aurum), W-tungsten (wolfram, Swedish). In these cases there is not an obviouscorrelation between the element's name and its symbol.

I will always give you a periodic table on exams and quizzes. A copy of what I will provide on theexams is at link mentioned previously. If you haven’t already printed yourself a copy, now is the time todo so. As you look at this information, remembering this is what I will give you on exams, it should beapparent that I will not require you to memorize element names and symbols. However, I very strongly

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suggest that you know the names and symbols of the first twenty elements, along with the names andsymbols of all other elements commonly used in this class. It will help you immensely and make yourwork go more quickly if you have committed these names and symbols to memory. You do not need tomemorize other information such as atomic numbers and atomic weights. If pertinent they will beprovided.

The top number in the periodic table entry for each element is the atomic number, thenumber of protons an atom has in its nucleus. The bottom number in the periodic table for eachelement is the atomic weight, the weighted average of the atomic masses of all of the isotopes ofa particular element. The atomic mass is the sum of the protons and neutrons in an isotope or aparticular atom. The masses of individual atoms is expressed in amu, or atomic mass units and isequal to 1/12 the mass of a single 12C isotope (or, 1.661 x 10-27 kg; don’t memorize this number. Ifyou need to use it, I will give it to you). We will discuss atomic number, atomic mass, atomicweight, and isotopes in more detail in Chapter 2 (will you be able to wait, or will the suspense justkill you?).

In the periodic table, groups are "chemical families," because they exhibit similarchemical behavior. We’ll explain why this is so in chapter 2. Some of the groups have specialnames, which you must know:

Group 1A - alkali metals.Group 2A - alkali earth metals.Group 6A - chalcogens.Group 7A - halogens.Group 8A - Noble Gases.Groups 1B-8B: transition metals. This includes the ten series of groups thecontains the elements from scandium to zinc (4th period), yttrium to cadmium (5th

period), lanthanum to mercury (6th period), and actinium to copernicum (element#112, 7th period).Groups 1A, 2A, 3A-8A: main block elements.

The "island" by itself beneath the main body of the periodic table consists of two periods,the lanthanides and actinides. Collectively these two periods are known as rare earth metals.Interestingly there are a number of rare earth metals far more common than some of the transitionmetals (the 1B-8B elements in the periodic table) such as gold, silver, platinum, and etc. Normally it isgroups that have special names. These two rows are something of an anomaly in this sense, that theyhave their own special names.

The periodic table can also be divided into three families of elements, metals, nonmetals,and metalloids.

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Metals are elements characterized by the tendency to give up electrons in reactions (i.e.form cations), are good thermal and electrical conductors, and are usually lustrous, malleable,and ductile. Metals are found to the left of the "staircase" line on the periodic table. Lustrousmeans shiny. Malleability is the ability to be pounded into a particular shape. Ductility is the ability tobe drawn into a fine wire. One other important note: it’s true that hydrogen is found to the left of thestaircase line. However, hydrogen is a nonmetal. In this sense, it’s position in the periodic table issomething of an anomaly. There are very good reasons for this anomalous position, but I will notexplain them. Take comfort in knowing the hydrogen is the only exception, i.e., the only nonmetal foundto the left of the staircase line.

Nonmetal elements are diametrically opposite to metallic elements in their primaryfeatures. They are characterized by the tendency to gain electrons in reactions (i.e. form anions)and are poor conductors of heat and electricity, dull, and brittle. Nonmetals are found to the rightof the "staircase" line on the periodic table.

Metalloids (or, semi-metals) are elements characterized by the some of the properties ofboth metals and nonmetals. Metalloids straddle the "staircase" line on the periodic table, i.e, theyare the elements found directly on either side of the staircase line. Having mentioned metalloidshere, you can promptly forget about them as they have no relevance to us as we study chemistryin this class. You will need to be able to classify each and every element as a metal or anonmetal, for reasons that will become apparent when we discuss chemical bonds in Chapter 3.In other words, unless I ask you very specifically about metalloids, for the purposes of this class you canforget they exist. This is probably the first, last, and only time we’ll use the word metalloid in class. Forour purposes in this class, every element is either a metal or a nonmetal. It’s that simple.

Examine the periodic table in the front inside cover of your textbook, or the one found atthe link I mention above that you’ll be using on the exams. Can you find the staircase line? Canyou tell which elements are metals? Which are nonmetals? This is an essential fundamental skillyou must develop.

Chemistry is an empirical science

Chemistry is an empirical science. This means it is based on observations andmeasurements made during experiments. This chapter will address measurements andcalculations in chemistry, although we do not actually make any measurements in this course.

Scientific notation and powers of 10

(Note: excluding the sections on “energy and heat” and “density,” nearly everything from this pointthrough the remainder of the chapter is a review of things you should have learned in math 0900,0950, and 0980/1010, the prerequisite math courses for this class. It is your responsibility to know

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these things. If you do not remember discussing them in math, it is your responsibility to do thenecessary review work to bring yourself up to speed! This includes reviewing information in the course'stext book, material in your old math books and notebooks, and online support material such as may befound at Kahn Academy online (https://www.khanacademy.org/). Math is the single biggest challenge tomost students in this class, often far more so than any chemical concepts we discuss. If your math skillsare weak or you are out of practice, you face a long, uphill, challenging climb in class. I’m not trying toscare you off. I do however want you to be fully aware of what awaits you this semester.

It is common to use either very large or very small numbers in science. The size of mostatoms is in the trillionths of an inch (0.000000000001 inch) range. The atoms in a few ounces ofa substance may number as many as one million billion billion(1,000,000,000,000,000,000,000,000) or more. How can we conveniently express numbers thatare either very large or very small?

Scientific notation is a scientific shorthand that makes the expression of large and smallnumbers easier. It is based on our use of a base ten number system. Each digit in a number is aplaceholder representing some multiple of ten. For example, a number greater than 1, such as4321, can also be written as 4321 = (4 x 1000) + ( 3 x 100) + (2 x 10) + (1 x 1), or, as (4 x 10 x10 x 10) + (3 x 10 x 10) + (2 x 10) + (1 x 1). A number less than 1, such as 0.5678, can also bewritten as 0.5678 = (5 x .1) + (6 x .01) + (7 x .001) + (8 x .0001).

4321 is written as 4.321 x 103 in scientific notation. Numbers written using scientificnotation consist of two pieces, the coefficient and the exponent. The coefficient is the part withthe decimal, in this case, 4.321. The exponent is the power to which 10 is raised, in this case, 3.In scientific notation the coefficient is always obtained by moving the decimal until is just oneplace to the right of the first non-zero digit. The magnitude of the exponent depends on howmany places the decimal is moved. The sign of the exponent depends on which direction thedecimal is moved. If the number is greater than one, the decimal will be moved to the left and theexponent will be a positive number. If the number is less than 1 (note: less than one is a smallnumber but not a negative number. Negative numbers are less than 0), the decimal will be movedto the right and the exponent will be negative. The number 0.5678 would be written as 5.678 x10-1.

A few examples:

54405 = 5.4405 x 104

0.000006036 = 6.036 x 10-6

-201,000 = -2.01 x 105

-0.023473 = -2.3473 x 10-2

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Being able to use scientific notation is an essential fundamental skill required in this class. I will oftenrequire you to report your answers on homework, quizzes, and exams using scientific notation. If youdo not remember how scientific notation works, you need to review and bring yourself up to speed.

Measurements and significant figures

First, some definitions. There are many words used in science that have slightly differentmeanings than in everyday use. To most people the words accuracy and precision aresynonymous. To a scientist they have similar but distinct and different meanings. Accuracy is anindication of how close to a correct value a measurement is. Precision is a function of thecloseness of the results of a series of measurements to each other in value.

To the layman this distinction may seem absurd. It is often wondered how the correctvalue of a measurement cannot be known. But in the real world, the absolute correctness of mostmeasurements is seldom known.

As an example, assume someone takes exactly 1.000 pound of finely powdered lead andscatters it evenly across a 10.00 acre field. Now assume a farmer plows the lead deep into theground. Is there still exactly 1.000 pound of lead, or is it possible that perhaps some of the leadstuck to the blades of the plow and the tires of the tractor? Now, assume that the field is inconstant use over a ten year period. Will the field still contain exactly 1.000 pound of lead? Whatprocesses might be responsible for a loss of lead in the 10 acres? The action of wind and water,the integration of lead into growing plants, and the transport of lead by either vehicles or animalsmoving through the field may all contribute to a loss of lead. Lead may also move into the field,especially if there is a copper smelter nearby, since lead is a common by-product of copper andsilver refining. The finely powdered lead that results from metal refining processes can becarried by the wind and deposited in the field, even if the refinery is many miles distant. Also,lead may have existed in the soil before we added our 1.000 pound. If we are asked how muchlead remains in the 10.00 acres, how accurate will our answer be if we state 1.000 pound?

How can we determine exactly how much lead remains in the field? If we want to beexact, we must remove all of the soil to a depth of several feet or more, over all of the ten acres,and measure the total amount of lead. Obviously, this is not a feasible approach.

Alternatively, we may take a number of soil samples scattered across the ten acres andthen extrapolate our test results as being representative for the entire field. This may result in aloss of accuracy, but it is more realistic in terms of time and money since testing can beexpensive. The results are treated statistically to help us estimate the accuracy of our testing. Intesting the samples we hope that the results of our tests are similar in value from one set ofsamples to the next. If the results are in fact similar, we say that the results are precise. Thegreater the difference between the results of various sets of samples, the more imprecise our dataare.

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One very important aspect of testing involves the measurement of standards. Standardsare samples in which the amount of the substance of interest is known exactly because it is addedunder very carefully controlled conditions in a laboratory. Standards are always analyzed alongwith samples from the real world. The results obtained from the analysis of similar standardsshould be very nearly the same, i.e., they should be precise. The greater the precision in themeasurement of a series of standards, the greater the likelihood of the measurements beingaccurate under normal conditions. In other words, precise measurements are also usuallyaccurate measurements, although this might not be entirely true for a number of reasons we willnot discuss here.

Realistically, can we ever know exactly how much lead remains in the field? No. Butwith careful testing we can provide a highly accurate estimation, one that is very close to theactual, correct vaue.

Whenever measurements are made in the real world, there are limitations in themeasurements. For example, if the pound of lead was initially measured on an electronicscientific balance before it was scattered through the field, it might be confidently said that itweighed exactly 1.000 pound. But what if the lead was initially weighed on a bathroom scale,which only indicates in one pound increments? Or worse, what if the lead was weighed on a feedstore scale which only registers 10 pound increments? How certain might we be that the poundof lead weighed exactly 1.000 pound?

Scientists are required to know the limitations of the equipment they use when makingmeasurements during experiments. These limitations are reflected in the manner in whichscientists report their observations. Whenever a scientist reports a measurement, he writes thenumber to as many places as he is certain are correct plus a last place, in which there isuncertainty due to the limitations of the instrument used to make the measurements.

If we define the distance one mile by measuring it with a yardstick, we might report that1 mile is equal to 5280 feet. However, if we measure that same distance using lasers and GPS(global positioning satellite) technology, we might find that 1 mile is equal to 5280.000 feet,since the distance can be measured more exactly with the more sophisticated equipment.

The number of significant figures in a measurement is equal to all of the places in ameasurement we are certain about plus the first uncertain place. In this class we do not makemeasurements. We must learn to recognize how many significant figures are in the numbers reportedto us. Nonzero digits are always significant. Zeros may or may not be significant, as describedbelow.

There are three types of zeros used in numbers.

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Leading zeros are found before (to the left of) the first non-zero digit in a number arenever significant. As an example, the zeros in 0.053 are leading zeros. This value has twosignificant figures.

Captive zeros are found between two non-zero digits and are always significant. Giventhe number 100.01, we have three captive zeros and five total significant figures.

Trailing zeros are found after (to the right of) the last non-zero digit. They may or maynot be significant, depending on the presence of a explicit decimal in a number. The zeros in1300 are trailing zeros but are not counted as significant figures because a decimal is not shownexplicitly in the value. But given 1300. with it’s explicit decimal shown, the two trailing zerosbecome significant and the overall value has four significant figures.

Here are some examples. How many significant figures are found in each value below?

10101 5 sf significant figures, or, sig figs as many lovingly call them; captivezeros are always significant

73,000 2 sf trailing zeros without a decimal are not significant73,000. 5 sf trailing zeros with a decimal are significant0.0006149 4 sf leading zeros are never significant0.00456000 6 sf leading zeros are never significant and trailing zeros with a

decimal are significant 0.002022300 7 sf leading zeros are never significant, captive zeros are always

significant, and trailing zeros with a decimal are significant4321 = 4 sf.5678 = 4 sf0.5678 = 4 sf why isn’t the zero counted as a sig fig?0.56078 5 sf why is the second zero counted as a sig fig when the first one isn’t?93,567,891 = 8 sf.123456789 = 9 sf-23,456 = 5 sf note that the sign of a number does not affect the way its sig figs arecounted-.456 = 3 sf-0.456 = 3 sf why isn’t the zero counted as a sig fig?

Do not confuse the relevance of placeholder zeros with significant figures. Leading ortrailing zeros are often essential in the reporting of a number, but they may or may not besignificant (i.e., they may or may not be counted as sig figs), depending on the type of zero theyare.

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Numbers may need to be rounded if they are to be reported to the correct number of sigfigs. Never round any numbers until all calculations have been completed. The rules forrounding may be found in your text if you are not familiar with them.

SI units

The International System (or SI system) is the measurement system used in science. It isalso known as the metric system. Its basic units are mass in kilogram (kg), length in meter (m),time in seconds (s), temperature in Kelvins (K; note: not degrees Kelvin), and quantity in moles(mol).

Prefixes are used to indicate how many or what part of a base unit is being described. Theprefixes and their abbreviations are as follows. Note: these abbreviations are case-sensitive.

unit abbreviation powers of ten examples

yotta Y 1024

zetta Z 1021

exa E 1018

peta P 1015

tera T 1012 trillions

Every year vegetation around the world emits about 500teragrams of isoprene, a hydrocarbon that plays an importantrole in atmospheric chemistry.

giga G 109

billionsA 32 gigabyte thumb drive can hold 32 billion bytes ofinformation.

mega M 106

millionsA one megaton nuclear bomb delivers the same amount ofdestructive energy as the detonation as one million tons of TNT.

kilo k 103

thousandsOne kilogram is exactly equal to one thousand grams.

deci d 10-1

tenthIn sound measurements one decibel is equal to one-tenth of abel, a unit of sound pressure.

centi c 10-2

hundredthsA centimeter is exactly one one-hundredth of a meter.

milli m 10-3

thousandthsA milligram is exactly one one-thousandth of a gram.

micro μ, u, or mc 10-6

millionthsFine human hair has a diameter of about 50 micrometers, 50one-millionths of a meter.

nano n 10-9

billionthsA nanosecond is exactly one one-billionth of a second.

pico p 10-12

trillionthsThe diameters of a single copper atom is about 157 picometers,or 157 trillionths of a meter.

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femto f 10-15

thousand-trillionthThere are some very fast chemical reactions that occur on afemtosecond time scale, in thousand-trillionths of seconds.

atto a 10-18

million-trillionthThe concentration of certain chemicals in individuals nerve cellshas been measured in the attomolar range.

zepto z 10-21

yocto y 10-24

Work at femto/atto scale is often at the cutting edge of chemistry today. You must know the baseunits and the prefixes from tera to pico in this class. You do not need to know from yotta to peta, orfrom femto to yocto.

Measurements and measured quantities

In chemistry length is commonly reported in meters (m), centimeters (cm), andmillimeters (mm). The sizes of individual atoms and molecules are reported in the picometer(pm) and nanometer (nm) range. A unit called the Angstrom (Å) is occasionally encountered andis equal to 0.1 nm (1 Å = 1 x 10-10m).

Mass is reported in kilograms, although the masses of individual atoms may beexceedingly small. Note that there is a difference between mass and weight. Mass is the amountof matter present in a substance and does not change with respect to local gravitational fields.Mass is constant. Weight depends on gravity. As an example, if a person weighs 180 pounds onearth and travels to the moon, the mass of the person will remain constant. How much of theperson is there will not change. However, since the gravity of the moon is about one-sixth that ofearth, the person's weight would change from 180 pounds to 30 pounds. This is why a trip to themoon might be appealing to some who perceive themselves as overweight.

Temperature is reported using the Fahrenheit scale in the United States but the centigrade(or Celsius) and Kelvin scales are used in science. The Fahrenheit scale is based on the boilingpoint and the melting point of water at 212°F and 32°F respectively. The centigrade scale is alsobased on the melting and boiling points of water, but these are stipulated as occurring at 100°Cand 0°C respectively. Note that 1°C = 1.8°F, i.e., the Celsius degree is larger by nearly a factorof 2. The Kelvin scale is based on 0 K at absolute zero, the point at which all molecular motionstops. On this scale we find the melting point of water at 273 K and the boiling point of water at373 K. Note that 1K = 1°C, i.e., they are the same size. Also note that we refer to degreesFahrenheit and degrees Celsius but never to degrees Kelvin. The word degree is implied in thename Kelvin. So, we state that the melting point of water occurs at 32 degrees Fahrenheit, 0degrees Celsius, and at 273 Kelvin.

The equations used to convert between the temperature scales is as follows:

F to C: (F - 32)/1.8

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C to F: (C x 1.8) + 32

C to K: C + 273

K to C: K - 273

K to F: convert K to C and then convert to F

F to K: convert F to C and then convert to K

Derived quantities-volume

Two of the most important derived quantities are volume and density. Derived quantitiesare based on the measurement of more than one attribute of a material or object.

Volume is the product of the length, width, and height of an object ( for a cube or box-shaped object, V = l x w x h). Since the standard unit of length is the meter, the standard unit ofvolume could be stated in cubic meters (m3), but this is too large a unit to be practical unless thevolume of large objects, such as houses, mountains, or planets, is being discussed. In chemistryvolume is commonly expressed in units of cubic centimeters (cm3). Since 1 m = 100 cm, then 1m3 = (100 cm)3 = 1 x 106 cm3. Another common unit of volume in chemistry is the liter (L),which is equivalent to one-one thousandth of a cubic meter, or, to one deciliter: 0.001 m3 = 1 dm3

= 1 L. There are one thousand milliliters in one liter (1000 mL = 1 L), and a milliliter and a cubiccentimeter are exactly the same size (1 mL = 1 cm3 ).

Derived quantities-density

The density (D) of any material is the ratio of the mass (m) of the material to the volume(V) occupied by that mass (D = m/V). Density is an important physical property of substances. Itgives information about a substance at both a macroscopic and a microscopic level. Givensamples of two substances, the more dense material either has more particles per unit volume,heavier particles, or both.

Density is determined by the careful measurement of a substance's mass and volume.Mass can be determined on a scale or balance. Volume can be determined in a variety of waysbut is commonly determined by displacement. To measure the volume of an object bydisplacement, water is placed in a device in which its volume can be carefully measured. Onesuch common device is a thin glass cylinder with volume increments marked on its side called agraduated cylinder (see, for example, Figure 2.3 in your text). After carefully noting the volumeoccupied by just the water, an object may be added. It will be noticed that the level of the waterin the cylinder moves upward, as water is displaced (moved out of the way) by the object. Thedifference between the starting volume and the final volume of water is equal to the volume ofthe object.

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The following are examples of how density may be used in chemical calculations.

Throughout these lecture notes I will work many problems as examples. An essential part of your studyin this class should consist of working these same example problems on your own, along with similarexamples in the text and also the recommended end-of-chapter problems. Like it or not, chemistry isvery problem based. If you don’t practice, practice, practice then you’re going to have some real troublewith the exams. But don’t worry. When you repeat the course because you failed to practice, maybeyou’ll have learned how important this really is to your success in this course.

• If 10.0 g of liquid occupies a volume of 13.5 mL, what is the density of the liquid?

density = m/V = 10.0g/13.5 mL = 0.741 g/mL

• Pure gold (24 K) has a density of 19.32 g/cm3 . A wedding band weighs 3.50 grams. Howmuch water must it displace (i.e., what must its volume be?) if the band is in fact 24 Kgold?

if D = m/V then V = m/D3.50 g/19.32 g/cm3 = 0.181 cm3

• Benzene has a density of 0.880 g/mL. If you need exactly 78.12 grams of benzene, whatvolume would you measure out?

if D = m/V then V = m/D 78.12 g / 0.880 g/mL = 88.77 mL .88.8 mL

Energy and heat

In science, energy is defined as the capacity to do useful work. We speak of potentialenergy, which is energy that is stored and based on position, and kinetic energy which is theenergy of moving objects. Energy can be transferred as heat, light, sound (acoustic), orelectricity by various chemical reactions. Reactions that give off energy are said to beexothermic, while reactions that absorb energy to make them happen are said to be endothermic.We’ll discuss exothermic and endothermic reactions in more detail in Chapter 7.

A few examples of exothermic reactions include fire, explosions, and the flow ofelectricity from a battery. Examples of endothermic reactions include cooking and recharging abattery.

Energy is often used to increase the temperature of a substance. The common unit ofenergy is the calorie (cal). One calorie is the amount of heat required to raise the temperature of

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1 g of water 1°C or 1 K since these temperature units are the same relative size, they are used moreor less interchangeably in this discussion. The SI unit of energy is the Joule (J). There is arelationship between calories and Joules: 4.184 J = 1 cal. We should note that the calories usedto calculate the nutrition value of food are actually equal to one thousand of the calories wediscuss in this chapter. In other words, in nutrition kilocalories are the basis for nutritionaldiscussions.

Let’s do a thought experiment. Imagine taking pieces of gold and iron that are equal inmass and immersing them in boiling water. After allowing them to reach the temperature of thewater, they are removed and dried. The gold will return to room temperature much quicker thanthe iron. This is because of an important physical property called specific heat. The specific heatof a substance is the amount of heat required to raise the temperature of one gram of thesubstance by 1°C. Comparatively speaking things with high specific heats heat more slowly andcool more slowly than things with lower specific heats.

Time for another thought experiment, although this can be attempted at home. Analuminum pan at room temperature is filled with a mass of water equal to the mass of the pan.The pan of water is placed on a stove. Which will heat more quickly, the pan or the water? Onceremoved from the burner, the water is poured out of the pan into another container at roomtemperature. Which will cool more rapidly, the empty pan or the water in its new container?

The pan will heat more rapidly than the water while on the burner, not just because it isin contact with burner, but also because aluminum has a lower specific heat than water. This isconfirmed by the second part of the experiment. The empty pan will return to room temperaturemuch quicker than the water. This is because aluminum in the pan has a lower specific heat thanwater.

Dimensional analysis and problem solving

It is often necessary to convert from one set of units to another while working scienceproblems. This is made possible by a problem-solving technique called dimensional analysis orthe factor label method.

At the risk of belaboring the point I just made above, you absolutely must learn how to use thistechnique. There are times I will require you to answer problems using dimensional analysis and will notaccept an answer, even if it is correct, obtained using other means. Chemistry is difficult, not because ofthe math, which is usually the sort of math most of us learned in junior high school and high school(although we may not remember it). Chemistry is difficult because the problems are nearly always storyproblems (oh damn, I hear you say) and can only be mastered through practice, practice, and morepractice. If you do not practice what follows below you will be extremely unhappy at exam time. Do nothesitate to spend time getting help from the Science Resource Center, other tutors, or your instructor.You have been warned!

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Dimensional analysis works by using conversion factors, which make the correct conversionfrom a value with one set of units to a value with a different set of units possible.

In mathematics there is something called the multiplicative identity, which states that anynumber multiplied by 1 will result in that number i.e., 1 x y = y. A conversion factor permits theconversion from one set of units to another without changing the true value of the numberassociated with the units, because a conversion factor is always exactly equal to one. Conversionfactors are created through equivalent ways of expressing things. As examples:

• 1 mile = 5280 feet If we divide both sides by 5280 feet we obtain (1 mile/5280 feet) = (5280 feet/5280 feet) = 1 The right side of the equation cancels, since the numerator and denominator are equal,and this leaves us with (1 mile/5280 feet) = 1

• 1 foot = 12 inchesIf we divide both sides by 12 inches we obtain (1 foot/ 12 inches) = (12 inches/12 inches) =1The right side of the equation cancels, since the numerator and denominator are equal,and this leaves us with (1 foot/12 inches) = 1

• 1 hour = 3600 secondsIf we divide both sides by 3600 seconds we obtain (1 hour/3600 seconds) = (3600 seconds/3600 seconds) = 1The right side of the equation cancels, since the numerator and denominator are equal,and this leaves us with (1 hour/3600 seconds) = 1

• 1 m = 1000 mm If we divide both sides by 1000 mm we obtain (1 m/1000 mm) = (1000 mm/1000 mm) = 1The right side of the equation cancels, since the numerator and denominator are equal,and this leaves us with (1 m/1000 mm) = 1

Dimensional analysis is the process of problem solving using conversion factors. Howare conversion factors used? While if you know another approach to this technique you’rewelcome to use it, I suggest the following approach, using as an example the conversion of 150lbs to kg:

1. Identify what is given and how should the problem wind up. In this case we are given150 pounds, and expect to wind up with a certain number of kg.

2. Identify which conversion factors are needed. In this example we need to know therelationship between pound and kilograms, which is 1 kg = 2.2 lbs or (1 kg/2.2 lbs) = 1

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3. Starting with what is given, we multiply it by our conversion factor. After using theconversion factor do the units cancel, or is another conversion factor needed?

150 lbs x (1 kg/2.2 lbs) = kg; the units are correct, no additional conversion factors areneeded

4. Do the arithmetic.

150 lbs x (1 kg/2.2 lbs) = 68.2 kg = 68 kg

Since there are only 2 significant figures in 150 lbs (why is the “0" not counted as a sigfig in this number?), we would report our answer as 68 kg.

Note: I strongly suggest that you always check to make sure that your units cancel each otherout and that you wind up with the correct units before you attempt any of the arithmetic in theproblem. I will never require you to do a dimensional analysis problem any one specific way, butI know from experience if you take care of the units in a problem first, then the arithmetic partnearly always turns out correct. If you rush through problems you will ultimately prove to beyour own worst enemy at exam time.

What does this mean physically? That there are 68 kg in 150 pounds of anything.

Some additional examples of dimensional analysis problems are as follows:

• Convert 62.5 cm to inches

conversion factors needed: 1 inch = 2.54 cm

62.5 cm x (1 inch/2.54 cm) = 24.6 inches

• Convert 3.62 ounces to mg

conversion factors needed: 1 lb = 16 oz; 1 lb = 453.6 g; 1 g = 1000 mg

(3.62 oz) x (1 lb/16 oz) x (453.6 g/1 lb) x (1000 mg/1 g) = 102,627 mg = 1.02 x 105 mg

Note that the solution of this problem required the use of several conversion factors. We can use asmany as we need to solve a problem. Also note that when we perform multiplication or division, thenumber of significant figures we report in our answer is determined by the value with the fewest sig figsin our calculations. Conversion factors are usually considered to be exact, and, as such, have an infinitenumber of significant figures. This means that, in this class, when we work problems, we report ouranswers with the correct number of sig figs, and the correct number of sig figs is determined by the

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information we are given in the problem, not the conversion factors. You should also be comforted toread that I will always provide you with all non-metric conversion factors in this class. The metricprefixes are very much your responsibility, and I require you to know them.

• Convert 125 m/s to mph (miles per hour)

conversion factors needed: 60 sec = 1 minute; 60 min = 1 hr; 1 m = 100 cm; 1 in = 2.54 cm; 1 ft = 12 in; 1 mile = 5280 ft

(125 m/s) x (60 s/1 min) x (60 min/1 hr) x (100 cm/m) x (1 in/2.54 cm) x (1 ft/12 in) x (1 mile/5280 ft) = 279.6 mile/hr = 280. mph

• Convert the area of an 8.5" by 11" piece of paper to square centimeters (cm2)

conversion factors needed: 1 in = 2.54 cm

step 1. 8.5 in x 11 in = 93.5 in2

step 2. (93.5 in2) x (2.54 cm/1 in)2 = 603.2 cm2 = 6.0 x 102 cm2

Note that we do not have a conversion factor that permits the conversion from square inches to squarecentimeters. We can, however, create our own, simply by squaring both sides of the conversion factorgiven i.e., if 1.00 in = 2.54 cm then (1.00 in)2 = (2.54 cm)2. If you need to do this you must rememberto square the number as well as the unit during the calculation. You should also note that there is morethan one way to solve this problem and come up with the correct answer. You should also note thatwhile normally we like to do dimensional analysis problems using one continuous string of conversionfactors, sometimes it is easier to break a problem down into pieces as we did here.

• Convert 6 yds3 of cement to cm3

conversion factors needed: 1 yd = 3 ft; 1 ft = 12 in; 1 in = 2.54 cm

(6 yds3) x (3 ft/1 yd)3 x (12 in/1 ft)3 x (2.54 cm/1 in)3 = 4,587,329.1 cm3 = 5 x 106 cm3

Note again the use of conversion factors created, this time, by cubing both sides of a relationship, e.g.(3 ft)3 = (1 yd)3

• Hexane is an organic liquid used in the manufacture of gasoline. If a railroad car contains2.5 x 104 gallons of hexane, how much does the liquid weigh in pounds? The density ofhexane is 0.6594 g/mL

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Ok, this problem is a bit of a tough one but we can do it. You must know that the density is needed toact as a bridging conversion factor between the conversion of mass to volume, or, visa versa, theconversion of volume to mass. It is not possible to correctly convert from mass to volume, or fromvolume to mass, without using the materials density as a conversion factor.

conversion factors needed: 1 gal = 4 qt; 1 L = 1.057 qt; 1000 mL = 1 L; 1 lb = 453.6 g

step 1. (2.5 x 104 gallons) x (4 qt/1 gal) x (1 L/1.057 qt) x (1000 mL/1 L) =94,607,379.376 mL

step 2. (94,607,379.376 mL) x (0.6594 g/mL) x (1 lb/453.6 g) = 137,531.1 lbs =1.4 x 105 lbs

Organic or inorganic - a postscript

I expect you to be able to differentiate between organic and inorganic compounds. You are goingto see questions on quizzes and exams in which you’ll be asked to identify a compound asorganic or inorganic, based strictly on a compound’s molecular formula. It is as simple as this:

If a compound's molecular formula includes both carbon (C) and hydrogen (H), it'sorganic. There are a small group of exceptions to this rule. In this class the only exceptions arethe bicarbonates (bicarbonates are also known as hydrogen carbonates; either the word"bicarbonate" or the words "hydrogen carbonate" will be part of the name.) Organic compoundsmay also include other types of atoms in addition to carbon and hydrogen atoms. The presence ofatoms other than C and H does not affect the compound’s designation as organic. Organiccompounds may include metal atoms, as well as nonmetal atoms in addition to carbon andhydrogen.

If a compound's molecular formula does not contain both carbon and hydrogen (C andH), it's inorganic - end of story.

As a few examples:

The molecular formula of water is H2O. This molecular formula contains H but not C.This is an inorganic compound.

The molecular formula of carbon monoxide is CO. This molecular formula contains Cbut not H. This is an inorganic compound.

The molecular formula of dinitrogen tetroxide is N2O4. This molecular formula does notcontain either C or H. This is an inorganic compound.

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The molecular formula of methane (natural gas) is CH4. This molecular formula containsboth C and H. This is an organic compound.

The molecular formula of strychnine is C21H22N2O2. This molecular formula containsboth C and H. This is an organic compound. The presence of atoms other than C and H does notaffect the designation of strychnine as an organic compound.

The molecular formula of sodium bicarbonate is NaHCO3. While this molecular formulacontains both C and H, the inclusion of the word "bicarbonate" in the name tells us this is anexception to the rule and is an inorganic compound.

The molecular formula of lead(II) acetate is Pb(C2H3O2)2.This molecular formulacontains both C and H. The presence of a metal atom, lead (Pb), does not change the status ofthis substance as an organic compound. A common misconception is that organic compoundscannot contain metal atoms. This is incorrect. As long as a compound contains both carbon andhydrogen in its molecular formula, it is an organic compound.

It is essential to remember that there are 11 elements with symbols that begin with the letter "C."These include chlorine (Cl), cesium (Cs), chromium (Cr), copper (Cu), and etc. None of theseelements are carbon, and so their presence in a molecular formula does not make the compoundorganic. As examples:

HCl - inorganic (not organic)

CsH - inorganic (not organic)

CuH - inorganic (not organic)

Please, be sure you use a periodic table and a list of the elements and their symbols whenyou're asked to determine if a compound is organic or inorganic. It's easy to inadvertently make amistake if you're not careful.

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Chapter 2: Atoms and the Periodic Table

The structure of the atom

Atoms are made of three types of subatomic particles which are summarized in the tablebelow.

charge relative mass absolute mass (kg)

proton +1 1 1.6726 x 10-27

neutron 0 1 1.6749 x 10-27

electron -1 1/1800 9.109 x 10-31

The Nucleus

Protons and neutrons are held together in the nucleus at the center of the atom. Undernormal circumstances particles with similar electrical charges repel each other, but protons areheld together in the nucleus despite this repulsion by the Strong Force, which is about 100 timesstronger than the electrostatic repulsion experienced by the similarly charged particles. TheStrong Force, while not a particularly clever name, is the strongest physical force in the knownuniverse. Neutrons also help to mitigate the electrical repulsion that arises when positivelycharged protons are jammed tightly together in an atomic nucleus. Electrons orbit the nucleusrelatively far away. As we stated above, the atom is mostly empty space.

To provide an illustration of this statement, the diameter of the nucleus of a hydrogenatom is reported as 1.75 fm (source: Atomic nucleus. (2016, July 12). In Wikipedia, The FreeEncyclopedia. Retrieved 16:53, August 12, 2016, fromhttps://en.wikipedia.org/w/index.php?title=Atomic_nucleus&oldid=729430288). However, thehydrogen atom, the combined size of the hydrogen atom’s nucleus and it’s single electron, havean approximate diameter of 0.106 nm, or 1.06 x 105 fm. In other words, the diameter of the atomis nearly one hundred thousand times larger than that of just the nucleus. To put this anotherway, if the nucleus of a hydrogen atom was represented by a ball exactly one meter in diameterin downtown Salt Lake City, the outer boundary of the hydrogen atom would be found inBrigham City (to the north, or Payson, UT, to the south), as the approximate distance fromdowntown Salt Lake City to Brigham City in the north, and to Payson in the south, is roughly100 km.

The notion that an atom is mostly just empty space becomes even more evident when we think ofatomic sizes in terms of volumes, rather than diameters. The nucleus of a hydrogen atom with adiameter of 1.75 fm occupies a volume of approximately 2.81 fm3. If a hydrogen atom has adiameter of 0.106 nm, the atom’s volume is 6.24 x 1014 fm3. In plain English, the overall volumeof the atom is 100 trillion times greater than the volume of the atom’s nucleus alone. What is it

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that occupies all of this space between the nucleus and the electron? Absolutely nothing, as far aswe know.

Hydrogen atoms are the smallest of any element. How much larger do the nuclei of atomsbecome? The nucleus of a typical uranium atom contains 92 protons and 146 neutrons. In spiteof this large number of protons and neutrons, the nucleus of a uranium atom is about 15 fm indiameter. In other words, even in atoms much larger than hydrogen, the atoms’ nuclei remainextremely small.

An electrostatic attraction, or, electromagnetic attraction, occurs between the positivelycharged protons in the nucleus and the negatively charged electrons as they move around it.While electrons are much smaller than protons, it is the number of electrons possessed by anatom, or, more correctly, the way that the electrons are arranged around the nucleus, thatdetermines that atom's chemical reactivity. This is one of the essential topics of chapter 2 andindeed, of this entire course.

Atoms and protons

It is the number of protons in an atomic nucleus that determines the identity of the atom.The number of protons in the nucleus is unique for the atoms of each element. The number ofprotons in an atom is invariable in chemical reactions. If the number of protons did change in achemical reaction, the atom would be converted into an atom of a different element. These typesof events do occur in nature but they are nuclear events, not chemical events. The process ofconverting an atom of one substance into an atom of another substance is (through the gain orloss of proton) is called transmutation and will be discussed in Chapter 11. To reiterate:transmutation never occurs during chemical reactions.

As we just said, and let us reiterate for emphasis: in chemical reactions, the number of protons inreacting atoms never changes. In nuclear reactions, which are not chemical reactions, the number ofprotons in the nuclei of reacting atoms may or may not change, through transmutation.

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If we examine a periodic table we find three pieces of information for every element.

The top number is the atomic number. In the above figure we see that this number, 29, and is thenumber of protons in the nucleus of all copper atoms. The atomic number is sometimes referredto as the"Z" number of an element although it is a bit archaic to do so. The atomic number of eachelement tells us the number of protons found in the nuclei of atoms of that element.

Atoms, neutrons, and isotopes

While the number of protons for a given element is invariable, the number of neutrons ina nucleus can differ. In other words, it is possible for atoms to have the same numbers of protonsbut different numbers of neutrons. We call these isotopes, atoms with the same atomic number(same number of protons) but with different numbers of neutrons. All elements have at least oneisotope. Most elements have two or more isotopes.

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The sum of the protons and neutrons in any isotope is known as its atomic mass. If wewant or need to be precise we can also include the masses of the electrons when calculating atomicmass but usually we ignore them because the are so small compared to the masses of the protons andneutrons. We can indicate the atomic number and the atomic mass (also called the mass number)of any isotope using it's elemental symbol:

The atomic mass is indicated as superscript to the left of the elemental symbol. The atomicnumber of an element is indicated as a subscript to the left of the elemental symbol. It is alsocommon to represent isotopes using only the elemental symbol and the mass number of theisotope.

As we know the element from its symbol, we must also know its atomic number as each elementhas a unique number of protons.

Isotopes are named by first stating the element and then the mass number. In thisillustration we are looking at the symbol for carbon-12.

If we know the atomic mass of an isotope we can calculate the number of neutrons foundin its nucleus. The number of neutrons in any isotope is equal to the difference between theisotope's mass number and its atomic number (number of protons). In other words,

atomic mass - atomic number = number of neutrons

As an example carbon has three common isotopes, carbon-12, carbon-13, and carbon-14. As allthree of these isotopes are of carbon, they all have six protons. The difference between thesethree isotopes is that each has a different number of neutrons from the other.

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126C

136C

146C

atomic mass 12 13 14

atomic number 6 6 6

number of neutrons 6 7 8

Students are sometimes confused as to how carbon-12 can be an isotope when it has an equalnumber of protons and neutrons. Isotopes are atoms with the same atomic number but withdifferent numbers of neutrons. This does not preclude substances from having isotopes with thesame number of protons and neutrons. The phrase “different numbers of neutrons” meansrelative to other atoms with the same numbers of protons. So carbon-12 has six protons and sixneutrons. It differs from carbon-13 in that while both have six protons, carbon-13 has sevenneutrons. And carbon-12 and carbon-13 both differ from carbon-14 in that while each isotopehas six protons, carbon-14 has eight neutrons.

As another example chlorine has two isotopes, chlorine-35 and chlorine-37. Theseisotopes have the following numbers of protons and neutrons.

3517Cl 37

17C

atomic mass 35 37

atomic number 17 17

number of neutrons 18 20

The mass of individual atoms is expressed in amu, atomic mass units. One amu is equalto 1/12 the mass of a single atom of carbon-12, and is also equal to 1.661 x 10-27 kg. In simplerterms, we can state that protons and neutrons each have a mass of 1 amu when calculatingatomic mass. This isn’t quite exact but it’s good enough for purposes in this course. This means thatthe three isotopes of carbon have atomic masses of 12 amu, 13 amu, and 14 amu respectively.The atomic masses of the two isotopes of chlorine have masses of 35 amu and 37 amurespectively.

The bottom number in the entry for each element in the Periodic Table is the atomicweight, which is the weighted average of the atomic masses of all of the isotopes of that element.In the figure above the atomic weight of carbon is given as 12.011 amu. A weighted average iscalculated by taking into consideration both the mass and the abundance of each of the isotopesof an element. We calculate the atomic weight of an element by multiplying the atomic mass ofeach isotope by its fractional abundance and by then adding all of the resulting numbers for all ofan element's isotopes together.

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The abundance of 12C is 98.9%, of 13C is 1.1%, and of 14C is around ~0%. This isotopicabundance information is not found in the periodic table. We have to turn to other sources ofinformation to learn the abundances of the isotopes of the elements. In other words, if we take amillion carbon atoms and examine them, we will find that roughly 989,000 of them are 12Cisotopes, 11,000 are 13C isotopes, and less than 1 of every trillion atoms is a 14C isotope. Eventhough 14C is not very abundant it is still useful, since it is the isotope examined when objects arecarbon dated in archeological studies. To calculate the atomic weight of carbon we use thefollowing set-up:

(0.989)(12) + (0.011)(13) + (0)(14) = 12.011 amu

which is the same as the atomic weight reported for carbon on your periodic table. Note that thepercent abundances must be converted to their decimal equivalents by dividing each by 100.Other examples of atomic weight calculations:

• The relative abundance of 35Cl is 75.78% and of 37Cl is 24.22%. The atomic weight ofchlorine is equal to:

(0.7578)(35) + (0.242)(37) = 35.48 amu

• The relative abundance of 50Cr is 4.345%, of 52Cr is 83.789%, of 53Cr is 9.501%, and of54Cr is 2.315%. The atomic weight of chromium is equal to:

(0.04345)(50) + (0.83789)(52) + (0.09501)(53) + (0.02315)(54) = 52.03 amu

Atoms that have two or more isotopes will always have an atomic weight that is not aninteger (i.e., an integer is not a fraction, i.e. there is no decimal place in the number). Check theperiodic table in your text and determine how many elements have an atomic weight that is awhole number. What does this suggest about the number of isotopes for each of these elements?Any element with a whole number for its atomic weight must only have one isotope.

What role do isotopes play in chemistry? In a class at this level, none whatsoever. Inother words, if we burn carbon in the presence of oxygen to form carbon dioxide gas

C + O2 => CO2

the carbon used in the reaction will be a mixture of all three common isotopes of carbon. Allthree of the isotopes will react with oxygen in exactly the same way, to form carbon dioxide.Most of the carbon dioxide molecules will have a carbon-12 atom, a few will have a carbon-13atom, and a small number of carbon dioxide molecules will have a carbon-14 atom, but all of thecarbon dioxide molecules formed in this reaction will behave exactly the same, both physicallyand chemically, for our purposes in this class. Similarly, if we react sodium with chlorine to formsodium chloride in the following reaction

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2 Na + Cl2 => 2 NaCl

we know that some of the chlorine atoms in the reaction will be chlorine-35 while others will bechlorine-37. However, they all react with sodium in exactly the same way. All of the sodiumchloride molecules formed, whether with chlorine-35 or chlorine-37 atoms, will have identicalphysical and chemical properties. Is there ever a time in chemistry when the isotopes of elements doplay a role in the outcome of a chemical reaction or influence the physical or chemical properties of acompound? Yes, but we will not ever see this happen in this particular class. So what does this mean toyou, as a student in this class? You should know what an isotope is, but you’re not going to ever worryabout them influencing the outcome of chemical reactions.

Atoms and electrons

In their elemental (natural) state elements are electrically neutral. In other words, in eachatom the number of electrons is equal to the number of protons. Remember, atoms never gain orlose protons in chemistry! However, it is possible for atoms to lose or gain electrons in chemicalreactions. If an atom gains or loses electrons, an electrical imbalance will be created and theatom will become electrically charged. Atoms that have electrical charge through the gain or lossof electrons are called ions. We speak of two categories of ions in chemistry.

• Cations have lost electrons, and since (#electrons < #protons), cations have a net positive(+) charge.

• Anions have gained electrons, and since (#electrons > #protons), anions have a netnegative (-) charge.

For ions, charge is represented as a superscript to the right of an element's symbol. Thesign tells us whether electrons have been lost or gained. Something that has lost electrons hasmore protons than electrons and becomes positively (+) charged. Something that has gainedelectrons has more electrons than protons and becomes negatively (-) charged. The number tellsus how many electrons have been lost or gained. This number is always a whole number. We donot lose or gain parts of electrons. As examples:

• Mg has 12 protons (how do we know? hint: what is its atomic number? You must learn how touse the periodic table to help you. I don’t expect you to memorize atomic numbers, atomicmasses, or atomic weights.) and therefore 12 electrons. Mg2+ has 12 protons but only 10electrons.

• As has 33 protons and therefore 33 electrons. As3- has 33 protons but 36 electrons.

• NH4+ and SO4

2- are examples of molecular ions. The "+" on the ammonium ion (NH4+)

tells us that the entire molecule has lost one electron, i.e., if we add up all of the protonsand all of the electrons in the molecule we will find that there is one fewer electron than

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protons. The "2-" charge on the sulfate ion (SO42-) tells us that the entire molecule has

gained two electrons.

Note that by convention we write the number and then the sign when representing charge. Mg2+ iscorrect, Mg+2 is incorrect. If an ion has lost or gained a single electron, we use only a "+" or "-" signand do not write "1+" or "1-". Na+ is correct, Na1+ is incorrect. Cl- is correct, Cl1- is incorrect. If an atomis electrically neutral there is no superscript to the right of the symbol. This absence tells us that thenumber of protons and electrons are equal in the atom or molecule.

Why is the loss or gain or electrons important? Because the resulting substance has verydifferent physical and chemical properties from that of the neutral atom or molecule. In otherwords, the physical and chemical properties of sodium, Na, and sodium ion, Na+, are verydifferent. If we compare the physical and chemical properties of fluorine, F, and fluoride ion, F-,we find that they are significantly different. Electrons may be physically small compared toprotons and neutrons, but the loss or gain of electrons profoundly affects the physical andchemical properties of the resulting ion.

An important thing to know about molecular ions is that we can never tell which atom ina molecular ion has lost or gained electrons. The symbol for ammonium ion, NH4

+, tells us thefive atoms in this molecule have collectively lost one electron. We do not know if it is thenitrogen atom or one of the hydrogen atoms that has lost the electron, nor do we care. It mattersonly that this collection of five atoms in an ammonium ion has lost an electron, i.e, the entiremolecule now has one fewer electrons than the total number of protons in the molecule. Whileone electron is very small compared to the overall size of the five atoms, it’s loss has a dramaticeffect in that both the physical and the chemical properties of these five atoms changes as aresult.

The same is true of sulfate ion, SO42-. We do not know if the two electrons gained go to

the sulfur atom or to one or more of the oxygen atoms. Quite simply, we don’t care where theygo within the molecule. We know that, collectively, there are two more electrons in the moleculethan the total number of protons in the molecule. We care very much that these five atoms have,collectively, gained two electrons because the resulting ion has completely different physical andchemical properties from an electrically neutral molecule consisting of one sulfur atom and fouroxygen atoms, i.e., a molecule with an equal number of protons and electrons.

The octet rule

Why would an atom want to gain or lose electrons? When two or more substances havethe same number of electrons, they are said to be isoelectronic. We will see that substances withthe same numbers of electrons also have the same arrangement of those electrons around theirnucleus. When ths occurs, the substances are said to be isoelectronic.

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It turns out that atoms are ultimately more stable when they are isoelectronic with thenearest Noble Gas. This is the basis for what is known as the octet rule. Atoms becomeisoelectronic with the nearest Noble Gas either by gaining or losing electrons, or by formingchemical bonds (which we will discuss later). The number of electrons an atom gains or losesdepends on how far away it is (on the Periodic Table) from the nearest Noble Gas. This behaviorof becoming isoelectronic with the nearest Noble Gas an important factor in the outcome ofmany chemical reactions.

For many elements we can predict how many electrons an atom may gain or lose basedon the element's position in the Periodic Table. Bear in mind it takes energy for each electronlost or gained, so the atoms of an element will be inclined to gain or lose the smallest possiblenumber of electrons necessary to become isoelectronic with the nearest Noble Gas. It is alsoessential to remember that, as a general rule, metals tend to lose electrons in chemical reactions,while non-metals tend to gain electrons in chemical reactions. This is a general rule, but a verygood and usually accurate statement, as far as general rules go.

All alkali metals (Group 1A elements) find it easiest to become isoelectronic with thenearest Noble Gas by losing one electron. When lithium loses an electron, it becomesisoelectronic with helium. When sodium loses an electron, it becomes isoelectronic with neon.When potassium loses an electron it becomes isoelectronic with argon. The same is true forrubidium, cesium, and francium. They lose an electron and become isoelectronic with the noblegas in the period beneath them. Be sure you’re looking at a periodic table as you read this section!

All alkali earth metals (Group 2A elements) find it easiest to become isoelectronic withthe nearest Noble Gas by losing two electrons. When beryllium loses two electrons it becomesisoelectronic with helium. When magnesium loses two electrons, it becomes isoelectronic withneon. When calcium loses two electrons it becomes isoelectronic with argon. And so on for therest of the Group 2A metals.

The Group 3A elements find it easiest to become isoelectronic with the nearest noble gasby losing three electrons. In other words, we often find Al3+, Ga3+, and so on. When it comes tothe octet rule, boron is a bit of an oddball for several reasons. We’ll talk a bit about boron’s oddbehavior in chapter 4.

We’re going to skip the Group 4A elements for the moment. They do obey the octet rule,but in a way slightly more complicated than we want to worry about at the moment.

The Group 5A elements find it easiest to become isoelectronic with the nearest noble gasby gaining three electrons. In other words, we often find the group’s non-metals gaining threeelectrons and forming N3-, P3-, As3-, and so on. How many electrons would you expect the metallic5A elements to lose? The answer is 5, the same as their group number.

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The non-metallic chalcogens (Group 6A elements) find it easiest to become isoelectronicwith the nearest noble gas by gaining two electrons. In other words, we often find O2-, S2-, Se2-,and so on. There is one metallic 6a element, polonium. How many electrons would you expect it tolose? If you guessed 6, you’re right!

The halogens (Group 7A elements) find it easiest to become isoelectronic with thenearest noble gas by gaining one electron. It is common to find F-, Cl-, Br-, and I- in the realworld.

As a rule, the transition metals and rare earth metals usually do not obey the octet rule.Trying to predict the charge of the transition metals and rare earth metals based on their positionin the periodic table is not easily done as rules other than the octet rule often govern theirbehavior. We will not discuss these other rules in this class. As I mentioned above, you shouldknow that, as a rule, metals typically lose electrons in chemical reactions and non-metalstypically gain electrons in chemical reactions.

You will find it incredibly useful to memorize the relationship between charge and position in thePeriodic Table as quickly as possible. This means the group 1A-8A elements, while ignoring thetransition metals and rare earth metals which cannot be predicted using the octet rule. You do notneed to worry about unusual atomic or chemical behavior in this class. If you can predict in general howthings will behave, you have achieved my expectations of you.

Fundamentals of electrons in atoms

In order to better understand why things behave the way they do in chemical reactions, itis often helpful to understand how the electrons of an atom are arranged with respect to thenucleus. Ultimately, our goal is to be able to predict the location of every electron in the atoms ofeach of the elements. If we can do this, we can correctly predict and understand the reason formany types of chemical behavior. Let’s begin with a few fundamental concepts about electrons.

• In neutral atoms the number of protons is equal to the number of electrons.

• Electrons do not orbit the nucleus in planetary fashion, as is commonly believed. Theposition of an electron is described in terms of statistical probability and may becalculated using quantum mechanics (we will not discuss quantum mechanics in thisclass).

• Electrons cannot exist just anywhere with respect to the nucleus. They can only be foundat certain discrete (specific) distances from the nucleus, which distances may becalculated using quantum mechanics.

• These distances correspond to energies. The further an electron is from the nucleus thegreater its energy.

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• Areas of high probability (90% probability) of finding an electron with a specific energyare called orbitals.

The discrete distances and locations at which electrons can be found with respect to thenucleus of any type of atom can be broken down into shells, subshells, and orbitals. These shells,subshells, and orbitals can be identified using quantum numbers which help us identify whereeach of the electrons in an atom is likely to be found.

The position of every electron in an atom can be described using four quantum numbers,the principal quantum number “n”, the angular momentum quantum number “l”, the magneticquantum number “ml”, and the spin quantum number “ms“. Every electron in an atom has its ownset of four quantum numbers unique from those of any other electron in the same atom. Thesefour quantum numbers serve essentially as an address system for electrons, just as your homeaddress helps others find you. If a friend from another country wanted to visit you, you might tellthem you lived in the United States, in Utah, in Salt Lake City, at 4600 South Redwood Road.These four bits of information identify your location to your friend.

The shell in which an electron is found describes its distance from the nucleus.

• Shell numbers, known as principal quantum numbers (n), range from 1 to 7 in chemistry.

• In other words, there are only seven average distances from the nucleus of any atom atwhich electrons may be found.

• The shell closest to the nucleus is n=1, the next nearest is n=2, and shell numbersincrease in integer values to the shell furthest away from the nucleus at n=7.

• As n increases the distance from the nucleus increases, as does the energy of the electronsfound in that shell.

• The spacing between shells is not linear. This has an important consequence, as we willshortly see.

You should also note that there are seven rows, or periods, in the periodic table. Eachperiod corresponds to one of the seven principal quantum numbers. The first period correspondsto n=1, the second period corresponds to n=2, the third period corresponds to n=3, and so on.This is because position in which an element is found in the periodic table is directly linked notonly to the number of its protons, but also to the number of its electrons and to the configurationof its electrons with respect to the nucleus. As we will see, the periodic table is a valuable tool tohelp us establish the electron configurations for the atoms of each of the 118 elements.

Each shell contains subshells, which correspond to the angular momentum quantumnumbers (l). A few important points about subshells are:

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• There are seven different types of subshells, s, p, d, f, g, h, and i subshells.

• Each shell contains as many subshells as its principal quantum number, i.e., the shell withn=1 has 1 subshell, the shell with n=2 has 2 subshells, the shell with n=3 has 3 subshells,and so on to the seventh shell (n=7) which is the only shell to have all seven subshells.

• While, in a given shell, all of the orbitals are at roughly the same distance from thenucleus, and therefore at roughly the same energy, the s subshells are the lowest inenergy, followed in order of increasing energy by the p, d, f, g, h, and i subshells.

• In other words, within a given shell, not all of the subshells are at exactly thesame distance from the nucleus. In the 2nd shell, the 2s subshell is a bit closer tothe nucleus and lower in energy, relative to the 2p subshell, which is a bit furtheraway from the nucleus and somewhat higher in energy. In the 7th shell, the 7ssubshell is the closest to the nucleus of all of the subshells found in the 7th shell.The 7i subshell is furthest from the nucleus and the highest in energy of all of thesubshells found in the 7th shell.

• These differences in energy and distances from the nucleus within a shell are notgreat but they are nonetheless significant.

• When assigning subshells to a shell, begin with the lowest energy subshell and work yourway up in terms of increasing energy.

• The first shell (n=1) has one subshell, s. It never has a p, d, f, or any othersubshell.

• The second shell (n=2) has two subshells, s and p. There are never d, f, g, etc.subshells in the second shell.

• The third shell (n=3) has three subshells, s, p, and d. There are never f, g, h, or isubshells in the third shell.

• And so on up to the seventh subshell (n=7), which has s, p, d, f, g, h, and isubshells.

• Each subshell differs from the others as to its distance from the nucleus, the number oforbitals it contains, the shape of its orbitals, and the orientation of its orbitals with respectto the nucleus.

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subshell #orbitals

s 1

p 3

d 5

f 7

g 9

h 11

i 13

The shapes and orientations of s, p, d, and f orbitals may be found at the following Wikipedialink:

http://en.wikipedia.org/wiki/Atomic_orbital#Orbitals_table.

Orbital shapes play an important role in the formation of chemical bonds, but I do not expect youto memorize orbital shapes.

The orbitals in the various subshells have magnetic quantum numbers (ml) whichdescribe their orientation with respect to a Cartesian coordinate system with its origin at thenucleus of the atom.

Any orbital, regardless of its shell and subshell, can only hold a maximum of twoelectrons. Electrons within the same orbital have opposite spin quantum numbers (ms). Electronscan have one of two spin quantum numbers, spin up or spin down.

We can summarize what we know about shells, subshells, and the total number ofelectrons in a shell in a table:

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shell (n) subshells (l) # orbitals/subshell #electrons/orbital

totalelectrons/shell

1 s 1 2 2

2 s,p 1+3=4 2 8

3 s, p, d 1+3+5=9 2 18

4 s, p, d, f 1+3+5+7=16 2 32

5 s, p, d, f, g 1+3+5+7+9=25 2 50

6 s, p, d, f, g, h 1+3+5+7+9+11=36 2 72

7 s, p, d, f, g, h, i 1+3+5+7+9+11+13=49 2 98

There are a few interesting features in the table we should point out. Notice that the total numberof orbitals in any shell is equal to n2. The first shell holds one orbital, the second shell holds fourorbitals, the third shell holds nine orbitals, and so on. Also note that the total number of electronsin any given shell is equal to 2n2.

Based on my years of experience as a teacher, I expect that some of you are on the verge of panicright about now. Peace, my children. Be calm. Much of this section is commentary that goes somewhatbeyond the scope of this class. You don’t need to memorize this table. I do however expect you tounderstand what it summarizes, which is the basic information about shells, subshells, and orbitals andhow many orbitals are in each shell and subshell. That's about it. Now, with that said, I offer a warning.It is painfully easy on quiz and exam questions to read a bit too quickly and to confuse shells, subshells,and orbitals. Treating seemingly “easy” questions too casually is one of the leading causes of studentsmissing quiz and exam questions on this material. When you encounter a question such as those you’llsee on the chapter 2 quiz, take a breath, read the question carefully, take a second breath, and thinkbefore you answer.

There’s one other thing I want to mention before we move on. At this very moment, there are severallaboratories around the world busily engaged in the attempted creation of 8th period elements. In otherwords, it is extremely likely that, at some future point, the periodic table will have eight periods ratherthan the seven we now see. While there are no known naturally existing 8th period elements, there iscontroversial experimental evidence in the world around us that such elements are likely to haveexisted at some point in the history of the universe. Given the extremely brief lifetime of most elementswith an atomic number of 112 or greater, why should we care? There are experimentalists who arguethat several of these potential 8th period elements may have long enough lifetimes and sufficiently

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unique physical and chemical properties that they could revolutionize certain aspects of life. Thefollowing link is optional but may be of interest to some of you:

http://www.rsc.org/chemistryworld/2016/01/beyond-new-elements-periodic-table-eighth-row-island-stability

Ground states and excited states

Regardless of how many electrons they may actually have, the atoms of all elements haveall of the possible shells and subshells described above available. In other words, even thoughhydrogen atoms only have one electron, they still have empty but available orbitals in the 2s, 2p,3s, 3p, 3d, and etc. shells and subshells up to and including the 7i shell and subshell. However,the majority of these shells and subshells are empty most of the time. Electrons always occupyorbitals in the shells and subshells with the lowest energy first. When all of the electrons in anatom are found in orbitals in the lowest energy shells and subshells they are said to be in theground state, i.e. electrons in their lowest energy state are said to be in the ground state.

When an electron absorbs energy, either thermal or electromagnetic, it will becomeexcited and can move to higher energy unoccupied orbitals. When an electron becomes excitedand moves to an orbital higher in energy than its ground state orbital the electron is said to be inan exited state. We mentioned g, h, and I subshells above, and we should point out that they arevirtual subshells, only populated by excited electrons. In the ground state electrons are neverfound in these subshells.

The difference in energy between the orbitals of the various shells and subshellscorrespond to specific amounts of energy. These, in turn, correspond to specific wavelengths oflight since there are mathematical relationships between the energy, frequency, and thewavelength of radiation.

We are only interested in ground state electrons in this class. While there are a number ofreasons to be interested in exited electrons, we will not discuss them in this class.

Electron configurations: full configurations and the periodic table

Of what practical value is all of this to us as we study chemistry? We are interested inbeing able to predict how things will behave in chemical reactions. Believe it or not, this is askill you will develop in this class. One of the most important predictive tools we have is theknowledge that the electron configuration of atoms has a direct bearing on their behavior inchemical reactions. In other words, if we know the electron configuration of an element, we canoften predict how it will react, and why it behaves the way it does.

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The electron configuration of an element is a description of the shells, subshells, andorbitals in which its electrons are found in the ground state. While this might seem like animposing task, it is made easier if we remember what we said above about the relationshipbetween shells, subshells, orbitals, and numbers of electrons. There are three rules that guide usas we write electron configurations.

• The aufbau principle: In the ground state electrons always occupy the lowest energyshells and subshells first. This means that the lowest energy electron in all elements isfound in the 1st shell, in its s subshell. Once we fill the orbitals in a subshell, the nextelectrons are placed in the orbitals of the next highest shell and subshell. In other words,the first two electrons of an atom are found in the 1s orbital. The 3rd electron must inhabitan orbital in the next available shell and subshell, which is 2s. Once the 2s orbital holds 2electrons, the next electron is placed in a 2p orbital. This is known as the Aufbau, or“building up,” principle which tells us that every electron configuration starts with thelowest energy shell and subshell (always 1s) and works its way up, in terms of energy.

• The Pauli exclusion principle: In the ground state no orbital can ever hold more than twoelectrons. Electrons paired up in the same orbital must have opposite spins (spin up andspin down). This is a restatement of the Pauli exclusion principle, which tells us that inthe ground state, no two electrons in the same atom can have the same set of fourquantum numbers. In other words, in the ground state in the same atom no two electrons canhave the same address. If you think about it, it makes sense, doesn’t it?.

• Hund’s rule: This pertains more to electron spin (arrow) diagrams than to the fullconfigurations and noble gas configurations we will initially discuss. When assigningelectrons to subshells in the ground state, all orbitals must hold at least one electronbefore any orbital may hold two electrons. In other words, electrons will never pair up inan orbital together if there is an empty orbital available (can you guess why?). This isknown as Hund’s rule.

We will discuss three types of electron configurations in this class, full configurations,Noble Gas configurations, and electron spin diagrams (also known as arrow diagrams or orbitaldiagrams). We will begin with full configurations because they serve as the basis for the othertwo types of configurations.

Full configurations are assigned to elements by determining how many electrons theyhave and then by placing them in orbitals in shells and subshells, starting with those orbitals ofthe lowest energy and working upward.

What we need is a tool to help us remember the order in which subshells are filled. Youmay be amazed to learn that the periodic table is just such a device. The periodic table not onlyprovides information about the number of protons and neutrons in each of the elements, it alsogives information about the highest energy electron for each element.

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The periodic table can be divided into four regions, the s block, p block, d block, and fblock. Groups 1A and 2A are known are the "s" block, because the highest energy electron ofeach element in these two groups is an "s" electron. Groups 3A, 4A, 5A, 6A, 7A, and 8A areknown as the "p" block, because the highest energy electron of each element in these six groupsis a"p" electron. The transition metals (Groups 1B - 8B) are found in the "d" block, because thehighest energy electron of each element in these ten groups is a"d" electron. The lanthanides andactinides are found in the "f" block, because the highest energy electron of each element in these14 groups is a"f" electron.

Do you notice that the "s" block contains two groups? Did you remember that an s subshell has oneorbital which can hold a maximum of two electrons? The "p" block contains six electrons in its threeorbitals. The "d" block holds 10 groups, and a d subshell can hold up to 10 electrons. (How manyorbitals are found in any d subshell?) The "f" block holds 14 groups, and a f subshell can hold 14electrons? (How many orbitals are found in an f subshell?) Coincidence? I think not. Remarkably clever?Absolutely!

How do we use the periodic table to assign electron configurations to elements? We mustfind the position of the element of interest, notice which period it is in and which block it is in,

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and observe how many groups to the right of the left-most group in the block the element is.Then we work backwards to account for the rest of the electrons. Let's choose a few examplesand see how this works:

Hydrogen has one electron. Hydrogen’s “address” in the periodic table is in the 1stperiod, in the s block, and in the first column, counting from the left within the block. Thereforethe full configuration of hydrogen is 1s1. The first number tell us the shell in which the subshellis found (1), the letter tells us which subshell contains the orbital in which the electron is found,and the superscripted "1" tells us that the orbital is occupied by a single electron. A handy tip: besure to have a periodic table handy and refer to it often when assigning electron configuration. I don’texpect you to memorize electron configurations. But plan on seeing them on quizzes and exams, forwhich you will always be provided a periodic table.

Helium has two electrons. Helium’s “address” in the periodic table is in the 1st period, inthe s block, and in the second column, counting from the left within the block. Therefore the fullconfiguration of helium is 1s2. In other words, there are two electrons in the single orbital of thes subshell of the 1st shell. At this point we have filled both the 1st shell and the s subshell.

Lithium has three electrons. Lithium’s “address” in the periodic table is in the 2nd period,in the s block, and in the first column, counting from the left within the block. Therefore theconfiguration of lithium’s highest energy electron is 2s1. But this only accounts for one oflithium’s three electrons. According to the Aufbau principle the first two electrons are assignedto orbitals in the lowest energy shells and subshell, which as we have just seen is the orbital inthe 1s shell and subshell. This means that the two lowest energy electrons in lithium have thesame configuration as helium, 1s2. The full configuration of lithium is 1s2 2s1.

Beryllium has four electrons. Beryllium’s “address” in the periodic table is in the 2ndperiod, in the s block, and in the second column, counting from the left within the block.Therefore the configuration of beryllium’s highest energy electron is 2s2. But this only accountsfor two of beryllium’s four electrons. According to the Aufbau principle the first two electronsare assigned to orbitals in the lowest energy shells and subshell, the 1s shell and subshell. Thismeans that the two lowest energy electrons in beryllium have the same configuration as helium,1s2. The full configuration of beryllium is 1s2 2s2. At this point we have filled the 2s subshell.

Boron has five electrons. Boron’s “address” in the periodic table is in the 2nd period, inthe p block, and in the first column, counting from the left within the block. The configuration ofboron’s highest energy electron is 2p1. But this only accounts for one of boron’s five electrons.According to the Aufbau principle the other four electrons are assigned to orbitals in the lowestenergy shells and subshells. This means that the two lowest energy electrons in boron have thesame configuration as helium, 1s2. The next two electrons are found in the orbital in the 2s shelland subshell and have the configuration 2s2. The full configuration of boron is 1s2 2s2 2p1.

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Carbon has six electrons. Carbon’s “address” in the periodic table is in the 2nd period, inthe p block, and in the second column, counting from the left within the block. The configurationof carbon’s highest energy electron is 2p2. But this only accounts for two of carbon’s sixelectrons. According to the Aufbau principle the other four electrons are assigned to orbitals inthe lowest energy shells and subshells. This means that the two lowest energy electrons in carbonhave the same configuration as helium, 1s2. The next two electrons are found in the orbital in the2s shell and subshell and have the configuration 2s2. The full configuration of carbon is 1s2 2s2

2p2. Note how the sum of the superscripts in the configuration tell us the total number of electrons forwhich the configuration accounts. In the case of carbon, 2+2+2=6.

Nitrogen has seven electrons. Nitrogen’s “address” in the periodic table is in the 2ndperiod, in the p block, and in the third column, counting from the left within the block. Theconfiguration of nitrogen’s highest energy electron is 2p3. But this only accounts for three ofnitrogen’s seven electrons. According to the Aufbau principle the other four electrons areassigned to orbitals in the lowest energy shells and subshells. This means that the two lowestenergy electrons in nitrogen have the same configuration as helium, 1s2. The next two electronsare found in the orbital in the 2s shell and subshell and have the configuration 2s2. The fullconfiguration of nitrogen is 1s2 2s2 2p3.

Oxygen has eight electrons. Oxygen’s “address” in the periodic table is in the 2nd period,in the p block, and in the fourth column, counting from the left within the block. Theconfiguration of oxygen’s highest energy electron is 2p4. But this only accounts for four ofoxygen’s eight electrons. According to the Aufbau principle the other four electrons are assignedto orbitals in the lowest energy shells and subshells. This means that the two lowest energyelectrons in oxygen have the same configuration as helium, 1s2. The next two electrons are foundin the orbital in the 2s shell and subshell and have the configuration 2s2. The full configuration ofoxygen is 1s2 2s2 2p4.

Fluorine has nine electrons. (note: fluorine is an element. Flourine - I don't know whatflourine is, perhaps an element made of ground-up wheat? Be *very* careful with spellings.Spelling mistakes will absolutely kill you on the Chapter 2 quiz!) Fluorine’s “address” in theperiodic table is in the 2nd period, in the p block, and in the fifth column, counting from the leftwithin the block. The configuration of fluorine’s highest energy electron is 2p5. But this onlyaccounts for five of fluorine’s nine electrons. According to the Aufbau principle the other fourelectrons are assigned to orbitals in the lowest energy shells and subshells. This means that thetwo lowest energy electrons in fluorine have the same configuration as helium, 1s2. The next twoelectrons are found in the orbital in the 2s shell and subshell and have the configuration 2s2. Thefull configuration of fluorine is 1s2 2s2 2p5.

Neon has ten electrons. Neon’s “address” in the periodic table is in the 2nd period, in thep block, and in the sixth column, counting from the left within the block. The configuration ofneon’s highest energy electron is 2p6. But this only accounts for six of neon’s ten electrons.According to the Aufbau principle the other four electrons are assigned to orbitals in the lowestenergy shells and subshells. This means that the two lowest energy electrons in neon have the

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same configuration as helium, 1s2. The next two electrons are found in the orbital in the 2s shelland subshell and have the configuration 2s2. The full configuration of neon is 1s2 2s2 2p6.

At this point we've filled all of the orbitals in all of the subshells in the first two shells.Where will we put additional electrons? If you guessed the third shell and it's subshells, you'redoing great!

Sodium is the first element in the third period (the 3rd row down). It has 11 electrons.Sodium’s “address” in the periodic table is in the 3rd period, in the s block, and in the firstcolumn, counting from the left within the block. The configuration of sodium’s highest energyelectron is 3s1. But this only accounts for one of sodium’s eleven electrons. According to theAufbau principle the other ten electrons are assigned to orbitals in the lowest energy shells andsubshells. This means that these ten electrons have the same configuration as neon, 1s2 2s2 2p6.The full configuration of sodium is 1s2 2s2 2p6 3s1.

Magnesium has 12 electrons. Magnesium’s “address” in the periodic table is in the 3rdperiod, in the s block, and in the second column, counting from the left within the block. Theconfiguration of magnesium’s highest energy electron is 3s2. But this only accounts for two ofmagnesium’s twelve electrons. According to the Aufbau principle the other ten electrons areassigned to orbitals in the lowest energy shells and subshells. This means that these ten electronshave the same configuration as neon, 1s2 2s2 2p6. The full configuration of magnesiumis 1s2 2s2 2p6 3s2. At this point the 3s shell and subshell are filled.

Aluminum has 13 electrons. Aluminum’s “address” in the periodic table is in the 3rdperiod, in the p block, and in the first column, counting from the left within the block. Theconfiguration of aluminum’s highest energy electron is 3p1. But this only accounts for one ofaluminum’s thirteen electrons. According to the Aufbau principle the other twelve electrons areassigned to orbitals in the lowest energy shells and subshells. This means that these twelveelectrons have the same configuration as magnesium, 1s2 2s2 2p6 3s2. The full configuration ofaluminum is 1s2 2s2 2p6 3s2 3p1.

The next five elements, silicon, phosphorus, sulfur, chlorine, and argon also haveelectrons that occupy the orbitals in the 3p subshell.

Silicon has 14 electrons. Silicon’s “address” in the periodic table is in the 3rd period, inthe p block, and in the second column, counting from the left within the block. The configurationof silicon’s highest energy electron is 3p2. But this only accounts for two of silicon’s fourteenelectrons. According to the Aufbau principle the other twelve electrons are assigned to orbitals inthe lowest energy shells and subshells. This means that these twelve electrons have the sameconfiguration as magnesium, 1s2 2s2 2p6 3s2. The full configuration of silicon is 1s2 2s2 2p6 3s2

3p2.

Phosphorus has 15 electrons. Phosphorus’s “address” in the periodic table is in the 3rdperiod, in the p block, and in the third column, counting from the left within the block. The

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configuration of phosphorus’s highest energy electron is 3p3. But this only accounts for three ofphosphorus’s fifteen electrons. According to the Aufbau principle the other twelve electrons areassigned to orbitals in the lowest energy shells and subshells. This means that these twelveelectrons have the same configuration as magnesium, 1s2 2s2 2p6 3s2. The full configuration ofphosphorus is 1s2 2s2 2p6 3s2 3p3.

Sulfur has 16 electrons. Sulfur’s “address” in the periodic table is in the 3rd period, in thep block, and in the fourth column, counting from the left within the block. The configuration ofsulfur’s highest energy electron is 3p4. But this only accounts for four of sulfur’s sixteenelectrons. According to the Aufbau principle the other twelve electrons are assigned to orbitals inthe lowest energy shells and subshells. This means that these twelve electrons have the sameconfiguration as magnesium, 1s2 2s2 2p6 3s2. The full configuration of sulfur is 1s2 2s2 2p6 3s2

3p4.

Chlorine has 17 electrons. Chlorine’s “address” in the periodic table is in the 3rd period,in the p block, and in the fifth column, counting from the left within the block. The configurationof chlorine’s highest energy electron is 3p5. But this only accounts for five of chlorine’sseventeen electrons. According to the Aufbau principle the other twelve electrons are assigned toorbitals in the lowest energy shells and subshells. This means that these twelve electrons havethe same configuration as magnesium, 1s2 2s2 2p6 3s2. The full configuration of chlorine is 1s2 2s2

2p6 3s2 3p5.

Argon has 18 electrons. Argon’s “address” in the periodic table is in the 3rd period, in thep block, and in the sixth column, counting from the left within the block. The configuration ofargon’s highest energy electron is 3p6. But this only accounts for six of argon’s eighteenelectrons. According to the Aufbau principle the other twelve electrons are assigned to orbitals inthe lowest energy shells and subshells. This means that these twelve electrons have the sameconfiguration as magnesium, 1s2 2s2 2p6 3s2. The full configuration of argon is 1s2 2s2 2p6 3s2 3p6.

Does each of these configurations account for the correct number of electrons? How doyou know? The sum of the superscripts should equal the total number of electrons.

The next element is potassium, which has 19 electrons (how do we know it has 19electrons?). We find potassium in the 4th period (the 4th row down). The first 18 electrons havethe same configuration as argon, 1s2 2s2 2p6 3s2 3p6 . How do we assign the 19th electron? On theone hand we need to remember that the 3rd shell has 3 subshells, 3s, 3p, and 3d. We have yet toplace any electrons in the orbitals in the 3d subshell. On the other hand, potassium is found in the4th period. Is the correct full configuration of potassium 1s2 2s2 2p6 3s2 3p6 3d1 or 1s2 2s2 2p6 3s2

3p6 4s1?

We use the periodic table to help us answer this question. We find potassium located inthe 4th period, and in the left-most column of the "s" block. This makes the configuration ofpotassium's highest energy electron 4s1. The other eighteen electrons have the same

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configuration as argon, 1s2 2s2 2p6 3s2 3p6. The full configuration of potassium is 1s2 2s2 2p6 3s2

3p6 4s1.

Calcium has 20 electrons. It is located in the 4th period, second to the left in the "s"block. Calcium's highest energy electron is 4s2. This accounts for two of calcium’s twentyelectrons. The other eighteen electrons have the same configuration as argon, 1s2 2s2 2p6 3s2 3p6.Calcium's full configuration is 1s2 2s2 2p6 3s2 3p6 4s2.

Next we come to scandium, which has 21 electrons. Scandium is located in the forthperiod, in the left-most group of the "d" block. Does this mean that scandium's highest energyelectron has a 4d1 configuration? Not quite. Remember that the 3rd shell has 3 subshells, s, p,and d. Recall as well that the spacing between shells is not linear. An important consequence ofthis non-linearity is that as shells move further from the nucleus, their subshells begin to overlapwith those in higher shells. The first d electrons we encounter in the periodic table are 3delectrons, meaning that scandium's highest energy electron has a 3d1 configuration. Scandium'sfull configuration is 1s2 2s2 2p6 3s2 3p6 4s2 3d1.

Because of overlap, the d electrons in every period are from one shell lower than theperiod in which they reside, while the s and p electrons are from the same shell as the period. Inother words, as we see in the figure near the beginning of this section, the subshells follow theorder 4s 3d 4p, 5s 4d 5p, 6s 5d 6p, and 7s 6d 7p.

Where do we find elements with f electrons in the periodic table? Note the order of theelements in the 6th period. The atomic number of cesium (Cs) is 55, the atomic number ofbarium (Ba) is 56, the atomic number of lanthanum (La) is 57, and the atomic number ofhafnium (Hf) is 72. Where are the elements with atomic numbers from 58 to 71? In the "f" block,in the lanthanides and actinides, which are found in a strip below the periodic table. Why arethese elements placed separately from the main body of the periodic table rather than integratedin their proper place? There is a reason, although it is not necessarily a very good one. Byincluding the 14 f-block groups in their proper place a periodic table we nearly double it's width.This means that it is difficult to fit the entire table on an regular-sized piece of paper in print thatis large enough to read clearly. A periodic table with the f-block in its proper place looks likethis:

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To which f-block do elements 58 through 71 belong? The first shell with an f subshell isthe 4th shell. Elements 58 through 71 have electrons with a 4f configuration. The f subshellsoverlap even more severely with higher level subshells than the d subshells.

In the 7th period there is a gap between actinium (Z=89) and Rutherfordium (Z=104),with the missing elements, those with atomic numbers from 90 to 103, again found in thef-block. These are 5f elements.

To summarize, s and p electrons always have the same number as the period in whichthey are found. d electrons always have a number one less than the period in which they reside,and f electrons, 2 less.

Do you feel like you can assign full electron configurations now? Let's do a fewexamples.

Arsenic (As) has 33 electrons. It is found in the 4th period, three groups from the left inthe p-block. Arsenic's highest energy electron has the configuration 4p3. By using the periodictable to remind us which subshells lie beneath 4p in energy, and their order, we assign arsenic afull electron configuration of 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p3. Note that 2 + 2 + 6 + 2 + 6 + 2 +10 + 3 = 33 a way to check that we have accounted for all of the electrons.

Palladium (Pd) has 46 electrons. It lies in the 5th period, 8 groups from the left.Palladium has the full electron configuration 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d8. Again, let’scheck to make sure that we have accounted for all of the electrons: 2 + 2 + 6 + 2 + 6 + 2 + 10 + 6+ 2 + 8 = 46

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Polonium (Po) has 84 electrons. It lies in the 6th period, four groups from the left in thep-block. Polonium has the full electron configuration 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6

6s2 4f14 5d10 6p4. Note that 2 + 2 + 6 + 2 + 6 + 2 + 10 + 6 + 2 + 10 + 6 + 2 + 14 + 10 + 4 = 84.That’s an awful lot of electrons, but we did it correctly.

A few final notes on full electron configurations.

• As I said above, you do not need to memorize the periodic table. I will always provideyou with one on exams and permit you to use one on quizzes.

• There are periodic tables available at campus bookstores which offer electronconfigurations that are correct but not in the order provided by the periodic table. As anexample, you might find the electron configuration of arsenic given as 1s2 2s2 2p6 3s2 3p6

3d10 4s2 4p3, rather than as we have written it above. I require you to be able to writeelectron configurations as they are found on the periodic table. You will not be permitted to useany information source that provides you with electron configurations on exams, other thanperiodic table such as the one found on the inside cover of your text. If you provide an electronconfiguration that is not congruent with the order found in the periodic table, as with ourarsenic example, even though it is correct, it will be marked wrong.

• The electron configurations of d and f-block elements can be tricky things. Sometimestheir full electron configurations differ slightly from those we might predict using theperiodic table. There are good reasons for these deviations, but we will not discuss themin this class and I do not require you to know them. If I ask for the electron configurationof a d-block element I will accept as correct the configuration derived from the periodictable, regardless of whether it is absolutely correct or not. You will not be asked aboutthe electron configuration of f-block elements. In summary, my goal for you in this section issimply to be able to provide the full configuration of any s-block, p-block, or d-block element asthey appear on any periodic table. I do not expect you to know and understand the exceptions.I do not expect you to provide configurations for f-block elements, or in the disputed areas ofthe periodic table where the lanthanides and actinides are found.

• Note there is a single space between each term in an electron configuration. The use of acomma, hyphen, or any form of punctuation rather than a single space will result in awrong answer on the chapter 2 quiz.

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Electron configurations: Noble Gas configurations

Ok, let's admit it. Writing full electron configurations, especially of elements with manyelectrons, is a real pain. Let's learn a short-cut. We can substitute the symbol of the noble gas inthe previous period for the configuration of lower energy electrons in a full electronconfiguration. We must place the symbol for the noble gas in square brackets to correctlyindicate what we're doing. Let's see how this works.

Element full configuration noble gas inprevious period

noble gasconfiguration

H 1s1 - 1s1

He 1s2 - 1s2

Li 1s2 2s1 He [He] 2s1

Be 1s2 2s2 He [He] 2s2

B 1s2 2s2 2p1 He [He] 2s2 2p1

C 1s2 2s2 2p2 He [He] 2s2 2p2

N 1s2 2s2 2p3 He [He] 2s2 2p3

O 1s2 2s2 2p4 He [He] 2s2 2p4

F 1s2 2s2 2p5 He [He] 2s2 2p5

Ne 1s2 2s2 2p6 He [He] 2s2 2p6 or [Ne]

Na 1s2 2s2 2p6 3s1 Ne [Ne] 3s1

Mg 1s2 2s2 2p6 3s2 Ne [Ne] 3s2

Al 1s2 2s2 2p6 3s2 3p1 Ne [Ne] 3s2 3p1

Si 1s2 2s2 2p6 3s2 3p2 Ne [Ne] 3s2 3p2

P 1s2 2s2 2p6 3s2 3p3 Ne [Ne] 3s2 3p3

S 1s2 2s2 2p6 3s2 3p4 Ne [Ne] 3s2 3p4

Cl 1s2 2s2 2p6 3s2 3p5 Ne [Ne] 3s2 3p5

Ar 1s2 2s2 2p6 3s2 3p6 Ne [Ne] 3s2 3p6 or [Ar]

Note that there are not noble gas configurations for the first period elements. All of the secondperiod elements have 1s2 as a part of their full configuration. Since this is also the fullconfiguration for helium, which is the noble gas in the period before the second period, we canrepresent these two electrons by using [He] in the configurations for the second period elements.

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And all of the third period elements have 1s2 2s2 2p6 as a part of their configuration, which is alsothe full configuration for neon which is the noble gas in the period before the third period. So weuse [Ne] to represent those electrons in their Noble gas configurations.

Let’s also look at the noble gas configurations of the three elements we did a moment ago,arsenic, palladium, and polonium.

element full configuration noble gasconfiguration

arsenic 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p3 [Ar] 4s2 3d10 4p3

palladium 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d8 [Kr] 5s2 4d8

polonium 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f14 5d10 6p4 [Xe] 6s2 4f14 5d10 6p4

We find arsenic in the fourth period. Argon is the noble gas in the third period with a fullconfiguration of 1s2 2s2 2p6 3s2 3p6. So we use [Ar] to represent those electrons in the noble gasconfiguration for arsenic.

We find palladium in the fifth period. Krypton is the noble gas in the fourth period with afull configuration of 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6. So we use [Kr] to represent those electronsin the noble gas configuration for palladium.

We find polonium in the sixth period. Xenon is the noble gas in the fifth period with afull configuration of 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6. So we use [Xe] to represent thoseelectrons in the noble gas configuration for polonium.

Electron configurations: electron spin diagrams (arrow diagrams)

Full electron configurations and Noble Gas configurations account for all of the electronsin an atom, but there are times when it is helpful to display electron configurations in a slightlydifferent way. Electron spin diagrams, also known as arrow diagrams, orbital diagrams, or spindiagrams use dashes to represent each of the orbitals in a subshell and arrows to representelectrons. Electron spin diagrams may be drawn vertically or horizontally. The energy ofelectrons in the diagram increases as we move from lower to higher levels or as we move fromleft to right, depending on if the electron spin diagrams is drawn vertically or horizontally.Remember that electrons repel each other (why?) and to minimize the repulsion, electrons thatshare an orbital have different spins, spin up or spin down. Of course, this is congruent with thePauli exclusion principle as two electrons in the same orbital and with the same spin would havethe same sets of four quantum numbers.

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Hydrogen has the full configuration 1s1. Its electron spin diagram is as follows.

Note the use of a single dash for the 1s orbital, the 2s orbital, and the 3s orbital, while the 2p and3p subshells are represented by three dashes because each has three orbitals. The shell andsubshell names are placed as a label beneath the appropriate dashes. There is only one dash witha single arrow in the electron spin diagram for hydrogen because there is only one orbital inhydrogen occupied by its lone electron. Really, we could have left the blank lines out if we hadwanted to and we usually do.

The electron spin diagram for helium with its 1s2 configuration is

Note that the 1s orbital is now filled, and the two electrons, represented by the arrows, point inopposite directions since they have opposite spins.

The electron spin diagram for lithium with its 1s2 2s1 configuration is

We now have two dashes occupied, one for the orbital in the 1s shell and subshell, and one forthe 2s orbital. Again, we always label the orbitals (dashes) as to which shell and subshell theybelong.

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The electron spin diagram for beryllium (1s2 2s2) is

At this point we have filled the 2s orbital. The next six elements will have electrons in the three2p orbitals. The first three elements with 2p electrons are boron (1s2 2s2 2p1 ),

carbon (1s2 2s2 2p2 ),

Did you notice that each 2p electron occupied an empty 2p orbital? You may have been temptedto place both of carbon’s 2p electrons in same 2p orbital:

The above electron spin diagram for carbon is wrong because it violates Hund’s rule. Allelectrons are negatively charged and are also therefore mutually repulsive. This is mitigated to aslight extent when electrons have opposite spins, but it is better still for them to stay as far apart

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as possible by occupying different orbitals and not pairing up until it becomes necessary. Thespin diagram for nitrogen is

Again, Hund’s rule requires an unpaired electron in each of the three 2p orbitals before electronsbegin to pair up in orbitals. As we look the electron spin diagrams for oxygen, fluorine, and neonwe see that it now becomes necessary to place pairs of electrons in the 2p orbitals. Here are theelectron spin diagrams for oxygen (1s2 2s2 2p4),

fluorine (1s2 2s2 2p5),

and neon (1s2 2s2 2p6).

And so it goes. We draw a single dash to represent the 3s orbital, and then, when necessary, threedashes to represent the three 3p orbitals. As examples, the electron spin diagram for the first fiveelements in the third period are as follow,

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sodium (1s2 2s2 2p6 3s1),

magnesium (1s2 2s2 2p6 3s2),

aluminum (1s2 2s2 2p6 3s2 3p1),

silicon (1s2 2s2 2p6 3s2 3p2),

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and phosphorus (1s2 2s2 2p6 3s2 3p3).

Have you noticed that each electron spin diagram begins with knowing the full electronconfiguration for the element in question?

Inner shell, outer shell, and valence electrons

Noble gas configurations provide us with a short-hand way of writing electronconfigurations. They also help us categorize an atom's electrons into one of two types, inner andouter shell electrons.

All electrons in periods prior to the period of the element of interest, i.e., all electronsthat can be accounted for by the noble gas portion of the noble gas configuration for an elementare called inner shell electrons. Inner shell electrons never participate in chemical reactions.

All electrons found in the same period as the element of interest are known as outer shellelectrons. Outer shell electrons are often called valence electrons, but there is actually a slightdistinction between the two. Valence electrons are those outer shell electrons available toparticipate in chemical reactions. Not all outer shell electrons are valence electrons. Some of youjust missed that and it’s extremely important, so let me re-iterate: not all outer shell electrons arevalence electrons. Chemists are often sloppy and use the terms interchangeably, the justificationbeing that outer shell electrons are usually, but not always, valence electrons. However, there arenumerous, common exceptions. You need to be able to make the distinction - when are outershell electrons valence electrons? when are outer shell electrons not valence electrons? - tocorrectly answer questions you’ll see on the quizzes and exams in this class.

Electrons found in completely filled p, d, and f subshells are extremely stable, to theextent that these electrons behave as though they are inner shell. In other words, the electrons offilled p, d, and f subshells do not participate in chemical reactions. This means that all of anatom's outer shell electrons may or may not be valence electrons.

For example, if we examine aluminum, we find it’s noble gas configuration to be [Ne] 3s2

3p1. This means aluminum has 13 total electrons, of which 10 are inner shell and three are outershell electrons, all three of which are valence electrons.

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Arsenic has the full configuration 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p3 and the noble gasconfiguration [Ar]4s2 3d10 4p3. Arsenic has 33 total electrons, of which there are 18 inner shellelectrons and 15 outer shell electrons. Of the 15 outer shell electrons there are only 5 valenceelectrons, the 4s and 4p electrons. The other 10 are tied up in the filled 3d subshell and areunreactive. The following table summarizes total electrons, inner shell electrons, outer shellelectrons, and valence electrons for the first 36 elements.

element noble gasconfiguration

inner shellelectrons

outer shellelectrons

valenceelectrons

commonlyformed ions

H 1s1 0 1 1 H+

He 1s2 0 2 0 none

Li [He] 2s1 2 1 1 Li+

Be [He] 2s2 2 2 2 Be2+

B [He] 2s2 2p1 2 3 3 B3+

C [He] 2s2 2p2 2 4 4 C4-

N [He] 2s2 2p3 2 5 5 N3-

O [He] 2s2 2p4 2 6 6 O2-

F [He] 2s2 2p5 2 7 7 F-

Ne [Ne] 2 8 0 none

Na [Ne] 3s1 10 1 1 Na+

Mg [Ne] 3s2 10 2 2 Mg2+

Al [Ne] 3s2 3p1 10 3 3 Al3+

Si [Ne] 3s2 3p2 10 4 4 Si4-

P [Ne] 3s2 3p3 10 5 5 P3-

S [Ne] 3s2 3p4 10 6 6 S2-

Cl [Ne] 3s2 3p5 10 7 7 Sl-

Ar [Ar] 10 8 0 none

K [Ar] 4s1 18 1 1 K+

Ca [Ar] 4s2 18 2 2 Ca2+

Sc [Ar] 4s2 3d1 18 3 3 Sc3+

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Ti [Ar] 4s2 3d2 18 4 4 Ti4+

V [Ar] 4s2 3d3 18 5 5 V5+

Cr [Ar] 4s2 3d4 18 6 6 Cr6+

Mn [Ar] 4s2 3d5 18 7 7 Mn7+

Fe [Ar] 4s2 3d6 18 8 8 Fe3+

Co [Ar] 4s2 3d7 18 9 9 Co2+

Ni [Ar] 4s2 3d8 18 10 10 Ni3+

Cu [Ar] 4s2 3d9 18 11 11 Cu2+

Zn [Ar] 4s2 3d10 18 12 2 Zn2+

Ga [Ar] 4s2 3d10 4p1 18 13 3 Ga3+

Ge [Ar] 4s2 3d10 4p2 18 14 4 Ge4+

As [Ar] 4s2 3d10 4p3 18 15 5 As3-

Se [Ar] 4s2 3d10 4p4 18 16 6 Se2-

Br [Ar] 4s2 3d10 4p5 18 17 7 Br-

Kr [Kr] 18 18 0 none

Generally there is a very good correlation between the number of valence electrons anelement has and the ion it forms to satisfy the octet rule. This is especially true of the s and pblock elements found in the first three periods. The behavior of the transition metals and thelarger (4th period through 7th period) p block elements is often not correctly described by theoctet rule. We will not discuss why they sometimes fail to obey the octet rule in this class.

You should be able to accurately predict for each s and p block element whether it formsa cation or an anion and its charge. This is based on position in the periodic table, as we haveseen. You are not responsible for remembering the various cations each of the transition metalsmay form (as well as the rare earth metals and p-block metals). However, you will find inChapter 3 that when we examine these metals compounds and their names, there is a fairlysimple way to determine the metal cation's charge.

One of the most important skills I expect you to develop in this chapter is the ability todetermine the number of inner shell electrons, outer shell electrons, and valence electrons theatoms of a given elements have. I guarantee there will be a series of three questions on MT1 andalso on the final exam in which I give you five elements and ask you which has the most innershell electrons, the most outer shell electrons, and the most valence electrons. You need to be

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sure you are prepared to respond. There is another way to approach this problem, using theperiodic table and an element’s noble gas configuration.

On the first midterm and also on the final exam, you will see a series of three questionsthat will look something like this:

1. Which has the most inner shell electrons? Si As Zr Te Cs

2. Which has the most outer shell electrons? Si As Zr Te Cs

3. Which has the most valence shell electrons? Si As Zr Te Cs

Normally, I don't care how you determine the answer for a particular question, as long asyou do it honestly. However, I very strongly recommend the following approach, and, shouldyou miss one of these questions on the exam, I will require you to use the following method ifyou wish to earn partial credit for the questions you missed if you choose to do the re-grade.

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To answer these questions (how many inner shell, outer shell or valence electrons) beginby making a table. There will be six rows, one for the column headings and one for eachelement. There will be six columns with the headings "element," "noble gas configuration(NGC)," "total electrons (TE)," "inner shell electrons (ISE)," "outer shell electrons (OSE)," and"valence electrons (VE)." Let's begin with the elements and their noble gas configurations.

element NGC TE ISE OSE VE

Si [Ne]3s2 3p2

As [Ar]4s2 3d10 4p3

Zr [Kr]5s2 4d2

Te [Kr]5s2 4d10 5p4

Cs [Xe]6s1

The total electrons can be determined by finding each element in the periodic table. The

atomic number of each element tells how many protons the atoms of each element have, but italso tells how many electrons the atoms of each element have (since un-reacted atoms areelectrically neutral, the number of protons and electrons in atoms must be equal):

element NGC TE ISE OSE VE

Si [Ne]3s2 3p2 14

As [Ar]4s2 3d10 4p3 33

Zr [Kr]5s2 4d2 40

Te [Kr]5s2 4d10 5p4 52

Cs [Xe]6s1 55

Each element has as many inner shell electrons as the noble gas in its noble gasconfiguration. We can determine this simply by looking at the periodic table and seeing howmany electrons the noble gas has.

element NGC TE ISE OSE VE

Si [Ne]3s2 3p2 14 10

As [Ar]4s2 3d10 4p3 33 18

Zr [Kr]5s2 4d2 40 36

Te [Kr]5s2 4d10 5p4 52 36

Cs [Xe]6s1 55 54

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The remaining electrons in the configuration are outer shell electrons. In other words, thedifference between the total electrons and the inner shell electrons is the number of outer shellelectrons (OSE = TE - ISE).

element NGC TE ISE OSE VE

Si [Ne]3s2 3p2 14 10 4

As [Ar]4s2 3d10 4p3 33 18 15

Zr [Kr]5s2 4d2 40 36 4

Te [Kr]5s2 4d10 5p4 52 36 16

Cs [Xe]6s1 55 54 1

However, outer shell electrons are not always valence electrons. Outer shell electrons in

filled d, f, or p subshells are not available to participate in reactions, so they are not counted asvalence electrons. Also, if all of the orbitals in a p subshell are filled, the electrons in a filleds subshell will also behave as though they're inner shell. As an example, the noble gasconfiguration of argon can be written [Ne] 3s2 3p6. Argon has 18 total electrons, 10 inner shellelectrons and 8 outer shell electrons. But, since the p subshell is filled, the 3p electrons behave asthough they are inner shell electrons. Since the 3p subshell is filled, the 3s electrons also behaveas though they are inner shell electrons. As a consequence, none of argon's 8 outer shellelectrons are valence electrons.

Only outer shell electrons in s subshells or in unfilled p, d, or f subshells can react andtherefore also be counted as valence electrons. The number of valence electrons for each elementin the question is:

element NGC TE ISE OSE VE

Si [Ne]3s2 3p2 14 10 4 4

As [Ar]4s2 3d10 4p3 33 18 15 5

Zr [Kr]5s2 4d2 40 36 4 4

Te [Kr]5s2 4d10 5p4 52 36 16 6

Cs [Xe]6s1 55 54 1 1

Early in these lecture notes we made the following statement: “In the periodic table, groups are"chemical families," because they exhibit similar chemical behavior.” In light of what you see inthe discussion and examples in this section, or more generally, in comparing the number of innershell, outer shell, and valence electrons for members of a chemical group, why is this true? Note:this is one of the most important questions you should be able to answer after mastering the materialin this chapter!

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The electron configurations of ions

When two substances have the same electron configuration (because they have the samenumber of electrons), they are said to be isoelectronic. The noble gas configuration of neon is[Ne] and of sodium is [Ne]3s1. When an atom loses electrons, it loses outer shell electrons. Thismeans that when sodium becomes sodium ion, Na+, it loses the 3s electron. This makes sodiumion isoelectronic with neon. Therefore, the noble gas configuration of Na+ can be represented as[Ne], or also as [Ne] 3s0. The latter is sometimes used to show that one or more electrons oncepresent have been lost. How do we know that sodium always tends to lose one electron? Where is Nafound in the periodic table? And remember: metals tend to lose electrons in chemical reactions,nonmetals tend to gain electrons in chemical reactions.

Magnesium has a noble gas configuration of [Ne] 3s2. Magnesium nearly always exists asMg2+ in chemical reactions. how do we know this? where is Mg found in the periodic table? Mg2+ haslost its two 3s electrons, and has become isoelectronic with both neon and Na+ in the process. Inother words, Mg2+ has the electron configuration of [Ne], or [Ne] 3s0.

Aluminum has a noble gas configuration of [Ne] 3s2 3p1. Aluminum nearly always existsas Al3+ in chemical reactions. This means it loses its two 3s electrons and also its 3p electrons.The Al3+ cation is also isoelectronic with neon, Na+ and Mg2+ . So Al3+ also has the electronconfiguration of [Ne] or [Ne] 3s0 3p0.

Fluorine has the electron configuration 1s2 2s2 2p5. It tends to gain an electron inreactions. how do we know this? where is F found in the periodic table? When an atom gainselectrons, they are always added to empty or half-filled orbitals in the outer shell. The electronconfiguration of F- is 1s2 2s2 2p6, which makes fluoride ion isoelectronic with neon.

Oxygen has the electron configuration 1s2 2s22p4. Oxygen tend to gain two electrons. howdo we know this? where is O found in the periodic table? The electron configuration of O2- is 1s2 2s2

2p6, which makes it isoelectronic with neon and F-.

Note that more than two substances can be isoelectronic. In our examples in this sectionwe find that Na+, Mg2+, Ne, F-, and O2- are all isoelectronic. What are examples of other ions thatare isoelectronic with neon? Al3+ and N3- are two examples.

As I mentioned above, it is important to remember that electrons in filled d and fsubshells, even when they are outer shell electrons, are no longer valence electrons. This effectsthe way we write electron configurations for cations in the 4th period and higher. Take gallium asan example. In its ground state (when the atom has an equal number of protons and electrons),gallium has the noble gas configuration [Ar] 4s2 3d10 4p1. Gallium commonly forms a 3+ cation,Ga3+. The noble gas configuration of this cation is [Ar] 4s0 3d10 4p0. The two 4s electrons and the

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4p electron are lost. None of the 3d electrons are available to react, even though they are outershell electrons. Once a d subshell contains 10 electrons, they no longer behave as valenceelectrons.

This is also true of the electrons in full f subshells. We mentioned above that the noblegas configuration of polonium is [Xe] 6s2 4f14 5d10 6p4. Polonium commonly forms a 6+ cation,Po6+. This cation has the noble gas configuration of [Xe] 6s0 4f14 5d10 6p0. The six electrons lostcome from the 6s and 6p subshells. None of the 4f or 5d electrons are available to participate inreactions. As these subshells are full, even though they contain outer shell electrons, none ofthem are available to participate in chemical reactions.

Atomic properties and periodic trends

There are four basic properties of elements, the extent of which can be predicted in arelative sense from position of the elements in the periodic table. These are properties are calledperiodic trends. They include atomic size, metallic character, ionization energy, andelectronegativity.

Atomic size is the actual size of the atoms. Within a group, atoms become larger as wemove down the group. Since moving down a group involves moving to electron shells that arefurther away from the nucleus, it should not surprise you that atomic size increases. As anexample, iodine is larger than bromine, which is larger than chlorine, which is larger thanfluorine. Atomic size also increases as we move from right to left within a period. This is a bitsurprising but not difficult to explain. Within a period, as we move from left to right, the numberof electrons increases. Lithium has fewer electrons (3 electrons) than beryllium (4 electrons),which has fewer electrons than boron (5 electrons), which has fewer electrons than carbon (6electrons), and so on. While there is a balance between the numbers of protons and electrons ineach of these elements, as the number of protons in the nucleus increases, electrons at the sameaverage distance from the nucleus (i.e., in the same period) are attracted more forcefully to thenucleus. As a result, the size of the atom becomes smaller. Atomic size increases as we movefrom nonmetals to metals and as we move down a group.

If we are interested in comparing the size of the atoms of two or more elements, withoutactually knowing any specific information about the size of the atoms, we can still qualitativelypredict which atoms are the largest and which are the smallest. Look at a periodic table andcompare the positions of barium (Ba) and chlorine. Based on position, which element has largeratoms? barium Using a periodic table compare the positions of atoms of cesium (Cs), ruthenium(Ru), and sulfur and arrange them in order of increasing size. S (smallest) < Ru < Cs (largest)

Metallic character involves the extent to which a substance behaves as a metal.Remember from chapter 1 that metals are elements characterized by good thermal and electricalconductivity and are usually lustrous, malleable, and ductile. Metals are also characterized by thetendency to give up electrons in reactions (i.e. form cations).

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It is the "looseness" with which valence electrons are held that determines how metallican element is. The further down a group, the further the valence electrons are found from thenucleus and the more loosely they are held. As we stated above, metallic character increases aswe move from nonmetals to metals within a period and as we move down a group.

We can qualitatively predict how metallic elements are relative to each based on thisinformation. While looking at a periodic table compare the atoms of cesium (Cs), ruthenium(Ru), and sulfur and arrange them in order of increasing metallic character. S (least metallic, infact a non-metal) < Ru < Cs (most metallic)

Ionization energy is the energy required to remove an electron from a neutral atom, thusmaking it a cation. Whenever an atom loses an electron it takes energy. The energy required toremove an electron increases as they are held more tightly. This means that the closer electronsare found to the nucleus (i.e., going up a group or from right to left within a period), the moreenergy required to remove them from an atom.

We can qualitatively predict how ionization energies are relative to each based on thisinformation. While looking at a periodic table compare the atoms of cesium (Cs), ruthenium(Ru), and sulfur and arrange them in order of increasing ionization energy. Cs (lowest ionizationenergy) < Ru < S (highest ionization energy)

There is a relationship between ionization energy and ion size. When an electron isremoved from an atom and a cation is formed, an imbalance between electrons and protonsresults. As a consequence, the remaining electrons are pulled nearer to the nucleus then theywere in the electrically neutral atom. In other words, cations are smaller than the neutral atomfrom which they originated. It is possible to remove more than one electron from an atom and toform cations with +2, +3 and higher charges. This loss can be thought of as occurring in astepwise process, meaning if an atom loses three electrons, it doesn't lose all threesimultaneously, but rather, in three steps that follow each other in rapid succession. Since eachloss of an electron results in the electrons being held more tightly by the nucleus, it should notsurprise you to read that it takes more energy to remove a second electron from an atom than itdid the first, far more still to remove a third electron than it did to remove the second, and so on.At the same time, with the loss of each electron, the cation becomes smaller and smaller as itholds its remaining electrons more and more tightly. So, as an example, if we compare the sizeof aluminum with that of its cations: Al > Al+ > Al2+ > Al3+

Electron affinity is the energy associated with the addition of a single electron to aneutral atom, thus making it an anion. Whenever an atom gains an electron, it may result in therelease of energy (i.e. an exothermic process) if a stable anion is formed, or it may require aninput of energy (i.e. an endothermic process) to make it happen if an unstable anion is formed.Electron affinity does not have a consistent periodic trend, but it is related to electronegativitywhich we shall soon discuss and which does have a consistent periodicity to it.

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As with cations, the size of anions differs from that of neutral atoms. When an electron isadded to an atom and an anion is formed, an imbalance between electrons and protons results. Asa consequence, the electrons are not held quite as tightly as in the neutral atom. In other words,anions are larger than the neutral atom from which they originated. It is possible to add morethan one electron to an atom and to form anions with -2, -3 and higher charges. This addition canbe thought of as occurring in a stepwise process, meaning if an atom gains three electrons, itdoesn't gain all three simultaneously, but rather, in three steps that follow each other in rapidsuccession. With the addition of each electron, the anion becomes larger and larger as it holds itselectrons more and more loosely. So, as an example, if we compare the size of nitrogen with thatof its anions: N < N– < N2- < N3-

Electronegativity is the tendency of an atom to attract electrons to itself. As we thinkabout the elements, we should recall that metals tend to lose electrons in reactions and thatnonmetals tend to gain electrons in chemical reactions. Therefore, as we move across a periodfrom left to right, from metals to nonmetals, we would expect that electronegativity shouldincrease, and it does. Electronegativity also increase as we move up a group. This is related tothe tendency of smaller atoms to cling to their electrons more tightly. The noble gases do nothave electronegativity, i.e, they have an electronegativity of zero. why?

We can qualitatively predict how electronegativities are relative to each based on thisinformation. Compare the atoms of cesium (Cs), ruthenium (Ru), and sulfur and arrange them inorder of increasing electronegativity. Cs (lowest electronegativity) < Ru < S (highestelectronegativity).

Electronegativity is commonly ranked on a scale from 0 to 4. The most electronegativeelement is fluorine, with an electronegativity of 4. Francium (Fr) is the least electronegativeelement with a value of 0.7. The noble gases have no tendency to attract electrons to themselvesand therefore have an electronegativity of “0.” This statement about the electronegativity of noblegases that I’ve just made is not something that all chemists agree upon. But still, we’re going to treat itthis way in this class. If you need to know the numerical of the electronegativity of a noble gas in thisclass, which you will in Chapter 4, use “0".

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Chapter 3: Ionic Compounds

Chemical bonds and chemical compounds

While there are substances in the world around us that consist of collections of individualatoms, most things are made up of molecules. Molecules consist of two or more atoms that havechemically bonded to each other.

A chemical bond occurs between two atoms. The bonding atoms may be the same typesof atoms or two atoms from different elements. Each atom donates one electron to the bond. Thebond is the result of the attraction between particles with opposite electrical charges. The typesof oppositely charged particles, the way the electrons are donated or shared, and the extent towhich they are shared is a function of the type of bond. The strength of the interactions betweentwo charged particles is described by Coulomb’s law, which is the following equation

in which F is the force of attraction or repulsion between two charged particles, k is a constant,q1 and q2 are the charges on the interacting particles, and r is the distance between the interactingparticles. Not to overstate the obvious, but this equation tells us that the force of attractionbetween two particles depends on the charges on the bonding particles. We use the word particle,rather than atom, because the equation describes the attractive and repulsive interactions between allcharged particles, not just atoms. While we will not use this equation quantitatively in other words,we’re not going to plug numbers into Coulomb’s law and solve it in this class you should understandthat it proves that as the magnitude of the charges on mutually-attracting particles increases, theforce of attraction also increases.

In other words: Let’s assume we’re trying to compare the strength of the interactionsbetween Na+ and Cl- as compared with the interactions between Ca2+ and O2-. As Na+ ispositively charged and Cl- is negatively charged, we know they will attract each other. The sameis true for Ca2+ and O2-. Let’s assume for a moment that these particles are all the same size.They’re not but if we make this assumption then the “r” term stays constant and it makes thiscomparison cleaner and simpler. If we use Coulomb’s law to calculate the force of attractionbetween the sodium and chloride particles, q1 would be the +1 charge on the sodium cation,while q2 would be the -1 charge on the chlorine anion. If we made a similar calculation for theforce of attraction between the calcium cation and the oxygen anion, q1 would be the +2 chargeon the calcium cation, while q2 would be the -2 charge on the oxygen anion. In each case, theattractive force between the ions depends on the product obtained from multiplying the ioncharges, q1 and q2. This means that, in relative terms, the force of attraction between the sodiumand chlorine ions is equal to ( +1 x -1 = -1), while the force of attraction between the calciumand chlorine ions is equal to ( +2 x -2 = -4). In other words, we can predict that the bonding

Fkq q

r 1 2

2

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(attractive) force between calcium and oxygen ions should be about 4 times greater than thebonding force between sodium and chlorine ions. What if we wanted to compare the strength ofthe bonding force between Al3+ and N3- to the bonding force between Na+ and Cl-. Assuming thedistance between the particles remains the same, how much stronger would the bonding force betweenthe aluminum cation and the nitrogen anion be? Why is “9 times stronger" the correct answer?

There are three types of chemical bonds. Each bond is found in a particular type ofcompound.

Metallic bonds occur between two metal atoms (M-M). The metal nuclei remain fixedwith a specific number of electrons around them proportional to the charge of the nuclei.However, the valence electrons are not fixed in place around the nuclei of the metal atoms butare free to move around them, or even to flow away under appropriate conditions, just as thewaters of a river flow around the islands that lie in mid-channel. The ratio of water to landaround the islands is fixed, even though the exact water molecules adjacent to the island changefrom one moment to the next. This model of metallic bonding is known as the "electron-seamodel." It is commonly offered as an analogy in which bonding metal nuclei are likened untoislands in an ocean, surrounded by a sea of electrons. In my opinion, the "electron-river" modelis a better analogy, but whatever works for you. The key thought in metallic bonding is that thevalence electrons involving in bonding between metal atoms have the ability to move from oneatom to another, provided valence electrons from other, adjacent metal atoms move into thesevacated positions.

Ionic bonds occur between ions. They are the types of bonds that form between metalcations and nonmetal anions (M-NM). The cations and anions may be either monatomic ions orpolyatomic ions. Monatomic ions are composed of one type of atom while polyatomic ions arecomposed of two more types of atoms. We will be discussing monatomic and polyatomic ions shortly.In ionic bonds between a metal atom and a nonmetal atom, the bonding electrons are not sharedequally. Instead, the nonmetal atom seizes the bonding electrons, making itself an anion, whilethe metal atom, having lost electrons, becomes a cation. This is not a perfectly correct explanationof what occurs but it is a good working description and will suffice for our purposes in this class.

Covalent bonds occur between two nonmetal atoms (NM-NM). In a covalent bond thetwo bonding electrons are shared more or less equally. This is a very good working generalization,but one that will require some future elaboration in Chapter 4.

As an essential reminder: elements are either metals or nonmetals, depending on their position in theperiodic table. For purposes of nomenclature metalloids do not exist. An element is either a metal or anonmetal. It’s that simple and there are no exceptions.

Being able to recognize the type of bond between atoms is a simple but essential skill,and for this reason: there are three common families of compounds, metallic compounds, ionic

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compounds, and covalent compounds. The family to which a compound belongs is determinedby the types of bonds between the atoms within the compound. This is relevant because compound names are assigned based on the compound family to which they belong.

We can determine the family to which a compound belongs simply by looking at itsmolecular formula. The molecular formula tells us the types of atoms found in molecules of thecompound. This in turn helps identify the types of bonds that occur between the atoms in thecompound. And it is the type of bonds between the atoms that determine the family of which thecompound is a member

Metallic compounds are compounds in which the bonding metal atoms are held togetherby metallic bonds. Pure metals and alloys (homogeneous metal mixtures such as brass, bronze,and 18 K gold, for example) are metallic compounds. We will not spend time discussing metalliccompounds further in this class, or how these compounds are named.

Ionic compounds are those in which the atoms are held together by ionic bonds. Thisfamily of compounds includes most of the rocks and minerals in the world around us, and allsalts. If ionic compounds dissolve in water they will also dissociate (ionize). The resultingsolutions of ionic compounds are called electrolyte solutions. The dissolved ions are capable ofcarrying charge and of conducting electricity. A few important clarifying notes here. Salt is not asingle chemical, but rather, a family of chemicals with hundreds of members. Salts are the ioniccompounds formed as products in neutralization reactions between acids and bases, a topic we’lldiscuss in chapter 5. All salts are ionic compounds, but not all ionic compounds are salts. Dissociation isthe process of ionizing, or if you’d rather, the process of a molecule breaking apart into ions. There is adifference between dissociation and dissolution, the process of dissolving, which is also known assolvation. We will elaborate on this in chapter 9.

Covalent compounds (sometimes known as molecular compounds) are those in which thebonding atoms are held together by covalent bonds. All organic compounds and biopolymers arecovalent compounds, as are all compounds that consist solely of nonmetal atoms. Polymers aremolecular chains; regularly repeating molecules are the links of the molecular chain. This includes all ofthe important large biological molecules such as proteins, carbohydrates, nucleic acids, and fats. Manymolecular compounds dissolve in water, but most of them that will dissolve do not dissociate.

An atom may satisfy the octet rule either through gaining or losing electrons in reactionsor through forming bonds. The alkali metals (1A) usually either form 1+ cations or one bond inchemical reactions. The alkali earth metals (2A) usually either form 2+ cations or two bonds inchemical reactions. The 3A elements usually either form 3+ cations or three bonds in chemicalreactions. The Group 4A elements usually form 4 bonds in chemical reactions. The Group 5Aelements usually either form 3- anions or three bonds in chemical reactions. The chalcogens (6A)usually either form 2- anions or a two bonds in chemical reactions. And the halogens (7A)

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usually either form 1- anions or a single bond in chemical reactions. (Based on what you learned inchapter 2 about valence electrons and electron configurations, why is none of this a surprise?)

You must keep in mind that much of what we have just said are generalizations. They are very goodworking generalizations, especially for this class, but there are many exceptions to thesegeneralizations. We will discuss a few exceptions, but they are mostly beyond the scope of this class. Itis not that we are trying to be untruthful as we teach you chemistry. It is that, in a class at this level,much of what you are taught is generalization. You would need to major in chemistry, and perhaps dograduate work as well, to begin to learn all of the exceptions and the reasons that they are exceptions.I hope this is not too upsetting to you. If it is, please let me know and we’ll talk about your decision tochange your major to chemistry.

Lewis electron dot structures (Lewis structures) for atoms

An important goal of chemistry is to understand why molecules behave the way they do,especially in biological systems. The actual three dimensional structure of molecules oftendetermines the behavior of molecules in chemical reactions. Although this is a first semesterchemistry class we can still learn how to predict the shapes of many small molecules and willlearn precisely how this is done in chapter 4.

Lewis electrons dot structures, also known as Lewis structures, are a first step in thisdirection. Lewis structures can be used to represent atoms, ions, and molecules. We will learn howto draw Lewis structures for atoms in this chapter and for ionic and covalent molecules in chapter 4.

Remember what you learned in chapter 2 about valence electrons. Valence electrons arethe reactive outer shell s and p electrons of atoms. Inner shell electrons are unreactive. Electronsin filled outer shell d and f subshells behave as inner shell electrons. In partially filled outer shelld subshells, the d electrons are both outer shell and valence electrons (transition metals). Mostoften we will only consider elements with filled d subshells. While this is also true about elementswith partially filled outer shell f subshells, we’re not going to worry about these elements and theirbonding behavior in this class.

The Lewis structures of atoms and ions consist of the elemental symbol surrounded byone dot for each valence electron of the substance. The Lewis structures of hydrogen and heliumare as follows:

The Lewis structures of the second period elements look like

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and the Lewis structures of the third period elements look nearly the same, except that theelement symbols differ

In some respects drawing a Lewis structure for an element is similar to writings it’s electronconfiguration, although with Lewis structures we’re interested exclusively with valenceelectrons. You may be wondering what the Lewis structures for elements with d subshell valenceelectrons look like. Don't bother worrying. We will only work with Lewis structures for elements with sand p valence electrons. You may also have noticed that your text does these Lewis structures in aslightly different way. They do not begin to pair electrons up until there is an electron on each of thefour sides of the element's symbol. Either way is acceptable and correct. What is ultimately important isnot where the dots are drawn with respect to the element symbol but that the correct number of dotsis drawn, i.e., that the correct number of valence electrons is represented.

Ionic compound nomenclature: naming ionic compounds

First, a definition. Nomenclature is a set of rules used in the naming of chemicalcompounds. To correctly name a compound we must first identify it as an ionic or a covalentcompound. This is because the rules for naming ionic and covalent compounds differ. We cannotcorrectly name ionic compounds using the rules of covalent nomenclature. And it is not possibleto correctly name covalent compounds using the rules of ionic nomenclature. We will not studymetallic compounds and their names in this class. Sorry, I know how disappointed you are.

The rules of nomenclature for all compounds have been established by IUPAC, theInternational Union of Pure and Applied Chemistry. However, chemistry is a very old science.Many of the compounds we will discuss have been known for decades, some for centuries. Manywere discovered before IUPAC was organized in 1919. As a consequence many compounds areknown by more than one name. We will focus on the systematic nomenclature advocated byIUPAC. It is this system that we will teach you and that you are required to learn for this class.Note: while there are usually several ways to name most compounds you are required to use the rulesoutlined in this chapter. No other names will be accepted.

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Ionic compounds: cation names

In the majority of the ionic compounds we’ll see in this class, the cations will be metalatoms that have lost electrons. All of the transition (d block) metals, the rare earth (f block)metals, and the p-block metals are theoretically capable of forming cations with two or moredifferent charge states. For example, iron commonly occurs as either Fe2+ or as Fe3+. Copper canoccur as Cu+ or as Cu2+. Chromium may be found as Cr2+, Cr3+, and Cr6+. And so on. As theyalmost never obey the octet rule, we cannot estimate the charge on these metal cations based ontheir position in the periodic table. This makes it necessary to convey charge information in thename of the cation. This is done by naming the element, including the cation charge in Romannumerals in parentheses, and by adding the word ion to the end of the name. As examples:

Fe2+ iron(II) ion

Fe3+ iron(III) ion

Cu+ copper(I) ion

Cu2+ copper(II) ion

Cr2+ chromium(II) ion

Cr3+ chromium(III) ion

Cr6+ chromium(VI) ion

To put this more plainly, to correctly name the metal cation in an ionic compound, we mustinclude the element name and the cation charge in Roman numerals in parentheses. For thepurposes of the chapter 3 quiz, it is essential to note that you must correctly spell the element name.The left parenthesis “(“ may directly follow the element name, or there may be a single space betweenthe element name and the left parenthesis although this latter method is much less popular than in thepast.

There are a small handful of metal cations that do not require the cation charge in Romannumerals as a part of their name. Group 1A and 2A metals and aluminum can only form one typeof cation, i.e.,a cation with one charge state. That is, all alkali metal cations are always 1+

cations, all alkali earth metal cations are always 2+ cations, and aluminum is always a 3+ cation.This is a generalization but an excellent one, especially for this class. The names of the 1A and 2Acations and of aluminum ion consists only of the element name plus the word ion. If you includethe charge information in Roman numerals for these elements on quizzes, exams, and MasteringChemistry, your answer will be marked wrong.

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Li+ lithium ion

Na+ sodium ion

K+ potassium ion

Rb+ rubidium ion

Cs+ cesium ion

Be2+ beryllium ion

Mg2+ magnesium ion

Ca2+ calcium ion

Sr2+ strontium ion

Ba2+ barium ion

Al3+ aluminum ion

Bear in mind that these are exceptions. For any of the other p-block, d-block, or f-block metals,if you neglect to include the cation charge in Roman numerals as part of its name, you haveincorrectly named the cation.

Ionic compounds: monatomic and polyatomic anion names

Anion names depend on the type of anion. As we mentioned above there are two types ofanions, monatomic anions and polyatomic anions.

Monatomic anion are anions which consist of only one type of atom. Each of thenonmetals is found as a monatomic anion in the natural world. The names of monatomic anionsconsist of the element name with its ending changed to "ide."

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neutral element name anion name symbol

fluorine fluoride F-

chlorine chloride Cl-

bromine bromide Br-

iodine iodide I-

oxygen oxide O2-

sulfur sulfide S2-

selenium selenide Se2-

tellurium telluride Te2-

nitrogen nitride N3-

phosphorus phosphide P3-

arsenic arsenide As3-

carbon carbide C4-

silicon silicide Si4-

boron boride doesn’t follow the rules

It is possible for an ionic compound to contain more than one monatomic anion. Toillustrate this, consider sodium chloride (NaCl), calcium chloride (CaCl2), aluminum chloride(AlCl3), titanium(IV) chloride (TiCl4), niobium(V) chloride (NbCl5), and tungsten(VI) chloride(WCl6). All of these contain M-NM bonds and are therefore ionic compounds. In each of thesecases chloride is a monatomic anion. It is true that for five of these compounds there is more thanone chloride anion, but since they are all the same type of anion, chloride, this makes chloride amonatomic anion. It is essential to note that these do not occur as clusters of anions. Forexample, in calcium chloride we do not have a Ca2+ cation and a Cl2

2- anion. Rather, we have aCa2+ cation and a two Cl- anions. In the case of (AlCl3) we have an Al3+ cation and three Cl-

anions, titanium(IV) chloride (TiCl4) contains a Ti4+ cations and four Cl- anions, niobium(V)chloride (NbCl5) has a Nb5+ cation and five Cl- anions, and there is a W6+ cation and six Cl-

anions found in tungsten(VI) chloride (WCl6).

So far we've only discussed ions consisting of a single type of atom. Many commonimportant ions are made up of two or more types of atoms. These are called polyatomic ions.While there are many thousands of polyatomic ions in the real world, there are two commonpolyatomic cations and 15 common polyatomic anions that you must know for this class.

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You must know the names, formulas, and charges of these polyatomic ions as soon as possible. Isuggest making flash cards and then drill yourself night and day until you know all of them by heart.The odds are that you may wind up repeating this class if you do not know them. Do yourself a favorand learn them now. Even though this is information you may include on the 3" x 5" card you bringwith you to exams, the better you know these polyatomic ions the more quickly and accurately you willbe able to solve problems in this class.

These common polyatomic cations and anions you must know are:

Hg22+ mercury (I) ion

NH4+ ammonium ion

SO42- sulfate ion

SO32- sulfite ion

NO3- nitrate ion

NO2- nitrite ion

OH- hydroxide ion

CO32- carbonate ion

HCO3- hydrogen carbonate (or bicarbonate) ion

CrO42- chromate ion

MnO4- permanganate ion

C2H3O2- acetate* ion

PO43- phosphate ion

CN- cyanide ion

ClO4- perchlorate ion

ClO3- chlorate ion

ClO2- chlorite ion

ClO- hypochlorite ion

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Let’s begin by taking a quick look at the two polyatomic cations. Ammonium ion isextremely common in the world around us. The mercury (I) ion consist of two mercury (I)cations that are chemically bonded to each other. Do not confuse the mercury (I) ion with themercury (II) ion (Hg2+), which consists of a single mercury cation with a 2+ charge.

Please observe that the HCO3- ion is known by two names and is correctly referred to as

either “bicarbonate” or as “hydrogen carbonate”. This is the lone exception to our rule about organiccompounds being those with molecular formulas that contain both C and H. We mentioned it back inchapter 1.

*Note: many texts, including the one for this course, will tell you that the formula for acetate ion isCH3COO-. This is not a molecular formula. It is a condensed structural formula, something we willdiscuss briefly in Chapter 4. It certainly is not wrong, but it conveys more information than we need inour study of nomenclature. As such, it will not be accepted when you take the online quizzes. I expectyou to use C2H3O2

- to represent acetate, and if you would like credit for your work you will heed thisadvice. However, be careful when doing Mastering Chemistry as its authors seem to prefer students touse the condensed structural formula.

It is worth mentioning that most of the polyatomic anions in our list are oxyanions,anions that contain one or more oxygen atoms. Oxyanion is a term you should know.

As I mentioned, this is not an exhaustive list of polyatomic ions. Many that are common in thereal world have not been included. But this is a good starting point, and I do hold youaccountable for all of these ions in this list for this class. There are a variety of ways to learnthese. I suggest you make a flash card for each of these polyatomic ions with the name on oneside and its structure on the other.

A digression: old ionic compound nomenclature systems

Before we proceed further into using IUPAC rules, I want to spend a moment discussingthe naming of ionic compounds in the past. Chemistry is an old science. Various approaches tonaming ionic compounds have been used during the several hundred year history of modernchemistry. Because of these older nomenclature systems, there are some compounds with asmany as 5-10 or more names. While you only need to know the IUPAC system, I will digress fora moment to explain some of the non-IUPAC names for ionic compounds you may see as youstudy chemistry.

IUPAC nomenclature replaced the Stock system for naming ionic compounds. You donot need to know the Stock system for this class. However, since you will almost certainlyencounter cations and anions that have been named using the system, you should be aware thatmany compounds are still known by their Stock system names as well as by their systematic(IUPAC) names.

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The Stock system used the suffixes "ic" and "ous" to indicate which of a pair of cationswas the higher and lower charged of the pair. The “ic” suffix was used with the higher chargedcation while the “ous” suffix was used with the lower charged cation.

Fe2+ iron (II) ion ferrous ion

Fe3+ iron (III) ion ferric ion

Cu+ copper (I) ion cuprous ion

Cu2+ copper (II) ion cupric ion

There is no intrinsic information about cation charge in this type of nomenclature. Rather, thesystem requires you to know beforehand something about the cations in question and providesrelative information about cation charges.

The naming of oxyanions utilized the suffixes "ate" and "ite" to differentiate between apair of oxyanions with different numbers of oxygen atoms. The oxyanion with more oxygenatoms bonded to the central atom (the first named atom in the molecular formula) received the“ate” suffix while the oxyanion with fewer oxygen atoms bonded to the central atom received the“ite” suffix.

SO42- sulfate

SO32- sulfite

NO3- nitrate

NO2- nitrite

Again, the problem this presents to those learning chemistry is that the “ate” and “ite” suffixesdo not convey absolute information about the number of oxygen atoms bonded to the centralatom. It is presupposed that you know more than you do at this point.

Those series of oxyanions with the same central atom (e.g., Cl in perchlorate) but withfour different oxyanions used prefixes in combination with the suffixes. "Per" and "ate" indicatedwhich of the four had the most oxygen atoms, "ate" the second most, "ite" the third most, and"hypo" and "ite" for the oxyanion with the fewest oxygen atoms of the four polyatomic anionswith the same central atom.

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ClO4- perchlorate

ClO3- chlorate

ClO2- chlorite

ClO- hypochlorite

Again, remember: you do not need to know how to use the Stock system. This is simply abit of background material that some people find helpful. If it confuses you, just ignore the lastfew paragraphs.

Note: all of the polyatomic anions I require you to know were named before IUPAC existed, and insome cases perhaps before the Stock system was in use. I do not expect you to know why the SO4

2-

anion is called sulfate ion. Very simply, I expect you to know that SO42- anion is called sulfate ion. This

is also true of the other monatomic and polyatomic anions in the lists above. Don’t worry about whythese ions have the names they have, or where these names come from. Quite simply, know the namesand the corresponding molecular formulas without worrying about where they come from, or whichsystem was used to name them.

Note: Stock system names are not acceptable answers on quizzes and exams for this class. As anexample, were I to ask you the correct name of CuCl on a quiz or exam, the IUPAC answer is copper(I)chloride. Cuprous chloride would not be acceptable. Mastering Chemistry may ask you a shortquestions about Stock system names, but I’ve tried to eliminate as many of these questions from yourhomework as possible.

It should also be pointed out that, in the past, it was acceptable to use prefixes to indicatehow many of a particular type of ion are found in a molecule of ionic compound. This is nolonger true with IUPAC rules. As examples, the compounds TiCl4, TiO2, and UF6 were knownhistorically as titanium tetrachloride, titanium dioxide, and uranium hexafluoride. The properIUPAC names of these three compounds are titanium(IV) chloride, titanium(IV) oxide, anduranium(VI) fluoride. It is never acceptable to use prefixes to indicate the number of atoms inthe IUPAC names of ionic compounds.

The names and molecular formulas of ionic compounds

Ionic compounds contain ionic bonds. Ionic bonds are usually found between metals andnonmetals, or when the name or molecular formula of a compound contains one or morepolyatomic ions. As examples, we know FeCl2 is an ionic compound because it’s molecularformula contains metal and nonmetal atoms. We know sodium permanganate is an ioniccompound, even without the compound’s molecular formula, because permanganate is apolyatomic ion, hence, the entire compound is an ionic compound. Even without a name we

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know (NH4)2S is an ionic compound because the compound’s molecular formula contains apolyatomic ion. This is obvious because the only time it is appropriate to use parentheses in themolecular formulas of compounds is to indicate the presence of more than one polyatomic ion.

In both the names and the molecular formulas of ionic compounds the cation is namedfirst and the anion is named second. There are no exceptions to this rule in IUPAC nomenclaturefor this class. As an example, we know the substance sodium chloride, NaCl, is an ioniccompound because one of the bonding atoms is a metal (Na) and the other bonding atom is anonmetal (Cl). Where do we look to establish whether an element is a metal or a nonmetal? Withoutknowing anything else about sodium chloride other than that it is an ionic compound, we knowthat sodium is a cation and that chloride is an anion because of the rules of ionic compoundnomenclature, i.e., sodium is listed first in both the name and the molecular formula so it must bea cation. Chlorine is listed second in both the name and the molecular formula, it must thereforebe an anion.

The following are a few examples of molecular formulas of ionic compounds and theircorresponding names:

• KBr: potassium bromide• Ca(NO3)2: calcium nitrate• (NH4)2S: ammonium sulfide• FeSO4: iron (II) sulfate• Zn3(PO4)2: zinc (II) phosphate

At this point and for the remainder of this chapter, as we got through this material you should have aperiodic table, a list of the monatomic and polyatomic ion, a pencil, and some paper besides you at alltimes. Don’t rush through this material! This is an excellent opportunity to practice, which is the onlyway you’re going to learn. If you go through the remainder of this chapter simply reading, withouttaking a more active role in the learning process, you’re doing yourself a tremendous disservice.

Here is a list of names of a few ionic compound, and their corresponding molecularformulas:

• gold (III) cyanide: Au(CN)3

• mercury (II) acetate: Hg(C2H3O2)2

• titanium (IV) oxide: TiO2

• silver (I) perchlorate: AgClO4

• sodium phosphate: Na3PO4

To reiterate something I mentioned above: parentheses are only used in molecularformulas when there is more than one polyatomic cation or anion in the compound. For example,KNO3 is correct, while K(NO3) is incorrect. Why? Because there’s only one polyatomic ion inthe molecular formula of this compound. Na3PO4 is correct, but (Na)3PO4 is incorrect. We never

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use parentheses for monatomic anions or - in this case - monatomic cations. The molecularformula NH4NO3 is correct, while (NH4)(NO3) is wrong. We never use parentheses if there is justone polyatomic cation or anion in the molecular formula of an ionic compound. In this case thereis one of each, but as there is only one polyatomic cation in the compound, and only onepolyatomic anion in the compound, the use of parentheses is inappropriate.

You're probably wondering how we know how many cations and anions go together in aparticular compound. We'll discuss this below.

Warning: the rules of nomenclature are very particular. If, in naming a compound, you forget any of thelittle things we have discussed or will be discussing in this chapter, your work is incorrect and will bemarked as such. You must practice, practice, practice! To help you practice, you should work all of theexamples we discuss in these lecture notes, all of the required problems in Mastering Chemistry and inthis chapter quiz, and you should also feel free to explore the problems at the end of chapter 3 in thetext to help you practice. I’ve also created an “Ionic Compound Nomenclature Self-Quiz” found at theSLCC-Science website to help you practice putting cations and anions together in their proper ratios. Ifyou click on a compound’s molecular formula, a Javascript box will open tell you the compound’scorrect name. You’ll find this at self-quiz athttp://www.slcc-science.org/chem/giddings/chem1110int/nomenclature/nomenclature.htm . While youshould feel free to print this table if you’d like, you’ll lose the ability to click on molecular formulas andsee compound names on the printed version.

How do we know how many cations and how many anions go together to form an ioniccompound? Let's look again at the examples of names and molecular formulas we listed above:

• KBr: potassium bromide• Ca(NO3)2: calcium nitrate• (NH4)2S: ammonium sulfide• FeSO4: iron (II) sulfate• Zn3(PO4)2: zinc (II) phosphate• gold (III) cyanide: Au(CN)3

• mercury (II) acetate: Hg(C2H3O2)2

• titanium (IV) oxide: TiO2

• silver (I) perchlorate: AgClO4

• sodium phosphate: Na3PO4

Are any of these compounds charged? The answer is no. Charged compounds have a superscriptto the right of the molecular formula indicating whether charge is positive or negative and towhat extent. As an example, CuCl4

2- has a 2- superscript, which indicates that the combination ofthis particular copper cation and the four chloride anions results in a compound with a net chargeof 2-. It is not electrically neutral. All ten of the compounds in our examples electrically neutral,i.e., are not charged, which is another way of saying they have no net charge. To many this

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seems to be a paradox. How is it that ionic compounds can be made up of charged particles andyet remain electrically neutral? (As an aside, there really is a CuCl4

2- polyatomic anion. However, it’snot on the list of polyatomic anions you’re required to know, so we’re not going to worry about it. I’monly using it as an example.)

For ionic compounds to be made up of charged particles and yet remain electricallyneutral, there must be a balance between the positive charge of the cations and the negativecharge of the anions. In other words, all of the electrons lost by all of the cations in an ioniccompound must be offset by all of the electrons gained by all of the anions in the same ioniccompound. We will not worry about where the electrons lost by the cations go, or where theelectrons gained by the anions come from. Often we not do not know, but no matter. We simplydon’t care. Our only concern in this class is: the molecular formulas we write for ioniccompounds must be electrically neutral.

Ionic compounds: going from names to molecular formulas

If we are given the name of an ionic compound and asked to write its molecular formula,how can we determine how many cations and how many anions are required to form a neutralcompound? The process can be stated in a series of four steps.

1. Ask yourself this question: is the compound ionic or covalent? If it’s covalent, don’t usethese rules. We’ll discuss covalent compounds below.

2. If the compound is ionic, write the symbols for the cation and the anion beneath the nameof the compound.

3. Find the lowest common denominator of the charges. Use this to determine how manycations and anions are needed to form a neutral compound.

4. Write the formula of the neutral compound.

The ratio of cations to anions in a neutral ionic compound is determined by the ratio oftheir charges. Take potassium bromide, which is made up of a potassium ion (K+) and a bromideion (Br-). The lowest common denominator (LCD) of the charges is often - but not always -found by multiplying them together.

(+1) x (-1) = |1|

I consistently use the phrase “lowest common denominator” in these notes but technically speaking itwould be more correct to use the phrase “least common multiple.” By definition, the least commonmultiple of two numbers, "a" and "b," is "the smallest positive integer that is a multiple both of a andof b. Since it is a multiple, it can be divided by a and b without a remainder." (See "Least commonmultiple." Wikipedia, The Free Encyclopedia. http://en.wikipedia.org/wiki/Least_common_multiple). This means that multiplying the charges may not always be the correct way to find the lowest common

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denominator in the problems in this chapter, as we will shortly see. Also bear in mind that finding thelowest common denominator of two numbers if a skill you are required to have learned in theprerequisite math classes for this course. If you don’t remember how, it’s time for you to do a bit ofreview! The lines on either side of |1| indicate we’re talking about the absolute value of thenumber and are not interested in its sign, only the number. In this example, how many cationsare needed to give us a total charge of 1+? How many anions are needed to give us a total chargeof 1-? This means that K+ and Br- go together in a 1:1 ratio.

Let's look at the rest of these examples. The way to do this is to write the symbols for thecation and the anion in the compound first, calculate the lowest common denominator of thecharges, and then write the molecular formula.

compound cation anion LCD cationcharge xLCD = ?

anioncharge xLCD = ?

formula(based on

cation:anionratio)

potassium bromide K+ Br- 1+ x 1- = |1| 1 x 1+ = 1 1 x 1- = 1 KBr

calcium nitrate Ca2+ NO3- 2+ x 1- = |2| 1 x 2+ = 2 2 x 1- = 2 Ca(NO3)2

ammonium sulfide NH4+ S2- 1+ x 2- = |2| 2 x 1+ = 2 1 x 2- = 2 (NH4)2S

iron(II) sulfate Fe2+ SO42- 2+ x 2- = |2| 1 x 2+ = 2 1 x 2- = 2 FeSO4

zinc(II) phosphate Zn2+ PO43- 2+ x 3- = |6| 3 x 2+ = 6 2 x 3-= 6 Zn3(PO4)2:

gold(III) cyanide Au3+ CN- 3+ x 1- = |3| 1 x 3+ = 3 3 x 1- = 3 Au(CN)3

mercury(II) acetate Hg2+ C2H3O2- 2+ x 1- = |2| 1 x 2+ = 2 2 x 1- = 2 Hg(C2H3O2)2

titanium(IV) oxide Ti4+ O2- 4+ x 2- = |4| 1 x 4+ = 4 2 x 2- = 4 TiO2

silver(I) perchlorate Ag+ ClO4- 1+ x 1- = |1| 1 x 1+ = 1 1 x 1- = 1 AgClO4

sodium phosphate Na+ PO43- 1+ x 3- = |3| 3 x 1+ = 3 1 x 3- = 3 Na3PO4

Let’s be sure we’re clear as to what’s going on in this table.

• The first three columns are hopefully self-evident, although you will need to know theinformation discussed previously in this chapter to be able to correctly associate ionnames and symbols.

• We start with the compound name. Based on the name we write the symbol for thecation, including it’s charge. We do the same for the anion.

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• It is essential to remember any time we have a Roman numeral in parentheses in thename of a compound, it always indicates two things. It guarantees we’re dealing with anionic compound, because we never use Roman numerals in parentheses in the names ofcovalent compounds. Also, any time you see a Roman numeral in parentheses it tells youthat the element to the left is a cation, and the Roman numeral tells you exactly howmany electrons that cation has lost. So in iron(II) phosphate, we know we have an ioniccompound in which there is an iron cation that has lost two electrons. Given the namegold(III) cyanide, we know we have an ionic compound in which we have a gold cationthat has lost three electrons. The compound titanium(IV) oxide is an ionic compoundwith a Ti4+ cation. And so on.

• In the LCD (lowest common denominator) column we’re using the cation charge and theanion charge to determine the lowest common denominator of the charges. We may ormay not actually be multiplying the two charges to find the lowest common denominator.In the case of iron (II) sulfate, obviously 2 x -2 does not equal 4. The lowest commondenominator of 2+ and 2- is 2, not 4.

• In the “cation charge x LCD = ?” column we ask and answer the question: what numbermultiplied by the cation charge will equal the lowest common denominator? In the caseof zinc (II) phosphate we see “3 x 2+ = 6" in the column. By working backwards wedetermine that we must multiply the cation’s “2+ ” charge by a factor of “3" to equal “6",the lowest common denominator.

• We do exactly the same thing in the “anion charge x LCD = ?” column. We ask andanswer the question: what number multiplied by the anion charge will equal the lowestcommon denominator? In the case of zinc (II) phosphate, given that phosphate has a 3-charge, we see “2 x 3- = 6" in the column. By working backwards we determine that wemust multiply the “3- ” charge by a factor of “2" to equal “6", the lowest commondenominator.

• In the “formula (based on cation:anion ratio)” column we combine the numberscalculated in the two previous columns. In the “cation charge x LCD = ?” column wedetermined we needed 3 cations for zinc (II) phosphate, while in the “anion charge xLCD = ?” column we calculated that it would require 2 phosphate anions. This tells usthat Zn2+ and PO4

3- must go together in a 3:2 ratio to give us the correct molecularformula. Zn3(PO4)2.

Yup, I’m going to nag you again. You can’t just stare at this stuff and have it make sense. You need toroll up your sleeves, get out a pencil and some paper, and work through these examples. Do it now! Ifyou don’t work through lots and lots of examples you may never figure it out.

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Ionic compounds: going from molecular formulas to names

You must also be able to state the name of an ionic compound when given its molecularformula. This too can be thought of as a process consisting of four steps.

• Ask yourself this question: is the compound ionic or covalent? If it’s covalent, don’t usethese rules. We’ll discuss covalent compounds below.

• If ionic, write the symbols for the cation and the anion beneath the formula of thecompound.

• Name the cation and anion.• Write the name of the neutral compound.

In many cases, the charge on the cation will not be immediately apparent since mostcations are capable of two or more different charge states. How do we determine the charge onthe cation? We use the anion - the anion charge and how many of them there are - to help usfigure out the cation charge. Anion charges are fixed. They never vary. This makes it possible toalways determine the charge on the cation in any ionic compound.

Let’s work a few examples and see how we do this. We’ll begin by naming thecompound FePO4.

1. Is this an ionic compound? How do we know? It has a polyatomic ion in its molecular formula.Any compound in this class with a polyatomic ion in its molecular formula is an ionic compound.

2. Write the symbols for the cation and the anion beneath the formula of the compound.

There are several important things to note. First, as iron (Fe) is a transition metal we can’tguess its charge from its position in the periodic table. We’re going to have to use the anioninformation in the molecular formula to help us determine the cation charge. Second, we have asingle PO4

3- group in the molecular formula of this compound. Remember, the subscripted “4"tells us that this polyatomic anion has four oxygen atoms in it, not that there are four polyatomicions. (Any time we have more than one polyatomic ion in the molecular formula of a compound itmust be enclosed in parentheses with a subscript to indicate how many of the polyatomic ions thereare in the compound. To have four of these groups in the example, the molecular formula would haveto be written Fe(PO4)4 and it’s not. We only have one polyatomic anion in the molecular formula of thiscompound. According to its molecular formula the polyatomic anion contains one phosphorus atom,four oxygen atoms, and three extra electrons.)

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We have one PO43- group, which has 3 extra electrons. We know this because we find it on

the list of polyatomic ions you were given earlier in the chapter. Do you have that list in front of youright now, as you’re reading this problem? If you don’t, why don’t you? It’s a waste of your time to try tostudy this material if you’re not properly prepared. We know this from the “3-“ charge on the anion.If we multiply the number of anions in the molecular formula by the number of extra electronsper anion, we calculate the total number of extra electrons the anions have brought to thecompound.

According to the molecular formula, we have one polyatomic ion with a 3- charge. Therefore,the total electrons brought to this compound by the anion is (1 x 3- = 3-).

Because the compound is electrically neutral (there is no net charge shown in themolecular formula for the compound), the number of electrons lost by the cations must equal thenumber of extra electrons gained by the anions. Note: this will always be true in this class. Thenumber of electrons lost by the cations will equal the number of electrons gained by the anions.Without exception. Ever.

So the cations have lost a total of 3 electrons. There is only one cation in the molecular formulaof this compound. By using a bit of simple arithmetic we find

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that the charge on the cation must be 3+.

3. Name the cation and anion. The Fe3+ cation is the iron(III) ion. The PO43- anion is the

phosphate ion.

4. To write the name of the neutral compound we combine the name of the cation with the nameof the anion, dropping the word “ion” from both the cation name and the anion name. So thecompound FePO4 is named iron(III) phosphate.

Let’s try another one, this time providing a name for UO3.

1. Is this an ionic compound? How do we know? As always, take a look at the periodic table. In themolecular formula of this compound we see that we have metal bonding with a nonmetal. It must bean ionic compound.

2. Write the symbols for the cation and the anion beneath the formula of the compound.

As “U” is a rare earth metal we can not deduce its charge from its position on the periodic table.But we can use the anions to help us figure out the charge on the cation. We have three O2-

groups, each of which has 2 extra electrons. We know this from the “2-“ charge on the anion. Ifwe multiply the number of anions by the number of extra electrons per anion, we calculate thetotal number of extra electrons the anions have brought to the compound.

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Because the compound is electrically neutral (there is no net charge shown in the molecularformula for the compound), the number of electrons lost by the cations must equal the number ofextra electrons gained by the anions.

So the cations have lost a total of 6 electrons. There is only one cation in the molecular formulaof this compound. By using a bit of simple arithmetic we find

that the charge on the cation must be 6+.

3. Name the cation and anion. The U6+ cation is the uranium(VI) ion. The O2- anion is the oxideion.

4. To write the name of the neutral compound we combine the name of the cation with the nameof the anion, dropping the word “ion.” So the compound UO3 is uranium(VI) oxide.

Remember what we said earlier in the chapter, we never have clusters of monatomic anions. In thiscompound we have three O2- anions, not one O3

6- anion. One molecule of UO3 contains one uraniumatom and three oxygen atoms.

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Let’s work through one more example and come up with the name for Sc2(CO3)3.

1. Is this an ionic compound? How do we know?

2. Write the symbols for the cation and the anion beneath the formula of the compound.

As “Sc” is a transition metal we can not deduce its charge from its position on the periodic table.But we can use the anions to help us figure out the charge on the cation. We have three CO3

2-

groups, each of which has 2 extra electrons. We know this from the “2-“ charge on the anion.And remember: the subscripted “3" immediately next to the O in the molecular formula tells usthat there are 3 oxygen atoms per carbon atom, while the subscripted “3" outside the parenthesesis the value that tells us we have three anions. If we multiply the number of anions by thenumber of extra electrons per anion, we calculate the total number of extra electrons the anionshave brought to the compound.

Because the compound is electrically neutral (there is no net charge shown in the molecularformula for the compound), the number of electrons lost by the cations must equal the number ofextra electrons gained by the anions.

So the cations have lost a total of 6 electrons. There are two cations in the molecular formula ofthis compound. By using a bit of simple arithmetic we find

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that the charge on the cation must be 3+.

3. Name the cation and anion. The Sc3+ cation is the scandium(III) ion. The CO32- anion is the

carbonate ion.

4. To write the name of the neutral compound we combine the name of the cation with the nameof the anion, dropping the word “ion.” So the compound Sc2(CO3)3 is scandium (III) carbonate.

Just as we never have clusters of monatomic anions, we never have clusters of metals cations (the onlyexception ever being the mercury(I) cation we mentioned earlier in this chapter). In this compound wehave two Sc3+ cations, not one Sc2

6+ cation. One molecule of scandium(III) carbonate contains twoscandium atoms, three carbon atoms, and nine oxygen atoms.

These examples will hopefully give you a much better idea of why it is so important toknow the names and charges of all of the polyatomic ions and how to find the charges of themonatomic anions based on their location in the periodic table. But to really learn and retain theseconcepts you must practice, practice, practice!

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The nomenclature of binary molecular compounds

Binary molecular compounds contain covalent bonds and are made up of two types ofnonmetal atoms. The order in which elements are named in binary molecular compounds isestablished by convention. It starts at the left of the nonmetal elements and works it way fromleft to right in periods, and from bottom to top in groups, with the exceptions of oxygen andhydrogen which fall out of order: B-Si-C-Sb-As-P-N-H-Te-Se-S-I-Br-Cl-O-F. This order isbased on the electronegativity of the elements. You do not need to know this order. It is presentedonly so that you understand that there is in fact a rhyme and reason to the order in which elements arelisted in molecular compounds. Having said this, the order of the elements in the names and themolecular formulas is consistent. As you will always be given one or the other, you will always knowwhich order should be followed.

The compound name states the elements in the same order as the molecular formula andvisa versa. In other words, if the name of a compound is carbon monoxide then the elements arelisted in that same order in the molecular formula, CO, not OC. If the molecular formula of acompound is NO2 then the name of the compound is nitrogen dioxide, not dioxygen nitride.

The name of the first element is stated as the first element's exact name. The name of thesecond element ends in the "ide" suffix. Prefixes are used to indicate the number of each type ofatom in the compound.

Take care! Remember, these are covalent compounds. None of the atoms in covalent compounds arecharged, i.e., there are no cations or anions in molecular compounds. This does mean that the suffix“ide” can mean one of two different things, depending on context. In the names of ionic compounds itindicates that the substance is a monatomic anion (except for hydroxide and cyanide). In the names ofcovalent compounds it simply means that element is the second named element, nothing more.

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1 mono

2 di

3 tri

4 tetra

5 penta

6 hexa

7 hepta

8 octa

9 nona

10 deca

There are two exceptions to the use of these prefixes when naming binary molecular compounds:

Exception 1: if the first element is by itself, "mono" is not used e.g. nitrogen dioxide (NO2), notmononitrogen dioxide. CO is carbon monoxide, not monocarbon monoxide.

Exception 2: chop "o"/"a" from the end of the prefix if the element name is oxygen, e.g. carbonmonoxide (CO), not carbon monooxide. N2O4 is dinitrogen tetroxide, not dinitrogen tetraoxide.

Some illustrations of the names of binary molecular compounds and their correspondingmolecular formulas are as follows.

• XeF6 : xenon hexafluoride• ICl5 : iodine pentachloride• N4S4 : tetranitrogen tetrasulfide• P2O5: diphophorus pentoxide• N2O: dinitrogen monoxide

It surprises many people to learn that noble gas compounds do in fact exist. Several hundred havebeen synthesized; I do not think that many, if any, are naturally occurring and if they do, they are rare.While their chemistry is interesting, we will not discuss them in any detail in this class although we willencounter a few more noble gas compounds in Chapter 4.

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A few examples of molecular formulas and their corresponding names:

• iodine heptafluoride: IF7

• dinitrogen pentoxide: N2O5

• tetraphosphorus decoxide: P4O10

• krypton difluoride: KrF2

• nitrogen monoxide: NO

You've probably noticed that the names of binary molecular compounds are easier than those ioniccompounds. Do not let this keep you from learning the rules and putting them into practice.

As I mentioned above, for many years the prefixes that are now only correctly used toname molecular compounds were also used in the naming of ionic compounds. For example,during World War II and the development of the atomic bomb, a key step in the processing ofuranium used in the bombs involved a compound known then as uranium hexafluoride, UF6.This compound is now called uranium(VI) fluoride. There are many other examples. Remember:the practice of using prefixes to indicate numbers of cations and anions in ionic compounds isnot acceptable as IUPAC nomenclature and is therefore also not acceptable in this class.

The nomenclature of acids and bases

There are three common acid-base theories. Each provides a slightly different butnonetheless distinct definition of what acids and bases are. We will examine these in greaterdetail in Chapter 10 but for the time being we will confine ourselves to the Arrhenius theorywhich defines acids as substances that can donate a hydrogen ion (H+), and bases as compoundsthat can donate a hydroxide ion (OH-). The molecular formulas of acids are usually written withhydrogen listed first, e.g., HCl, HBr, HC2H3O2 and so on. In this course the names of acids willalways include the word "acid." Some common acids include:

carbonic acid H2CO3

acetic acid HC2H3O2

nitric acid HNO3

nitrous acid HNO2

sulfuric acid H2SO4

sulfurous acid H2SO3

phosphoric acid H3PO4

hypochlorous acid HOCl

chlorous acid HClO2

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chloric acid HClO3

perchloric acid HClO4

hydrofluoric acid HF

hydrochloric acid HCl

hydrobromic acid HBr

hydroiodic acid HI

The names of acids are common names but are still accepted by IUPAC. Note that the acidsnamed above are all covalent compounds, not ionic compounds. This is a bit deceptive in that,while they consist entirely of nonmetal atoms, they must behave to some extent as ioniccompounds to act as acids. Most of them are also oxyacids, i.e., they are acids which contain oneor more oxygen atoms.

The common bases are the Group 1A and 2A hydroxides, ammonia, and ammoniumhydroxide.

lithium hydroxide LiOH

sodium hydroxide NaOH

potassium hydroxide KOH

rubidium hydroxide RbOH

cesium hydroxide CsOH

beryllium hydroxide Be(OH)2

magnesiumhydroxide

Mg(OH)2

calcium hydroxide Ca(OH)2

strontium hydroxide Sr(OH)2

barium hydroxide Ba(OH)2

ammonium hydroxide NH4OH

It is important to note that not every substance that contains hydroxide can behave as a base.There are many ionic compounds which contain hydroxide ion but which do not behave as bases.Iron (III) hydroxide, Fe(OH)3, is only one of many dozens of examples.

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Ammonia has the molecular formula NH3. Its ability to act as a base is not obvious asthere is no hydroxide ion in its molecular formula. The bases we have listed in the table aboveare Arrhenius bases. Ammonia is a Brønsted-Lowry base, not an Arrhenius base. You do not atpresent need to understand these distinctions. They will be made more clear in Chapter 10.

Again, you must memorize the names and structures of these compounds if you have any seriousexpectations of doing well in this class. I really do try to keep the amount of memorization I require ofyou to a minimum, but it is not completely unavoidable. And if you refuse to learn these concepts andnames or cannot learn them, your grade will suffer accordingly. This is not a threat. I’m trying to helpyou understand the nature of the beast. You must master the rules of nomenclature in this chapter or itwill continue to haunt you through out the remainder of the semester.

Common nomenclature problems - some troubleshooting help

If you’re having trouble on the chapter quiz or Mastering Chemistry, the following is a brief listof some of the more common sources of frustration. All of these are mentioned in the chapterabove, but are easy to overlook or forget.

1. When naming the metal cations in ionic compounds, you must include the elementname and the cation charge in Roman numerals in parentheses. There are only a small number ofmetals that do not require the Roman numeral/charge information as part of the name? Whichmetals are they, and why are they exceptions? Is it wrong to include Roman numeral/chargeinformation in the names of these specific cations?

2. It is never acceptable to use prefixes to indicate the number of atoms/polyatomic ions in theIUPAC names of ionic compounds. Prefixes can only be used in the names of covalentcompounds.

3. It's easy to confuse the atomic symbols for some of the elements. Be dead certain you're usingthe correct elemental symbols when you write the molecular formulas for compounds.

4. When given an ionic compound's molecular formula, it's easy to miscalculate charge.Sometimes students calculate the correct charge but then inadvertently used the wrong set ofRoman numerals to indicate cation charge. Either way, this can cause trouble.

5. Consistent success depends on practice. In addition to the lecture notes, your text and it'sexamples, and Mastering chemistry, I've provided an "Ionic Compound NomenclatureSelf-Quiz" I created for this class. You'll find the link to it on the slcc-science home page, or athttp://www.slcc-science.org/chem/giddings/chem1110int/nomenclature/nomenclature.htm(Links to an external site.) . This consists of rows of cations and columns of anons. You canpractice putting various cations and anions together in their proper ratios, based on charge. If youclick on the molecular formula for any one of the compounds in the matrix, it will tell you thecompound's name; this provides you with naming practice.

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Chapter 4: Molecular Compounds

Covalent bonds and molecular compounds

Covalent bonds are the typical bond formed when both bonding atoms are nonmetals(NM-NM). Typically each atom donates one electron to the bond. The bonding electrons areshared more or less equally. Atoms participating in covalent bonds gain an octet of electronsthrough sharing electrons, i.e. through forming chemical bonds, rather than through ionizationand gaining or losing electrons. Covalent bonds result from the overlap of orbitals. Atoms mustbe close enough to attract each other and share electrons (through the overlap of the orbitals inwhich the bonding electrons are found), but not so close that the positively-charged nuclei of thebonding atoms repel each other.

There is a strong correlation between the length of a covalent bond (i.e., the distancebetween the bonding atoms) and the strength of the bond. Remember Coulomb’s law, theequation that describes the strength of electrostatic interactions as presented in Chapter 4:

The closer bonding atoms are to each other (i.e., the shorter the bond length), the smaller thevalue of "r" and the stronger the attractive force between them. The further apart bonding atomsare from each other (i.e., the greater the bond length), the larger the value of "r" and the weakerthe attractive force between them. Not all covalent bonds are equally long, so it stands to reasonthat not all covalent bonds are equally strong either. This is more than just theory. This has beenexperimentally corroborated using instrumentation and techniques that make it possible toexamine lengths and strengths of bonds in molecules. Why do bond lengths and bond strengthsvary? They are determined by a number of different factors at an atomic level that we will notconsider here.

In Chapter 3 we reviewed some of the properties of covalent compounds. They are heldtogether by covalent bonds. All organic compounds and polymers are covalent compounds, asare most binary covalent compounds (as the name suggests!). Many covalent compounds dissolvein water, but most of them that will dissolve do not dissociate. Remember, there is a differencebetween dissolution (dissolving) and dissociation (ionizing).

Generally speaking, covalent bonds are not usually as strong as ionic bonds. As aconsequence, the melting points and boiling points of covalent compounds are usually lowerthan those of ionic compounds with a similar mass. Why would this be? Think about what has tohappen for any compound, ionic or covalent, to change from a solid to a liquid. The attractiveinteractions between the compound's particles must be overcome so that the distance betweenthem can increase. If the attractive interactions are stronger in one substance than in another,

Fkq q

r 1 2

2

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more energy must be added (i.e., the temperature must be increased to an even greater extent) toconvert that substance from a solid to a liquid. This is the same reason that the boiling points ofsubstances increases as the strength of the bonds between their particles increases. This meansthat by comparing the melting points or the boiling points of two or more compounds, we canobtain an approximate relative idea as to the strengths of the bonding interactions between theparticles in the substances.

Covalent bonds and the Periodic Table

As mentioned in Chapter 3 the number of bonds a nonmetal is likely to form can bepredicted from its position in the periodic table. As a rule, the behavior of nonmetals in chemicalreactions tends to be well explained and predicted by the octet rule. Atoms are more stable whenthey are surrounded by an octet of electrons, whether that octet is obtained through bondformation or through ionization. Actually, it is more correct to say that atoms are more stablewhen their outer shell is completely filled and they are isoelectronic with the nearest noble gas, butsaying they require an octet of electrons is very much the same thing in most cases. There are anumber of elements with behavior that does not follow the octet rule that will be discussed laterin this chapter.

Multiple covalent bonds

It is possible for two bonding nonmetal atoms to share more than one pair of electrons.When two bonding nonmetal atoms share two pairs of electrons rather than a single pair (i.e.,when each of the bonding nonmetal atoms donates two electrons to the bond, rather than one), adouble covalent bond consisting of two shared pairs of electrons is formed (that’s a total of 4electrons being shared in the bond). When two bonding nonmetal atoms share three pairs ofelectrons rather than a single pair (i.e, when each of the bonding nonmetal atoms donates threeelectrons to the bond), a triple covalent bond is formed. This triple covalent bond consists ofthree shared pairs of electrons (In a triple bond between two nonmetal atoms, A and B, atom Adonates 3 electrons to the bond, while bonding atom B also donates three electrons to the bond. Thismeans that A & B have a bond between them consisting of 6 electrons, or, if you’d rather, 3 pairs ofelectrons. Make sense?)

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These double covalent bonds and triple covalent bonds are shorter and stronger than singlebonds. (No, you do not need to memorize the values in this chart. They’re only here for the purposesof illustrating differences in the lengths and strengths of single, double and triple covalent bonds)

the strengths of various covalent bonds (kJ/mol)

single bond double bond triple bond

C-C 348 612 820

N-N 163 409 945

O-O 146 497 -

C-N 305 613 890

Covalent compounds containing double or triple covalent bonds are common in the real world.When we draw the structures of these compounds, we indicate single bonds with a dash betweenthe bonding atoms. Double bonds are represented with two parallel dashes and triple bonds withthree parallel dashes.

So, in the real world, are there such things as quadruple, quintuple, or even sextuple bonds. Yes, thereare, although they’re not found in covalent compounds. Are we going to worry about them in this class,or say anything more about them? Nope, not a word. Just thought I’d mention it in case you werecurious.

Coordinate covalent bonds

Covalent bonds result from the nuclei of two nonmetal atoms sharing a pair of electrons.Most commonly each of the bonding nuclei donate one electron to the shared pair. An example isthe covalent bond between two hydrogen atoms in molecular hydrogen.

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Each hydrogen atom donates one electron to the bond and the bonding electrons are sharedequally between the two bonding atoms. Why is hydrogen satisfied with two electrons, rather thanrequiring an octet of electrons around it? Hopefully you can answer this question after what youlearned in chapter 3.

There are occasions when one of the bonding atoms donates both of the bondingelectrons to the bond and the other bonding atom does not donate any electrons to the bond.Since the electrons are shared between the bonding atoms the bond is still a covalent bond. Butthis type of covalent bond is called a coordinate covalent bond. Valence electrons always occurin pairs in molecules in this class. Either the pairs are involved in bond formation and are calledbonding pairs, or they are not involved in bond formation and are called nonbonding pairs, orlone pairs. In the reaction of ammonia (NH3) and hydrogen ion

the nitrogen atom of ammonia has three bonding pairs of electrons and one nonbonding pair ofelectrons. In this reaction nitrogen shares its non-bonding pair of electrons with the hydrogenion. A fourth N-H bond is formed and the entire compound, ammonium ion, assumes the positivecharge formerly carried by just the hydrogen ion. This fourth bond, formed in this reaction, is acoordinate covalent bond because nitrogen donated both of the bonding electrons and hydrogenion did not contribute any electrons to the bond.

Coordinate covalent bonds are identical in length and strength to regular covalent bonds.Their existence can only be predicted from an understanding of how reactant molecules interactwith each other in the formation of a product compound. Still, the formation of coordinatecovalent bonds is often of great importance in many organic and inorganic reactions.

Molecular formulas and structural formulas

Why do we care about the shapes of molecules? Because often, especially in biologicalsystems, the chemical behavior is determined by molecular shape. The shapes of molecules areoften wildly complex, especially when we examine the shapes of large biological molecules. Asyou look at the structure of large biological molecules in your textbooks or on the internet theirshape might not make much sense to you but it makes perfect sense to various enzymes andhormone receptors in your body. And if the shape of these large biological molecules is changedby even a tiny amount, they may become completely unrecognizable to those enzymes andhormone receptors with which they need to interact. In other words, changes in shape to things

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as tiny as molecules can result in serious illness and even death. It is very much in the bestinterests of our good health for our molecules to keep their proper shapes. In a first semesterchemistry class we will not learn about the shapes of large molecules, but we will take a step inthat direction by learning how to predict the shapes of small molecules.

Molecules can be represented in a variety of ways. The molecular formulas we have beenusing most of the semester tell us the types of atoms in a compound as well as the number ofeach type of atom. The molecular formula of water is H2O, which tells us that a single watermolecule is made up of two hydrogen atoms and a single oxygen atom. The molecular formulaof glucose, C6H12O6, informs us there are 6 carbon atoms, 12 hydrogen atoms, and 6 oxygenatoms in a single glucose molecule.

It is common for two or more compounds to have the same molecular formula, especiallyin the study of organic chemistry. (This is much less common in inorganic chemistry. In fact, it’sdownright infrequent, especially for the compounds we’ll discuss in this class) We can differentiatebetween these compounds with the same molecular formulas using various types of structuralformulas (or, molecular structural formulas), which show the types and numbers of atoms in acompound and also how the atoms are connected to each other within the molecule. Structuralformulas show all of the bonds between all of the atoms, as in the following illustration ofethanol and dimethyl ether, both of which have the molecular formula C2H6O.

Compounds with the same molecular formula but with the atoms connected differently are calledconstitutional isomers. Compounds that are constitutional isomers have different physical andchemical properties, even though they have the same types and numbers of atoms. This isbecause the physical and chemical properties of compounds are determined both by the numbersand types of atoms in the compound as well as by how the atoms are connected, one to another.

Chemists often use an abbreviated type of structural formula, called a condensedstructural formula, in which all of the bonds have been collapsed. A carbon atom bonded to threehydrogen atoms is called a methyl group and is shown as a CH3 group. A carbon atom bonded totwo hydrogens called a methylene group and is shown as a CH2 group. And the O-H group isshown as an OH group. The condensed structure of ethanol is CH3CH2OH while that of dimethylether is CH3OCH3.

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The most abbreviated representation of molecular structure is achieved with a linestructure. These are also sometimes known as backbone or skeletal structures. The line structuresof ethanol and dimethyl ether are

in which the end of a line is a methyl (CH3) group and the vertex of a line is a methylene (CH2)group. The structural formula, condensed structure, backbone structure, and line structure ofoctane, a compound with the molecular formula C8H18 appear as follows.

The various types of structural formulas are of greatest use in the study of organic compounds. Weseldom use anything other than the molecular formulas of inorganic compounds to represent them.Constitutional isomerism is uncommon for inorganic compounds, particularly in this class.

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The Lewis structures of ionic compounds

We learned in Chapter 3 that the Lewis structures of atoms and ions consist of theelemental symbol surrounded by one dot for each valence electron of the substance. The Lewisstructures of ions are drawn with the element symbol surrounded by the correct number ofvalence electrons, including those it has lost or gained.

Since oppositely charged particles attract each other, cations and anions form ionic bonds as theresulting ionic compounds are produced. Ionic compounds can be represented using Lewisstructures. The reactions of sodium ion and fluoride, magnesium ion and oxide, and of sodiumion and oxide, and the resulting ionic compounds, appear as follows:

Note that in sodium fluoride, fluoride is surrounded by one bonding pair and three nonbondingpairs of electrons. In magnesium oxide, oxide is also surrounded by one bonding pair and threenonbonding pairs of electrons. And in sodium oxide, oxide is surrounded by two bonding andtwo nonbonding pairs of electrons.

The Lewis structures of covalent compounds

Lewis structures can also show covalent bonds in molecular compounds. As wementioned a moment ago, valence electrons always occur as pairs in molecules. These electronpairs may be found in bonds, as bonding pairs, or they may occur as nonbonding pairs (or, lonepairs) that do not participate in bonding. In small covalent compounds both bonding pairs andnonbonding pairs of electrons affect the shapes of the molecules. We will be using the Lewisstructures of covalent compounds to determine their molecular geometry, or molecular shape, ifyou’d rather. This statement about unpaired electrons is only true for the molecules we will discuss inthis class. In the real world we encounter molecules with unpaired valence electrons, usually on the

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central atom, but we will neither discuss nor worry about these compounds in this class as we discussLewis structures.

There is a procedure we must follow when drawing the Lewis structures of covalentmolecules. We’ll go through the rules in this procedure rather quickly, but we will then spend asubstantial fraction of this chapter discussing what the rules mean and how we apply them. Touse your time wisely as you proceed through this section (and the remainder of the chapter aswell), you should have a periodic table, a pencil, and paper at hand so you can practice as wework our way through examples. Bear in mind that, the octet rule almost always plays animportant rule in correctly drawing Lewis structures.

1. Find the sum of all of the valence electrons of all of the atoms in the molecule.2. Draw the skeleton structure of the molecule by placing the central atom (usually the first

named atom in the molecular formula, except for hydrogen) in the middle and all of theother atoms (terminal atoms) around it. Draw a line (which represents a single covalentbond) between the central atom and each of the terminal atoms.

3. Subtract two electrons for each bond formed in Step 2 from the total valence electronsfound in Step 1. Divide the remainder by two. This tells us the number of nonbondingpairs of electrons left to assign to atoms in the molecule.

4. Assign nonbonding pairs to terminal atoms, until each is surrounded by an octet ofelectrons.

5. Assign nonbonding pairs to the central atom so that it also has an octet. In some cases thecentral atom may wind up with more than an octet. We'll be discussing what this means in amoment.

6. If there are not enough nonbonding pairs to give the central atom an octet, borrownonbonding pairs from terminal atoms and form multiple bonds with the central atom.

7. Draw resonance structures if appropriate. We'll also be discussing this in one of theexamples below.

Let's see how to use these rules. We'll work a series of examples to illustrate their use.Before we begin, let me point out that drawing Lewis structures is much like cooking. If youfollow the recipe (i.e., the rules I’ve just given you) exactly, it’s very hard to go wrong.However, I’ll warn you what I’ve learned from years of experience watching students work withthis material: the moment you begin to deviate from these rules, the more likely you are to comeup with the wrong Lewis structure. In other words, there are dozens of way to draw Lewisstructures wrong, but one way to do them correctly. Note: I will not give you the rules for drawing Lewis structures on exams. They are your responsibility.This means you should have a copy of them on the 3" x 5" card you bring with you to the exam. It alsomeans that you should never attempt to draw a Lewis structure without a copy of these rules directly infront of you, right next to your periodic table you should always have in front of you. This applies tochapter quizzes and Mastering Chemistry as well as exams.. Please, don’t make life harder for yourselfthan it needs to be. Take these suggestions to heart!

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Example 1: methane has the molecular formula CH4.

While I am providing you with both the names and the molecular formulas in these examples, I mayrequire you on quizzes and examinations to be able to write to correct molecular formula simply givena compound's name. Be sure you know your Chapter 3 nomenclature material!

1. Total valence electrons: 4 + (1 x 4) = 8 Carbon has four valence electrons, each of the four hydrogen atoms has one valence electron.We learn this from their position on the periodic table. Carbon is a group 4A element, whilehydrogen is a group 1A element. The group number for elements in groups 1A - 8A tell us howmany valence electrons they have when we’re drawing Lewis structures. This doesn’t work fromelements in groups 1B-8B, but we’re not going to ask you to draw Lewis structures forcompounds containing transition metal and rare earth metal atoms.

2. Skeletal structure:

Carbon is named first and is the central atom; therefore the hydrogens are terminal atoms.

3. Four covalent bonds were formed. 8 - (4 x 2) = 0. 8 is the number of valence electrons we started with. We formed four bonds and each bondcontains 2 electrons. There are no electrons left to act as nonbonding pairs.

This “0" is the number of valence electrons remaining. Since there are no more electronsto assign, and since both carbon and all of the hydrogens are satisfied with respect to the octetrule, we're done! The skeletal structure we drew in Step 2 and the Lewis structure of methane arethe same.

In other words we know we're done drawing a Lewis structure when (1.) all of the nonbonding pairshave been assigned (i.e., we've run out of electrons), and (2.) all of the atoms in the molecule aresatisfied with respect to the octet rule. Take note of this. Sometimes students have a hard timeknowing when to quit. Again, we’re done when every atom is satisfied with respect to the octet rule. Butmethane was rather too easy. Now let's try one a little more challenging.

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Example 2: phosphorus trichloride has the molecular formula PCl3.

1. Total valence electrons: 5 + (3 x 7) = 26 total valence electrons.As I mentioned above, we can tell how many valence electrons an element has by virtue of itsposition in the periodic table. Phosphorus is a group 5A element and has 5 valence electrons.Chlorine is a group 7A element, so each chlorine atom has 7 valence electrons.

2. Skeletal structure:

Phosphorus is named first in the molecular formula and is therefore the central atom, whichleaves the chlorine atoms as terminal atoms. We draw a single covalent bond from the centralatom to each of the terminal atoms.

3. 26 total valence electrons - (3 x 2 electrons used per bond) = 20 unassigned valence electronremaining; 20/2 = 10 nonbonding pairs of electrons left to assign.

26 total valence electrons - (3 bonds x 2 electrons per bond) = 20 electrons left; 20 electrons left/2 electrons per nonbonding pair = 10 nonbonding pairs of electrons.

4. Assign nonbonding pairs such that all terminal atoms are satisfied with respect to the octetrule:

5. Of the 10 nonbonding pairs, it took 9 of the 10 to give each terminal chlorine atom an octet.The nonbonding pair is assigned to the central phosphorus atom:

6. Since all of the terminal and central atoms are now surrounded by an octet of electrons, andsince there are no more electrons left, this is the Lewis structure for phosphorus trichloride.

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Example 3: hydrogen cyanide has the molecular formula HCN.

1. Total valence electrons: 1 + 4 + 5 = 10 1 valence electron for hydrogen, 4 valence electrons for carbon, and 5 valence electrons fornitrogen. How do we know this? In which group do we find each element?

2. Skeletal structure:

While H is listed first it is never a central atom because it can only form one bond. In this casethe next listed atom - carbon - is the central atom.

3. 10 total valence electrons - (2 x 2 electrons used per bond) = 6 unassigned valence electronsremaining; 6/2 = 3 nonbonding pairs remain to be assigned.

10 total valence electrons - (2 bonds x 2 electrons per bond) = 6 valence electrons left; 6 valence electrons left/2 electrons per nonbonding pair = 3 nonbonding pairs of electrons.

4. Assign nonbonding pairs such that all terminal atoms are satisfied with respect to the octetrule:

5. All three nonbonding pairs were assigned to the nitrogen atom. H and N are satisfied withrespect to the octet rule, but carbon is not. As stated above, a Lewis structure is not completeuntil (1) all of the valence electrons in the molecule have been assigned, and (2) every atom inthe Lewis structure is surrounded by an octet of electrons. So, what do we do, now that we’verun out of electrons but carbon is only surrounded by 4 electrons, rather than the required 8?

6. Since there are no more nonbonding pairs to assign, one of nitrogen's nonbonding pairs isconverted into a bonding pair and a double bond is formed between nitrogen and carbon.

Making this nonbonding pair of electrons to a bonding pair obviously does not affecthydrogen. It remains satisfied with respect to the octet rule. Nitrogen also remains surrounded by8 electrons (four pairs of electrons), two of these pairs being bonding pairs, while the other twoare nonbonding pairs. Forming a double bond between nitrogen and carbon does help a bit, ascarbon is now surrounded by six electrons (three bonding pairs of electrons). However, six isinsufficient. Carbon needs to be surrounded by an octet of electrons. So, another of nitrogen'snonbonding pairs is converted into a bonding pair, and a triple bond is formed between nitrogenand carbon.

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In this last Lewis structure, all three atoms are now satisfied with respect to the octet rule.Therefore, this last structure is the correct structure for hydrogen cyanide. It is possible to confirmexperimentally that the carbon-nitrogen bond in hydrogen cyanide is in fact a triple bond. Lewisstructures often do a remarkable job of predicting how atoms are arranged at a molecular level. Whilethey don’t always work well in the real world, in this class they will prove more than adequate for ourneeds.

Example 4: bromate ion has the molecular formula BrO3-.

1. Total valence electrons: 7 + (3 x 6) + 1 = 26 total valence electrons. 7 valence electrons for bromine, 6 for each of the 3 oxygen atoms, and since the compound isa 1- anion, we must add an extra electron to the total number of valence electrons in amolecule of he compound to account for the charge of the ion.

2. Skeletal structure:

3. 26 total valence electrons - (3 x 2 electrons used per bond) = 20 unassigned valence electronsremaining; 20/2 = 10 nonbonding pairs of electrons left to assign.

26 total valence electrons - (3 bonds x 2 electrons per bond) = 20 electrons left; 20 electrons left/2 electrons per nonbonding pair = 10 nonbonding pairs of electrons

4. Assign nonbonding pairs such that all terminal atoms are satisfied with respect to the octetrule:

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5. Of the 10 nonbonding pairs, it took 9 of the 10 to give each terminal oxygen atom an octet.The nonbonding pair is assigned to the central bromine atom:

6. In this last Lewis structure, all of the electrons have been used and all four atoms are satisfiedwith respect to the octet rule. Therefore, this last structure is the correct structure for bromateion. We’re done, but for this: the Lewis structures of ions must be enclosed in square bracketswith the ion charge indicated as a superscript to the left of the structure.

Note that the square brackets are mandatory for the correct Lewis structures of ions. In thisparticular example, it means the four atoms that constitute a bromate ion have collectivelygained one extra electron. We do not know which electron in bromate’s Lewis structure is theextra electron, and we frankly don’t care, because we couldn’t identify “the extra electron” if ourlives depended on it. It is sufficient to know that this collection of four atoms has one moreelectron than the sum of all of the protons found in the nuclei of the four atoms in this molecule.The simple act of gaining a solitary electron changes the physical and chemical properties of thismolecule. In other words, the physical and chemical properties of BrO3 and BrO3

- are notablydifferent. Some find it hard to believe that something as small as the addition of one extraelectron can cause these differences, but it does.

Example 5: ammonium ion has the molecular formula NH4+.

1. Total valence electrons: 5 + (4 x 1) - 1 = 8 total valence electrons.5 valence electrons for nitrogen, 1 for each of the 4 hydrogen atoms, and since the compoundis a 1+ cation, we must subtract an electron from the total number of valence electrons in themolecule to account for the charge of the ion.

2. Skeletal structure:

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3. 8 total valence electrons - (4 x 2 electrons used per bond) = 0 valence electrons left to assign.

There are no valence electrons remaining, and so there are no nonbonding pairs to assign.Since there are no more electrons to assign, and since nitrogen and all of the hydrogens aresatisfied with respect to the octet rule, this is the correct Lewis structure except, since it is an ion,it must be enclosed in square brackets and show the charge.

What this means is the five atoms that constitute an ammonium ion have collectively lost onevalence electron. Again, we do not know which atom in ammonium’s Lewis structure lost anelectron. Again, we simply don’t care, because we couldn’t possibly identify the atom which“lost” the electron. It is enough to know that this collection of five atoms has one fewerelectrons than the sum of all of the protons found in the nuclei of the five atoms in this molecule.

Example 6: sulfur trioxide has the molecular formula SO3.

1. Total valence electrons: 6 + (3 x 6) = 24 total valence electrons

2. Skeletal structure:

3. 24 total valence electrons - (3 x 2 electrons used per bond) = 18 unassigned valence electronsremaining; 18/2 = 9 nonbonding pairs of electrons left to assign

4. Assign nonbonding pairs such that all terminal atoms are satisfied with respect to the octetrule:

5. Of the 9 nonbonding pairs, it took all 9 to give each terminal oxygen atom an octet. Whilethey are satisfied with respect to the octet rule, the central sulfur atom is not. We will borrow a

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pair of nonbonding electrons from one of the three oxygen atoms and form a double bond to givethe sulfur atom an octet.

6. Wait a minute. Aren't all three oxygen atoms identical to each other? How do we know thatsulfur might not borrow a pair of nonbonding electrons from one of the other oxygen atoms?

The answer is: we don't know. In actual fact there are three equivalent Lewis structures thatdescribe sulfur trioxide. All are equally probable and equally correct. When a molecule can onlycorrectly be described by two or more equivalent Lewis structures, we say that it is showingresonance. The Lewis structures of molecules exhibiting resonance are also known as resonancestructures.

There are two very important consequences to resonance in molecules. First, for anymolecule in which resonance occurs, the measurable physical attributes of the molecule are theinstantaneous average of the physical attributes of all of it's resonance structures. In real life, themolecule does not rapidly switch from one resonance form to another. A molecule withresonance is always the instantaneous average of all its resonance structures. This has someinteresting implications that can be proven experimentally. We will discuss two of theseimplications

The S-O bond lengths in sulfur trioxide are shorter than S-O single bond lengths but longer thanS=O double bond lengths. They are, in fact, intermediate between the two. The S-O bondstrengths in sulfur trioxide are also intermediate between those of S-O single and S=O doublebond strengths. These bits of information corroborate our belief that sulfur trioxide cannot beaccurately described by any one of its Lewis structures, or by assuming that it rapidly fluctuatesfrom one structure to the next. The sulfur trioxide molecule is, in verifiable fact, theinstantaneous average of its three resonance structures. We represent resonance by drawing all ofthe correct Lewis structures, and then by drawing two-headed arrows between them to denoteresonance.

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More generally, this is true for every molecule in which we find resonance in its Lewisstructures: the physical features (bond length and bond strength) of the molecule are always theinstantaneous average of all of it’s resonance structures.

Second, molecules are stabilized by resonance. This is to say they are more stable, andless reactive, than we might otherwise predict. This stabilization is related to something calledelectron delocalization, which is beyond the scope of this class. Suffice it to say that resonance isa naturally-occurring event that is energetically favorable and which stabilizes those moleculesin which it occurs.

Example 7: carbonate ion has the molecular formula CO32-.

1. Total valence electrons: 4 + (3 x 6) + 2 = 24 total valence electrons. Remember, we add two electrons to the total because this is a 2- anion.

2. Skeletal structure:

3. 24 total valence electrons - (3 x 2 electrons used per bond) = 18 unassigned valence electronsremaining; 18/2 = 9 nonbonding pairs of electrons left to assign.

4. Assign nonbonding pairs such that all terminal atoms are satisfied with respect to the octetrule:

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5. Of the 9 remaining nonbonding pairs, it took all 9 pairs to give each terminal oxygen atom anoctet. While these terminal oxygen atoms are satisfied with respect to the octet rule, the centralcarbon atom is not. We will borrow a pair of nonbonding electrons from one of the oxygen atomsand form a double bond to give the carbon atom an octet. It is obvious that the double bondrequired to give carbon an octet can be drawn to any of the three oxygen atoms, all of which areequivalent to each other.

6. Since we have three equivalent Lewis structures, we know that we have resonance and mustrepresent all three resonance structures. We also know that, since the structures are of ions, eachmust be enclosed in square brackets and show its charge.

Exceptions to the octet rule

There are a number of atoms and molecules that can show behavior that deviates from thepredictions made using the octet rule.

There are numerous molecules with an odd number of valence electrons, such as ClO2,NO, and NO2. As I mentioned above, it is almost always the central atom that winds up with theunpaired electron. While Lewis structures can be drawn for all of these compounds, we will notconsider them in this class. This is reiteration, but it’s important. If you're drawing a Lewis structurefor this class and wind up with an odd number of valence electrons in Step 1, you've made a mistake.These types of molecules do exist and they do play important roles in many aspects of the worldaround us. But, quite simply, you will never be asked to draw Lewis structures for these compounds inthis class. Never.

There are molecules in which an atom might actually be satisfied with less than an octet.Hydrogen is an example we've already seen. When it has two electrons it is isoelectronic withhelium. Beryllium and boron are two more atoms that are satisfied with less than an octet.Beryllium forms ionic bonds and is satisfied with a total of four valence electrons around it, as inBeH2. Boron only forms three covalent bonds and is satisfied with a total of six electrons around

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it, as in BF3. This is because beryllium and boron are such small atoms that they do not haveroom to accommodate an entire octet around them. I do expect you to remember these threeexceptions.

There are also molecules with a central atom with more than an octet of electrons aroundit. Only central atoms will ever have more than an octet. Terminal atoms may never have moreor less than exactly an octet of electrons. The central atom may only have more than an octet if itis a nonmetal found in the third period or higher. This is because elements in the 3rd Period andhigher may take advantage of empty orbitals in their d subshells to accommodate the extraelectrons. In other words, if an atom fills its s and p subshells it can store the extra electrons in emptyorbitals in its d subshell. But since d subshells only occur in the 3rd period and higher, elements in the1st and 2nd period do not have this option. We see this most commonly in phosphorus, which mayhave 10 valence electrons, and in sulfur, which may have 12 valence electrons. Let's look at twoexamples. These are only two examples, but before you're done with this class you will see othermolecules in which the central atom is surrounded by more than an octet of electrons.

It is important to remember that we never use the formation of double or triple covalentbonds to violate the octet rule. These violations only occurs when there are left-overnon-bonding pairs and all of the terminal atoms have been satisfied with respect to the octet rule.In this class if you draw a Lewis structure in which the central atom violates the octet ruledue to the formation of double or triple bonds, you have made a mistake! This is one of themost common mistakes I see students make with this material. Heed this paragraph well! Terminalatoms never violate the octet rule. And in this class, central atoms only violate the octet rule whenthere are an excess of nonbonding pairs and rule 5 requires them to be “given” to the central atom.There, are of course, compounds in the real world for which the preferred Lewis structure is one inwhich the central atom violates the octet rule due to the formation of multiple bonds. Sulfate ion is agood example. But the compounds in which this occurs will not be considered in this class - withoutexception!

Iodine tetrachloride ion has the molecular formula ICl4-.

1. Total valence electrons: 7 + (4 x 7) + 1 = 36 total valence electrons

2. Skeletal structure:

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3. 36 total valence electron- (4 x 2 electrons used per bond) = 28 unassigned valence electronsremaining; 28/2 = 14 nonbonding pairs of valence electrons left to assign

4. Assign nonbonding pairs such that all terminal atoms are satisfied with respect to the octetrule:

5. Of the 14 nonbonding pairs, it took 12 to give each terminal chlorine atom an octet. Whilethey are satisfied with respect to the octet rule, and the central iodine atom is satisfied withrespect to the octet rule, there are still two nonbonding pairs left to assign. According to rule 5above, they are assigned to (placed around) the central atom.

6. Since iodine tetrachloride is an ion, it must be placed in square brackets and show its charge.

Note: this is the correct, expected Lewis structure for this ion. We do not “rid ourselves” of those twoextra nonbonding pairs around the central iodine atom by drawing double bonds with terminal atoms.Some of you are going to be irresistibly tempted to do what I’ve just told you not to do. Fight thetemptation!

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Phosphorus pentachloride has the molecular formula PCl5.

1. Total valence electrons: 5 + (5 x 7) = 40 total valence electrons

2. Skeletal structure:

3. 40 total valence electrons - (5 x 2 valence electrons used per bond) = 30 unassigned valenceelectrons remain; 30/2 = 15 nonbonding pairs of electrons left to assign.

4. Assign nonbonding pairs such that all terminal atoms are satisfied with respect to the octetrule:

5. Of the 15 nonbonding pairs, it took all 15 to give each terminal chlorine atom an octet. Theyare satisfied with respect to the octet rule, as is the central phosphorus atom, which is surroundedby five bonding pairs of electrons, or, by a total of ten electrons. This is of course a violation ofthe octet rule. But, it is possible for phosphorus to do this, and phosphorus is a stable, naturallyoccurring element when it behaves this way, unexpected though it may be.

The shapes of molecules and VSEPR - Valence Shell Electron-Pair Repulsion

The shape, or, the geometry of a molecule is determined by the number of bonding andnonbonding sets of electrons that surround its central atom. Electrons are repelled by otherelectrons, since they have the same negative charge. Pairs of electrons in the same orbital arerepelled by pairs of electrons in other orbitals. As a consequence, pairs of electrons will orientthemselves as far apart from each other as is possible in three dimensional space. The theory ofmolecular geometry based on this repulsion behavior is called Valence Shell Electron-PairRepulsion, or VSEPR. Some books prefer VSPER, Valence Shell Paired Electron Repulsion. By eithername, they’re exactly the same theory and employ the same principles and logic.

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To use VSEPR to determine the shape of a molecule we begin by drawing it's Lewisstructure. The molecule's geometry depends on the total number of sets of electrons thatsurround the central atom, by how many of those are bonding sets of electrons, and by how manyare nonbonding sets of electrons.

Two further notes to keep in mind. All of the electrons in multiple bonds count as one setof electrons. In other words, the presence of multiple covalent bonds does not affect ourdetermination of the geometry of a molecule. Strangely enough, while double bonds and triplebonds are much shorter and stronger than single bonds, they seem to play exactly the same rolein the determination of molecular geometry as single bonds.

Nonbonding pairs of electrons take up more space than bonding pairs of electrons.Bonding electrons are generally confined between the two bonding nuclei. Nonbonding electronsare not constrained by the presence of a second nucleus, and hence can physically take up morespace. As a result, nonbonding pairs of electrons typically result in the distortion of moleculargeometry.

Below is a summary of commonly encountered geometries. You do not need to memorizethis table. I will always provide you with a copy on quizzes and exams. You simply need toknow how to use the table. Correctly using the table will always start with the correct Lewisstructure for the compound of interest. This means if you can’t correctly draw Lewis structures, you’rein a world of hurt when it comes to determining molecular geometries.

I advise you in the strongest possible terms to refer back to the geometry table on the following page aswe work our examples. In other words, as you work the examples in these notes, and problems on thechapter quiz and Mastering Chemistry, a copy of this table should always be in front of you, along witha periodic table and a list of the rules for correctly drawing Lewis structures. When I ask you moleculargeometry questions on quizzes or exams you will be permitted to use a copy of this geometry table. Iwill provide you with a copy of this table on exams covering this material.

There is one curious and rather unexpected aspect to molecular geometry I want to point out. Todetermine a compound’s molecular geometry, you begin by drawing it’s Lewis structure. Youthen count the “sets of electrons” around the central atom and use the geometry table on thefollowing page to determine molecular geometry. By and large, geometry is determined withreference to the central atom in a molecule. We are rarely interested in the geometry with respect toany atom in a molecule other than the central atom. A bonding pairs of electrons counts as one“set” of electrons. A nonbonding pair of electrons counts as one “set” of electrons. So far, nogreat surprises. But, it is rather astonishing to learn that the two pairs of electrons in a doublebonds are counted as only one “set” of electrons. The three pairs of electrons in a triple bond arealso counted as just one “set” of electrons. The immediate implication of this is that doublebonds and triple bonds may be shorter and stronger than single bonds, but they all have the same

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effect on molecular geometry. In the real world this isn’t always completely true, but it’s generallyvery close. It’s certainly an adequate assumption for our purposes in this class.

To reiterate, in this geometry table, a "set" may refer either to single pair of bonding electrons, asingle pair of nonbonding electrons, the two pairs of bonding electrons in a double bond, or tothe three pairs of bonding electrons in a triple bond. Each are counted as one set of electrons.“Total sets” is the total number of electron sets around the central atom. “Bonding sets” are thosesets of electrons involved in the formation of a single, double, or triple bond. “Nonbonding sets”are the number of lone pairs (nonbonding electron pairs) around the central atom.

total sets bonding setsnonbonding

setsgeometry bond angles examples

2 2 0 linear 180o BeH2

3 3 0trigonalplanar

120o BF3

3 2 1 bent - NO2-

4 4 0 tetrahedral 109.5o CH4

4 3 1 pyramidal - NH3

4 2 2 bent - H2O

5 5 0trigonal

bipyramidal120o, 180o PCl5

5 4 1 seesaw - SCl4

5 3 2 t-shaped - ClF3

5 2 3 linear - XeF2

6 6 0 octahedral 90o, 180o SF6

6 5 1square

pyramidal- BrF5

6 4 2 square planar - XeF4

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Two Sets of Electrons

When the central atom of a molecule is surrounded by two bonding sets of electrons, asin beryllium hydride (BeH2, which is also an ionic compound)

the optimal geometry places the bonding sets of electrons at an 180° angle with respect to eachother.

Ask yourself this question: what angle with respect to the central atom places the two terminal atomsas physically far apart from each other as possible? I suggest that you take a bag of marshmallows orgumdrops and some toothpicks and use them to make 3-D models of the molecules we discuss. It willhelp you visualize the correctness of what I tell you in this section. The marshmallows (gumdrops)represent atoms. The toothpicks represent bonds (sets of electrons between the atoms. Try to optimizethe distance between the terminal marshmallows with respect to the central marshmallow by changingthe angles between them. Which arrangement maximizes the distance between the terminalmarshmallows? As a bonus, you may eat your models when your are finished. You won't get any extrapoints but it's still a bonus all the same.

This geometry is called linear geometry. There are no nonbonding pairs of electrons around thecentral atom, and it is thus an ideal geometry.

“Ideal geometries” are molecular geometries which are not distorted by the presence of nonbondingpairs of electrons around the central atom. This concept of “ideal geometry” is one you will not find inany text book. It is, as far as I know, a device of my own creation. But you will find it remarkably usefulwhen we reach our discussion of molecular polarity at the end of this chapter so pay very closeattention to which geometries are ideal and which are not.

Three Sets of Electrons

When the central atom of a molecule is surrounded by three sets of electrons there are twodifferent common geometries.

When the central atom of a molecule is surrounded by three sets of electrons and they are allbonding sets of electrons, as in boron trifluoride,

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the optimal geometry places the bonding sets of electrons at an 120° angle with respect to eachother. This geometry is called trigonal planar geometry. It is an ideal geometry. The threeterminal atoms are found at the corners of a equilateral triangle with the central atom in themiddle. All four atoms are found in the same plane.

When the central atom of a molecule is surrounded by three sets of electrons, two of which arebonding sets of electrons and one of which is a nonbonding set of electrons, as in nitrite ion,

the optimal geometry ideally places the sets of electrons at an 120° angle with respect to eachother. Note that the double bond only counts as one set. You should also note that nitrite ion has tworesonance structures, which I have omitted, as well as the square brackets and charge informationnormally required for the Lewis structures of ions. This is in the interest of trying to keep things simple.However, the nonbonding pair distorts the trigonal planar geometry and forces the two bondingpairs closer together than a 120° angle. This type of distorted trigonal planar geometry is calledbent geometry, or angular geometry. All three atoms are found in the same plane.

An unexpected consequence of nonbonding sets is distortion of molecular geometry.Nonbonding pairs of electrons take up more physical space than bonding pairs of electrons. Thisis because you have a pair of electrons being attracted by the nucleus of a single atom, ratherthan a pair of electrons being attracted by a pair of atomic nuclei as one typically finds in asingle covalent bond. As a consequence, nonbonding pairs are typically more electricallyrepulsive to each other and to bonding sets of electrons, than bonding sets of electrons are toeach other. This causes distortion in molecular geometry. The bonding sets of electrons arepushed closer to each other than they would be under “ideal” circumstances (i.e. the absence ofthe nonbonding pair of electrons). The amount of geometric distortion introduced by nonbondingpairs depends on the terminal atoms in the molecule. In other words, the O-N=O bond angle willnot be 120o but something that puts the terminal oxygen atoms a bit closer to each other, about115o. If we were to investigate the geometries of the NS2

- and NSe2- molecules, we would expect

S-N=S and Se-N=Se to have bent geometry, like the NS2- ion seen above, but we could not begin

to predict the extent of the distortion caused by the nonbonding pair on the central nitrogen atomin each molecule. We can simply guess that whatever the bond angle may be, it will not be 120o.

As a consequence, I expect you to remember the bond angles of the ideal geometries, those undistortedby the presence of nonbonding pairs. I do not expect you to know or remember the bond angles of thedistorted geometries. It is sufficient that you know the names of these distorted geometries and whythey have been distorted.

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Four Sets of Electrons

When the central atom of a molecule is surrounded by four sets of electrons, there are threedifferent common geometries.

When the central atom of a molecule is surrounded by four sets of electrons and they are allbonding sets of electrons, as in methane,

the optimal geometry places the bonding sets of electrons at an 109.5° angle with respect to eachother. This geometry is called tetrahedral geometry. It is an ideal geometry (i.e. there are nononbonding pairs of electrons causing distortion of the molecular geometry). The four terminalatoms are found at the corners of a tetrahedron, with the central atom in the middle of thetetrahedron.

Since it is difficult to show three dimensionality with two dimensional figures we usethree different bond representations to help achieve the effect. Line bonds, as are the C-H bondsto H1a and H1b, are bonds in the plane of the surface on which the drawing is rendered. Wedgebonds, like the C-H bond to H1c, are bond that project out of the surface toward the viewer.Dash bonds, like the C-H bond to H1d, are bonds that pass behind the surface away from theviewer.

When the central atom of a molecule is surrounded by four sets of electrons, three ofwhich are bonding sets of electrons and one of which is a nonbonding set of electrons, as inammonia (NH3),

the optimal geometry ideally places the sets of electrons at an 109.5° angle with respect to eachother. However, the nonbonding pair of electrons distorts the tetrahedral geometry and forces thethree bonding pairs closer together than an 109.5° angle. This distorted tetrahedral geometry iscalled pyramidal geometry. Note that the bond to H1a is in the plane of the paper, the bond toH1b comes out of the plane toward us, and the bond to H1c passes behind the plane, away fromus. The amount of distortion caused by the nonbonding pair depends on both the central atomand the terminal atoms in the molecule. In the case of ammonia, the H-N-H bond angles are all107o. Some sources claim the H-N-H bond angle is closer to 107.5o but either way, we’re not really

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interested, other than to note that the bond angle is no longer 109.5o. In the cases of NF3 and NCl3,which also have pyramidal geometries, the bond F-N-F and Cl-N-Cl bond angles are 102.3o and107.1o respectively.

When the central atom of a molecule is surrounded by four sets of electrons, two ofwhich are bonding sets of electrons and two of which are nonbonding sets of electrons, as inwater,

the optimal geometry ideally places the sets of electrons at an 109.5° angle with respect to eachother. However, the nonbonding pairs on the central oxygen atom distort the ideal tetrahedralgeometry and forces the two bonding pairs closer together than an 109.5° angle . This distortedtetrahedral geometry is called bent geometry, or angular geometry. As was the case withpyramidal geometry, the amount of distortion depends on the central and terminal atoms in themolecule. In water the H-O-H bond angle is around 104.5o, rather than the 109.5o we’d see ifnonbonding pairs had not distorted the geometry.

Ok, so there's a certain poverty of imagination in the name of this geometry seeing as how we’vealready got a bent geometry. Nonetheless, bent geometry may be either due to the effect of onenonbonding pair on two bonding pairs or due to the effect of two nonbonding pairs on two bondingpairs. I should also point out that there is not universal agreement on the names of some of thegeometries in the table. This means you need to take care. When you’re doing Mastering Chemistry, besure to use the geometry names the text provides you. When you’re doing my quizzes and exams,you’re required to use the geometry names I provide in the geometry table in theses lecture notes.

Five Sets of Electrons

When the central atom of a molecule is surrounded by five sets of electrons there are fourdifferent common geometries.

When the central atom of a molecule is surrounded by five sets of electrons and they areall bonding sets of electrons, as in phosphorus pentachloride,

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the optimal geometry places the bonding sets of electrons at either an 180° angle or an 120°angle with respect to each other. This geometry is called trigonal bipyramidal geometry. It is anideal geometry. Atoms or groups in the 1, 2, and 3 positions are said to be in an equatorialposition with respect to the central atom. Equatorial groups are found at an 120° angle withrespect to each other. The three equatorial groups and the central atom are in the same plane.Atoms or groups in the 4 and 5 positions are said to be in an axial position with respect to thecentral atom. Axial groups are found at an 180° angle with respect to each other and areperpendicular to the plane containing the central atom and the three equatorial atoms. Equatorialand axial, as in the equator and the axis of the earth.

When the central atom of a molecule is surrounded by five sets of electrons, four ofwhich are bonding sets of electrons and one of which is a nonbonding set of electrons, as insulfur tetrachloride,

the nonbonding pair distorts the ideal trigonal bipyramidal geometry and forces the bondingpairs closer together. When the total number of sets of electrons is five, nonbonding pairs arealways found in the equatorial position. This distorted trigonal bipyramidal geometry is calledsee-saw geometry.

Did you notice something curious with sulfur tetrachloride? Given its molecular formula,SCl4, you might be tempted to leap to the conclusion that it has tetrahedral geometry. But you’dbe wrong. The problem with molecular formulas is that while they do tell us about the numbersand types of atoms we find in a compound they don’t tell us if there are nonbonding pairs ofelectrons around the central atom and if so, how many. Don’t take short cuts when determiningmolecular geometry. You must begin with the correct Lewis structure for the compound.Molecular formulas alone are not adequate. Notice how often this is the case in the remainingexamples we work.

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When the central atom of a molecule is surrounded by five sets of electrons, three ofwhich are bonding sets of electrons and two of which are nonbonding sets of electrons, as inchlorine trifluoride,

the nonbonding pairs distort the ideal trigonal bipyramidal geometry and force the bonding pairscloser together. Both nonbonding pairs of electrons are found in equatorial positions. Thisdistorted trigonal bipyramidal geometry is called t-shaped geometry. I've never been quite sure how many beers whoever named the see-saw and t-shaped geometries musthave had to drink but I'm assuming it was more than one. As you try to figure out how in the worldthese geometries look like a see-saw or the letter "T," remember that when we "look" at moleculeswith instruments capable of doing so we cannot see the nonbonding pairs of electrons. We only see theatoms that are present and the effects of nonbonding pairs on the arrangement of the terminal atomswith respect to the central atom. Nonbonding pairs are invisible. We only see their effects.

When the central atom of a molecule is surrounded by five sets of electrons, three ofwhich are nonbonding sets of electrons and two of which are bonding sets of electrons, as inxenon difluoride,

the nonbonding pairs do not distort the ideal trigonal bipyramidal geometry because they offseteach other. The terminal atoms are found in the axial positions, 180° apart from each other. Thisversion of trigonal bipyramidal geometry is called linear geometry. This geometry, although ithas nonbonding pairs of electrons around the central atom, is an ideal geometry. As all of thenonbonding pairs are in the equatorial position they cancel each other out and in effect it is as ifthey are not present. Yes, I know, we've already got a linear geometry. Now we've got another. Don't complain to me, I justwork here. Linear geometry can be found either when a central atom is surrounded by two bonding

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pairs of electrons, or when a central atom is surrounded by five pairs of electrons and three of the fiveare nonbonding pairs.

Six Sets of Electrons

When the central atom of a molecule is surrounded by six sets of electrons there are threedifferent common geometries.

When the central atom of a molecule is surrounded by six sets of electrons and they areall bonding sets of electrons, as in sulfur hexafluoride,the optimal geometry places the bondingsets of electrons at either an 180° angle or a 90° angle with respect to each other. This geometryis called octahedral geometry. It is an ideal geometry. Atoms or groups in the 1, 2, 3, and 4positions are said to be in an equatorial position with respect to the central atom. Equatorialgroups are found at a 90° angle with respect to each other. The four equatorial groups and thecentral atom are in the same plane. Atoms or groups in the 5 and 6 positions are said to be in anaxial position with respect to the central atom. Axial groups are found at an 180° angle withrespect to each other and are perpendicular to the plane containing the central atom and the fourequatorial atoms.

When the central atom of a molecule is surrounded by six sets of electrons, five of whichare bonding sets of electrons and one of which is a nonbonding set of electrons, as in brominepentafluoride,

the nonbonding pair distorts the ideal octahedral geometry and forces the bonding pairs closertogether. When the total number of sets of electrons is six, nonbonding pairs are always found inthe axial position. This distorted octahedral geometry is called square pyramidal geometry.

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When the central atom of a molecule is surrounded by six sets of electrons, four of whichare bonding sets of electrons and two of which are nonbonding sets of electrons, as in xenontetrafluoride, the nonbonding pairs do not distort the ideal octahedral geometry because theyoffset each other (i.e., their effects on molecular geometry negate each other). The terminalatoms are found in the equatorial positions, 90° apart from each other. This version of octahedralgeometry is called square planar geometry. This geometry, although it has nonbonding pairs ofelectrons around the central atom, is an ideal geometry. As both of the nonbonding pairs are inthe axial position they cancel each other out and in effect it is as if they are not present.

Let me reiterate the warning I gave you above. You usually cannot look at the molecular formula of acompound and guess it's geometry. Molecular formulas do not account for the presence of nonbondingpairs found around the central atom. You have to first correctly draw its Lewis structure, and then usethe geometry table. If we did not do this above, we might have thought that bromine pentafluoride hastrigonal bipyramidal geometry, since the central atom forms five bonds. We might have thought thatxenon tetrafluoride has tetrahedral geometry, since the central atom forms four bonds. Take the timeto learn how to do this correctly and practice, practice, practice!

Polar covalent bonds and polar molecules

In covalent bonds each bonding atom donates one electron to the bond and the bondingelectrons are shared. These bonding electrons are shared equally when the bonding atoms are thesame, as for example in H-H, N-N, C-C, O-O, and F-F bonds. However, if the bonding atoms arenot the same, the electronegativity of the bonding atoms will affect the way in which the bondingelectrons are shared. Electronegativity, as we learned in Chapter 3, is the tendency of an atom toattract electrons to itself.

When a covalent bond is formed between two nonmetal atoms with differentelectronegativities, the bonding electrons will still be shared but they will spend adisproportionate amount of time in the vicinity of the more electronegative atom. Since the moreelectronegative atom has an extra electron around it part of the time, the atom develops a partialnegative charge because it has one more electron than it has protons for part of the time. On theother end of the bond, the less electronegative atom is missing one of its electrons part of thetime; the atom develops a partial positive charge because it has one more proton than it haselectrons part of the time. When this occurs, we say the bond is polarized, or that it is a polarcovalent bond. To reiterate: in polar covalent bonds the bonding electrons are shared butunequally, due to differences in the electronegativities of the bonding atoms.

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Polarity in bonds is indicated using a lowercase Greek letter delta (δ). The partiallynegatively charged atom is indicated with a δ-, and the partially positively charged atom isindicated with a δ+. This is illustrated with hydrochloric acid:

How do we know when the difference in electronegativity is sufficient to cause a bond tobe polar? If you have access to a table of electronegativity values (as, for example, the one foundin Chapter 4 of your text, more specifically, Figure 4.6 on p.120 of the 8th edition of the text) wecan simply compare the values of the bonding atoms. If the difference in electronegativity isabout 0.7 or greater, the bond is considered to be a polar covalent bond. In hydrochloric acid, theelectronegativity value of hydrogen is 2.1 and of chlorine is 3.0. The difference is 0.9, whichindicates that the bonding electron pair will not be shared equally and that the bond will be polar.

The value 0.7 is not one that is universally agreed upon. Some texts give one value and others another.Rather, 0.7 is a commonly cited value in many texts. It lies roughly in the middle of differences inelectronegativity from 0.5 to 1.0 that are thought to be the point at which a covalent bond becomespolar. It is important to note that the values I use do not agree with those found in your text, in whichthe authors assert that “0.5" is the “magic number” at which a difference in electronegativity betweenbonding atoms is sufficient to result in a polar covalent bond. It’s not a matter of who’s right and who’swrong. This is simply one of those areas in chemistry in which there are various opinions .

There are those who believe if the difference in electronegativities between bondingatoms is around 1.7 to 2.0 or greater, the bonding electrons are shared so unequally that the bondceases to be covalent and becomes an ionic bond in nature. There are also those, myselfincluded, who disagree with this theory. We will disregard it in this class and adhere to what wehave said previously: a bond between a metal and a nonmetal is always an ionic bond, while abond between two nonmetals is always a covalent bond regardless of the difference in theelectronegativity of the bonding atoms.

This is important to know, and it does place you, as a student, in a bit of a tricky position. When you’redoing Mastering Chemistry problems, you must use the text’s electronegativity values and go by theauthor’s definitions (p.120 in the 8th edition). However, on my quizzes and exams, you must use theelectronegativity values I give you and use the definitions I’ve provided you with on this page. You’re notfree to mix and match, and if you do so, you’re going to wind up making mistakes. You’ve beenwarned. Proceed with caution. Ask questions if you’re unclear.

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Bond polarity and molecular polarity

As a general rule, a molecule with one or more polar bonds is itself a polar molecule. Inother words, polar bonds in a molecule can affect the distribution of electrons in the molecule tothe extent that a portion of the molecule develops a partial negative charge while another portionof the molecule develops a partial positive charge. This molecular polarity affects the strength ofinteractions with neighboring molecules (which we will discuss in Chapter 8). Molecularpolarity is measured in Debye units (D). Values can range from 0.0 D for nonpolar molecules tovalues of 2-5 D or more for very polar molecules.

As a general rule, if a molecule has no polar bonds, it will be a nonpolar molecule. If amolecule has one or more polar bonds, the molecule will be a polar molecule. There is oneexception to this statement. If a molecule has one or more polar bonds but meets both of thefollowing conditions (1.) the central atom has one of the ideal geometries - linear, trigonalplanar, tetrahedral, trigonal bipyramidal, octahedral, or square planar and, (2.) all of the terminalatoms are the same, then the molecule will be a nonpolar molecule.

An example of this is phosphorus pentachloride. The P-Cl bonds are polar, but sincethere are five chlorine atoms pulling equally on the central phosphorus atom, they cancel eachother out and the molecule as a whole is nonpolar. However, if we do something to disrupt thedistribution of electrons within the molecule, like, for example, substitute a bromine atom forone of the five chlorine atoms, the molecule will itself become a polar molecule.

We are treating this concept of molecular polarity as though it is a black-and-white affair when, inreality, there is a great deal of gray in the real world. As we have done on previous occasions and as wewill do in the future in this class, we are simplifying the situation in an attempt to reduce it tosomething one can reasonably understand and learn in the space of a single semester.

Summary of determining molecular polarity

Is a molecule polar? In this class any molecule with no polar bonds itself is always anonpolar molecule. If a molecule has one or more polar bonds then the molecule itself is alwayspolar unless it meets both of the following conditions: (a.) it has an ideal geometry, and (b.) allof the terminal atoms are the same.

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Chapter 5: Classification and Balancing of ChemicalReactions

Chemical equations

Chemical equations are shorthand representations of chemical reactions. For example,take the following verbal description of an experiment:

If we take solid silver(I) nitrate and add it to a beaker containing water, take solid ammoniumchloride and add it to a second beaker also containing water, and then mix the contents of thetwo beakers together, a chemical reaction will occur in which we observe the formation of solidsilver(I) chloride. An aqueous solution of ammonium nitrate is also form.

Describing this reaction verbally is rather tedious. Alternatively, we can describe it usinga chemical equation:

This particular chemical equation tells us that one molecule of silver(I) nitrate will reactwith one molecule of ammonium chloride to form one molecule of silver(I) chloride and onemolecule of ammonium nitrate as products.

The substances involved in the reaction are represented by their molecular formulas. Ifone of the reactants is an element we use its symbol to represent the element.

The chemicals found to the left of the arrow are called reactants. Reactants are theingredients from which the products of the chemical reaction are made. Reactants are alwaysfound to the left of the arrow in a chemical equation. Products are the chemicals produced duringa reaction and are always found on the right side of the arrow. Normally we do not labelreactants or products as such, as it is understood that anything to the left of the arrow in achemical equation is always a reactant, and anything to the right of the arrow in a chemicalequation is always a product.

The physical state of each chemical is indicated with a subscript which is enclosed inparentheses, to the right of the substance’s molecular formula. The four states in which we mayfind reactants and products are:

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• the gaseous state, indicated (g)• as a pure liquid, indicated (l)• as a solid, indicated (s)• in aqueous solution, indicated (aq), which means “dissolved in water.” Note that

“aqueous” is not a state of matter as such. It is simply a reflection of the nature of things wecommonly find in life. In the real world, and particularly in living creatures, be they bacteria orgreat blue whales and giant sequoia trees, much of the chemistry of life depends on the abilityof essential substances to dissolve in water, i.e., to be found in the aqueous (aq) state.

An important attribute of all correctly written molecular formulas is mass balance.Remember that mass is never created or destroyed in chemical reactions. If we have “x” numberof a given atom on the reactant side of an equation we must always have exactly “x” of that sameatom on the product side of the equation, no matter how small or large the number “x” may be. Ifwe have exactly one billion “x” atoms on the reactant side of a chemical equation, we will alsohave exactly one billion “x” atoms on the product side of the chemical equation. This is aninviolable law of the universe. This does not however preclude them from being found indifferent physical states or in different substances on the two sides of the chemical equation.However, they must be present in equal numbers on both the reactant and the product side of achemical equation .

In the above reaction we have one silver (Ag) atom on each side of the equation. It doesnot matter that it is found in silver(I) nitrate on the reactant side and in silver(I) chloride on theproduct side. All that matters is that there is one silver atom on both sides of the equation.

We also have exactly one chlorine atom on each side of the equation, one atom inammonium chloride on the reactant side, one atom in silver(I) chloride on the product side.

When it comes to counting the atoms in polyatomic ions we can do one of two things. Wecan count the individual atoms in each polyatomic ion which is longer and harder. Or if thepolyatomic ions pass unchanged from the reactant side to the product side we can simply countpolyatomic ions themselves, which is quicker and easier when it is appropriate to do so.

In the above reaction we have one nitrate ion on each side of the equation, and oneammonium ion on each side of the equation. This means we have mass balance with respect tothese two polyatomic ions.

Before we move one, we should note that the order of reactants on the reactant side of theequation, and the order of products on the product side of the equation is unimportant. It is,however, essential to keep reactants on the reactant side of the equation, and products on theproduct side of the equation. What exactly do I mean by all of this?

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The chemical equation for the reaction we have been discussing was given as

AgNO3 (aq) + NH4Cl(aq) => AgCl(s) + NH4NO3 (aq)

It would also be appropriate to write this equation as

NH4Cl(aq) + AgNO3 (aq) => NH4NO3 (aq) + AgCl(s)

In other words, it doesn’t matter if we listed silver(I) nitrate 1st and ammonium chloride 2nd, orvisa versa, as long as they’re both on the reactant side of the equation. Similarly, on the productside of the equation, if doesn’t matter if silver(I) chloride is listed 1st and ammonium nitrate islisted 2nd, or visa versa. All that matters is that, as products, both are found on the product side ofthe equation. It’s a bit like math:

A + B = C + D

can also be correctly written as

B + A = D + C

Let's take a look at the reaction of aqueous barium nitrate and aqueous lithium phosphate:

3 Ba(NO3)2 (aq) + 2 Li3PO4 (aq) => Ba3(PO4)2 (s) + 6 LiNO3 (aq)

Aqueous solutions of barium nitrate and lithium phosphate are the reactants in this reaction.Solid barium phosphate and aqueous lithium nitrate are the products of this reaction. Thenumbers in front of the various molecular formulas are called coefficients. A coefficient tells ushow many of each molecule is needed for the reaction to proceed properly. If there is nocoefficient shown explicitly in front of a substance in a chemical equation, it should be inferredthat the number "one" is present. In the above reaction, the equation tells us that three moleculesof aqueous barium nitrate react with two molecules of aqueous lithium phosphate to form asingle molecule of solid barium phosphate and six molecules of aqueous lithium nitrate.

There is a very important difference between the coefficients in chemical equations andthe subscripts in molecular formulas.

The subscripts in molecular formulas tell us how many of each atom are found in a singlemolecule of the compound. These numbers are unique for each compound. If we change thesubscripts in a molecular formula we change the compound. We learned about this, particularly forionic compounds, in chapter 3. If you don’t remember chapter 3, or if your understanding of thematerial is shaky, you must master the chapter 3 material or this chapter will be an exercise infrustration and abject misery for you. There’s simply no easy way round it, and it won’t go away if you

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try ignore it. Much weeping, wailing, gnashing of teeth, hair-pulling, and swearing will result as youattempt to work your work though this material. Please, get help if you need it!

Regardless of the type of compound, how the compound is involved in a reaction, or thetype of reaction in which a compound is formed, the subscripts in a compound's molecularformula are fixed by nature and they never vary.

In ionic compounds, the subscripts in the molecular formula of a given compound alwaysdepend exclusively on the charges of the cations and anions in the compound. As we learned inChapter 3, cations and anions go together in such ratios as are necessary to form a neutralcompound.

In covalent compounds the subscripts in the molecular formula of a given compound arealso fixed and invariant. We are told how many of each atom occurs in each molecule ofcompound by the compound’s name.

As examples, water is always H2O regardless of whether it is a reactant or a product, andalso regardless of the reaction in which it is involved. Barium nitrate is always Ba(NO3)2, lithiumphosphate is always Li3PO4, barium phosphate is always Ba3(PO4)2 , and lithium nitrate is alwaysLiNO3, even if they are involved in reactions other than the one shown above.

Coefficients, on the other hand, always depend on the reaction. A compound may haveone coefficient in one reaction, and a different coefficient in another. As we see in the followingreaction

when barium nitrate reacts with lithium carbonate only a single molecule of barium nitrate isrequired as a reactant and only two molecules of lithium nitrate are produced. However, themolecular formulas of the barium nitrate and lithium nitrate are exactly the same in both of thepreceding reactions.

Balancing chemical equations

Let’s repeat some of the important points we just learned about chemical reactions. Massis always conserved in chemical reactions. If we have "x" number of a certain type of atoms in areactant compound (or compounds) there must also be exactly "x" of those atoms in productmolecules as well. Atoms are never lost or gained in chemical reactions. As a consequence,chemical equations must be balanced. There must be equal numbers of all atoms on both sides ofthe equation.

Let's take a second look at the reaction of barium nitrate and lithium carbonate. Theabove equation is balanced. We can tell whether or not an equation is balanced by counting thenumbers of each type of atom on both sides of the equation.

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In the balanced equation there is one barium atom and one carbon atom on each side of theequation.

As we said above, when balancing equations with polyatomic ions, it is usually easier tobalance the ions if they pass unchanged from one side of the reaction to the other. This willusually be the case in the reactions we examine in this chapter.

A chemical equation must be balanced to be of any use. When writing chemicalequations you must keep this in mind or your work will be incorrect. When balancing chemicalequations be sure to address the following points:

• Remember mass balance!• Use the Periodic Table and a knowledge of the formulas and charges of the common

polyatomic ions to decide the ratios of ions and how they go together to form ioniccompounds

• There is a difference between the subscripts in molecular formulas and the coefficients ofbalanced chemical equations.

• Coefficients are used to balance an equation.

We're almost to the point where we can begin to balance equations but we first need todiscuss one more topic. There are families of chemical reactions (or classes of chemicalreactions), just as there are families of elements and families of compounds. If we knowsomething about the general properties of these families of reactions we can easily predict theoutcomes for these types of reactions.

The two most common families of reactions we discuss in this chapter and class aredouble displacement reactions and combustion reactions.

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Double displacement reactions are the common type of reaction that occurs between twoionic compounds in aqueous solution (i.e., when both ionic compounds are dissolved in water).In a double displacement reaction the cations swap their anions. Generically, if AB and CD areionic compounds in aqueous solution, then in a double displacement reaction between them

AB + CD => AD + CB

AD and CB will be formed as a result. Note: given that these are ionic compounds, theirmolecular formulas tell us A and C are cations and B and D are anions. Remember the rule forionic compounds: cations are listed 1st, anions are listed 2nd. It would be incorrect to write theoutcome of this reaction as

AB + CD => AD + BC

because in the molecular formula BC, we’ve listed the anion 1st and the cation 2nd, which quitesimply is inappropriate.

If you look back at the three reactions we have just discussed in this chapter, you will be able tonotice this is exactly what happened in each case, i.e., all three reactions have been doubledisplacement reactions. Keep the constraining conditions in mind: both reacting compoundsmust be ionic compounds, and both must be in aqueous solution (dissolved in water).

Combustion reactions involve the interaction of some material with oxygen and result inthe formation of light and heat energy along with other products. Any time you see somethingburn you’re seeing a combustion reaction. All non-nuclear explosions are very fast combustionreactions. In the combustion of organic materials (materials containing both carbon andhydrogen in their molecular formulas) two products will always be formed, carbon dioxide gasand water vapor. In this class these are the only two products that will ever be formed in a reactioninvolving the combustion of an organic material. A generic version of the chemical equationdescribing the combustion of organic compounds is

"organic substance" + O2 (g) => CO2 (g) + H2O(l)

When balancing a double displacement reaction, it is helpful to remember the followingsteps:

1. Write the molecular formulas for the reactant compounds.2. Write the symbols for the cations and anions in each reactant compound beneath them,

since this will help predict the products that are formed.3. The charges of the ions determine the ratios in which they combine to form product

molecules; write the formulas of the product compounds.4. Place the products on the right side of the arrow.5. Use coefficients to balance the equation.

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As an example let's consider the double displacement reaction of potassium chloride andlead(II) nitrate. Hmm, do you think a knowledge of nomenclature might become useful at this point? Imean, it's not like this is the sort of thing I would put on an exam, is it? Or is it? You've got a 50/50chance here. Do you think there is any possibility of seeing problems like these on your exams for thisclass that one cannot begin to answer unless one is familiar with nomenclature?

Step 1 - write the molecular formulas for the reactants:

Step 2: Write the symbols for the cations and anions in each reactant compound beneath themsince this will help predict the products that are formed.

Step 3 - The charges of the ions determine the ratios in which they combine to form productmolecules; write the formulas of the product compounds. In this reaction the products will beKNO3 and PbCl2. If you have trouble with this step you need to re-visit Chapter 3.

Step 4 - Place the products on the right side of the arrow.

Step 5 - Use coefficients to balance the equation.

And that's all there is to it! When balancing the equations of double displacement reactions, itreally doesn't matter where you begin, but you will find it easiest if you start by balancing thehighest charged ions first and ions with the lowest charges last.

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When balancing organic combustion reactions you will find it easiest if you begin withcarbon, then balance hydrogen, and balance oxygen last. The number of moles of carbon dioxideformed is equal to the number of moles of carbon in the organic reactant. The number of molesof water vapor formed is equal to one-half the number of moles of hydrogen in the organicreactant. Consider the combustion of pentane, C5H12:

Step 1 - use the generic equation describing the combustion of an organic compound:

C5H12 + O2 => CO2 + H2O

Step 2 - balance carbon: As there are five carbon atoms in pentane we must have five carbondioxide molecules as product:

C5H12 + O2 => 5 CO2 + H2O

Step 3 - balance hydrogen: the twelve hydrogen atoms in pentane will form six water molecules,as

(6 water molecules) x (2 hydrogen atoms/water molecule) = 12 hydrogen atoms

C5H12 + O2 => 5 CO2 + 6 H2O

Step 4 - balance oxygen: on the product side we have (5 x 2) + (6 x 1) = 16 oxygen atoms. Theoxygen gas used in combustion reactions always occurs as a diatomic molecule consisting of twooxygen atoms. We need 8 oxygen molecules, as

(8 oxygen molecules) x (2 oxygen atoms/oxygen molecule) = 16 oxygen atoms

so the balanced equation describing the combustion of pentane is

C5H12 + 8 O2 => 5 CO2 + 6 H2O

Some combustion reaction require a 5th step. Let's look at the combustion of ethane,C2H6.

Step 1: use the generic equation describing the combustion of an organic compound:

C2H6 + O2 => CO2 + H2O

Step 2 - balance carbon: As there are two carbon atoms in ethane we must have two carbondioxide molecules as product:

C2H6 + O2 => 2 CO2 + H2O

Step 3 - balance hydrogen: the six hydrogen atoms in ethane will form three water molecules, as

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(3 water molecules) x (2 hydrogen atoms/water molecule) = 6 hydrogen atoms

C2H6 + O2 => 2 CO2 + 3 H2O

Step 4 - balance oxygen: on the product side we have (2 x 2) + (3 x 1) = 7 oxygen atoms. Theoxygen gas used in combustion reactions always occurs as a diatomic molecule consisting of twooxygen atoms. We need 3.5 oxygen molecules, as

(3.5 oxygen molecules) x (2 oxygen atoms/oxygen molecule) = 7 oxygen atoms

so the balanced equation describing the combustion of ethane is

C2H6 + 3½ O2 => 2 CO2 + 3 H2O

Step 5 - the problem with this balanced equation is that coefficients must always be wholenumbers. Fractional coefficients are not permitted in balanced equations (except under somevery special circumstances we see in chapter 7 when we discuss thermochemical equations). Wemust multiply all of the coefficients in the equation by two to convert the coefficient of 3½ to aninteger value of 7:

2 C2H6 + 7 O2 => 4 CO2 + 6 H2O

Is the equation still balanced? Count the atoms and see. In this class the only time you willencounter fractional coefficients is while balancing combustion reactions equations and then,only occasionally. You will never balance the equation of a double displacement reaction in thisclass and wind up with a fractional coefficient. Unless you’ve made a mistake.

Ok, let’s get some practice. Write a balanced chemical equation for each of the followingreactions. No peeking until you've solved the problem yourself first! At this point you have not yetbeen introduced to the rules that permit you to predict the state of a compound, so don't worry aboutthem - yet. Patience, my children, patience. This will be covered a little later in this chapter.

1. the double displacement reaction of sodium bromide and iron (III) nitrate2. the double displacement reaction of aluminum chloride and barium sulfate3. the double displacement reaction of ammonium phosphate and calcium sulfite4. the double displacement reaction of sodium cyanide and gold (III) nitrate5. the combustion of methane gas (CH4) 6. the combustion of propanol (C3H7OH)

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Answers:• 3 NaBr + Fe(NO3)3 => FeBr3 + 3 NaNO3

• 2 AlCl3 + 3 BaSO4 => Al2(SO4)3 + 3 BaCl2

• 2 (NH4)3PO4 + 3 CaSO3 => Ca3(PO4)2 + 3 (NH4)2SO3

• 3 NaCN + Au(NO3)3 => Au(CN)3 + 3 NaNO3

• CH4 + 2 O2 => CO2 + 2 H2O• 2 C3H7OH+ 9 O2 => 6 CO2 + 8 H2O

How did you do? Do you need more practice?

Strong electrolytes, weak electrolytes, and non-electrolytes

All compounds are classified as strong electrolytes, weak electrolytes, ornon-electrolytes. This is true of all ionic compounds and also of all covalent compounds.

Electrolytes are substances that form aqueous solutions capable of conducting electricity.A substance does this by dissociating (ionizing is a synonym) when in aqueous solution, whichmeans it breaks apart into ions when it dissolves in water. The ions serve as charge carriers. Purewater is non-conductive unless it has ions in it. Of course pure water with ions in it is no longer purewater, is it?

Strong electrolytes are substances that dissociate completely in aqueous solution. Nearly100% of all of the molecules of a sample of strong electrolyte will dissociate in aqueous solutionif they dissolve. All ionic compounds are strong electrolytes, as are all strong acids and bases.

Weak electrolytes are substances that dissociate incompletely in aqueous solution.Typically only1-10% of a sample of weak electrolytes will dissociate when they dissolve inwater (although weak electrolytes may also dissociate more than 10% or far less than 1%, dependingon the substance and on experimental conditions such as temperature pressure, concentration, andetc., things we’re not going to worry about in this class). The most common weak electrolytes are theweak acids and bases, which comprise a far more extensive list than that of strong acids andbases.

Organic acids (also known as carboxylic acids) are weak acids, and there are thousands of them(remember: how do we recognize a substance as an acid? A compound that has as molecular formulabeginning with hydrogen is typically an acid, for example, H3PO4, HCl, H2SO4, H2O, etc. Or, thecompound’s name includes the word “acid” in it).

Ammonia and a class of compounds known as amines are the most common weak bases. In otherwords, if a compound’s name includes “amine” it’s probably a weak base. Examples includemethlyamine, triethanolamine, and histamine.

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It is also common for organic compounds that contain nitrogen to be weak bases. Although thisis an extremely general rule, it is not a bad one to use in this class. This means that compoundssuch as pyridine (C5H5N), indole (C8H7N), and adenosine (C10H13N5O4) can be recognized asweak bases, simply by virtue of the presence of “N” in the compounds’ molecular formulas.

Non-electrolytes do not ionize at all in aqueous solution, even though they may dissolve.Nearly all organic compounds are non-electrolytes, with the exceptions noted in the previousparagraph, organic acids and organic bases. The majority of the binary covalent compounds weencounter in this class are also non-electrolytes.

strong electrolytes weak electrolytes non-electrolytes

general all ionic compounds

nearly all organiccompounds exceptorganic acids andbases; most binary

covalent compounds

acidsstrong acids: HCl,HBr, HI, HNO3,H2SO4, HClO4

weak acids: H2O, HF,H3PO4, H2CO3,H2SO3, H2S, all

organic acids, usuallyany acid not named

as a strong acid

basesstrong bases: all

Group I and Group IIhydroxides

weak bases:ammonia (NH3),NH4OH, amines,

organic bases(usually contain C,H, N in molecular

formula), usually anybase not named as a

strong base

This table is not absolutely correct but it does provide us with a very good set of generally accurateguidelines that are more than adequate for a class at this level. You would be very well advised tocommit the contents of this table to memory or a 3" x 5" card that you keep handy as you workthrough this chapter. In fact, I’ll tell you right now: when you take the quizzes and exams that coverthis material, plan on including the information in this table on your 3" x 5" card. You’ll need it. I canjust about promise it. And - I know this would never happen, but just in case you may have forgottenabout acids and bases: we talked about them near the end of Chapter 3. Do yourself a favor and brushup.

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Note that the distinction between strong and weak electrolytes lies in the extent to whichthey dissociate. There is absolutely no correlation between an acid or base being strong or weakand how dangerous or reactive it might be. Many of the most dangerous chemicals you mayencounter in a chemistry laboratory are weak acids and bases. I am confident that the number ofannual injuries and deaths caused by weak acids and bases far exceed those caused by strongacids and bases. In over 30 years of chemistry the most serious injury I have received was caused bythe mishandling of a weak acid. This is also true for most of my friends who have been injured by acidsor bases. The weak acids and bases can be remarkably corrosive to flesh. Handle them with a greatdeal of care!

We should also note that dissolution, or, the process of dissolving, and dissociation, theprocess of ionization, are treated as two different things in this class. Before a substance candissociate it must first be able to dissolve. We will explain this in a later chapter. If a substance cannot dissolve it cannot dissociate, even if it is a strong (or weak) electrolyte. Having said this,there are many ionic compounds that do not dissolve in aqueous solution but which are stillclassed as strong electrolytes, because if they could dissolve, they would also dissociatecompletely. There are many non-electrolytes that do dissolve in water but which still do notdissociate. Table sugar is a good example. Molecules of sugar dissolve readily in water, butbecause of their chemical nature they do not ionize at all. Compounds that dissolve in water aresaid to be soluble, and are indicated with the (aq) subscript in chemical equations. Compoundsthat do not dissolve in water are said to be insoluble and are indicated with the (s) subscript inchemical equations. They are sometimes referred to as precipitates, if they are formed asproducts in a double displacement reaction. This information will be valuable to you very shortly.Again, for your own sake: please keep careful, accurate notes of this material!

Molecular, ionic, and net ionic equations

We learned above that chemical equations are a shorthand way of describing a chemicalreaction. There are different types of chemical equations in common use.

Molecular equations are equations in which all reactants and products are written ascomplete molecules even though they may exist as ions in solution. In other words, in amolecular equation all compounds are represented with their molecular formulas, and allelements are represented with their elemental symbols. To this point in this course nearly all ofthe chemical equations you have seen in this class have been molecular equations.

Complete ionic equations, or, ionic equations as they are sometimes called, differ frommolecular equations in that soluble strong electrolytes are written as ions if they are in aqueoussolution. Ionic equations reflect the real state of things in aqueous solution. I’ll explain thisbelow.

Net ionic equations are ionic equations in which the spectator ions have been canceledout. The actual reaction that takes place is what remains. A net ionic equation often represents

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what we would see happening if we watched the reaction happen. Spectator ions are those ionswhich are needed to balance the mass and charge of a reaction but which do not actually takeplace in the reaction. Spectator ions are ions which are exactly the same in number and form onboth the reactant side and the product side of the equation.

It is important to note that not all ionic equations have spectator ions. In these cases, theionic equation and the net ionic equations are the same. It is equally noteworthy to observe thatthere may be ionic equations in which all of the ions are spectator ions. In these cases, there is nonet ionic equation, which may indicate there is no reaction that occurs when the reactants aremixed.

Let's do some examples of these three types of equations, beginning with the doubledisplacement reaction of sodium chloride and silver (I) nitrate.

• molecular equation:

NaCl(aq) + AgNO3 (aq) => AgCl(s) + NaNO3 (aq)

Note that all four compounds are written in their molecular form, i.e., represented using theirmolecular formulas. Further, note that we always begin with a correctly balanced molecularequation, although this one was pretty easy to balance as all of the coefficients are “one.”

• ionic equation:

Na+(aq) + Cl-

(aq) + Ag+(aq) + NO3

-(aq) => AgCl(s) + Na+

(aq) + NO3-(aq)

When writing ionic equations, in deciding whether to write compounds using their molecularformula or as ions, we must answer yes to two questions: (1.) is the compound a strongelectrolyte? and (2.) is it soluble? If we answer yes to both questions the compound is written asions. If we answer no to one or both questions, the compound is represented with its molecularformula. (These two rules are definitely something you want to remember and/or include on your3" x 5" card!)

As NaCl, AgNO3, and NaNO3 are all strong electrolytes and are all also soluble (asindicated by the (aq) subscript), they are written in ionic form since, as strong electrolytes, this isthe form in which they really exist in aqueous solution. Soluble strong electrolytes don’t floataround as intact molecules. They move about as ions. While AgCl is also a strong electrolyte, itis insoluble and as such it is left in its molecular form. While AgCl is a strong electrolyte, it doesn’tdissolve in water (at room temperature). And, if it can’t dissolve, that means it also can’t dissociate.How do we know that AgCl is insoluble? That's what the (s) subscript means. (s) does not stand forsoluble. It stands for solid, and compounds classed as solids are insoluble in aqueous solution. Butagain, how do we know? This is what we’ll learn in the next section of this chapter.

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• net ionic equation:

Cl-(aq) + Ag+

(aq) => AgCl(s)

In the ionic equation, Cl- and Ag+ combine to form solid AgCl while Na+ and NO3- do

nothing in the actual reaction. They are exactly the same in number and form on both thereactant side and the product side of the equation. This makes them spectator ions, and as suchthey are canceled from the ionic equation to leave the net ionic equation.

As discussed above, does it matter which order the ions are written in? Would it also becorrect to write

Ag+(aq) + Cl-

(aq) => AgCl(s)

The answer is yes, this is also correct. The order of reactants and products in a chemical equationdoes not matter as long as reactants are found on the reactant side and products are found on theproduct side.

Now let's write the molecular, complete ionic, and net ionic equations for thedouble-displacement reaction of barium nitrate and ammonium sulfide.

• molecular equation:

Ba(NO3)2 (aq) + (NH4)2S(aq) => BaS(s) + 2 NH4NO3 (aq)

We start with a properly balanced equation.

• ionic equation:

Ba2+(aq) + 2 NO3

-(aq) + 2 NH4

+(aq) + S2-

(aq) => BaS(s) + 2 NH4+(aq) + 2 NO3

-(aq)

Are Ba(NO3)2, (NH4)2S, and NH4NO3 strong electrolytes? Yes, they are ionic compounds. Are theysoluble? Yes, their (aq) subscript indicates that they are soluble. They are therefore represented as ionsin an ionic equation. While its true, BaS is a strong electrolyte, it is insoluble and so we leave it in itsmolecular form. If it can’t dissolve neither can it ionize.

• net ionic equation:

Ba2+(aq) + S2-

(aq) => BaS(s)

In the ionic equation we notice that NO3-(aq) and NH4

+(aq) occur in exactly the same number and form

on both the reactant side and the product side of the equation. They are therefore spectator ions andare canceled, leaving us with our net ionic equation.

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As a final example, let's write the molecular, complete ionic, and net ionic equations forthe double-displacement reaction of lithium sulfate and lead(II) acetate. See if you can do thisone by yourself and then check your answer below.

• molecular equation:

Li2SO4 (aq) + Pb(C2H3O2)2 (aq) => PbSO4 (s) + 2 LiC2H3O2 (aq)

• ionic equation:

2 Li+(aq) + SO4

2-(aq) + Pb2+

(aq) + 2 C2H3O2-(aq) => PbSO4 (s) + 2 Li+

(aq) + 2 C2H3O2-(aq)

• net ionic equation:

SO42-

(aq) + Pb2+(aq) => PbSO4 (s)

So, how did you do? Do you “get it,” or do you need more practice?

Double displacement reactions: precipitation formation and solubility rules

The mixing of two aqueous solutions of ionic compounds may result in the formation of anew insoluble ionic compound called a precipitate. Remember: insoluble means it will not dissolve inwater. This type of reaction is known generally as a precipitation reaction.

We can predict the solubility of ionic compounds using a set of solubility rules. Whilethese rules are limited in that they only describe the solubility of certain ionic compounds inwater, they still provide us with a surprisingly broad and accurate tool. Note: different text booksstate these rules in slightly different ways, even though they’re mostly similar. Regardless of what yourtext says, I hold you accountable for these rules in these notes only. You don’t need to memorize themor include them on your 3" x 5" card. I”ll give them to you on exams. You simply need to know howand when to use them. Also note: you cannot use these rules to predict the solubility of covalentcompounds in water. They only work for a limited number of ionic compounds.

Rule 1: ionic compounds with Group I cations and ammonium ion are always soluble. This ruletakes precedence over all other rules. Regardless of anion, if an ionic compound has one of thesecations, it will *always* be soluble.

Rule 2: ionic compounds with acetate, nitrate, and perchlorate as anions are always soluble.

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Rule 3: ionic compounds with halogens as anions are always soluble unless the cation is Ag+,Hg22+, Hg2+, or Pb2+ Don’t forget these exceptions! And be sure you know and remember thedifferences between mercury (I) ion and mercury (II) ion.

Rule 4: ionic compounds with sulfate as an anions are always soluble unless the cation is Ag+,Hg22+, Hg2+, Pb2+, Ca2+, Sr2+, or Ba2+ These are the same exceptions as those for the halogen anionsin Rule 3 plus the addition of the larger Group 2 cations.

Rule 5: ionic compounds with carbonate, phosphate, sulfide, and hydroxide as anions are alwaysinsoluble unless the cation is a Group 1 cation or ammonium ion or unless the compound is astrong base - and as I’m sure you recall, earlier in this chapter we told you that the strong bases arethe Group 1 and Group 2 hydroxides.

While I really do try to keep those things you should memorize to a minimum, you must be familiarwith and memorize these rules (or include them on your 3" x 5" card), and also those used to assignoxidation numbers, which we will discuss below. I will of course provide you with all of this informationon exams, but you really, really should know it. It will make life simpler for you, and it will make it morelikely that you will work problems quickly and correctly.

Will a reaction occur if the aqueous solutions of the following ionic compounds aremixed? Assume that if a reaction occurs it is a double-displacement reaction. Remember, thetypical reactions of two ionic compounds in aqueous solution is a double displacement reaction. And inthis chapter, the only other type of reaction we talk about is combustion reactions. So don’t makethings harder for yourself than is necessary. When I ask you about a reaction in this chapter, you’ve gota 50-50 chance of guessing what type of reaction it will be, either double displacement or combustion.If you really haven’t picked up on the differences between double displacement and combustionreactions, pull a coin out of your pocket and flip it. You’ve going to be right 50% of the time simply byguessing and with a bit of luck. We begin by writing the balanced molecular equation for thereaction. Then, using the solubility rules, we assign the state of each reactant and product. A veryimportant caveat: if one or both of the reactants is insoluble, the reaction will not occur aswritten because the ions are not free to mix and recombine as is required in a doubledisplacement reaction. This, however, does not prevent us from predicting the outcome of thereaction if it could occur.

! Ammonium phosphate and iron(III) nitrate " write the balanced equation: (NH4)3PO4 + Fe(NO3)3 => FePO4 + 3 NH4NO3

" is it soluble or solid? (NH4)3PO4 (aq) + Fe(NO3)3 (aq) => FePO4(s) + 3 NH4NO3 (aq)

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! Sodium iodide and mercury(II) acetate" write the balanced equation: 2 NaI + Hg(C2H3O2)2 => HgI2 + 2 NaC2H3O2

" is it soluble or solid? 2 NaI(aq) + Hg(C2H3O2)2 (aq) => HgI2 (s) + 2 NaC2H3O2 (aq)

! Ammonium iodide and potassium hydroxide" NH4I + KOH => KI + NH4OH" NH4I(aq) + KOH(aq) => KI(aq) + NH4OH(aq)

! Silver(I) acetate and cesium sulfate" 2 AgC2H3O2 + Cs2SO4 => Ag2SO4 + 2 CsC2H3O2

" 2 AgC2H3O2 (aq) + Cs2SO4 (aq) => Ag2SO4 (s) + 2 CsC2H3O2 (aq)

! Rubidium carbonate and chromium(III) chloride" 3 Rb2CO3 + 2 CrCl3 => Cr2(CO3)3 + 6 RbCl" 3 Rb2CO3 (aq) + 2 CrCl3 (aq) => Cr2(CO3)3 (s) + 6 RbCl(aq)

! Gold(III) carbonate and nickel(IV) sulfide" 2 Au2(CO3)3 + 3 NiS2 => 2 Au2S3 + 3 Ni(CO3)2

" 2 Au2(CO3)3 (s) + 3 NiS2 (s) => 2 Au2S3 (s) + 3 Ni(CO3)2 (s)

Double displacement reactions and water formation

Water can be formed in two common types of double displacement reactions, the reactionof an acid with a base, or in the reaction of an acid with an ionic hydroxide compound. Whilemost of the acids we discuss in this class are covalent compounds, in aqueous solution theybehave like ionic compounds in that they do dissociate to some extent, depending on whetherthey are strong acids or weak acids. You should also know that acids are always soluble in water,unless you are told expressly otherwise.

We can define what acids and bases are in a variety of ways and have mentioned thispreviously. You’ll find this in Chapter 3 as well as earlier in this chapter. We will discuss threedifferent acid-base theories in detail in Chapter 10, but for the time being we will contentourselves with the definitions provided by the Arrhenius theory. In aqueous solution, an acid is asubstance that donates a hydrogen ion (H+), and a base is a substance that donates a hydroxideion (OH-). All of the strong and weak acids and all of the strong bases described in theelectrolyte table we gave you earlier in the chapter are Arrhenius acids and bases. Ammonia andamines are not Arrhenius bases, but we won't worry about that until Chapter 10.

When Arrhenius acids and bases in aqueous solution are mixed, the resulting reaction iscalled a neutralization reaction. It is a double-displacement reaction in which water is formed asa pure liquid. An ionic compound, known as a salt, is also formed.

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Salts are a very large family of chemical compounds. By definition salts are ioniccompounds which may be formed as the result of neutralization reactions. All salts are ioniccompounds; however, not all ionic compounds are salts. Table salt, sodium chloride, is of coursea member of the salt family, but so are thousands of other ionic compounds. That a compound isclassed as a salt does not necessarily mean that it was formed by a neutralization reaction. Itmeans that it can be formed by a neutralization reaction.

The generic representation of a neutralization reaction looks something like this:

acid + base => H2O(l) + salt

An example of a neutralization reaction is the reaction of hydrochloric acid and sodiumhydroxide:

• molecular equation: HCl(aq) + NaOH(aq) => H2O(l) + NaCl(aq)

• ionic equation: H+(aq) + Cl-

(aq) + Na+(aq) + OH-

(aq) => H2O(l) + Na+(aq) + Cl-

(aq)

• net ionic equation: H+(aq) + OH-

(aq) => H2O(l)

Note in the molecular equation that the salt formed in this reaction is our old friend sodium chloride,without whom french fries or popcorn would never taste the same. However, not all salts are this tasty.Some are actually quite foul. There are even those that, when ingested, cause vomiting or more seriousillnesses, like - death.

Hydrochloric acid can donate a single proton to a neutralization reaction. We categorizeall acids, strong or weak, that can donate one hydrogen ion (i.e., one proton) to a reaction asmonoprotic acids.

Acids with two hydrogen ions available to participate in reactions are categorized asdiprotic acids. Sulfuric acid (H2SO4) is an example of a diprotic acid.

Acids with three available hydrogen ions are categorized as triprotic acids. Phosphoricacid (H3PO4) is an example of a triprotic acid.

Let's look at examples of the neutralization reactions of diprotic and triprotic acids. Whensulfuric acid is mixed with potassium hydroxide, we observe the following:

• molecular equation: H2SO4 (aq) + 2 KOH(aq) => 2 H2O(l) + K2SO4(aq)

• ionic equation: 2 H+(aq) + SO4

2-(aq) + 2 K+

(aq) + 2 OH-(aq) => 2 H2O(l) + 2 K+

(aq) + SO42-

(aq)

• net ionic equation: 2 H+(aq) + 2 OH-

(aq) => 2 H2O(l) *or* H+(aq) + OH-

(aq) => H2O(l) For net ionic equations it is common to reduce the coefficients of the equation to the lowestcommon denominator. This is the only time is it permissible to do this. You never adjust thecoefficients in molecular equations in this class.

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And when phosphoric acid reacts with calcium hydroxide:

• molecular equation: 2 H3PO4 (aq) + 3 Ca(OH)2 (aq) => 6 H2O(l) + Ca3(PO4)2 (s)

• ionic equation: 2 H3PO4 (aq) + 3 Ca2+(aq) + 6 OH-

(aq) => 6 H2O(l) + Ca3(PO4)2 (s)

• net ionic equation: 2 H3PO4 (aq) + 3 Ca2+(aq) + 6 OH-

(aq) => 6 H2O(l) + Ca3(PO4)2 (s)

There are no spectator ions, so the ionic equation and the net ionic equation are the same.

You may be wondering why phosphoric acid is left in its molecular form in the ionic andnet ionic equations. Remember, in an ionic equation, we only write a compound as ions if itpasses the two tests: (1.) is it a strong electrolyte? and (2.) is it soluble? Phosphoric acid is aweak acid, which means it is also a weak electrolyte. Since, in aqueous solution, far morephosphoric acid remains in its molecular form than dissociated as ions, it is more correct torepresent it in its molecular form. This is always true when representing weak electrolytes inionic and net ionic equations: leave them in their molecular form (represent them using theirmolecular formulas).

Finally, let's consider the reaction of acetic acid, HC2H3O2, a weak acid, and copper(II)hydroxide. This is not a neutralization reaction. Copper(II) hydroxide is not a base. How do weknow phosphoric and acetic acids are weak acids? Very simply, because they're not on the list of strongacids I gave you earlier. And in the case of acetic acid, it’s an organic acid; organic acids are alwaysweak acids. How do we know copper(II) hydroxide is not a base? Because it’s not on the list of strongbases you were given earlier in this chapter. Resist the temptation to over-complicate things and makethis harder than it is!

• molecular equation: 2 HC2H3O2 (aq) + Cu(OH)2 (s) => 2 H2O(l) + Cu(C2H3O2)2 (aq)

• ionic equation: 2 HC2H3O2 (aq) + Cu(OH)2 (s) => 2 H2O(l) + Cu2+(aq) + 2 C2H3O2

-(aq)

• net ionic equation: 2 HC2H3O2 (aq) + Cu(OH)2 (s) => 2 H2O(l) + Cu2+(aq) + 2 C2H3O2

-(aq)

Again there are no spectator ions, and the ionic and net ionic equations are the same.

It is important to note that in these double displacement reactions between an acid and abase or an ionic hydroxide, it is possible for one of the reactants to be a solid and for a reactionto still occur. The previous reaction is an example. Here’s another. Let's go to the kitchen and tryan experiment to prove this to ourselves. Pour a small amount of baking soda in a dish. Poursome vinegar onto the baking soda. Does a reaction take place? Vinegar is a dilute aqueoussolution of acetic acid. Baking soda is solid sodium hydrogen carbonate (NaHCO3), a base. Whatyou see when you perform this experiment is proof that a reaction can occur between a solid baseand a liquid acid. These reactions can also occur liquid or aqueous bases and solid acids.

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Double displacement reactions and gas formation

Double displacement reactions in aqueous solution which result in gas formation onlytake place under a very limited set of circumstances. There are many reactions other than thosethat we will discuss in this section which may result in the formation of a gas, but nearly all ofthem are electron transfer reactions and will be considered after this section. For gas formation tooccur in a double displacement reaction two conditions must be met:

• one of the reactants must be an acid• the anion of the other reactant must be either a carbonate, sulfite, sulfide, or cyanide.

Let's look at the behavior of each of these four anions in turn.

Consider the reaction of hydrobromic acid and sodium carbonate. It doesn't really matterwhich acid we select. All acids will react in essentially the same way with carbonate ion, although theremight be slight differences in the resulting net ionic equations. At first glance it might seem that themolecular equation should appear as

2 HBr(aq) + Na2CO3 (aq) => H2CO3 (aq) + 2 NaBr(aq)

but carbonic acid, H2CO3, is unstable when formed by double displacement reactions. Itspontaneously decomposes to form liquid water and carbon dioxide gas

H2CO3 (aq) => H2O(l) + CO2 (g)

The resulting equations appears as:

• molecular equation: 2 HBr(aq) + Na2CO3 (aq) => H2O(l) + CO2 (g) + 2 NaBr(aq)

• ionic equation: 2 H+(aq) + 2 Br-

(aq) + 2 Na+(aq) + CO3

2-(aq) => H2O(l) + CO2 (g) + 2 Na+

(aq) + 2Br-

(aq)

• net ionic equation: 2 H+(aq) + CO3

2-(aq) => H2O(l) + CO2 (g)

Sulfite ion behaves similarly to carbonate ion when it reacts with an acid. The result isthe formation of sulfurous acid, H2SO3, which is also unstable when formed in aqueous solutionas the product of a double displacement reaction. Sulfurous acid generated in a doubledisplacement reaction will spontaneously decompose to form water and sulfur dioxide gas

H2SO3 (aq) => H2O(l) + SO2 (g)

Sulfur dioxide is a very sharp-smelling irritating gas used to preserve fruits and vegetables. It isproduced by the combustion of sulfur and is an important contributor to the production of acid rain, as

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when rain water combines with atmospheric sulfur dioxide, sulfurous acid is formed. And unfortunately,when sulfurous acid is formed this way it is a bit more stable. So it falls to the ground and attacks andcorrodes stone and kills living things such as trees. To illustrate the behavior of sulfite ion in thepresence of an acid, let's examine the reaction of perchloric acid and ammonium sulfite.

• molecular equation: 2 HClO4 (aq) + (NH4)2SO3(aq) => H2O(l) + SO2 (g) + 2 NH4ClO4 (aq)

• ionic equation: 2 H+(aq) + 2 ClO4

-(aq) + 2 NH4

+(aq) + SO3

2-(aq) => H2O(l) + SO2 (g) + 2 NH4

+(aq)

+ 2 ClO4-(aq)

• net ionic equation: 2 H+(aq) + SO3

2-(aq) => H2O(l) + SO2 (g)

Sulfide ion, S2-, reacts directly with hydrogen ion in double displacement reactions toform hydrogen sulfide gas, H2S(g). I realize that, as a covalent compound, the correct name of H2Sshould be dihydrogen sulfide, but it's historic name, while systematically incorrect, is the accepted nameof the compound. Hydrogen sulfide is also known as rotten egg gas. You often smell it when passing oilrefineries and around geothermal features such as geysers and hot springs. Our noses are extremelysensitive to it and can detect it in concentrations of around 1 ppb (part per billion!) or less. At levels of10-50 ppm (parts per million), it can cause serious injury, and even quick - or immediate - death atsufficiently high concentrations. This is a serious problem on oil rigs when pockets of hydrogen sulfidegas are hit while drilling for petroleum or natural gas. Hydrogen sulfide gas is naturally produced by thedecay of dead plant and animal life, as are oil and natural gas. Fortunately, the stench of hydrogensulfide is so intolerable at toxic concentrations that most people don't wait around to be injured by it.One very nasty attribute of the gas is its ability to induce nasal tetany, paralysis of the olfactory nerves.Sometimes people exposed to it get one good sniff, lose their sense of smell, think that the gas hasdissipated, and continue to work in a dangerous environment until they collapse and die. You are highlyunlikely to ever be in this position unless you work on an oil rig. It is also fascinating to learn that ourbodies produce very, very small amounts of hydrogen sulfide naturally, and that it serves as an essentialneurotransmitter molecule. Without the H2S our bodies produce naturally, we’d die. And yet, if we’reexposed to too much of it from a source eternal to our bodies, the compound will kill us. Chemistry inthe real world is full of truly amazing paradoxes!

When, for example, sulfuric acid reacts with cesium sulfide, we observe the following:

• molecular equation: H2SO4 (aq) + Cs2S(aq) => H2S(g) + Cs2SO4 (aq)

• ionic equation: 2 H+(aq) + 2 SO4

2-(aq) + 2 Cs+

(aq) + S2-(aq) => H2S(g) + 2 Cs+

(aq) + SO42-

(aq)

• net ionic equation: 2 H+(aq) + S2-

(aq) => H2S(g)

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Cyanide ion behaves like sulfide ion in that it reacts directly with hydrogen ion in doubledisplacement reactions to form hydrogen cyanide gas, HCN(g). This is another very nasty anddangerous gas. It inhibits respiration at a cellular level. This is, in fact, the gas that was used in the gaschamber to execute people. There is an interesting article on gas chambers at Wikipedia, if you’reinterested, but obviously it is not required reading. Hydrogen cyanide gas is also naturally produced bythe combustion of organic compounds that also contain nitrogen - and many do. Hydrogen cyanideoccurs in very small concentrations in tobacco smoke, and also in camp fire smoke. But as long as youdon’t inhale too much camp fire smoke you don’t have anything to worry about.

When, for example, hydrochloric acid reacts with potassium cyanide, we observe the following:

• molecular equation: HCl (aq) + KCN(s) => HCN(g) + KCl(aq)

• ionic equation: H+(aq) + Cl-

(aq) + K+(aq) + CN-

(aq) => HCN(g) + K+(aq) + Cl-

(aq)

• net ionic equation: H+(aq)+ CN-

(aq) => HCN(g)

As is the case with neutralization reactions, it is possible for one of the reactants to be asolid and for a gas-forming reaction to occur. We see this in the previous example. Andremember our example of the reaction of baking soda and vinegar? The bubbles formed arecarbon dioxide gas. In this case the acetic acid reacts with sodium hydrogen carbonate accordingto the equation

HC2H3O2 (aq) + NaHCO3 (aq) => H2O(l) + CO2 (g) + NaC2H3O2 (aq)

Electron transfer (redox) reactions

Electron transfer reactions are chemical reactions in which one or more electrons istransferred from one atom to another. Electron transfer reactions are also known asoxidation-reduction reactions, or as redox reactions. pronounced REE-dox.

In the real world, redox reactions are far more common than double displacementreactions which result in precipitation, neutralization, or gas formation. Redox reactions are notconstrained to occur in aqueous solution, nor do the reactants have to be ionic compounds.Electron transfer lies at the heart of many important processes, from respiration andphotosynthesis and the reactions in batteries to corrosion and combustion and making thingspretty and shiny by applying a thin layer of metal coating to them.

Let's begin with some definitions.

• Oxidation (oxidized): the loss of one or more electrons• Reduction (reduced): the gain of one or more electrons oxidation and reduction are

inextricably tied to each other. One never occurs without the other also taking place inthe same reaction. Absolutely never!

• Oxidizing agent: a chemical that oxidizes something else and reduces itself

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• Reducing agent: a chemical that reduces something else and oxidizes itself

A commonly used mnemonic device for remembering oxidation and reduction is "OIL RIG"which stands for “oxidation is loss, reduction is gain.” Of course, we’re talking about oxidationbeing the loss of electrons, and reduction is the gain of electrons.

How can we tell when electron transfer occurs? By using a concept known as oxidationnumbers. Oxidation numbers are a theoretical book-keeping method devised as a way of keepingtrack of electrons in reactions, as well as the potential for certain types of atoms to lose or gainelectrons during redox reactions. With the rules described below, we assign an oxidation numberto every atom in every atom in the reaction, on the both the reactant and the product side of thechemical reaction equation. If the oxidation number of the same atom is different on the productside than it was on the reactant side, then it has either been oxidized, if the atom has lostelectrons, or reduced, if the atom has gained electrons. If the oxidation number for a specifictype of atom remains exactly the same on both the reactant and the product side of the chemicalequation., that atom has not played a role in a redox reaction.

Rules for determining oxidation numbers:

Rule 1: the oxidation number of atoms in their elemental state is zero. A substance is in itselemental state when it has not chemically combined with any other element, e.g., Fe, O2, andetc.

Rule 2: the oxidation number of a monatomic ion is equal to its charge. As examples, the Na+

cation also has an oxidation number of +1. The O2- anion has an oxidation number of -2.

Rule 3: the oxidation number of oxygen is always equal to -2 unless in it's molecular form (i.e, itappears as O2 and is not in a compound with another type of atom; see Rule 1) or in a peroxide(we will not discuss peroxides in this class, but O in peroxides has an oxidation number of -1).

Rule 4: the oxidation number of hydrogen is always +1 unless in it's molecular form (i.e, itappears as H2 and is not in a compound with another type of atom; see Rule 1) or in a hydride(we will not discuss hydrides in this class, but H in hydrides has an oxidation number of -1).

Rule 5: Fluorine always has an oxidation number of -1. The other halogens always have anoxidation of -1 as anions in ionic compounds or as the second named atom in binary molecularcompounds. Halogens listed as the first member of a binary molecular compound or involved inoxyanions have positive oxidation numbers and must be determined using Rule 6 below.

Rule 6: for either a neutral compound or for any polyatomic ion, the sum of the oxidationnumbers of the atoms in the molecule is equal to the net charge of the compound or ion.

You must be familiar with and able to use these rules. As with the solubility rules, I will provide them foryou on exams.

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Let's practice assigning oxidation numbers to a few compounds before we use it inequations.

! NO3-

" oxygen is always -2, and there are three of them" the net charge on the molecule is -1" we do not have a rule specifically for nitrogen, but we can figure out it's oxidation

number using Rule 6" assume the oxidation number of the one nitrogen atom is X" using Rule 6: (1)(X) + (3)(-2) = -1" X = +5, meaning the oxidation number of the nitrogen atom in nitrate ion is +5

! NO2-

" oxygen is always -2, and there are two of them" the net charge on the molecule is -1" we do not have a rule specifically for nitrogen, but we can figure out it's oxidation

number using Rule 6. That the nitrogen atom in nitrate ion has an oxidation numberof +5 doesn’t mean a thing here. We have to figure it out. Perhaps it will be same.Perhaps not. Let’s see.

" assume the oxidation number of the one nitrogen atom is X" using Rule 6: (1)(X) + (2)(-2) = -1" X = +3, meaning the oxidation number of the nitrogen atom in nitrite ion is +3

! H2SO4

" oxygen is always -2, and there are four of them" hydrogen is always +1, and there are two of them" the molecule is neutral, i.e., there is no net charge" we do not have a rule specifically for sulfur, but we can figure out it's oxidation

number using Rule 6" assume the oxidation number of the one sulfur atom is X" using Rule 6: (2)(+1) + (1)(X) + (4)(-2) = 0" X = +6, meaning the oxidation number of the sulfur atom is +6

! Fe(OH)2

" oxygen is always -2, and there are two of them" hydrogen is always +1, and there are two of them" the molecule is neutral, i.e., there is no net charge" we do not have a rule specifically for iron, but we can figure out it's oxidation

number using Rule 6" assume the oxidation number of iron is X" using Rule 6: X + (2)(+1) + (2)(-2) = 0" X = +2, meaning the oxidation number of the iron atom is +2

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Alternatively, remember that this is an ionic compound. When the compound dissociates it formsFe2+ ion and two OH- ions, but we already know how to figure out the oxidation numbers ofhydrogen and oxygen According to Rule 2, the oxidation number of the iron cation is equal to it'scharge, +2. This works for all ionic compounds, and in fact it is often simpler to break an ioniccompound down into its component ions before assigning oxidation numbers to its atoms.

! Li3PO4

" oxygen is always -2, and there are four of them" the molecule is neutral, i.e., there is no net charge" we do not have rules specifically for lithium or phosphorus, but we can figure out

their oxidation numbers using Rule 6 and by observing that this is an ioniccompound consisting of lithium ion and phosphate ion

" the cation is Li+, so the oxidation number of lithium is +1" assume the oxidation number of phosphorus is X" the net charge on phosphate is -3" using Rule 6: (1)(X) + (4)(-2) = -3" X = +5, meaning the oxidation number of the phosphorus atom is +5

! HClO3

" oxygen is always -2, and there are three of them" hydrogen is always +1, and there is one of them" the molecule is neutral, i.e., there is no net charge" we do not have a rule specifically for chlorine in a molecule made up of three

atoms, but we can figure out it's oxidation number using Rule 6" assume the oxidation number of chlorine is X" using Rule 6: (1)(+1) +(1)(X) + (3)(-2) = 0" X = +5, meaning the oxidation number of the chlorine atom is +5

! W2(SO3)3

" the molecule is neutral, i.e., there is no net charge" we do not have rules specifically for tungsten or sulfur, but we can figure out their

oxidation numbers using Rule 6 and by observing that this is an ionic compoundconsisting of a tungsten ion and sulfite ion

" sulfite ion has a 2- charge, and since there are three of them the tungsten ion mustbe W3+, which tells us that its oxidation number is also 3+

" the net charge on sulfite is -2" oxygen is always -2, and there are three of them in a single sulfite ion" assume the oxidation number of sulfur is X" using Rule 6: (1)(X) + (3)(-2) = -2" X = +4, meaning the oxidation number of the sulfur atom is +4

It is important to remember the distinction between charge and oxidation number. Chargeis real and is caused by the loss or gain of electrons. Oxidation number is imaginary and is adevice used to help track the movements of electrons from one atom to another in redox

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reactions. This is how it is possible to find positive and negative oxidation numbers for the atomsin covalent compounds. We know that atoms are electrically neutral in covalent compounds, butthey can still participate in electron transfer. Sometimes the charge and the oxidation number ofan atom are numerically the same, but do not forget that charge and oxidation number areconceptually two different things.

The formal IUPAC definition of oxidation number is: “Oxidation number is an empiricalconcept; it is not synonymous with the number of bonds to an atom. The oxidation number of anelement in any chemical entity is the charge which would be present on an atom of the element ifthe electrons in each bond to that atom were assigned to the more electronegative atom...”

In other words, what does an oxidation number actually mean? That can be difficult tosay. Fortunately, if we understand the rules we can uses the concept to help us ascertain whenelectron transfer reactions occur, even without really understanding what an oxidation numbermeans.

There are many different electron transfer reactions in the world around us, but there arefour common classes of reactions that are often - not always, but often - electron transferreactions. Prove to yourself that each of the example reactions is in fact an electron transferreaction by assigning oxidation numbers to all of the atoms in all of the compounds andobserving which atoms are oxidized and which atoms are reduced.

Combination reactions are those in which two smaller substances combine to form a thirdlarger substance. An example of a combination reaction that is also a redox reaction is thereaction of sodium and chlorine to form sodium chloride:

2 Na(s) + Cl2 (g) => 2 NaCl(s)

The oxidation numbers of sodium metal and chlorine gas are on the reactant side of the equationare 0, since both are in their elemental state. On the product side of the equation, the oxidationnumber of sodium ion is +1 and the oxidation number of chloride ion is -1. In this reactionsodium is oxidized and chlorine is reduced. Sodium is the reducing agent and chlorine gas is theoxidizing agent. Note that oxidizing agents and reducing agents are only ever found on the reactantside of the equation.

In decomposition reactions a single (relatively) larger compound reacts to form two ormore (relatively) smaller substances. The formation of hydrogen and oxygen from water is anexample of a decomposition reaction that is also an electron transfer reaction:

2 H2O(l) => 2 H2 (g) + O2 (g)

On the reactant side of the equation, the oxidation numbers of the hydrogen and oxygen in waterare +1 and -2 respectively. On the product side of the equation, the oxidation numbers ofhydrogen gas and oxygen gas are both 0, since each is in it's elemental state. In this reaction

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hydrogen is reduced and oxygen is oxidized, and water is both the oxidizing agent and thereducing agent.

Displacement reactions, or single displacement reactions, are reactions in which anelement reacts with a compound and replaces an element in the compound. They have thegeneral form

AB + C => A + BC

The reaction of chlorine and hydrogen sulfide is a single displacement reaction in whichoxidation and reduction occur:

8 H2S(g) + 8 Cl2 (g) => 16 HCl(g) + S8 (s)

On the reactant side of the equation, in hydrogen sulfide the oxidation number of hydrogen is +1and of sulfur is -2, while the oxidation number of chlorine gas is 0 since it is in it's elementalstate. On the product side of the equation, the oxidation number of hydrogen in hydrogenchloride gas is +1 while that of chloride ion in -1. The oxidation number of sulfur in S8 is 0,since this is the elemental state of sulfur (yes, we did mention this, but it was a long time ago.Check your Chapter 1 notes). Sulfur is oxidized and chlorine is reduced. Chlorine gas is theoxidizing agent and hydrogen sulfide is the reducing agent.

We've already learned a little bit about combustion reactions, in which a substance reactswith oxygen, resulting in a rapid release of heat and light energy. While many inorganicelements and compounds can be oxidized, we have learned specifically about organic oxidationreactions, in which carbon dioxide gas and water vapor are formed. Oxygen is reduced inorganic combustion reactions while carbon is oxidized. Take, for example, the combustion ofmethane (CH4):

CH4 (g) + 2 O2 (g) => CO2 (g) + 2 H2O(g)

The carbon atom in methane has an oxidation number of -4 while the hydrogen in methane hasan oxidation number of +1. The oxidation number of oxygen gas is 0 since it is in it's elementalstate. The oxidation number of carbon in carbon dioxide is +4 while that of oxygen is -2. Theoxidation numbers of the hydrogen and oxygen in water are +1 and -2 respectively. Carbon isoxidized in this reaction and oxygen is reduced. Oxygen gas is the oxidizing agent and methaneis the reducing agent.

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Post-script: responses to student questions about Chapter 5 and 6 concepts

Below are several notes I have sent to students over the years in response to material inthis chapter. While the questions were asked quite some time ago, there is useful information inthese notes for every student taking this class, no matter how well they may think they know thismaterial.

*************************This note addresses a question about balancing equations.

hi,

I would follow the sequence of steps suggested in the chapter 5 notes.

When balancing a double displacement reaction:

Step 1: write the molecular formulas for the reactant compounds

Step 2: write the cations and anions in each reactant compound beneath them, since this will helppredict the products that are formed

Step 3: the charges of the ions determine the ratios in which they combine to form productmolecules; write the formulas of the product compounds

Step 4: place the products on the right side of the arrow

Step 5: use coefficients to balance the equation

There is one example of how to employ these steps in the lecture notes. Here's another.

Provide a balanced molecular formula for the double displacement reaction of barium hydroxideand phosphoric acid.

Step 1: Ba(OH)2 and H3PO4

Step 2: Ba(OH)2 consists of Ba2+ and OH-; H3PO4 consists of H+ and PO43-

Step 3: Ba2+ and PO43- combine to form Ba3(PO4)2; H

+ and OH- combine to form HOH, or H2O asit is correctly written

Step 4: Ba(OH)2 + H3PO4 -> Ba3(PO4)2 + H2O

Step 5: 3 Ba(OH)2 + 2 H3PO4 -> Ba3(PO4)2 + 6 H2O

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*************************In this note I discuss how to use solubility rules.

hi,

regarding the reaction between lithium sulfate and lead (II) acetate on page 148 of the lecturenotes. You’ve asked how we can tell whether or not each of the reactant and product compoundsis soluble. The solubilities are determined by the solubility rules. Although these are generalrules, they're still quite useful. We need to look at each of the four compounds in the reaction andcompare it to what the solubility rules tell us. Note that these rules depend on the particularcompound and are independent of the coefficient in the balanced equation or even the reaction.In other words, according to the rules, lead (II) sulfate will always be a solid in aqueous solution,regardless of the reaction, regardless of whether it's a reactant or a product, regardless of itscoefficient in the balanced equation.

Ok, according to the rules:

Li2SO4 - according to Rule 1, compounds with a group 1 cation are always soluble in water. Ifwe check the periodic table, we see that Li+ is a group 1 cation. So this compound receives an(aq) subscript.

Pb(C2H3O2)2 - according to Rule 2, compounds with acetate as an anion are always soluble inwater, regardless of the cation. This compound also receives an (aq) subscript.

PbSO4 - according to Rule 4, compounds with sulfate as an anion are always soluble in waterexcept when the cation is Ag+, Pb2+, Hg2

2+, Hg2+, Ca2+, Sr2+, or Ba2+. Sulfate compounds with oneof these seven cations are insoluble in water and receive an (s) subscript. Hence PbSO4 islabeled (s).

LiC2H3O2 - this compound has a rule 1 cation and is therefore (aq). It also has a rule 2 anion,which also means it's (aq).

Remember, you don't need to memorize the rules. I'll give you a copy on the exam. You justneed to know how and when to use them.*************************To this student I attempted to address a question about the essential concepts in Chapter 5 and 6.

hi,

It is important to master this material. You will see several sets of questions exactly like theseQuiz 5 and 6 questions on the midterm and on the final as well. If you're going to make mistakes- and nearly everyone does on this material - it's better to make them on the quiz where it doesn'tcount against your grade as much as it does when an exam question is missed.

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Since you have asked general questions, I can only respond in a general manner. But it's astarting point.

First, there are a number of concepts addressed in Chapters 5 & 6. This is part of what makes itsuch a challenging chapter. Overall, the two most important concepts are those of stoichiometryand predicting the outcome of chemical reactions. But there are other concepts such as the moleconcept, strong and weak electrolytes, different types of chemical equations, and etc. that arealso important.

Despite how it may seem, I have only asked you about two types of reactions, combustionreactions and double displacement reactions. Don't make things harder for yourself than italready is. If a reaction you are asked about is not a combustion reaction, it must be a doubledisplacement reaction. Yes, it is as simple as that.

Double displacement reactions take place between two ionic compounds in aqueous solution.The acid-base reactions described in the notes are all double displacement reactions. The basesare ionic compounds. The acids are most commonly covalent compounds but acids alwaysbehave as though they are ionic compounds in that they ionize to a greater or lesser extent inaqueous solution.

You absolutely must know the Chapter 3 nomenclature stuff - as has been pointed out by others -or you haven't got a prayer at doing these problems. That is in part why I choose to emphasizethis material, as well as for its relevance in its own right.

I would strongly suggest that anyone having trouble with the Chapter 5 or 6 quiz to go back, as astarting point, to my lecture notes and very carefully re-read the sections entitled "Strongelectrolytes, weak electrolytes, and non-electrolytes," "Molecular, ionic, and net ionicequations," "Predicting the likelihood of chemical reactions," " Precipitation reactions andsolubility rules," "Acids, bases, and neutralization reactions," and "Gas formation." There is nofluff here. You must know everything in these sections to do these problems. If you skipanything, even a seemingly small detail, it will probably come back to cause trouble. As a part ofyour study of these sections, you really must work the examples I provide. Don't just read them.That will not help. You must sit down with a paper and pencil and work through each and everyexample with a great deal of care.

I don't know if this is helpful or not. This is exactly the same information I provide my lectureclasses, and these are exactly the same instructions I give my classroom students. This materialis extremely demanding and the only way to master it is to roll up your sleeves and get dirty withit. I know it's hard but don't get discouraged. Every semester students who persevere eventuallyfigure it out. Remember that the tutors in the Student Resource Center (SI 359) may not answerquiz questions but they are permitted to help you master the concepts. Please continue to postnotes here on the discussion board when you get stuck. And many, many thanks to those of youwho read these notes and who respond with helpful hints of your own.

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*************************In this final note I discuss redox reactions.

hi,

to determine whether or not a reaction is an electron transfer reaction you must assign anoxidation number to every atom in every substance in the reaction and see if they change as anatom passes from the reactant side to the product side of the equation. If none of the atoms haveoxidation numbers that change, it is not a redox reaction. If any of the atoms have oxidationnumbers that change, it is a redox reaction.

Oxidation is the loss of electrons. If an oxidation number becomes more positive, oxidation hastaken place. If an oxidation number becomes less negative, oxidation has taken place. As anexample, if in a reaction hydrogen has an oxidation number of 0 on the reactant side and anoxidation number of +1 on the product side, this means that hydrogen has been oxidized, i.e., ithas lost electrons.

Reduction is the gain of electrons. If an oxidation number becomes more negative, reduction hastaken place. If an oxidation number becomes less positive, reduction has taken place. As anexample, if in a reaction chlorine has an oxidation number of 0 on the reactant side and anoxidation number of -1 on the product side, this means that chlorine has been reduced, i.e., it hasgained electrons.

Oxidation and reduction are a paired event. You never have one without the other.

The word "agent" conveys a sense of what one compound does to another. An oxidizing agentoxidizes something else and in the process reduces itself. A reducing agent reduces somethingelse and in the process oxidizes itself. The oxidizing agent is the entire compound that containsthe atom that is reduced in the reaction. The reducing agent is the entire compound that containsthe atom that is oxidizes in the reaction. Oxidizing and reducing agents are always on thereactant side of the equation. Products are never oxidizing or reducing agents - at least, not inthis class.

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Chapter 6: Chemical Reactions - Mole and MassRelationships

Molecular formulas and empirical formulas

Let's review briefly. The molecular formula of a compound tells us what types of atomsare found in the compound and how many of each type of atom. It is not uncommon for two ormore covalent compounds to have the same molecular formula and in fact this happensfrequently. When this is the case we may use one of the four different types of structuralformulas to help differentiate between the compounds. See chapter 5 for a review of structuralformulas. There is usually only one ionic compound for a given molecular formula.

The empirical formula of a compound is the simplest whole number ratio of atoms in thatcompound. If we multiply the empirical formula of a compound by some integer we will obtainits molecular formula. The empirical formula and the molecular formula of many simplecompounds are the same.

compound molecular formula empirical formula

sodium chloride NaCl NaCl

ethene C2H4 CH2

1-octene C8H16 CH2

glucose C6H12O6 CH2O

formaldehyde CH2O CH2O

imaginary compound C25H75O15N5S10 C5H15O3NS2

Historically, the empirical formula of a compound was found using a technique calledelemental analysis which showed the proportions of the various elements by mass in a specificamount of the compound. This knowledge, coupled with the molar mass of the compound (whichis determined using other techniques) revealed the molecular formula of the compound. We willbe discussing molar masses below. This method of determining the molecular formula ofcompounds is seldom used any more. Powerful electronic instruments such as massspectrometers can determine the molecular formula of most compounds quickly and accurately,often in just a few minutes. We mention empirical formulas not because they have any great importance to us in class but becauseof their historical relevance. You should know what the definition of an empirical formula is and shouldalso be able to identify the empirical formula of a compound given it's molecular formula. However, you

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will never need to reduce the molecular formula of a compound to its empirical formula for anythingwe do in this class.

Molecular formulas, molecular weights, and formula weights

The molecular weight of a compound is the combined mass of all of the atoms in a singlemolecule of the compound. The molecular weight of a compound is found by finding the sum ofthe masses of all of the atoms in the compound. Molecular weight is abbreviated mw. The units ofmolecular weight are amu, atomic mass units. As we said in chapter 3 the mass of individual atomsis expressed in amu; one amu is equal to 1/12 the mass of single a 12C isotope, or 1.661 x 10-27 kg.The mass of individual atoms is found in the periodic table. The stated atomic weight for eachelement is the average mass of a single atom of the substance in amu. Since the mass ofindividual atoms is very small, it is more convenient to express that mass in amu than in gramsor kilograms.

compound molecularformula

molecular weight

sodiumchloride

NaCl1 Na atom x 22.989 amu/atom + 1 Cl atom x 35.453amu/atom = 58.442 amu

ethene C2H42 C atoms x 12.011 amu/atom + 4 H atoms x 1.008amu/atom = 28.054 amu

1-octene C8H168 C atoms x 12.011 amu/atom + 16 H atoms x 1.008amu/atom = 112.216 amu

glucose C6H12O6

6 C atoms x 12.011 amu/atom + 12 H atoms x 1.008amu/atom + 6 O atoms x 15.999 amu/atom =180.156 amu

formaldehyde

CH2O1 C atom x 12.011 amu/atom + 2 H atoms x 1.008amu/atom + 1 O atom x 15.999 amu/atom =30.026 amu

imaginarycompound

C25H75O15N5S10

25 C atoms x 12.011 amu/atom + 75 H atoms x 1.008amu/atom + 15 O atoms x 15.999 amu/atom + 5 Natoms x 14.0067 amu/atom + 10 S atoms x 32.06amu/atom = 1006.494 amu

The rules for calculating sig figs for addition and subtraction differ from those used to determine thenumber of sig figs when multiplying or dividing. However, I will not hold you responsible for correctlydetermining sig figs for addition and subtraction. The molecular weight of a compound may be

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small, as it is for elemental hydrogen (H2, mw = 2.016 amu). The molecular weight of largemolecules, such as proteins and strands of nucleic acids, may exceed 1 x 106 amu.

The molecular weights of ionic compounds are often called formula weights. Most ioniccompounds do not exist as discrete molecules but rather as interconnected lattices of ions. So,strictly speaking, it is not correct to think of ionic compounds as consisting of individualmolecules. Be this as it may, without going into any further discussion, the formula weight of anionic compound is numerically exactly the same as it's molecular weight. In this class we willalways refer to the molecular weight of ionic compounds and will not mention formula weightsagain.

The mole and Avogadro's number

By looking at the periodic table we can find the mass of an individual atom of anyelement. The units of these masses are given in amu (1.00 amu = 1.661 x 10-27 kg). As one amuis roughly equal to one-billionth of one-billlionth of one-billionth of a kilogram it is obvious thatthe mass of an individual atom is extremely small. Since it is difficult to measure the mass ofindividual atoms, is there a way to determine the mass of a large number of atoms or even better,to calculate the number of atoms in a given amount of a substance?

We can count atoms by weighing a substance. Then, using dimensional analysis, we candetermine the number of atoms in the mass of the substance we weighed.

A single carbon atom weighs 12.01 amu. If we take 12.01 amu of carbon, we have onecarbon atom. How many carbon atoms will we have if we take 12.01 grams of carbon rather than12.01 amu of carbon? Using dimensional analysis, we find

(12.01 g C) x (1 kg C/1000 g C) x (1 amu C/1.6605402 x 10-27 kg C) x (1 C atom/12.01 amu C) =6.02 x 1023 C atoms

This calculation means that in 12.01 grams of carbon, there are 6.02 x 1023 C atoms, or,there are about six hundred billion trillion carbon atoms. That means we have a trillion carbonatoms six hundred billion times That's rather a large number of carbon atoms.

Let's repeat this calculation for another element, this time for gold. From the periodictable we find the mass of a single gold atom is 196.97 amu. How many gold atoms are there in196.97 grams of gold?

(196.97 g Au) x (1 kg Au/1000 g Au) x (1 amu Au/1.6605402 x 10-27 kg Au) x (1 Auatom/196.97 amu Au) = 6.02 x 1023 Au atoms

Notice that the number of gold atoms in 196.97 grams of gold is exactly the same as thenumber of carbon atoms in 12.01 grams of carbon. How strange! Is this a coincidence?

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It is not a coincidence. We find that whenever we take the mass in grams of an elementequal to the mass in amu of one of it's atoms, the number of atoms in that mass is always6.02 x 1023 atoms.

This is also true of molecules. Whenever we take the mass in grams of a compound equalto the mass in amu of a single molecule (i.e. the mass in grams equal to the molecular weight ofthe compound in amu) there are always 6.02 x 1023 molecules present. There are no exceptions tothis, either for atoms or for molecules.

We calculated above that the mass of a single glucose molecule is 180.156 amu (from thesum of the atomic weights of the elements from the periodic table). How many glucosemolecules are there in 180.156 g of glucose?

(180.156 g glucose) x (1 kg glucose/1000 g glucose) x (1 amu glucose/1.6605402 x 10-27 kgglucose) x (1 glucose molecule/180.156 amu glucose) = 6.02 x 1023 glucose molecules

The mass in grams of any atom or of any molecule equal to its weight in amu will alwayscontain 6.02 x 1023 of that atom or molecule. The number 6.02 x 1023 is called Avogadro'snumber. Whenever we have Avogadro's number of anything, large or small, we have 6.02 x 1023

of that thing. We can have Avogadro's number of atoms and we can also have Avogadro'snumber of stars. And when we have Avogadro's number of anything, we say that we have 1 moleof that thing.

6.02 x 1023 of (x) = 1.00 mole of (x).

The mole is a unit of quantity. When we have a pair of things - whatever those thingsmight be - we have two of those things. When we have a dozen objects we have twelve of theobject. When we have a gross of something we have 144 of the thing. And when we have a moleof something we have 6.02 x 1023 of whatever. Given the small size of atoms, can you see anyadvantages in having a very large measure of quantity during the study of chemistry?

The mole is an unimaginably large number. If one does a Google search on the phrase“how big is a mole” various numbers can be found. While I cannot vouch for the accuracy ofthese calculations here are a few examples I found on the Internet that sound more or lesscorrect:

• one mole of glass marbles would cover the entire surface of the earth (oceans included) toa depth of several miles.

• a mole of beer cans (or Coke cans, if you prefer) would cover the entire surface of theearth (oceans included) to a depth in excess of 200 miles.

• If you count out loud starting with the number "one" at the rate of one count every secondit would take you about 2,000 trillion years to finish.

• If you covered the entire state of California with a mole of sand it would bury the state toa depth of ten feet.

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Molecular weight and molar mass

There is an important implication to all of this. The periodic table gives us information attwo levels. The atomic weight on the periodic table is both the mass of a single atom in amu andalso the mass of a mole of the substance in grams. We refer to the mass of one mole of anysubstance as it's molar mass. The molar mass of a compound is equal to the sum of the weightsof the moles of atoms in one mole of the compound.

compound molecularformula

molar mass

sodiumchloride

NaCl1 mole Na atoms x 22.989 g/mole + 1 mole Cl atomsx 35.453 g/mole = 58.442 g/mole

ethene C2H42 moles C atoms x 12.011 g/mole + 4 moles H atomsx 1.008 g/mole = 28.054 g/mole

1-octene C8H168 moles C atoms x 12.011 g/mole + 16 moles Hatoms x 1.008 g/mole = 112.216 g/mole

glucose C6H12O6

6 moles C atoms x 12.011 g/mole + 12 moles Hatoms x 1.008 g/mole + 6 moles O atoms x 15.999g/mole = 180.156 g/mole

formaldehyde

CH2O1 mole C atoms x 12.011 g/mole + 2 moles H atomsx 1.008 g/mole + 1 mole O atoms x 15.999 g/mole =30.026 g/mole

imaginarycompound

C25H75O15N5S10

25 moles C atoms x 12.011 g/mole + 75 moles Hatoms x 1.008 g/mole + 15 moles O atoms x 15.999g/mole + 5 moles N atoms x 14.0067 g/mole + 10moles S atoms x 32.06 g/mole = 1006.494 g/mole

As with the periodic table molecular formulas also give us information at two levels. Themolecular formula of a compound tells us how many of each atom it takes to make a singlemolecule of the compound. The molecular formula also tells us how many moles of each type ofatom it takes to make up one mole of molecules of the compound.

Let’s reiterate: the numbers for the molecular weight of a compound and the molar massof a compound are exactly the same. Only the units differ. The unit of molecular weight is theamu. The units of molar mass are grams per mole (g/mol). The abbreviation for mole is mol.Doesn't that save lots of time and ink?

water, H2OThe mass of a single molecule: 18.02 amu

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The mass of a mole of molecules: 18.02 g

carbon dioxide, CO2

The mass of a single molecule: 44.01 amuThe mass of a mole of molecules: 44.01 g

glucose, C6H12O6

The mass of a single molecule: 180 amuThe mass of a mole of molecules: 180 g

Diazinon is a pesticide with the molecular formula C12H21N2O3PSThe mass of a single molecule: 304.34 amuThe mass of a mole of molecules: 304.34 g

While molar mass and molecular weight are two different things the molar mass of acompound is frequently referred to as its molecular weight by people who use the termscarelessly. But as we have seen above, while they are conceptually similar in actual fact theyaren’t the same thing at all.

The linking relationship between the microscopic and the macroscopic is the mole.Microscopic: at an atomic or molecular level; macroscopic: on a level we can perceive with our unaidedsenses. If we know how many moles of a substance we have we also know how many atoms ormolecules of the substance we have. Conversely, if we know how many atoms or molecules ofthe substance we have we know how many moles of the substance we have. The followingexamples demonstrate how we may convert back and forth using the relationship betweenAvogadro’s number and the mole as a conversion factor.

! How many moles of carbon are in one mole of diazinon? how many carbon atoms? " there are 12 moles of carbon in every mole of diazinon We know this from

diazinon's molecular formula C12H21N2O3PS" (12 moles of carbon) x (6.02 x 1023 C atoms/1 mole carbon) = 7.22 x 1024 carbon

atoms! How many molecules are there in 1.4 moles of ethanol? The condensed structural formula

of ethanol is C2H5OH and it is commonly referred to as EtOH " (1.4 moles of EtOH) x (6.02 x 1023 molecules of EtOH/1 mole EtOH) = 8.43 x

1023 molecules EtOH ! A sample contains 7.75 x 1015 molecules of ethanol, how many moles is this?

" (7.75 x 1015 molecules EtOH ) x (1 mole EtOH/6.02 x 1023 molecules EtOH) =1.29 x 10-8 mole EtOH

! A sample of magnesium phosphate weighs 2.50 g, how many molecules is this? To do thisproblem you must first begin with the formula of magnesium phosphate (MP)." (2.50 g MP) x (1 mole MP/262.87 g MP) x (6.02 x 1023 MP molecules/1 mole

MP) = 5.73 x 1021 MP molecules

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! If a chemical assay based on the detection of phosphorus can detect 5.65 pg ofmagnesium phosphate, how many phosphorus atoms are present? An assay is a chemicaltest used to detect the presence of something, usually something present in very small amounts.

" (5.65 pg MP) x (1 g MP/1012 pg MP) x (1 mole MP/262.87 g MP) x (6.02 x 1023

MP molecules/1 mole MP ) x (2 P atoms/1 MP molecule) = 2.58 x 1010 P atomsDo you notice how we used relationships in the molecular formula of the compound asa conversion factor in this problem, 2 phosphorus atoms per MP molecule. Is it properto use the information in a molecular formula as a conversion factor? Absolutely.

! How many mg will 6.1 x 1020 butane molecules weigh? Butane has the molecular formulaC4H10 " (6.1 x 1020 molecules butane) x (1 mole butane/6.02 x 1023 molecules butane) x

(58.12 g butane/1 mole butane) x (1000 mg butane/1 g butane) = 58.9 mg butane

Stoichiometry

Let’s examine the combustion of liquid ethanol, C2H5OH(l):

C2H5OH(l) + 3 O2 (g) => 2 CO2 (g) + 3 H2O(g)

Chemical equations also provide us with information at two levels. Since molecularformulas provide both microscopic and macroscopic information, and since chemical equations consistof molecular formulas, this should come as no great surprise.

A chemical equation tells us how many molecules of each reactant are required toproduce a specific quantity of product molecules.

A chemical equation also indicates how many moles of reactants are required to producethe indicated number of moles of products.

In the above reaction, the equation shows that a single molecule of ethanol may reactwith a three molecules of oxygen gas, and one molecule of carbon dioxide and three moleculesof water will be produced. But it also tells us that when one mole of ethanol reacts with threemoles of oxygen gas, one mole of carbon dioxide gas and three mole of water vapor areproduced.

The balanced equation conveys information at two levels. It tells us that, on amicroscopic level, one ethanol molecule will react with three oxygen molecules to form twocarbon dioxide molecules and three water molecules. On a macroscopic level the balancedequation tells us that if one mole of ethanol reacts with three moles of oxygen, two moles ofcarbon dioxide and three moles of water vapor will form. But the balanced equation does not

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convey information about mass amounts. It would be utterly incorrect to assume from thebalanced equation that if we take one gram of ethanol and react it with three grams of oxygenthat two grams of carbon dioxide and three grams of water would form. No chemical equation,even a balanced one, ever provides that sort of information directly.

And yet mass to mass relationships are of a great deal of interest to chemists in the realworld. A common question one might ask is how much product will be formed during the courseof a reaction based on the starting amount of reactants. Or, how much reactant must one beginwith to guarantee the production of a certain amount of a compound. Using the above example ofthe combustion of ethanol, we might ask if 100.0 grams of ethanol are burned how many gramsof carbon dioxide will be produced? Or, in order to produce 100.0 grams of water vapor, howmany grams of ethanol must be used?

The answers to questions of this sort rely on a technique known as stoichiometry, whichhas been defined as the "calculation of the quantities of reactants and products involved in achemical reaction." I prefer to say that stoichiometry is the calculation of the quantities ofreactants and products based on their mole-to-mole relationships in a balanced chemicalequation. We will see that a balanced chemical equation is essential to stoichiometry, as is aknowledge of molar masses of the compounds of interest.

The calculations in stoichiometry problems are based on the mole-to-mole relationshipsderived from a balanced chemical equation. There is an equivalence between all of the reactantsand all of the products in every balanced chemical equation. These equivalences are used asconversion factors to set up and solve stoichiometric questions.

From this balanced chemical equation above we obtain the following sets ofequivalences. We use the equal sign (=) to indicate equivalence. It does not mean that the twocompounds are the same.

• 1 mol C2H5OH(l) = 3 mol O2 (g) • 3 mol O2 (g) = 2 mol CO2 (g) • 2 mol CO2 (g) = 3 mol H2O(g) • 1 mol C2H5OH(l) = 2 mol CO2 (g) • 3 mol O2 (g) = 3 mol H2O(g) • 1 mol C2H5OH(l) = 3 mol H2O(g)

These equivalences depend on the particular reaction and generally vary from onereaction to the next. While there is a 2:3 relationship between carbon dioxide gas and watervapor in this reaction, we found in the combustion of pentane (which is discussed above) thatthey existed in a 5:6 relationship to each other.

As I said, we can use these relationships as conversion factors. They are useful whencalculating amounts of reactants needed and/or amounts of product formed. Pay close attentionto the following examples. All are similar and yet each a little different from the next. Note that

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when mass to mass conversions must be made that they always go through mole to moleconversions first. They absolutely cannot ever be made directly. In other words, before we cando anything with a mass given to us in grams (or any other unit of mass) we must first convert itto moles using the molar mass of the substance.

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• How many moles of ethanol (EtOH) must be burned to produce 16.7 moles of carbondioxide?

(16.7 mol CO2) x (1 mol EtOH/2 mol CO2) = 8.35 mol EtOH• The combustion of 2.78 moles of ethanol will produce how many moles of water vapor?

(2.78 mol EtOH) x (3 mol H2O/1 mol EtOH) = 8.34 mol H2O

• How many moles of oxygen are required for the complete combustion of 33.6 moles ofethanol?

(33.6 mol EtOH) x (3 mol O2/1 mol EtOH) = 100.8 mol O2

• The combustion of ethanol produces 16.62 moles of carbon dioxide. How many moles ofwater vapor are also produced?

(16.62 mol CO2) x (3 mol H2O/2 mol CO2) = 24.93 mol H2O

• How many grams of ethanol must be burned to produce 125 grams of carbon dioxide?

(125 g CO2) x (1 mol CO2/44.01 g CO2) x (1 mol EtOH/2 mol CO2) x (46.07 g EtOH/1 molEtOH) = 65.4 g EtOH

• How many grams of oxygen are required for the complete combustion of 1500 grams ofethanol?

(1500 g EtOH) x (1 mol EtOH/46.07 g EtOH) x (3 mol O2/1 mol EtOH) x (32.0 g O2/1 mol O2) =3125.7 grams of oxygen

• The combustion of a certain amount of ethanol produces 6.5 milligrams of carbondioxide. How many milligrams of water vapor are also formed?

(6.5 mg CO2) x (1 g CO2/1000 mg CO2) x (1 mol CO2/44.01 g CO2) x (3 mol H2O/2 mol CO2) x(18.02 g H2O/1 mol H2O) x (1000 mg H2O/1 g H2O) = 3.99 mg H2O

Theoretical yield and percent yield

The masses calculated in stoichiometry problems represent theoretical possibilities. In thereal world the actual amount of a substance produced might differ significantly from what ispredicted through calculations. There are a number of reasons for this ranging fromunanticipated chemical effects to equipment problems to physical errors made during the actualexperiment. If, for example, during the course of a reaction you accidentally spilled some of yourproduct on the lab bench, do you think your final results would exactly match the amount youpredicted you would have? What if the substance clings to the interior wall of the reaction vessel

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so that you cannot transfer all of it? What if not all of the reactants react? What if there areimpurities in the reactants? What if a product changes into another chemical before you have theopportunity to assess the results of your experiment? And so on.

The amount of a chemical calculated using stoichiometry is called the theoretical yield.The amount of chemical that is actually measured in the lab is called the actual yield. The ratioof the actual yield to the theoretical yield, multiplied by 100, is called the percent yield.

(actual yield/theoretical yield) x 100 = percent yield

In the real world, chemists hope for high percent yields from their reactions. Of course a 100%percent yield is ideal and cannot be surpassed unless mistakes have been made. There are anumber of ways to end up with a percent yield greater than 100% when doing an experiment, butnone of them is correct. Remember that mass is always conserved in chemical reactions and is nevercreated or destroyed. Actual yields are often much lower. Percent yields of 70-90% or better arecommon for some types of reactions. For others, the percent yields might be much less. For somereactions, especially those that take place in a series of steps, the overall percent yield might beonly 10-15% or less.

• We calculated above that 65.4 g of EtOH must be burned to produce 125 g of CO2. Ifonly 88.6 g of CO2 are produced, what is the percent yield of the reaction?

(88.6 g CO2/125 g CO2) x 100 = 70.9% percent yield

• We calculated above that 3125.7 g of O2 are required for the complete combustion of1500 g of EtOH. If the percent yield of CO2 in this reaction is only 65.0%, how manygrams of CO2 are produced?

since (actual yield/theoretical yield) x 100 = percent yield, then actual yield = (percent yield x theoretical yield)/100

since we are not given the theoretical yield of carbon dioxide it must be calculated

(1500 g EtOH) x (1 mol EtOH/46.07 g EtOH) x (2 mol CO2/1 mol EtOH) x (44.01 g CO2/1 molCO2) = 2865.86 g CO2

(65.0 x 2865.86)/100 = 1862.8 g CO2

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Chapter 7: Chemical Reactions - Energy, Rates, andEquilibrium

An introduction to thermodynamics and kinetics

Two of the most important questions in the study of any chemical reaction are (1) is thereaction spontaneous (or non-spontaneous), and (2) if the reaction is spontaneous, how quickly(or slowly) will the reaction proceed? For each reaction, the answers to these questions aredetermined by changes in energy and in the molecular orderliness that occur during the reaction.

Many naturally occurring reactions are spontaneous. The double displacement reactionsof ionic compounds in aqueous solution in which there is the formation of a precipitate, a pureliquid, or a gas are usually spontaneous reactions. Many of the common reactions in whichelectron transfer takes place are also spontaneous reactions. However, there are also numerousimportant non-spontaneous reactions essential to life as we know it.

Thermodynamics is the branch of chemistry in which we study the changes in energy thatoccur during chemical reactions and changes in physical state. Thermodynamics is defined as thestudy of the transformations of energy, especially the transformation of heat into work and workinto heat. By observing whether a chemical reaction emits or absorbs energy, and whether itresults in the formation of substances that are more orderly at a molecular level (such as solids)or less orderly (such as gases) we can predict with perfect correctness whether the reaction willbe spontaneous or non-spontaneous. We also turn to thermodynamics to help us predict howmuch energy will be released by a spontaneous reaction, and how much energy we will need toput into a non-spontaneous reaction to make it happen.

In addition to predicting whether or not reactions are spontaneous or non-spontaneous, itis often important to have some idea of the time frame of a reaction. Comparatively speaking,different reactions occur at different rates. Some reactions are fast, and some are slow. Whether areaction is spontaneous or non-spontaneous does not have any apparent correlation with how fastor slow it might be. Some reactions, such as those that generate electricity in batteries or thattake place during combustion, are both spontaneous and fast. Some of the bond-breaking andbond-making events that occur during an explosion might take place on a picosecond or even afemtosecond time scale. Remember that a picosecond is a trillionth of a second and a femtosecond isa thousand-trillionth of a second. Other reactions, such as the oxidation of iron or the weathering ofrocks and other geological processes are spontaneous reactions that might take place over days,weeks, months, years, or even longer. Kinetics is the branch of chemistry dedicated to the studyof the rates at which reactions occur and those factors that affect the rates of reactions.

Energy, work, and other thermodynamic definitions

One of the quirky aspects of thermodynamics is that words we use one way in oureveryday communications sometimes have a somewhat different and very precise meaning in

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thermodynamics. The study of thermodynamics began during the early 19th century as scientistsattempted to make steam engines perform more efficiently. Surprisingly, their work on steamengines is both relevant and accurate in describing events that take place in a collection of atomsor molecules. It sometimes helps to keep this history in mind as we encounter the definitions ofthermodynamics terms.

In thermodynamics we divide the universe into two regions, the system and thesurroundings. The system is the part of the world in which we have an interest. The system iswhat we measure or observe. We make our measurements or observations from thesurroundings. In chemistry the system might be an event taking place in a test tube or a beaker.A system can be as small as a cell or bacterium (or even smaller), or as large as a planet or staror galaxy (or even larger). The system in chemistry is where ever the reaction of interest istaking place, however big or small that reaction might be. The surroundings are everything else.In essence, the surroundings are the rest of the universe.

We also need to say a word or two on sign conventions in thermodynamics. If a quantitysuch as heat or work is transferred from the system to its surroundings, the quantity is given anegative (-) sign. If, on the other hand, a thermodynamic property such as work or heat istransferred from the surroundings to the system, it is given a positive (+) sign.

from system to surrounds: (-)

from surroundings to system: (+)

As a part of our discussion of thermodynamics we must spend a few moments talkingabout energy, work, and heat.

Energy (commonly represented with the symbol “E”) is defined as the capacity to dowork.

Work (represented with the symbol “w”) is defined as a transfer of energy that can beused to the change the height of a weight somewhere in the surroundings. Remember that steamengines were not only used to propel trains and boats. They were also used to lift coal out of mines innorthern England and Wales and to help drain water in mines that would otherwise have flooded.While we typically could care less about the height of weights, scientists trying to improve theefficiencies of steam engines found this a compelling topic.

Heat (represented with the symbol “q”) is defined as a transfer of energy as a result of atemperature difference between the system and the surroundings.

Energy is measured in Joules (J) in the SI system and in calories (cal) in the Britishsystem. The relationship between Joules and calories is 4.184 J = 1 cal. Dietary calories areactually equal to 1,000 calories as we discuss them in this chapter. I'm not entirely sure as to why

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things are done this way, but can you imagine the shock and panic you might experience when youdiscover that the Snickers bar you just ate contained 280,000 calories? Perhaps there is some merit todoing things this way after all.

Energy governs everything that happens in chemistry. The lower the energy of a system,the greater its stability. The higher the energy of a system, the greater its instability. If eventsresult in the system having a lower energy at the end than it did in the beginning, the process willbe spontaneous. Events that result in the system having more energy in the end than it startedwith are non-spontaneous. This is true both of chemical reactions and also of physical changes.

There are two types of energy, potential energy and kinetic energy.

Potential energy is stored energy based on position. Most commonly we think ofpotential energy in terms of gravitational potential energy. If we pick an object up and hold itabove the surface upon which it rested, we have given it gravitational potential energy. If werelease the object, gravity will pull it back to the surface as the object's potential energy isconverted to kinetic energy. This occurs spontaneously not only because gravity is acting on theobject, but also because the object returns to a lower energy state as it falls. While gravity playsan essential role in the behavior of large objects, it has no effect on the outcome of chemicalreactions. This is true as far as we know. However, recent experiments on the Space Shuttle and inthe International Space Station suggest that gravity may in fact play a small role in the chemicalbehavior of atoms and molecules that we do not yet understand. Stay tuned over the next few decades.

In chemistry we are interested in chemical potential energy. This is the energy stored inchemical bonds, which is based on the strength of the attractive interactions between the bondingatoms, which in turn is related to their charges and to their positions with respect to each other.Remember the equation that describes bonding forces given in the "Chemical bonds and chemicalcompounds" section of Chapter 4 to understand this statement. According to this equation the closertwo oppositely charged particles are to each other the stronger the attractive interactions betweenthem. The further two oppositely charged particles are from each other, the weaker the attractiveinteractions they experience.

Kinetic energy is energy associated with motion and movement. We are most familiarwith translational motion, which is the energy of any object moving in a straight line from onepoint in space to another. Translational motion is possible in the gaseous and liquid states but notin the solid state, as particles are more or less locked into place.

There are also three other important types of motion that may impart kinetic energy toatoms and molecules.

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Bonded atoms vibrate with respect to each other in molecules. The behavior of twobonded atoms is similar to that of two masses at opposite ends of a spring. Vibrations are asource of kinetic energy to molecules in all three states of matter.

Molecules in the liquid and gas phases are free to rotate, or tumble, as they move. Thereis kinetic energy associated with rotational motion.

And, within molecules, atoms and their nuclei and electrons are free to spinindependently of the motion of the molecule in which they are found. This rotation is also asource of kinetic energy to particles in the solid, liquid and gaseous states.

The total kinetic energy of a molecule is the sum of all of its translational, vibrational,rotational, and spin motions. All molecules always have some kinetic energy, no matter how lowthe temperature. And the higher the temperature of a molecule, the greater its total kineticenergy.

It is important to remember that energy is neither created nor destroyed during chemicalreactions. This principle is called the conservation of energy. While energy is not created ordestroyed, it can be transferred (as heat, light, or electrical energy) or transformed (from heat tolight, etc.) during chemical reactions.

As stated above, work (w) is a transfer of energy that can be used to the change theheight of a weight somewhere in the surroundings. In chemical reactions it is most common forwork to be done by a gas which expands against an external force, such as steam driving a pistonagainst atmospheric pressure. This sort of work is called expansion work, or PV work. Insituations in which a system does work on its surroundings through the expansion of a gascreated in the system

w = - PΔV

where P is the applied pressure and ΔV is the change in volume that results from the expansionof the gas. While chemical reactions are capable of performing other types of work, such aselectrical work, these are not considered during fundamental discussions of thermodynamics.

Ok, I think that's it. Now that we're equipped with the language of thermodynamics, we can begin todiscuss it.

Internal energy

There are four important fundamental quantities used in thermodynamics to predict thespontaneity of reactions. These are internal energy (U), enthalpy (H), entropy (S), and Gibb'sfree energy (G).

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Internal energy is the total energy of a system, the sum of all of the potential and kineticenergies of all of the reactant particles and product particles in the system. During the course of achemical reaction the internal energy always changes.

ΔU = Ufinal - Uinitial

Remember that the Greek letter ) is used to indicate a change in the value that it precedes. As anychemical reaction proceeds from reactants to products, bonds are broken in the reactant particlesand new bonds are formed in the product particles. As bonds are storehouses of chemicalpotential energy, it should not surprise you to learn that the amount of chemical potential energystored in the bonds of the reactants is always different from the amount of chemical potentialenergy stored in the bonds of the products. If the temperature of the system changes as thereaction progresses because the reaction emits or absorbs heat, the kinetic energy of the reactantand product particles changes as well.

In other words, as a reaction proceeds from its initial state to its final state (i.e., fromreactants to products), the change in free energy in the system is equal to the difference betweenthe free energy of the system in its initial state and in its final state. This change in energy occursas heat and (or) work are transferred from the system to its surroundings, or from thesurroundings to the system. Work and heat are equivalent ways of changing the internal energyof the system. If the only interactions of a system with its surroundings consist of transfers ofheat to the system and the performance of work on the system, then

ΔU = q + w

Enthalpy

In a reaction in a system at constant pressure, the heat-associated change in internalenergy is called enthalpy. Enthalpy comes from the Greek word "enthalpein" which means "towarm." Since many important reactions, including a substantial fraction of all chemical reactionsthat occur in living systems, occur at atmospheric pressure (which is more or less constant)enthalpy is an important property of substances.

H = U + PV

We cannot directly measure the enthalpy of a substance, but we can measure changes inenthalpy so we are interested in ΔH, which is equal to

ΔH = Hfinal - Hinitial

or, alternatively

ΔH = ΔU + PΔV

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Let me interject two notes at this point. First, we will not prove this equation or demonstrate its origin,but you can find a proof of it in any thermodynamics textbook if you're really interested. Ok, I didn'tthink so. Second, we’re not going to use these equations in this class. But, as mathematics is thelanguage of chemistry, we actually find ourselves handcuffed if we attempt to discuss some of theseprinciples without the use of a few equations. So take a deep breath and stop hyperventilating! Thechange in enthalpy during a chemical reaction is equal to the difference between the enthalpy ofthe final state and the enthalpy of the initial state. We can also say that the change in enthalpy ofa system during a reaction is equal to its change in internal energy and any PV work done to orby the system.

Different types of enthalpy changes can be encountered during chemical reactions. Thisis to say, there are different events that, at constant pressure, lead to changes in internal energydue to releases or absorbing of heat. Some of the more common processes which result inchanges in enthalpy are:

• the enthalpy of reaction, or ΔHrxn which is also known as the "heat of reaction"• the standard enthalpy of formation, associated with the enthalpy changes of forming

compounds from the elements in their standard elemental state, ΔH°f

• there are enthalpies associated with physical changes of state (phase change), such as theenthalpy of fusion and melting, and of vaporization and condensation

• the processes of gaining or losing electrons also affect the enthalpy of compounds

In this class we are most interested in the enthalpy of reaction, ΔHrxn. The units of ΔHrxn

are usually expressed in kJ/mol. The enthalpy of reaction is the net change in heat-related energythat results from the breaking and making of chemical bonds that occurs during chemicalreactions.

We categorize all chemical reactions based on whether they emit or absorb heat.

In exothermic reactions, heat is given off by the reaction and it flows from the system toits surroundings. The products of exothermic reactions have less chemical potential energy thanthe reactants. Heat is essentially a product of the reaction. The sign of ΔHrxn for exothermicreactions is negative, -ΔHrxn

In endothermic reactions, heat is absorbed as it flows from the surroundings to thesystem. The products of endothermic reactions have more chemical potential energy than thereactants, so heat is essentially a reactant. The sign of ΔHrxn is positive for endothermic reactions,+ΔHrxn.

If a chemical reaction is exothermic it is usually spontaneous. If a reaction is endothermicit is usually non-spontaneous. However, there are numerous exceptions, and we will discuss thisa bit later in this chapter.

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In chemistry, properties can be classified as intensive or extensive. Intensive propertiesare independent of amount, in other words, intensive properties are the same whether we have alot or only a little bit of a substance. A few common intensive properties are color, odor, state ofmatter, density, temperature, melting point, and boiling point. Most of the properties we discussin this class are intensive properties. Extensive properties depend on the amount of substancepresent. Extensive properties include mass and volume.

Enthalpy is an extensive property. In other words, the more matter we have in a reaction,the greater the enthalpy change. Take, for example, the reaction of hydrogen gas and oxygen gasas they form liquid water:

2 H2 (g) + O2 (g) => 2 H2O(l) ΔHrxn = -571.6 kJ

A balanced chemical equation that also includes thermodynamic information is called athermochemical equation. The above thermochemical equation tells us that when two moles ofhydrogen gas and one mole of oxygen gas react to form two moles of liquid water vapor, 571.6kilojoules of energy is released. So, is this an exothermic or endothermic reaction? If we cut theamount of reactants by one-half, we find that the amount of energy released during the reactiondiminishes proportionally as well.

H2 (g) + ½O2 (g) => H2O(l) ΔHrxn = -285.8 kJ

Note that while we never use fractional coefficients in chemical equations, it is both correct andnecessary on occasion to use them in thermochemical equations. This is the only time in thisclass that the use of fractional coefficients is acceptable, as we study and discussthermodynamics. This is true because enthalpy is extensive. If we multiply through by two to eliminatethe fractional coefficient of oxygen, we must also multiply the enthalpy of reaction by a factor of two aswell. And if we increase the amount of reactants and products by a factor of ten, the enthalpy ofreaction also increases by a factor of ten:

20 H2 (g) + 10 O2 (g) => 20 H2O(l) ΔHrxn = -5716 kJ

But you already knew the enthalpy was extensive although you may not have realized it. Think about it.If enthalpy was not extensive, which would give off more heat, a very large fire, a very small fire, orwould they in fact give off equal amounts of heat? (Answer: if enthalpy was intensive, not extensive, abig fire and a small fire would give off equal amounts of heat) If enthalpy was not extensive, whatwould happen when you stepped on the gas pedal in your car? (Answer: nothing) Giving the enginemore fuel would not produce any more energy than giving it a little bit of fuel. And so on. Be gratefulthat enthalpy is extensive!

Enthalpies of reaction depend on the physical states of the reactants and products. If weexamine the following reaction in which water vapor is formed rather than liquid water:

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2 H2 (g) + O2 (g) => 2 H2O(g) ΔHrxn = -483.6 kJ

The reaction of hydrogen and oxygen to form water vapor is an exothermic reaction. Ifwe want to decompose water vapor to form hydrogen gas and oxygen gas, will it be anexothermic or endothermic process? And, will the process be spontaneous or non-spontaneous?To write the thermochemical equation for this reverse reaction, we simply place the products onthe reactant side of the equation, and the reactants on the products side:

2 H2O(g) => 2 H2 (g) + O2 (g) ΔHrxn = +483.6 kJ

This reversal of the reaction involves the same amount of energy as the forward reaction. Notethat the value of the heat of reaction has remained the same. However, the sign has changed. Tobreak two moles of water vapor down into two moles of hydrogen gas and one mole of oxygengas will require us to put 483.6 kilojoules of energy into the system.

This illustrates an important point. In theory, nearly all chemical reactions are reversible,although there may be practical considerations that make it unlikely. Given any exothermicreaction, it's reverse reaction is always endothermic. The reverse reaction for any endothermicreaction is always exothermic. The magnitude of the enthalpy change is constant. Only the signchanges.

We can use the relationships in thermochemical equations as conversion factors instoichiometry problems. As an example, we can ask how much energy will be released by thereaction of 100.0 g of hydrogen gas to form water vapor.

(100.0 g H2) x (1 mole H2/2.016 g H2) x (-483.6 kJ/2 moles H2) = -11,994 kJ

We can also predict how much energy will be required to break 100.0 grams of water vapordown into hydrogen and oxygen gases:

(100.0 g H2O) x (1 mole H2O/18.02 g H2O) x (+483.6 kJ/2 moles H2O) = +1,342 kJ

I probably will not require you to do this sort of problem. Probably. I’d still advise you to be sure youunderstand conceptually how do to these sorts of problems, just in case.

Entropy

We said above that if a chemical reaction is exothermic it is usually spontaneous, and thatif a reaction is endothermic it is usually non-spontaneous. This is a good general rule but thereare some exceptions. This is because changes in the order of the system also affect thespontaneity of chemical reactions.

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One of the fundamental laws of thermodynamics is that the order of a system decreasesduring spontaneous reactions. Although, as we will see in a moment, there are a few exceptions tothis. In the world around us there are many spontaneous processes in which order decreases.Melting is one example. Vaporization is another. Dissolving table salt or sugar, either one, inwater is yet another example. And this is also true of many chemical reactions.

So what do we mean by order and randomness in thermodynamics? Order andrandomness are measures of the structure in a system at an atomic level. An ordered system is ahighly structured system, much like a very large orchard. If you stand in the middle of a well laidout orchard the view looks the same in every direction, no matter which way you look. If youcould shrink yourself down to an atomic level and then stand in the middle of an orderedsubstance the view would look the same in every direction, no matter which way you look,including overhead and beneath your feet. In other words, in ordered substances there is a highdegree of repetition of particles and spacing between the particles in all three dimensions, formany hundreds (or thousands or even millions) of particle diameters.

We can make distinctions between the three common states of matter based on thedifferences in their internal order.

Solids are highly ordered (structured) substances. They have both short range and longrange order. Solids are, at an atomic level, like a very large three dimensional orchard. No matterwhere you look in a solid, the view is exactly the same in all directions for many hundreds orthousands (or millions, billions, or more) of atoms or molecules. This is true of crystalline solids. Allionic compounds form crystalline solids, as do many other substances in the solid state. Amorphoussolids, such as glass, are structurally more like liquids. The atoms and molecules in different types ofglass are not nearly as well-ordered and highly structured as the ions in a salt (sodium chloride) crystal,although this is not apparent to the naked eye. But you don’t need to worry about it. In this class, inthis chapter, we’ll assume all solids are crystalline solids.

Liquids are much less ordered (structured) at the atomic level than solids. Liquids haveshort range order but do not have long range order. This is similar to standing in an orchardwhere no matter which way you look the view is the same for three or four trees, but then thelines of trees become crooked and perhaps the spacing between the trees changes and you findthem closer together or further apart than the trees in your immediate vicinity. Because liquidshave less order than solids, we say that they are more disordered, or more random, or morechaotic. We use disorder, randomness, and chaos more or less synonymously in thermodynamics. Themore disordered (or random or chaotic) something is, the less structure (order) it has.

Gases have absolutely no order (structure) at the atomic level. Gases have neither shortrange order or long range order. At an atomic level gases are like an orchard that was laid out bysome one walking randomly around the grounds and throwing a handful of seeds here and there

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every now and then. It is like an orchard that has been hit by a bomb. No matter where you stand,nothing looks the same in any direction.

Entropy is a measure of the randomness (or disorder or chaos) of a substance. Arelationship exists between free energy and entropy, and we will discuss it in a moment.

As is the case with enthalpy, we cannot directly measure entropy. We can only measurethe changes in entropy that occur within a system as the system changes. We are particularlyinterested in the changes in entropy that result from chemical reactions. The change in entropyduring a chemical reaction is equal to the difference between the entropy of the final state andthe entropy of the initial state of the system

ΔSrxn = Sfinal - Sinitial

Any process which results in an increase in the entropy of the system has a value of +ΔS, whileany process the results in a decrease in the entropy of the system has a value of -ΔS. These signconventions are typically something of a challenge to reconcile with what we have said above aboutsigns in thermodynamics. So I’m going to let you take it on faith that when an entropy change results ingreater chaos (less order or structure) the sign of )S is positive, and when an entropy change results inless chaos (more order or structure) the sign of )S is negative.

Entropy increases when there are changes from more ordered states to less ordered states,such as the formation of a gas from solid reactants. Entropy also increases when the number ofmoles of products is greater than the number of moles of reactants. Entropy decreases when, as aresult of a chemical reaction, products are formed that are more ordered than the reactants orwhen the number of moles of products formed are less than the moles of reactants initiallypresent. The units of ΔSrxn are usually in J/molAK. We could discuss entropy in a far morequantitative manner, i.e., with math and equations and all that fun stuff, but it will be sufficient for ourpurposes if you understand entropy in a more general, qualitative sense.

Gibbs free energy and the spontaneity of reactions

The ultimate indicator of the spontaneity or non-spontaneity of any (and all) chemicalreactions is Gibb's free energy. The Gibbs free energy of a reaction, indicated with the symbol G,is the energy available to do useful work in a chemical reaction. As is true with enthalpy, wecannot measure Gibbs free energy directly, but we can observe how it changes during the courseof a chemical reaction.

ΔGrxn = Gfinal - Ginitial

The sign of ΔGrxn for spontaneous reactions is always negative, -ΔGrxn. The sign is positive fornon-spontaneous reactions, +ΔGrxn . The units of ΔGrxn are generally in kJ/mol. In other words, ifwe know the sign of ΔGrxn for any reaction, we also know with absolute certainty whether the

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reaction is spontaneous or non-spontaneous as written. The magnitude of ΔGrxn tells us howmuch energy released by a spontaneous reaction is available to do useful work, or how muchenergy must be put into a non-spontaneous reaction to make it happen.

It is important to note that non-spontaneous reactions can and do occur. It’s just that theydon’t happen to any great extent. Reactions with negative values of Gibbs free energy willtypically have a great deal of product and relatively little reactant remaining. Reactions withpositive values of Gibbs free energy will form very little (but not necessarily zero) product andhave a great deal of reactant that remains.

The Gibbs free energy of reaction ΔGrxn is always dependant on the enthalpy of thereaction ΔHrxn, the entropy of the reaction ΔSrxn, and the temperature at which the reaction isperformed. This relationship is described by the equation

ΔGrxn = ΔHrxn - TΔSrxn

Let's examine the implications of this equation.

If a reaction is exothermic and if the entropy of the system increases during the reaction,the reaction will always be spontaneous. Think about why this is true. If entropy increases as aconsequence of the reaction, ΔSrxn is positive. Temperature is always a positive number. We'reworking with the Kelvin scale during our study of thermodynamics. The lowest possible temperature onthe Kelvin scale is 0 K. It is not possible to have a negative Kelvin temperature. A positive numbermultiplied by a positive number is always a positive number, i.e., TΔSrxn is a positive number. Ifwe multiply a positive number by (-1), as we do when using this equation, the quantity TΔSrxn

becomes a negative number. If we add this negative number to the enthalpy of reaction, whichfor an exothermic reaction is always a negative number (-ΔHrxn), ΔGrxn cannot be other than anegative value, which means the reaction is spontaneous. Remember, the sum of two negativenumbers is always itself a negative number: a negative plus a negative equals a negative.

On the other hand, if a reaction is endothermic and if the entropy of the system decreasesduring the reaction, the reaction will always be non-spontaneous. If entropy decreases as aconsequence of the reaction, ΔSrxn is negative. Temperature is, as we said above, always apositive number . A positive number multiplied by a negative number is always a negativenumber, i.e., TΔSrxn is a negative number. If we multiply TΔSrxn by (-1), the quantity TΔSrxn

becomes a positive number. If we add this positive number to the enthalpy of reaction, which foran endothermic reaction is always a positive number, ΔGrxn will always have a positive value,and the reaction is non-spontaneous. Remember, the sum of two positive numbers is always itself apositive number: a positive plus a positive equals a positive.

This leaves us to discuss those reactions in which the enthalpy and entropy of thereaction are both either positive or negative. Let's consider an exothermic reaction in which theentropy decreases. This means that ΔHrxn is negative, ΔSrxn is negative, and -TΔSrxn is positive.

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Whether ΔGrxn is negative or positive depends on the temperature of the reaction. If the reactionis carried out at a high temperature, the -TΔSrxn term will be large enough to override thenegative enthalpy term and ΔGrxn will be positive. On the other hand, if the reaction is carried outat a low temperature, the -TΔSrxn term will be too small to override the negative enthalpy termand ΔGrxn will be negative.

Finally, let's examine an endothermic reaction in which the entropy increases. This meansthat ΔHrxn is positive, ΔSrxn is positive, and -TΔSrxn is negative. Again, whether ΔGrxn is negativeor positive depends on the temperature of the reaction. If the reaction is carried out at a hightemperature, the -TΔSrxn term will be large enough to override the positive enthalpy term andΔGrxn will be negative. If the reaction is carried out at a low temperature, the -TΔSrxn term will betoo small to override the positive enthalpy term and ΔGrxn will be positive. The following tablesummarizes the relationship between the signs of ΔHrxn, ΔSrxn, temperature, and ΔGrxn.

ΔHrxn ΔSrxn T ΔGrxn

- + any always -

+ - any always +

- - low/high -/+

+ + low/high +/-

Do you need to memorize this table? No. But you should be able to use it to help you answerquestions. You need to know what this stuff means and how it is used. An example of how it may beused is found on the Chapter 7 quiz..

So how do we use this information? In the conversion of ozone gas to oxygen gas

2 O3 (g) => 3 O2 (g)

the sign of ΔHrxn is negative and the sign of ΔSrxn is positive. This means, according to the table,that the sign of ΔGrxn is negative at all temperatures (i.e., at any temperature) and that thereaction is therefore spontaneous at all temperatures.

As another example, the process of dissolving solid sodium chloride in water has anenthalpy of reaction of +3.6 kJ/mol and an entropy of reaction of +43.2 J/mol.K. given that thesigns of both of these are positive and based on the above table, we can surmise that the sign offor ΔGrxn this process is positive at low temperatures and negative at high temperatures. In otherwords, sodium chloride will spontaneously dissolve in water at high temperatures but not at lowtemperatures. "High" temperature and "low" temperature are very much relative terms in thissituation. Exactly how high or low a temperature must be for a reaction of this type to be spontaneousor non-spontaneous depends on the magnitudes of the enthalpy and entropy terms which depend, in

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turn , on the particular reaction of interest. In some cases “high” temperature might be manythousands of degrees. The combustion of methane is an example of reaction that is exothermic andwhich also results in a decrease in entropy. For this particular combustion reaction it will remainspontaneous as long as the reaction occurs at a temperature less than 160,000 K. This is good newsunless you plan on using methane to heat your home in some place very warm, like, say for example,the surface of the sun. In other reactions “the “high” temperature may be only a few degrees aboveabsolute zero.

The rates of chemical reactions

The rates of chemical reactions can be expressed several different ways. It is common toexpress the rate of a reaction in terms of how fast a reactant disappears as a function of time, orin terms of how quickly a product appears as a function of time. Mathematically, this can beexpressed as

rate = Δ[reactant]/Δt or

rate = Δ[product]/Δt

where the square brackets [] represent the molar concentration of the substance contained withinthe brackets and "Δ" represents "the change in." Molar concentration is the number of moles of asubstance per liter of solution. It is discussed in Chapter 9 of your text. As an example, if we arediscussing the molar concentration of sodium chloride in a solution, we would it express it as [NaCl].This means that the mathematical statement "Δ[reactant]/Δt " means “the change in reactantconcentration with the change in time.” The rate of any chemical reaction depends on thesechanges in concentration as time progresses.

Let's examine the data from a fairly simple reaction. Hydrogen peroxide decomposes toform water and oxygen according to the equation

2 H2O2 (aq) => 2 H2O(l) + O2 (g)

Out of curiosity, is this a redox reaction? If you know the answer is yes without guessing - but becauseyou took the time to assign oxidation numbers, give yourself a big pat on the back!

If we begin with roughly 0.9 moles of hydrogen peroxide in 1.00 L of water, we can measure thedecrease in it's concentration with the passage of time. The concentration of hydrogen peroxidedecreases because it is being consumed as the reaction progresses. If we plot this datagraphically, we see the following:

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There are several important things to note in this plot. You can see that the trend line describedby the data is not linear, but gently curving. This behavior is not unique to this particular reactionbut is in fact characteristic of many reactions. An important implication of this non-linearbehavior is that the concentration of hydrogen peroxide will never reach zero, i.e., all of thehydrogen peroxide will never be completely consumed. We will discuss why this is so a littlelater.

At any given time, the average rate of the reaction is equal to the slope of the linedescribed by the data. We can see in the graph that the rate of the reaction changes as thereaction proceeds. We can calculate the slope between any two of the adjacent data points, solet's choose the first two data points. Remember your algebra: given two points, (x1, y1,) and (x2, y2)the slope m between the points is equal to m = (y2-y1)/(x2-x1). And a further word about slopes: thisequation is the way we calculate the average slope. This is fine for our purposes in this class. In trulyserious studies of kinetics we are usually more interested in the instantaneous slope, the slope of thecurve at any given instant, but since it requires a bit of calculus to determine it, we simply will not worryabout it.

m = (0.70 - 0.88)/(60 - 0) = -0.00300

If we do this for all of the pairs of data we will find that the reaction has slowed to about 15% ofit's initial rate by the time 10 minutes (600 seconds) has passed. This is consistent with the slopeof the line decreasing as the reaction progresses.

Reaction rates and molecular collisions

Why does the rate of the decomposition of hydrogen peroxide slow as it is consumed?Why doesn't the reaction continue on at the same rate until all of the peroxide is consumed?

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In most reactions particles must be able to collide if they are to react. Remember, we usethe word “particle” to generically represent atoms, molecules, and ions. Various factors affectreaction rates. Nearly all are factors that pertain to collisions between particles and include:

• reactant concentration: how many particles there are in a given volume of space. Themore particles in a given volume, the more frequently they will collide and the faster thereaction.

• reactant pressure (if a reactant is in the gas phase): as pressure increases in the gas phase,the number of reacting particles per unit volume of space will also increase, and thereforeso will the frequency of collisions and the rate of the reaction.

• spatial orientation: how the particles are aligned with respect to one another when theycollide. You can think of the interactions between particles in many reactions as beingsimilar to the interaction between a key and a lock. While there are numerous ways for akey and a lock to interact with each other, only when the key interacts with the lock inprecisely the right place - the key hole - will something happen. If the key strikes the lockanywhere but the keyhole, nothing will happen.

• system temperature affects both the speed of the particles, and therefore how often theycollide, but also the kinetic energy involved in these collisions. As a general rule, formost common reactions the rate increases as temperature increases, and reaction rateslows as temperature decreases.

• the presence of catalysts, which make reactions proceed more rapidly. For reasons we willsoon explain.

Chemical reactions are, at the atomic level, events in which chemical bonds are brokenand made. Bonds must be broken for the reactant molecules to react. New bonds are formed asthe “broken” reactant molecules recombine and product molecules are created. The energy usedto break chemical bonds is often the kinetic energy of the colliding molecules. This means thatduring most chemical reactions, the reacting molecules must collide with enough energy to resultin the breaking of bonds.

The reacting molecules must also be oriented properly, positioned in such a way that thecollisions between the molecules happen with the parts of the molecules that have the ability toreact. Why this is so may not be apparent but we will not elaborate further on this topic in this class.Suffice it to say that, tiny as they are, not all parts of most molecules are equally reactive. As I havedescribed above you can think of reacting particles as very tiny locks and keys.

Concentration has a significant effect on the number of effective collisions. As a reactantis consumed the number of reactant molecules decreases so there are fewer available to collideand react.

Temperature also has a profound affect on the rate of chemical reactions. As temperatureincreases, the kinetic energy of the molecules also increases and they collide with more energy,making the breaking of chemical bonds easier. As a general rule, reaction rates double with

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every 10°C increase in temperature, because the fraction of molecules in the system withsufficient energy to break bonds when they collide with other molecules increases with theincrease in temperature.

Rate Laws

Rates of reaction can also be expressed using rate laws. Rate laws are mathematicalexpressions of the relationship between reaction rate, temperature, and reactant concentration.For the reaction wA + xB <=> yC + zD, where w, x, y, and z are the stoichiometric coefficientsof the reactants A and B and the products C and D, the reaction’s rate law is written

rate = k[A]m[B]n

Rate laws are independent of time. They can be used to calculate the rate of a reaction at anypoint during the reaction as long as reactant concentrations are known. The value k is a rateconstant that depends on the particular reaction and on the temperature at which the reaction isperformed. The exponents m and n are called the orders of the reactants A and B respectively.They describe exactly how the concentrations of A and B affect the reaction rate and they mustbe determined experimentally. Which we will not do in this class but if you feel cheated, plan ontaking Chem 1220 from me some time and I'll show you how this is done. I should also tell you thatthe lack of a connection between rate and time in rate laws can be something of a nuisance. But fearnot. Using calculus we can integrate rate laws and establish a relationship between reaction rate,reactant concentrations, and time, something wonderfully useful in chemistry but sadly, something wewill not discuss in this class.

Activation energy and activated complexes

The process of breaking old bonds and making new bonds is integral to all chemicalreactions. The process is generally rapid, but it is not instantaneous.

During a reaction a reacting substance will typically find itself very briefly in anabnormally high energy state. We're talking milliseconds to femtoseconds here and in some caseseven faster, although in many cases it may also be substantially slower. There are very good reasonsfor this temporary increase in energy. Some of the reactant's atoms may briefly have more orfewer bonds than are required for them to obey the octet rule. Or, one or more of the reactantatoms may gain or lose an electron (or, electrons) temporarily, again causing these reactantatoms to violate the octet rule. Whatever the reason, when atoms violate the octet rule theirenergy increases. This in turn raises the energy of the entire reactant molecule in which theatoms are found. There are of course other ways for reactant molecule energy to increase besidesviolations of the octet rule but we will not consider them here. This temporary high energy state isalways an unstable one, as it is higher in chemical potential energy than either the starting

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potential energy state of the reactant molecules or the final potential energy state of the productmolecules. Remember that molecules have chemical potential energy, the energy stored in the bondsbetween the atoms. And also remember the correlation between energy and stability: in relative terms,high energy equals unstable and low energy equals stable. This temporary high energy state is calledthe transition state. The unstable substance in the transition state is called an activated complex.The increased potential energy of the activated complex is called activation energy and isrepresented with the symbol Ea. By definition the activation energy of a reaction is the differencein energy between the energy of the reactants and the energy of the activated complex in thetransition state.

Activation energy can be thought of as an energy barrier between the reactant andproduct states. Spontaneous reactions often require a bit of a "nudge" to get them started, usuallyin the form of a small amount of energy. For example, combustion reactions are spontaneousreactions, yet they seldom begin without a small initial input of energy to get them started. Youwill never enjoy a campfire without initially striking a match. And yet, this is still classified as aspontaneous reaction. Spontaneous reactions are reactions that result in a lower overall system energyin the product state than in the reactant state. The small amount of energy a spontaneous reactionmay require to get it started is generally insignificant compared to the overall changes in energy in thesystem as the reactants become products.

A car parked at the very edge of a steep hill will spontaneously roll down the hill if it isnot in gear and the emergency brake is not engaged. Some of you have probably learned this thehard way, through experience. All it takes is the slightest of pushes to start it rolling. If there is arock in front of one of the tires one must push a little harder to make the spontaneous rollingoccur. The larger the rock, the harder one must push to make something that is ostensiblyspontaneous happen. This rock is like the activation energy of a reaction. And if one wishes tomake the reverse reaction occur, one must not only supply enough energy to push the car backup the hill, but also enough to get the tires over the rock at the top of the hill.

Let's look at the change in the energy of a system during an exothermic reaction. In anexothermic reaction (a reaction that emits heat) the chemical potential energy of the reactants isgreater than the chemical potential energy of the products. The difference in potential energybetween the two is emitted as heat as reactants are converted to products. This emitted heat is theenthalpy of reaction, or, the heat of reaction. Even though exothermic reactions are generallyspontaneous (this also depends on entropy changes in the reaction), reactant molecules mustcollide with sufficient energy to be able to break bonds and form the activated complex thatexists in the transition state. Here we have a plot of the changes in the energy of a system as anexothermic reaction progresses from reactant to products. The “progress of reaction” axis is, really,just a measure of time as it increases.

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During an endothermic reaction (a reaction that absorbs heat) the starting chemicalpotential energy of the reactants is less than the final chemical potential energy of the products.The difference in potential energy between the two is absorbed, usually as heat, as reactants areconverted to products. This absorbed heat is the enthalpy of reaction. Endothermic reactions aregenerally non-spontaneous, because reactant molecules must collide with sufficient energy to beable to break bonds and form the activated complex that exists in the transition state.

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The reversibility of chemical reactions

At this point we need to learn a rather unexpected but important truth: nearly all chemicalreactions are reversible. This is true no matter how improbable the likelihood of the reservereaction occurring may seem. And there is excellent experimental evidence to confirm thisstatement.

For any exothermic reaction, it's reverse reaction will always be endothermic. Themagnitude of the enthalpy change is constant. Only the sign changes. In other words, if areaction has an enthalpy of reaction of “-X” then the enthalpy of reaction for the reverse reactionwill be exactly “+X.” Remember that exothermic reactions have negative enthalpies of reaction andendothermic reactions have positive enthalpies of reaction. The reason that this observation is true isdue to the extensive nature of enthalpy. Do you remember what extensive properties are and whythey’re important? For the reverse reaction of an exothermic process to occur, enough energy mustbe added to the products to (a.) restore them to the higher potential energy state of the reactants,and (b.) overcome the activation energy of the transition state.

Endothermic reactions are also reversible. For any endothermic reaction, it's reversereaction is always exothermic. The magnitude of the enthalpy change is constant. Only the signchanges. For the reverse reaction of an endothermic reaction to occur, enough energy must beadded to the products to overcome the activation energy of the transition state. Energy will be

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emitted as the higher energy product molecules are converted to lower energy reactantmolecules.

It is extremely important to realize that in saying that, in theory, all reactions arereversible, we are *not* saying that reverse reactions are equally as probable - or, if you’d rather,as energetically favorable - as the forward reaction. Far from it.

As we established above, all spontaneous reactions have a negative Gibbs free energyassociated with them, i.e., a -ΔG value. This means that the Gibbs free energy change associatedwith the reverse reaction will be equal in magnitude but opposite in sign. This is, as withenthalpy, due to the extensive nature of Gibbs free energy. In other words, if a reaction has anΔG value of “-X” then the value of ΔG value for the reverse reaction will be exactly “+X.” Inother words, the reverse reaction of a spontaneous reaction is nonspontaneous, meaning, it willtypically only happen to a slight extent.

Also, as we learned above, all nonspontaneous reactions have a positive Gibbs freeenergy associated with them, i.e., a +ΔG value. The value of the Gibbs free energy changeassociated with the reverse reaction will be equal in magnitude but opposite in sign. In otherwords, the reverse reaction of a nonspontaneous reaction is itself spontaneous, meaning, it willtypically happen to a far greater extent than the forward reaction.

Catalysts

A catalyst is a substance that increases the rate of a reaction without any net change in itsconcentration. In other words, catalysts make chemical reactions happen faster and are usuallynot consumed during the course of the reaction. Or if a catalyst is consumed at one point in areaction, it is immediately regenerated during another portion of the same reaction so that thereis no net change in its concentration.

Catalysts make reactions rates increase by lowering the activation energy of the reaction.This can be accomplished in one of two ways. Some catalysts work by providing an alternativereaction pathway with a lower energy transition state. Perhaps more common, catalysts provide asurface on which the reaction takes place. Molecules adsorp to the surface of the catalyst."Adsorp" means to chemically bond to the surface; this is different from "absorb", which is what occurswhen one material incorporates another material into itself, as when one absorbs spilled water with apaper towel. Features of the catalyst's surface result in the stretching of chemical bonds in theadsorped molecules. This results in weakening of the bonds which makes them easier to breakwhen other reactant molecules collide with them.

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In looking at this image it is hopefully apparent that enzymes catalyze both the forwardand the reverse reactions equally. If a catalyst lowers the activation energy for a given reactionby “X” kilojoules, it makes both the forward and the reverse reactions that much easier.

Enzymes are biological catalysts. They play an essential role in making biologicalreactions happen that would otherwise not occur, or that would happen so slowly as to causeserious health problems. Like, for example, death. Enzymes are very specific as to the substancesupon which they act. In some cases, simply changing one or two of the atoms of a large moleculecan change its shape to the extent that an enzyme can no longer act upon it. This specificity ofenzymes for their target molecules often results in the use of a "lock and key" model whendiscussing them. Enzymes are likened unto a lock, and the molecules that they act on, calledsubstrates, are analogous to a key. If you change the shape of a key ever so slightly, it will oftenno longer unlock the lock for which it was intended. Conversely, if you change the configurationof the tumblers in the lock (in other words, if you do something to change the shape of theenzyme), the key will no longer work properly. Although the “lock and key” model has beenreplaced with the “induced fit” model, I do not think the lock and key model is so far from correct thatit will hurt you to think of things this way. See the wikipedia entry for enzymes athttp://en.wikipedia.org/wiki/Enzyme for more information, if you’re interested.

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Chemical equilibrium

Let's return to our discussion of the decomposition of hydrogen peroxide. We saw thatthe rate of the reaction slows as hydrogen peroxide is consumed. This is indicated by a levelingof the trend line that connects the data points. If we plot the change in the oxygen concentrationof the system as a function of time, we eventually see similar leveling behavior in the trend line.

At first the rate of the production of oxygen is relatively fast and the slope of the line is relativelysteep. But the rate of production of oxygen slows as the reaction progresses, and the slope of theline begins to level off. The graph shows that rate of oxygen production begins to level off atabout the same time in the experiment that the rate of the decomposition of hydrogen peroxidebegins to slow.

Why is it that the rate of hydrogen peroxide decomposition and the rate of oxygenproduction both slow as the reaction proceeds? When we discussed this behavior above, we saidthat as the concentration of hydrogen peroxide decreases the number of molecules available toparticipate in collisions decreases and that the reaction slows as a consequence. This is true, butit is only part of the story. It also does not explain why the hydrogen peroxide is never entirelyconsumed.

The complete answer to this question lies in the reversibility of chemical reactions. Ashydrogen peroxide decomposes, liquid water and oxygen gas are produced. We know this fromthe balanced chemical equation for the reaction. As molecules of water and oxygen areproduced, collisions between them will sometimes result in the reverse reaction taking place, i.e.,in the production of hydrogen peroxide. As the concentrations of water and oxygen increase, thenumber of collisions between water and oxygen molecules also increases, and the rate of thereverse reaction increases as well.

In every reaction a point is reached at which the rate of the forward reaction is equal tothe rate of the reverse reaction. Remember that the slopes of the trend lines in the plot of the data

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are related to the rates of the reactions. When the slopes of the trend lines are parallel, the ratesof the two reactions are equal. In the above plot we do not have quite enough data to show thisleveling of the trend lines. However, if we extrapolate from the data that we have, we can guess thatboth trend lines will level out and become parallel between 600 and 1200 seconds. When we reachthis point we say we have reached equilibrium in the reaction. By definition, in any reaction astate of dynamic equilibrium exists when the rate of a forward reaction is equal to the rate of itsreverse reaction. A reaction at equilibrium is indicated with a two-headed arrow between thereactants and the products. All chemical reactions always proceed toward a state of equilibrium.For example, once the decomposition of hydrogen peroxide is at equilibrium, the chemicalequation can be written

2 H2O2 (aq) W2 H2O(l) + O2 (g)

Equilibrium is as a state of dynamic equilibrium. A reaction does not simply "shut off"when equilibrium is reached. At equilibrium both the forward and the reverse reaction continueto occur, but at the same rate. As a consequence, the concentrations of reactants and productsbecome constant, because as quickly as reactant molecules are consumed in the forward reactionto form products, product molecules are consumed in the reverse reaction in the production ofthe original reactants.

How long does it take a reaction to reach equilibrium? It varies, depending on thereaction and on the temperature at which the reaction takes place. Some reactions proceed toequilibrium rapidly, reaching it in seconds or even quicker. Other reactions might require aperiod of hours, days, or longer before equilibrium is attained.

How does a catalyst affect equilibrium? Catalysts speed up the rate of both the forwardand the reverse reactions. A catalyst will help the system reach equilibrium more rapidly but theydo not cause a shift in the ratio of the concentrations of the products to the concentrations of thereactants. A catalyst will help a slow reaction reach equilibrium more quickly, and a fast reactionreach equilibrium faster still.

The equilibrium constant

Equilibrium can be expressed mathematically using rate laws. Let's assume we have thefollowing reaction

wA + xB W yC + zD

and that the rate laws for the forward and reverse reactions are

ratef = kf[A]w[B]x

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and

rater = kr[C]y[D]z where ratef and rater are the rates of the forward and reverse reactions, respectively, kf and kr arethe rate constants of the forward and reverse reactions, respectively, and [A]w, [B]x, [C]y and [D]z

are the concentrations of the reactants and products raised to the powers of their respectivecoefficients. At equilibrium the rates of the forward and reverse reactions are equal to each other.Therefore, at equilibrium,

ratef = rater

which can be re-written as

kf[A]w[B]x = kr[C]y[D]z

and rearranged as

There is, actually, an advantage to expressing things this way. This last equation tells usthat for any chemical reaction at equilibrium, the ratio of the concentrations of the products andthe reactants is always a constant value. We call this constant the equilibrium constant, Kc. Forany reaction at equilibrium we can write an equilibrium expression or, equilibrium constantexpression that describes the relationship between the equilibrium constant and the balancedchemical equation for the reaction. As an example, for the reaction wA + xB W yC + zD, theequilibrium constant expression is

Why is this useful? Because for any given reaction Kc will only change if we change thetemperature at which the reaction is performed or if we change to a completely differentreaction. This means that regardless of how much reactant and product with which we start areaction, when equilibrium is reached, the ratio of the concentrations of the products andreactants will always equal the equilibrium constant.

To demonstrate this, let's look at the results of the reaction of dinitrogen tetroxide andnitrogen dioxide. The balanced chemical equation for this reaction is

N2O4 (g) W 2 NO2 (g)

k

k

C D

A B

f

r

y z

w x

[ ] [ ]

[ ] [ ]

Kk

k

C D

A Bc

f

r

y z

w x

[ ] [ ]

[ ] [ ]

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The equilibrium constant expression for this reaction is

Note that we are calculating Kp in these experiments rather than Kc. There are a number ofdifferent type of equilibrium constants. Kp is used when we are studying equilibrium in gaseoussystems. Kc is used when we are studying equilibrium in aqueous solution. The "p" subscriptmeans that the equilibrium constant was calculated using partial pressures. The "c" subscriptmeans that the equilibrium constant was calculated using molar concentrations. We’ll explainthese concepts of partial pressure and molar concentration in Chapters 8 and 9 respectively. I shouldalso tell you that in addition to equilibrium constants calculated using molar concentrations, Kc, andequilibrium constants calculated using partial pressures, Kp, that there are equilibrium constants thatdescribe the dissociation behavior of acids in water, Ka, bases in water, Kb, and ionic compounds inwater, Ksp, and etc. The point is that there are a variety of ways in which we find it useful to applyequilibrium theory in chemistry. But we don’t have a chance to discuss them in this class, and that’struly a shame.

Let’s look at the data from three sets of experiments. In the first set of experiments we’llstart with different amounts of dinitrogen tetroxide and with no nitrogen dioxide. In the secondset of experiments we’ll start with different amounts of nitrogen dioxide and no dinitrogentetroxide. And in the third set of experiments we’ll start with varying amounts of bothcompounds.

Here we have the data for five experiments. In each of these we have an initial amount ofdinitrogen tetroxide (in units of atmospheres, N2O4 I, the “subscript “I” stands for “initial.”) andno nitrogen dioxide is initially present (again in units of atmospheres, NO2 I).

Once the reaction begins we let it progress until it reaches equilibrium. When we measure theamount of each compound at equilibrium we find that the concentration of dinitrogen tetroxidehas decreased, as we would expect of a reactant (still in units of atmospheres, N2O4 eqb in whichthe “eqb” subscript stands for “equilibrium”). We also find that at equilibrium the concentrationof nitrogen dioxide has increased, which is consistent behavior for a product (also in units ofatmospheres, NO2 eqb). To calculate the numerical value of the equilibrium constant we take theequilibrium

KNO

N Op [ ]

[ ]2

2

2 4

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concentrations of the reactant and the product from each of the five experiments and plug theminto the equilibrium constant expression for this reaction. Here we use the data from the firstexperiment:

The amazing thing is that in each of the five experiments, regardless of the amount of dinitrogentetroxide with which we began, the ratio of products to reactants at equilibrium is always thesame value, 6.45. Note that there are no units associated with this equilibrium constant, as byconvention we always express equilibrium constants as unitless numbers.

In our next set of five experiments we begin with no dinitrogen tetroxide and withdiffering amounts of nitrogen dioxide.

In these experiments it is not initially possible for the forward reaction to occur, as there is nodinitrogen tetroxide. We would expect the reverse reaction to be the prominent reaction, and thisis what we see in the results. In each experiment the concentration of nitrogen dioxide decreasesas we progress from our initial conditions to equilibrium, and the concentration of dinitrogentetroxide increases as we move from initial to equilibrium conditions. And, as with the first setsof experiments, again we see that if we take the equilibrium amounts of product and reactant andplug them into the equilibrium constant expression, we wind up with exactly the same ratio ofproducts to reactants in each case, independent of starting amounts.

In our last five experiments we begin with equal amounts each of reactant and product.

It’s interesting to note that for experiments 11, 12, and 13 dinitrogen tetroxide is consumed andnitrogen dioxide is produced as the experiment proceeds from initial to equilibrium conditions.In experiments 14 and 15 we see the reverse behavior: nitrogen dioxide is consumed and

KNO

N Op [ ]

[ ]

( . )

..2

2

2 4

2140

0 306 45

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dinitrogen tetroxide is produced as the experiments move from initial to equilibrium conditions.But in all five experiments the ratio of products to reactants as calculated using the equilibriumconstant expression remains the same as in our first ten experiments.

So what’s the point of all of these experiments? To underline what we said above:“regardless of how much reactant and product with which we start a reaction, when equilibriumis reached, the ratio of the concentrations of the products and reactants will always equal theequilibrium constant.” The only two things that will change the equilibrium constant for areaction is a change in temperature or a change to a completely different reaction.

Writing equilibrium constant expressions

I expect you to be able to write equilibrium constant expressions for reactions and tocalculate the value of equilibrium constants. Let’s look at a few more examples.

Let’s begin with a reaction called the Haber process in which nitrogen gas reacts withhydrogen gas to form ammonia gas according to the following chemical equation.

N2 (g) + 3 H2 (g) W 2 NH3 (g)

Based on this equation the equilibrium constant expression is

If we’re told that the equilibrium concentrations of nitrogen, hydrogen, and ammonia are 1M,3M, and 5M respectively, then the equilibrium constant is equal to

The unit “M” indicates that we are using molar concentration (molarity) for these values, again, a topicwe will discuss in Chapter 9. I should point out that this is not the real value of the equilibrium constantfor this reaction. I have arbitrarily selected some numbers to illustrate how they are used in thesecalculations. But even though these are not actual values, the way in which they are used is preciselythe same as though they are in fact, real and correct values.

Another example would be the reaction of hydrogen gas and iodine vapor to formhydrogen iodide gas.

K

NH

N Hc

32

2 23[ ][ ]

Kc

5

1 30 96

2

3[ ][ ].

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H2 (g) + I2 (g) W 2 HI(g)

Based on this chemical equation the equilibrium constant expression would be

If, at equilibrium, the molar concentrations of hydrogen, iodine, and hydrogen iodide are 0.1 M,0.25 M, and 3.7 M respectively, then the value of the equilibrium constant would be

Again, I have arbitrarily selected numbers at random. This is not the actual value of the equilibriumconstant for this reaction.

It is important to note that when writing equilibrium constant expressions we onlyinclude terms for those substances in the gaseous or aqueous state. We do not include terms forsubstances in the liquid or the solid state as the concentrations of materials in these states areconstant and, as such, unchanging during the course of a chemical reaction. This undoubtedlyseems an odd statement to you but it is one I will not justify in these notes. If you have questions,please contact me privately. I am, quite simply, going to ask you to trust me on this one. Let’s look ata few examples of how this affects the way in which we write equilibrium constant expressions.

Given the reaction

CO2 (g) + H2 (g) WCO(g) + H2O(l)

the equilibrium constant expression is

For the reaction

SnO2 (s) + 2 CO(g) W Sn(s) + 2 CO2 (g)

the equilibrium constant expression is

K

HI

H Ic

2

2 2[ ][ ]

Kc

37

01 0 25548

2.

[ . ][ . ]

KCO

CO Hc [ ]

[ ][ ]2 2

KCO

COc

[ ]

[ ]2

2

2

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For the reaction

3 Fe(s) + 4 H2O(g) W Fe3O4 (s) + 4 H2 (g)

the equilibrium constant expression is

And finally, for the reaction

2 Al(s) + Fe2O3 (s) W 2 Fe(s) + Al2O3 (s)

there is no equilibrium constant expression, as all of the substances in the reaction are in thesolid state.

Equilibrium constants and Gibbs free energy

Why is it that an equilibrium constant represents a seemingly “magic number” for areaction? Because the total energy of a system is at a minimum at equilibrium, and energygoverns everything that happens in chemistry. All physical changes and chemical reactions movespontaneously from states of higher energy to states of lower energy. And if the energy of asystem is at its lowest possible point at equilibrium, then that system will spontaneously progressto equilibrium.

Equilibrium constants provide insight into the favorability of chemical reactions. This isbecause for any chemical reaction (or physical change) there is an inverse relationship betweenthe Gibbs free energy of a reaction and the equilibrium constant for that reaction. Any reactionwith a negative Gibbs free energy of reaction will have an equilibrium constant greater than 1.And any reaction with a positive Gibbs free energy of reaction will have an equilibrium constantless than 1.

If the equilibrium constant for a reaction is much greater than 1 (Kc > 1), this means thatmore products are formed in the forward reaction than reactants are formed in the reversereaction. For a fraction to be greater than 1 the value of the numerator must be greater than thevalue of the denominator, as is the case in 2/1. The equilibrium mixture will contain a greateramount of products than reactants. Reactions in which there are more products formed thanreactants are said to be product-favored reactions. The forward reaction is the energeticallyfavorable reaction in product-favored reactions.

KH

H Oc

[ ]

[ ]2

4

24

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The larger the value of Kc the more favored the forward reaction must be. If we comparetwo reactions, one with an equilibrium constant of 1 x 105 and another with an equilibriumconstant of 1 x 1010, we know that the reaction with the larger equilibrium constant is the moreproduct-favored reaction.

If the equilibrium constant for a reaction is much less than 1 (Kc < 1), this means thatmore reactants are formed in the reverse reaction than products are formed in the forwardreaction. For a fraction to be less than 1, the value of the numerator must be less than the value ofthe denominator, as is the case in ½. The equilibrium mixture will contain a greater amount ofreactants than products. Reactions in which there are more reactants formed than products aresaid to be reactant-favored reactions. The reverse reaction is the energetically favorable reactionin reactant-favored reactions.

The smaller the value of Kc the more favored the reverse reaction must be. If we comparetwo reactions, one with an equilibrium constant of 1 x 10-5 and another with an equilibriumconstant of 1 x 10-10, we know that the reaction with the smaller equilibrium constant is the morereactant-favored reaction.

If the equilibrium constant for a reaction is around 1 (Kc ~ 1), this means that there areroughly equal amounts of reactants and products in the equilibrium mixture. Values of Kc and Kp

are never negative. Why is that? (Hint: is it possible to have a negative concentration or a negativepartial pressure? Answer: no!)

LeChatelier's principle

Why do all reactions always proceed toward equilibrium? As I said above, it is becausesystems at equilibrium are in a relatively low energy state, compared to the energy of the systemwhen it is in any state other than a state of equilibrium. In other words, a system at equilibrium ismore stable than the same system when it is not at equilibrium.

Knowing this, it should not surprise us to learn that if a system is at equilibrium and isdisturbed, i.e., the equilibrium is disrupted or, some books refer to this as "stressing" the system, thesystem will spontaneously react so as to restore its state of equilibrium. This is calledLeChatelier's le SHAT-lee-ay principle. LeChatelier's principle states that a system at equilibrium,if disturbed, will shift so as to minimize the effects of the disturbance. Or in other words, thesystem will do everything it can to return to a state of equilibrium.

We can disturb a system at equilibrium by adding or removing reactants or products, bychanging the temperature of the system, or by changing the pressure of the system. Tounderstand the effect of changing the concentration of reactants or products in an equilibrium,remember that at equilibrium the ratio of the products and reactants is a constant number (i.e.,the equilibrium constant).

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Let's say we have a system at equilibrium and we add more reactants to it. The ratio ofproducts to reactants is no longer equal to the equilibrium constant. To restore things to a state inwhich the ratio of products to reactants is equal to the equilibrium constant, the forward reactionmust consume some of the added reactant and form more products. This will continue until theratio of products and reactants is equal to the equilibrium constant and equilibrium has beenrestored.

Assume we have a system at equilibrium and we add more products to it. The ratio ofproducts to reactants is no longer equal to the equilibrium constant. To restore things to a state inwhich the ratio of products to reactants is equal to the equilibrium constant, the reverse reactionmust consume some of the added product and form more reactants. This will continue until theratio of products and reactants is equal to the equilibrium constant and equilibrium has beenrestored.

What will happen if we somehow remove reactant from a system at equilibrium? Again,the ratio of products to reactants is no longer equal to the equilibrium constant. To restore thingsto a state in which the ratio of products to reactants is equal to the equilibrium constant, thereverse reaction must consume some of the product and form more reactants. This will continueuntil the ratio of products and reactants is equal to the equilibrium constant and equilibrium hasbeen restored.

Finally, how will the equilibrium system respond if we remove some of the product? Asin the three cases above, the ratio of products to reactants is no longer equal to the equilibriumconstant. To restore things to a state in which the ratio of products to reactants is equal to theequilibrium constant, the forward reaction must consume some of the reactant and form moreproducts. This will continue until the ratio of products and reactants is equal to the equilibriumconstant and equilibrium has been restored.

Consider the reaction

A + B W C + D

with the equilibrium constant expression

The effects of various disturbances to this system at equilibrium are summarized in the followingtable:

if we the system will shift by and form more through the

add A and/or B consuming A and B C and D forward reaction

KC D

A Bc

[ ][ ]

[ ][ ]

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add C and/or D consuming C and D A and B reverse reaction

remove A and/or B consuming C and D A and B reverse reaction

remove C and/or D consuming A and B C and D forward reaction

How will changes in temperature affect a system at equilibrium? Of course, a change intemperature will result in a change in the equilibrium constant for a reaction in nearly all cases,but let's ignore that effect during this discussion. In fact, let's just forget about it all together. Wecan actually ignore it safely in this case. If you really want to learn more about this, you should take aproper thermodynamics class. We learned above that exothermic reactions give off heat andendothermic reactions absorb heat. To understand the effect of temperature change on a systemat equilibrium, let's consider heat as a product in exothermic reactions and as a reactant inendothermic reactions. In other words, we can write a chemical equation for an exothermicreaction

A + B W C + D + heatand an endothermic reaction as

heat + A + B W C + D

It is easy to appreciate that we can add heat to a system by doing something to increase it'stemperature. We remove heat from a system by doing something to lower it's temperature, i.e.,by cooling it. The effects of temperature changes to this system at equilibrium are summarized inthe following table:

if we to/from an ________ reaction the system will shifttoward

through the

add heat exothermic reactants reverse reaction

remove heat exothermic products forward reaction

add heat endothermic products forward reaction

remove heat endothermic reactants reverse reaction

How will changes in pressure affect a system at equilibrium? If the system is in aqueoussolution and an inert (unreactive) gas is added to the system, there will be no effect. This is alsotrue for gaseous systems: the addition of an inert gas will have no effect on the equilibrium.However, if a gas that participates in the reaction is added to or removed from the system, thesystem will behave exactly the same as a system in aqueous solution. We will ignore changes inpressure that result from changing the volume of the system in this class.

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How will addition of a catalyst affect a system at equilibrium? Quite simply, it will notaffect it at all. A catalyst speed up the rates of both the forward and the reverse reactions of asystem, so adding a catalyst could help a system reach equilibrium more quickly. But, once atequilibrium the catalyst will no longer have any effect.

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Chapter 8: Gases, Liquids, and Solids

General gas properties

There are three common states of matter, solids, liquids, and gases. At a molecular level therelative differences between these three states of matter generally depend on the distancebetween the particles and on the strengths of the attractive interactions between the particles(remember, we use the word "particle" in a generic sense to describe a substance that at an atomiclevel may consist of individual atoms, ions, or molecules):

• in gases the particles are far apart and there are no attractive interactions between the gasparticles

• in liquids the particles are close together and there are relatively weak attractiveinteractions between the liquid particles

• in solids the particles are very close together and there are relatively strong attractiveinteractions between the gas particles

Substances in a gas or liquid state are often referred to as fluids, substances with no fixedvolume or shape but which take the shape and volume of their container. The words vapor andgas are used synonymously in general vocabulary but there is a distinction in science, eventhough they have reference to the same physical state. Substances that are usually in a liquid orsolid state at room temperature and pressure but which have been converted to a gaseous stateare called vapors. Substances which are found in the gaseous state at room temperature andpressure are called gases. This is why we talk about hydrogen gas and nitrogen gas but watervapor.

To better understand the states of matter at a molecular level, particularly gases, we mustdiscuss temperature and pressure. Temperature, T, can be defined as a measure of how hot orcold something is. There is a direct correlation between the temperature of a substance and thekinetic energy of its particles. The kinetic energy (K.E.) of atoms and molecules can bedescribed mathematically as

K.E. = 3/2 kT

where k is Boltzmann's constant and is equal to 1.38 x 10-23 JAK-1 We won’t be solving thisequation, so you don't need to remember this constant in this class. Kinetic energy can also bedescribed as

K.E. = ½ mv2

where m is the mass of the particle and v is its velocity. Since both of these equations are equalto the same thing, kinetic energy, they are also equal to each other

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½ mv2 = 3/2 kTIf we multiply both sides by 2 we obtain

mv2 = 3 kT

and if we rearrange the equations in terms of velocity we see that

v = (3 kT/m)½

This equation tells us that the velocity of atoms and molecules is directly dependent on theirmass and on the system temperature. As temperature increases the velocity of particles will alsoincrease. And as temperature deceases the velocity of the particles will also decrease. Given therelationship between velocity and kinetic energy, this means that as particles move faster theyhave greater kinetic energy. Therefore, as the temperature of particles increases, they havegreater kinetic energy. Conversely, as the temperature of particles decreases, their kinetic energyalso declines. We will not be using these equations quantitatively in this class, but it will help some ofyou to more clearly understand the relationship between temperature and kinetic energy.

In our study of gases, temperature is always expressed in Kelvin. While we commonlywork with the Celsius scale in chemistry, especially when making measurements, you will neverbe able to correctly make calculations when studying gases without first converting Celsiustemperatures to the Kelvin scale. This is a common source of error for students in this chapter.Always think Kelvin!

Pressure, P, is defined as the ratio of a force applied to a surface divided by the area overwhich the force is applied, i.e.,

Pressure = Force/Area over which the force is applied

At a molecular level, when molecules collide with a surface they exert a force upon it. The forceexerted by molecular collisions depends on the kinetic energy of the collisions. It also dependson the frequency of the collisions. The more frequent the collisions, the greater the pressure. Wecan re-write our equation describing pressure as

Pressure = (K.E. of collisions) x (# of collisions per unit time)/Area over which collisions occur

The role of temperature in the behavior of gases becomes more apparent here. We provedabove the relationship between temperature, velocity, and kinetic energy. If pressure isdependent on both the kinetic energy of collisions and on the frequency of collisions, then itshould be apparent that as temperature increases, pressure will also increase. This is because astemperature increases, the particles move more rapidly and can collide more frequently. It is alsobecause as the particles move more rapidly, they have greater kinetic energy. Knowing thesethings, can you explain why pressure decreases as the temperature decreases?

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The units in which pressure may be expressed are a bit more varied than those oftemperature. Pressure is commonly expressed in units called atmospheres (atm), but there areother units also used.

Pressure may be expressed in pounds per square inch (psi, or lbs/in2 ).

1 atm = 14.7 psi

This is because the air contained in a column with a surface area of one inch by one inch thatstretches from sea level to the upper reaches of the atmosphere will weigh exactly 14.7 pounds.

Pressure may be expressed in units of bars:

1 bar = 14.5 psi

This unit is not typically used in elementary chemistry and will not be used in this class.However, it is commonly used in meteorology. If you are watching the weather on television,you will commonly hear barometric pressure expressed in units of millibars (thousandths ofbars).

Pressure may be expressed in units of torr, T, or in millimeters of mercury, mm Hg. Sincetemperature is also indicated with the symbol T, this is sometimes confusing. You must establishwhether temperature or the pressure of a gas in units of torr is being discussed by paying closeattention to the context of the discussion or problem in which the symbol is used. The unit torr takesits name from the Italian physicist Evangelista Torricelli who invented the mercury barometer. Asimple barometer consists of a sealed glass tube filled with mercury. The tube is inverted in adish of liquid mercury with the end of the tube submerged beneath the surface of the mercury inthe dish. Surprisingly, when the tube is inverted, not all of the mercury runs out of the tube. Thepressure of air on the surface of the mercury in the dish is sufficient to force much of themercury back up into the tube. In other words, the pressure exerted by air on the surface of themercury in the dish is sufficient to offset the force of gravity pulling the mercury down the tube.Whenever air pressure changes the height of the column of mercury in the tube changes as well.If atmospheric pressure increases, the height of the column of mercury in the tube also increasesas air pressure forces the column of mercury slightly further up the tube. If atmospheric pressuredecreases, the height of the column of mercury in the tube also decreases. The relationshipbetween atmospheres, torr, and mm Hg is

1 atm = 760 T = 760 mm Hg

Note that 1 torr is exactly equal to 1 mm Hg. You can see photographs and other images and learnmore about barometers at the Wikipedia address http://en.wikipedia.org/wiki/Barometer.

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The SI unit of pressure is the Pascal (Pa). We will not use this unit in this class but youshould still know what it is, as you may bump into it at some future point outside of class.

1 atm - 101,325 Pa

In summary, you should be familiar with the following relationships between the various units ofpressure:

1 atm = 14.7 psi = 760 T = 760 mm Hg = 101,325 Pa

Finally. When studying gases, experiments are commonly performed at standardtemperature and pressure (STP). For gases, these conditions are 1.00 atm and 273 K.

Ideal gases and the Ideal Gas Law (Universal Gas Law)

It will be easier to study gases if we make two assumptions about them. We will assumethat gas particles do not occupy any volume. Which is, of course, completely untrue. Gases aremostly empty space. But while in any sample of gas the volume occupied by the actual molecules of gasis extremely small, it is still a finite number. However, this assumption does make it possible to makesome remarkable calculations that are generally quite accurate as we study most gases. And we willassume that gas particles do not interact with each other. Again, the absolute correctness of thisassumption is rather poor in most cases, but the short-term benefits we derive from making it make itwell worth our while to do so. And surprisingly, in most cases, the error introduced is small. When wemake these two assumptions about a gas, about any gas, we state that the gas exhibits idealbehavior. In the real world nearly all gases show ideal behavior under appropriate conditions,usually at low pressures and at high temperatures. The implication of this is that, regardless ofcomposition, all gases behave roughly the same under the same set of conditions. This is a trulyremarkable implication, if you think about it!

Historically, the study of gases lies at the heart of the science of modern chemistry. Thesestudies began in the 17th century and continued to shape the development of chemistry duringthe 18th and 19th centuries. It is common to acknowledge the contributions of Robert Boyle,Jacques Charles, Joseph Gay-Lussac, and Amedeo Avogadro by using the laws and relationshipsthey discovered to explain the behavior of gases. But, in actual fact, it is not necessary to knowor remember their laws individually. They can be synthesized into one relationship, the ideal gaslaw. This law explains the relationship between pressure, volume, the amount (number of moles)of gas, and temperature as follows:

PV = nRT

• P is the pressure of the gas, units = atmospheres (atm)• V is the volume the gas occupies, units = liters (L)

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• n is the number of moles of gas in the sample under study, units = moles (mol)• T is the temperature of the gas, units = Kelvin (K)• R is a constant, called the gas constant and is equal to 0.0821 LAatm/molAK. This is rather

an odd set of units, but it derives from the origin of R which is calculated using the equationR = PV/nT.

How do we use the ideal gas law in the study of chemistry? This is an equation with fourvariables and one constant. In problems in which we study gases using the ideal gas law, we willbe given three of the four variables. Using algebra, we will re-arrange the equation and solve forthe variable of interest. Some examples of ideal gas law problems are as follows.

• If 1.00 mole of a gas occupies a volume of 1.00 L at 298 K, what is the pressure insidethe container?

Start by using a bit of algebra to rearrange the ideal gas law:

P = nRT/V

P = (1.00 mol) x (.0821 LAatm/molAK) x (298 K)/1.00 L = 24.46 atm

Be sure to check your units carefully when you set up and work ideal gas law problems! These are,really, just dimensional analysis problems. And smile - that’s supposed to make you feel better aboutthings.

• A 5.00 L container holds a gas at 810 T and 62oC. How many moles of gas are present?

Start by using a bit of algebra to rearrange the ideal gas law

n = PV/RT

convert 810 T to atm: 810 T/760 T/atm = 1.066 atmconvert 62oC to K: 62oC + 273 = 335 K

n = [(1.066 atm) x (5.00 L)]/[(.0821 LAatm/molAK) x (335 K)] = 0.194 moles

• If you blow exactly 2.00 moles of air into a balloon at a pressure of 780 mm Hg and at atemperature of 35oC, what volume must the balloon hold to keep from bursting? Note: an11" balloon holds a volume of 11.4 L, and an 18" balloon holds a volume of 50.0 L.

Start by using a bit of algebra to rearrange the ideal gas law

V = nRT/P

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convert 780 T to atm: 780 mm Hg/760 mm Hg/atm = 1.026 atmconvert 35oC to K: 35oC + 273 = 308 K

V = (2.00 mol) x (.0821 LAatm/molAK) x (308 K)/(1.026 atm) = 49.3 L

You're probably thinking to yourself, why in the world didn't we just use the 18" balloon and saveourselves the trouble of these calculations? It's true, we could have done things that way, but just thinkof the fun you would have cheated yourself out of!

• 6.75 moles of an unknown gas occupy a volume of 13.3 L at a pressure of 17.6 atm. Whatis the temperature of the gas?

Start by using a bit of algebra to rearrange the ideal gas law

T = PV/nR

T = [(17.6 atm) x (13.3 L)]/[(6.75 mol) x (.0821 LAatm/molAK)] = 422.4 K

If you're having problems using algebra to rearrange the equation in these examples, I can't help you.You're going to need to review your basic algebra on your own.

Let me point something out that you may not have noticed. We just worked fourexamples using the ideal gas law. In any one of the problems, did we ever give any considerationto what the gas in the problem was? Looking back, you can see that in three of the fourproblems, the gas isn't even identified. Why can we do this? Because the ideal gas law operateson the assumption that all gases behave ideally. This means that they all behave the same,regardless of their chemical composition, under the same physical conditions.

The ideal gas law and molar mass calculations

The Ideal Gas Law can be used to find the molar mass of substances. This is based on thevariable n in the ideal gas law. If we have a certain amount of a substance, we can calculate thenumber of moles of the substance if we know how much of it (i.e. it's mass in grams) we haveand if we know it's molar mass. In equation form

n = number of moles = mass of substance (grams)/molar mass (g/mol)

For example, if we have 36 grams of water, and given that the molar mass of water is 18 g/mol,then we have 2 moles of water.

Given that

n = g/Mm

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where g is the grams of substance and Mm is it molar mass, we can substitute this into the idealgas law

PV = gRT/Mm

and by re-arranging the ideal gas law and solving for molar mass, we have

Mm = gRT/PV

Here are two examples of problems that use this form of the ideal gas law:

• 44.9 g of an unknown gas exert a pressure of 2.55 atm within a 10.2 L container at 325 K.Is the unknown gas nitrogen dioxide or sulfur dioxide?

To answer this particular question we need to know the molar masses of the two gases ofinterest. The molar masses are 46.01 g/mol for NO2 and 64.06 g/mol for SO2.

Mm = gRT/PV

Mm = [(44.9 g) x (.0821 LAatm/molAK) x (325 K)]/[(2.55 atm) x (10.2 L)] = 46.06 g/mol

Based on the similarities of their molar masses, the unknown gas must be nitrogen dioxide.

• A damaged unmarked cylinder of compressed gas is found in the ruins of a burnedbuilding. You suspect that the gas is either argon, nitrogen, helium, or carbon dioxide. A5.00 L mylar bag is filled with the gas to a pressure of 1900 T at 298 K. When weighed,the mass of the gas is 20.42 g. What is it?

Again, to answer this particular question you'll need to know the molar masses of the fournamed gases that the unknown gas may possibly be.

Mm = gRT/PV

convert 1900 T to atm: 1900 T/760 T/atm = 2.50 atm

Mm = [(20.42 g) x (.0821 LAatm/molAK) x (298 K)]/[(2.50 atm) x (5.00 L)] = 39.97 g/mol

The unknown gas is argon.

The ideal gas law and stoichiometry

We can also use the ideal gas law to help in the solution of stoichiometry problems. Twoexamples follow.

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• An 18" balloon has a volume of 50.0 L. How many grams of water must be decomposedvia electrolysis to fill the balloon with hydrogen to a pressure of 1.05 atm at 298 K?By rearranging the ideal gas law and using the information given find n

n = PV/RT = [(1.05 atm) x (50.0 L)]/[(.0821 LAatm/molAK) x (298 K)] = 2.146 moles hydrogen

Now we need a balanced chemical equation for the reaction

2 H2O(l) => 2 H2 (g) + O2 (g)

Use the balanced chemical equation and the information used from solving the ideal gaslaw to answer the question

(2.146 moles H2) x (2 moles H2O/2 moles H2) x (18.02 g H2O/1 mol H2O) = 38.7 grams of water

• The reaction of 75.0 g of iron (III) sulfide with excess hydrochloric acid will producewhat volume of gas at 755 T and 293 K?

In this problem we need to begin with a balanced chemical equation but then, we alwaysneed a balanced equation to do stoichiometry problems:

Fe2S3 (s) + 6 HCl(aq) => 2 FeCl3 (aq) + 3 H2S(g)

Use the balanced chemical equation to determine the number of moles of gas that will beproduced.

(75.0 g Fe2S3) x (1 mol Fe2S3/207.87 g Fe2S3) x (3 mol H2/1 mol Fe2S3) = 1.082 moles H2S

By rearranging the ideal gas law and using the information given, find V:

convert 755 T to atm: 755 T/760 T/atm = 0.993 atm

V = nRT/P = [(1.082 mol H2S) x (.0821 LAatm/molAK) x (293 K)]/(0.993 atm) = 26.2 L

The combined gas law

We can use the ideal gas law to help us solve problems in which conditions are constant.But how do we solve problems in which one or more conditions change during the course of theproblem?

Let's say that we have two sets of conditions, our initial conditions and our finalconditions. We can describe both the initial and final conditions using the ideal gas law:

initial: PiVi = niRTi

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and final: PfVf = nfRTf

If we express these equations in terms of the gas constant R

R = (PiVi)/(niTi) and

R = (PfVf)/(nfTf)

Since R is a constant and therefore the same in both cases, we can say that

(PiVi)/(niTi) = (PfVf)/(nfTf)

This form of the ideal gas law is called the combined gas law. I’ve also heard students callit other things, but none of them should be repeated in polite company. This is the equation we usewhen we solve gas problems in which one or more conditions change during the course of theproblem. This provides for Boyle's Law, Charles's Law, Gay-Lussac's Law, and Avogadro's Lawall in the same equation. If you remember this relationship you don't need to remember all of theothers.

Below are five examples of problems solved using the combined gas law.

• 7.5 moles of a gas at 2.25 atm of pressure occupy a volume of 2.00 L. If the gas iscompressed to a volume of 1.00 L what is the new pressure of the gas?

To solve problems using the combined gas law, I suggest setting up a table in which youlist the information given to you in the problem. This will often help clarify exactlywhich variable we need to solve for

Pi 2.25 atm Pf ?

Vi 2.00 L Vf 1.00 L

ni 7.5 mol nf

Ti Tf

Note that temperature is not mentioned in the problem. We infer from this thattemperature does not change and therefore that the initial and final temperatures are thesame. Also note that the initial number of moles is mentioned, but it is not mentionedagain. We infer from this that this value also does not change during the course of theproblem. We begin with

(PiVi)/(niTi) = (PfVf)/(nfTf)

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Since the temperatures and numbers of moles are unchanging and are therefore constant,they can be cancelled and

PiVi = PfVf

We're interested in finding Pf so we rearrange the equation and arrive at

Pf = PiVi/Vf

[(2.25 atm) x ( 2.00 L)]/(1.00 L ) = 4.50 atm

Does it make sense, physically, for the pressure to increase when the volume decreases? Why? Think interms of collisions if you have difficulty answering this question.

This is actually a Boyle's Law problem. Boyle's law is the equation PiVi = PfVf, and yet we were able tosolve this problem without even realizing we needed Boyle's law, simply because we know how to usethe combined gas law. The next few problems we work will also be problems that could be solved usingthe appropriate law - Charles's law, Avogadro's relationship, Gay-Lussac's law - if we knew them. Butwe don't have to know them if we know how to use the combined gas law. We don't even have toworry about them.

• A sample of gas at 373 K exerts a pressure of 1500 T. If the temperature drops to 273 K,what will the new pressure be?

Let's begin with our table of values, remembering to convert our pressures from torr toatm:

Pi 1.974 atm Pf ?

Vi Vf

ni nf

Ti 373 K Tf 273 K

(PiVi)/(niTi) = (PfVf)/(nfTf)

Since the volumes and numbers of moles are unchanging and are therefore constant, theycan be cancelled and:

Pi/Ti = Pf/Tf

Once again we're interested in solving for Pf so we rearrange the equation and arrive at

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Pf = PiTf/Ti

[(1.974 atm) x ( 273 K)]/(373 K) = 1.44 atmIs the answer physically consistent with what we know about the relationship between temperature andpressure to see the pressure decrease with a decrease in temperature?

• 10.0 moles of helium at 298 K occupy a volume of 20.5 L at a pressure of 11.93 atm. Ifthe temperature of the gas increases to 328 K, how must the container volume change forthe pressure to remain the same?

Pi 11.93 atm Pf 11.93 atm

Vi 20.5 L Vf ?

ni 10.0 mol nf

Ti 298 K Tf 328 K

(PiVi)/(niTi) = (PfVf)/(nfTf)

If the pressure is to stay the same, this means that the initial and final pressure must bethe same. The number of moles is constant. Our equation becomes

Vi/Ti = Vf/Tf

We want to solve for Vf and so the equation becomes

Vf = ViTf/Ti

[(20.5 L) x ( 328 K)]/(298 K) = 22.56 L

We know that as temperature increases, pressure will also increase. The only way to prevent thepressure from increasing with an increase in temperature is to increase the size of the containerholding the gas, which will prevent an increase in the number of collisions by spreading the gas particlesfurther apart.

• A 3.6 mole sample of gas in a sealed vessel has a pressure of 4.25 atm. If the number ofmoles of gas is increased to 4.9 moles, what will the new pressure be?

Pi 4.25 atm Pf ?

Vi Vf

ni 3.6 mol nf 4.9 mol

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Ti Tf

(PiVi)/(niTi) = (PfVf)/(nfTf)

The volume and temperature are constant in this problem. Our equation becomes:

Pi/ni = Pf/nf

To solve for Pf the equation becomes

Pf = Pinf/ni

[(4.25 atm) x ( 4.9 mol)]/(3.6 mol) = 5.78 atm

Why should pressure increase if we add more particles at the same temperature to a gas constrainedby a constant volume?

• 10.0 moles of a gas at 26.0 atm and 303 K occupy a volume of 9.57 L. If the number ofmoles of gas is suddenly changed to 15.0 moles while the temperature increases to 373 Kand the volume decreases to 5.00 L, what will the new pressure in the container be?

Pi 26.0 atm Pf ?

Vi 9.57 L Vf 5.00 L

ni 10.0 mol nf 15.0 mol

Ti 303 K Tf 373 K

(PiVi)/(niTi) = (PfVf)/(nfTf)

Nearly everything changes in this problem. Can we still solve it? Of course we can. Thecombined gas law has eight variables. As long as we can account for seven of the eight,we can solve the problem. To solve for Pf the equation becomes

Pf = PiVinfTf/VfniTi

[(26.0 atm) x ( 9.57 L) x (15.0 mol) x (373 K)]/[(5.00 L) x (10.0 mol) x (303 K)] = 91.9 atm

Increasing the temperature increased both collision frequency and the kinetic energy of the collisions.Decreasing the volume increased the collision frequency, as did increasing the number of moles of gas.On the basis of these things we would expect the pressure to increase, and it did.

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Dalton's Law of Partial Pressures and mole fractions

Given a system that contains two or more gases, the total pressure of the system is equalto the sum of pressures of all of the gases present. This is described by the equation

PT = P1 + P2 + P3 + . . . .

where PT is the total pressure of the system and P1, P2, and P3 are the pressures of the gasespresent in the system. In other words, the pressures of multiple gases in a system are additive.They do not cancel each other out or affect each other in any other way. This equation is trueregardless of whether a system contains only two gases or one hundred or more gases. Afundamental assumption is being made here, that none of the gases react with each other. If they do,the things we are about to discuss do not work.

The pressure that each of the individual gases contributes to the total pressure is called itspartial pressure. The partial pressure of a gas is in no way incomplete for that gas, but rather, it issimply a part of the total pressure of the system.

In other words, any time you hear someone refer to the pressure of a gas as a partial pressure, they’reimplying that it is not the only gas in the system. As an example, when you study physiology you read ofthe partial pressure of oxygen in the blood. This means, quite simply, that there are other gasesdissolved in the blood besides oxygen.

The partial pressures of the gases in a system can be determined using the ideal gas law.As examples:

• A sealed flask contains nitrogen at a pressure of 500 T and oxygen at a pressure of 250 T.What is the total pressure in the flask?

PT = 500 T + 250 T = 750 T

You may have noticed that in this case we did not convert from torr to atmospheres. In problems inwhich the ideal gas law is not used, we can get away with it. If you don't believe me, convert the twovalues from torr to atm, add them together, and then convert them back to torr. And to think youdoubted me.

• A 10.0 L flask contains 1.0 mole of carbon dioxide, 2.0 moles of methane, and 4.0 molesof water vapor. If the temperature inside the flask is 303 K, what is the total pressure? Inthis problem we're going to need to use the ideal gas law to find the pressure of each ofthe three gases.

PCO2 = nRT/V = [(1.0 mol) x (.0821 LAatm/molAK) x (303 K)]/(10.0 L) = 2.49 atm

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Pmethane = nRT/V = [(2.0 mol) x (.0821 LAatm/molAK) x (303 K)]/(10.0 L) = 4.98 atm

PH2O = nRT/V = [(4.0 mol) x (.0821 LAatm/molAK) x (303 K)]/(10.0 L) = 9.95 atm

PT = PCO2 + Pmethane + PH2O = 2.49 atm + 4.98 atm + 9.95 atm = 17.4 atmThere's one final concept we need to mention while we're discussing partial pressures. In

a system containing two or more gases, the mole fraction of a gas can be described as either theratio of that gas's partial pressure divided by the total pressure of the system, or as the ratio of thenumber of moles of that gas to the total number of moles of gas in the system. In a system of twoor more gases, the mole fraction X1 of gas 1 is equal to

X1 = (P1/PT) = (n1/nT)

where P1 is the partial pressure of gas 1, PT is the total pressure in the system, n1 is the number ofmoles of gas 1, and nT is the total number of moles of gas in the system. The same would holdtrue for the second gas in the system, and the third, and so on. For gas 2,

X2 = (P2/PT) = (n2/nT)

In the example we just worked there is 1.0 mole of carbon dioxide gas, 2.0 moles of methanegas, and 4.0 moles of water vapor. There are therefore 7.0 (1.0 + 2.0 + 4.0) total moles of gas inthe system. The mole fraction of carbon dioxide XCO2 is therefore 0.143 (1.0/7.0), the molefraction of methane Xmethane is 0.286 (2.0/7.0), and the mole fraction of water vapor XH2O is 0.571(4.0/7.0).

The sum of the mole fractions of a system will always be equal to exactly 1. In a systemwith “n” different gases,

X1 + X2 + X3 + ... Xn = 1

In our example, 1/7 + 2/7 + 4/7 = 1. Mole fractions are always unitless values.

The concept of mole fractions becomes useful in the calculation of partial pressures. In a systemwith two or more gases, the partial pressure of any gas is equal to the product of its mole fractionand the total system pressure. In other words, for gas 1,

P1 = X1PT

and for gas 2

P2 = X2PT

Now let's apply this concept to the problem we just worked. Let's suppose we're told thatwe have a system containing 1.0 mole of carbon dioxide, 2.0 moles of methane, and 4.0 moles of

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water vapor and that the total pressure of the system is 17.4 atm. Having already calculated themole fractions of each of these three compounds, then

PCO2 = XCO2 PT = (0.143) x (17.4 atm) = 2.49 atm

Pmethane = Xmethane PT = (0.286) x (17.4 atm) = 4.98 atm

PH2O = XH2OPT = (0.571) x (17.4 atm) = 9.95 atm

This concept is often a useful and easy way to calculate partial pressures, given the properinformation.

States of matter, phase transitions, and enthalpies of phase transitions

We have discussed the three common states of matter in Chapter 1, in Chapter 7, andagain at the beginning of this chapter. Let's summarize the things we’ve discussed throughoutthis course thus far:

Gases• are fluids• have no fixed volume or shape - a gas will fill and take the shape of its container• have no intermolecular potential energy: there are no attractive or repulsive interactions

between gas particles• gas particles are very far apart compared to their size• the total energy of a perfect gas is the sum of its translational and internal energy

(vibrations, rotations, spins, etc.)• there is no short-range order or long-range in gases

Liquids• are fluids• have a fixed volume but no fixed shape - a liquid will take the shape of its container• intermolecular potential energy: there are attractive or repulsive interactions between

liquid particles• liquid molecules are close together compared to their size, which is why liquids are

largely incompressible• the translational energy of liquid particles is much lower than in gas (i.e., it's harder for

them to move around than in the gas phase)• in liquids there is short-range order but no long-range order

Solids• fixed volume and shape• intermolecular potential energy: there are attractive or repulsive interactions between

solid particles

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• solid particles are, on average, closer together than in particles in the liquid phase, whichis why solids are incompressible

• the translational energy of solids is much lower than in liquids (i.e., it's harder for them tomove around than in the liquid phase)

• in solids there is both short-range order and long-range order (especially in crystallinesolids)

We also learned in Chapter 1 that there are specific names for the transition of one stateof matter to another.

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In Chapter 7 we learned that there are enthalpies associated with physical changes ofstate (phase changes, or phase transitions), such as the enthalpy of fusion, melting, vaporization,and condensation. Since enthalpy is associated with processes that result in a system gaining orlosing energy, it should not surprise us to learn that we can also think of phase transitions interms of enthalpy. Below is a plot of the changes in the state of water with the addition of heat.

The plot begins with solid water (ice) at a temperature less than -50°C.

As a side question, is it possible to cool water to a temperature less than it’s freezing point? Moregenerally speaking, is it possible to cool any substance to a temperature below its freezing point? Theanswer is yes. We can cool any thing and all things to a temperature slightly above 0 K, absolute zero,regardless of their freezing points. So water at a temperature less than its freezing point should notcome as much of a surprise to you.

As heat is added to the ice its temperature increases rapidly. However, once we reach 0°C we seesomething curious. The temperature of the ice remains constant, even though heat is added to thesystem. This is represented by the first plateau, or leveling, on the left side of the line. Why doesthis occur? Because before the solid can be converted to a liquid we must add enough energy tothe system to help the water molecules overcome the strong attractive interactions that hold themtogether in the solid state. The energy that must be put into the system before the solid cantransform to a liquid is called the enthalpy of melting. Since we must put energy into the systemto convert a solid into a liquid, we say that melting is an endothermic process.

Once the solid has been converted to a liquid, the temperature of the liquid increases asheat is added to the system. The slope of this line is different from the slope of the solid line becauseice and liquid water have different heat capacities. That's the explanation, but you needn't worry about

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it in this class. The increase in temperature continues until the liquid water reaches a temperatureof 100°C. At this point we reach a second leveling of the line. This plateau is much longer thanthe first. It represents the addition of heat to the system but without an increase in thetemperature of the system. This occurs because energy must be added to the system to break thestrong interactions that take place between the water molecules in the liquid state. Theseinteractions must be broken before the liquid can transform to a vapor (gas). The energy thatmust be put into the system before the liquid can transform to a vapor is called the enthalpy ofvaporization. Since we must put energy into the system to convert a liquid into a vapor, we saythat vaporization is also an endothermic process.

Once the liquid has been converted to a vapor, the temperature of the vapor continues toincrease as heat is added to the system. The slope of this line is different from the slope of the liquidline because liquid water and water vapor have different heat capacities. Again, you don't need to worryabout it in this class.

What about the reverse processes, condensation and freezing?

Ok, here’s another question. Is it possible to heat water above its boiling point at 100oC? And again,more generally speaking, is it possible to heat any substance to a temperature in excess of its boilingpoint? Again, the answer is yes. Exactly how hot can water be heated? Any compound, regardless ofwhether it is ionic, covalent, or metallic, will eventually decompose into the elements of which it is madeif you heat it to a temperature hot enough to break the bonds holding the atoms together. At atemperature in excess of 2000oC water becomes so hot that the covalent bonds holding the hydrogenatoms and oxygen atoms together break and water decomposes. This decomposition temperature willdiffer for each compound.

Assume we begin a second experiment with water vapor heated to a temperature inexcess of 150°C. You can follow this experiment by following the line on the plot from right to left,rather than from left to right as we did just a moment ago. As we remove heat from the system (i.ecool the system) the temperature of the system decreases. This continues until our vapor reachesa temperature of 100°C. At this point, we reach a leveling of the line. We continue to removeheat from the system but the system temperature does not decrease. This is because the relativelyhigh energy vapor molecules must release some of their kinetic energy before they are able tosettle into the (relatively speaking, much more sedate, stable, lower energy) liquid phase. Theenergy the system must lose before the vapor can condense into a liquid is called the enthalpy ofcondensation. Thus we see that condensation is an exothermic process. The enthalpies ofvaporization and condensation are related. They are equal in magnitude but opposite in sign.

Once the vapor has been converted to a liquid, the temperature of the liquid decreases asheat is removed from the system. The decrease in temperature continues until the liquid waterreaches a temperature of 0°C. At this point we reach a second leveling of the line. It representsthe removal of heat from the system but without a decrease in the temperature of the system.

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This occurs because energy must be removed from the system before it is sufficiently stable forthe strong interactions that take place between water molecules in the solid phase to form. Theenergy the system must lose before the liquid can freeze into a solid is called the enthalpy offusion. Fusion (freezing) is also an exothermic process. The enthalpies of fusion and melting arealso related. They too are equal in magnitude but opposite in sign. Note that fusion can mean oneof two things. In thermodynamics fusion means freezing. But outside of thermodynamics, whichfortunately is most of the time, fusion refers to nuclear reactions as described in Chapter 11, and not tochemical reactions or physical changes

Intermolecular forces: attractive interactions between molecules

To this point in this class, when we've discussed chemical bonds we have been talkingabout intramolecular bonds (or, intramolecular forces), although you did not know it.Intramolecular bonds are the bonding interactions that occur between atoms within a molecule.Intramolecular forces include metallic, ionic, and covalent bonds.

There are attractive interactions besides intramolecular forces. These are the interactionsthat take place between molecules, rather than within molecules. Intermolecular forces are thebonding interactions between molecules. They are found primarily between particles in theliquid and solid phases. If there were no intermolecular forces, all substances would behave asideal gases. Liquids and solids are held together by intermolecular forces.

Intramolecular forces and intermolecular forces are similar in that they are both based onthe attraction of oppositely charged particles. In Chapter 4 we first saw the equation, which iscalled Coulomb's law, that describes the force with which oppositely charged particles attracteach other:

Intramolecular forces result from the attraction of ions for one another (ionic bonds) or for theattractions between electrons and protons (covalent bonds, metallic bonds). These attractiveinteractions are based on full integer charges. Intermolecular forces result from molecularpolarity. We discussed this in Chapter 5. If you don’t remember now would be an extremely good timeto do a bit of review, as you’re going to be terribly lost if you don’t recall the things you hopefullylearned back then. Molecular polarity result from a molecule having one or more polar bonds.The charges of polar molecules are only partial charges. As a result, intermolecular bonds aremuch weaker than intramolecular bonds. The strongest intermolecular bonds are only about10-25% as strong as a typical covalent bond.

We will discuss four types of intermolecular forces in this chapter, London (or,dispersion) forces, dipole-dipole interactions, hydrogen bonds, and ion-dipole interactions,

Fkq q

r 1 2

2

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although there are other types of intermolecular forces besides these. These first three of theseintermolecular interactions are known collectively as van der Waal's forces (interactions).

London (dispersion) forces

London forces, or dispersion forces, are also known as induced dipole-induced dipoleinteractions. They take their name from Fritz London, who identified their role in chemicalbonding in 1930. London forces are generally the weakest type of intermolecular interaction,although they can be significant, especially in large, nonpolar molecules. London forces are theonly intermolecular bonding force that occur between nonpolar molecules. However, allsubstances are capable of London forces. Even if other stronger forces predominate theinteractions between molecules, London forces may still be a major contributor to theinteractions between polar molecules. They are the most common intermolecular forces. Londonforces are one of the principle intermolecular forces in living systems.

London forces are weak because (a.) partial charges are involved and (b.) because theyare temporary. They arise from the random movement of electrons in atoms and molecules. Let'sconsider two nonpolar molecules. This we see in the below image.

Note: nonpolar molecules are not necessarily round. I have simply represented them this way for easeof illustration. While electrons are constrained by the laws of quantum mechanics to specificshells, subshells, and orbitals, depending on their energy, they do have a certain amount offreedom to move about within atoms and molecules. This movement is extremely rapid and, innonpolar molecules, it is also more or less random. In general, we can describe the electrondistribution in a neutral, nonpolar molecule as a " 'cloud' of negative charge" surrounding thenucleus, which is positively charged. ("Physical Chemistry;" Walter J. Moore; Prentice-Hall, Inc.4th Ed., p. 914.) While this electron cloud is, on average, distributed symmetrically with respectto the nucleus, distortions of this cloud may arise and a nonpolar molecule may temporarily havea slight excess of electrons in one portion of the molecule. We see this in the image below.

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This means that, for the instant in which this charge imbalance exists, the molecule becomespolar, with a partial negative charge at the end of the molecule with the excess of electrons andwith a partial positive charge at the end of the molecule that is temporarily deficient in electrons.The polarity of this molecule, temporary though it is, has an effect on those molecules that are itsneighbors. This is shown in this image:

Electrons in the neighboring molecules that are adjacent to the negative end of the temporarilypolar molecule are slightly repelled, causing the neighboring molecule to also becometemporarily polar. Electrons in molecules that are adjacent to the positive end of the temporarilypolar molecule are slightly attracted, which also causes these neighboring molecules to becometemporarily polar. During the brief interval that these temporarily polar molecules exist, they areattracted to each other. This is a London force, as represented by the dashed lines between thespheres in the above image.

Then, as the electrons in the initial molecule return to a random distribution

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the polarity disappears, the bond is broken, and the electron distribution in the neighboringmolecule returns to the random state in which its electrons were initially found.

As we said, London forces are weak because the charges are only partial charges and alsobecause these partial charges are only temporary. The duration of the temporary polarity isgenerally described as "instantaneous" in most books, although in actuality this instantaneouslifetime must be finite albeit very brief. Let's assume for the sake of illustration that thesetemporarily polar molecules only last for a few microseconds. In actual fact they may last for alonger or shorter period of time, but humor me for just a moment, if you please. If this is the case,given the very high velocity of electrons in atoms, every nonpolar molecule in a system may becapable of becoming temporarily polar and forming these short-lived interactions with theirneighbors many thousands of times every second. In a collection of a mole of molecules, thiswould result in an unimaginably large number of these interactions, even in a short period oftime. By themselves London forces are weak, but when you consider the enormous number ofinteractions that can take place between a large number of molecules in a few seconds, you cansee that collectively these interactions are not inconsequential.

The magnitude of London forces increases with molecular size and surface area

As molecules (and individual atoms as well) become larger, they have more electrons.We know from previous discussions that the further we find electrons from the nucleus the moreloosely they are held. This is also true of molecules. We say that large molecules are more

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polarizable than smaller molecules. Polarizability is a measure of the ease of distortion of theelectron cloud of atoms and molecules. Molecules that are more easily polarized can developlarger partial charges during the brief interval in which they have asymmetric electrondistributions.

Surface area also plays a role in the magnitude of London forces. We will only describethis qualitatively. Imagine two sets of refrigerator magnets. Imagine trying to stick two of therefrigerator magnets together by their ends. Now imagine sticking two of the magnets togetherface to face. Which will stick together more forcefully? This is analogous to the effect of surfacearea on intermolecular bond strength.

Dipole-dipole interactions

With very few exceptions, molecules that have one or more polar bonds are polarmolecules. Again, if you need to review this information you'll find it in Chapter 5. One importantproperty of polar molecules is called the dipole moment, which is a measure of the polarity ofpolar molecules. The dipole moment of a polar molecule gives information about the magnitudeof the partial positive and negative charges within it. The larger the dipole moment of a polarmolecule, the more polar the molecule. The units of dipole moment are Debye units (D). Someexamples of substances and their dipole moments follows:

compound dipole moment (D)

HCN 2.98

H2O 1.84

NH3 1.47

HCl 1.08

SF4 0.63

CO2 0

SF6 0

Ne 0

Note that the last three named compounds are nonpolar compounds, either due to a lack of polarcovalent bonds or due to symmetry. Symmetry, as explained at the end of Chapter 5 in these notes.

Dipole-dipole interactions occur when a molecule with a permanent dipole moment forms a bondwith another molecule with a permanent dipole moment.

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Dipole-dipole interactions only occur between permanently polar molecules. Compared withLondon forces, dipole-dipole interactions are relatively strong because while polar moleculesonly bear partial charges, they are permanently polar, as compared to the temporary polarityfound in London forces. Note: polar molecules are not necessarily oval shaped. I have simplyrepresented them this way for ease of illustration.

Hydrogen bonds - a special case of dipole-dipole interactions

Within a molecule, when a hydrogen atom is bonded directly to either a nitrogen, oxygen,or fluorine atom, the resulting covalent bond is so polar that a very strong dipole-dipole bond isformed when these molecules interact with other similar molecules. This type of dipole-dipoleinteraction between molecules is called a hydrogen bond.

Let's use an alcohol as an example. Alcohols are a very large family of organiccompounds with hundreds of family members. All alcohols are characterized by the presence ofat least one hydroxyl group. A hydroxyl group is an O-H group and differs from a hydroxide ion,OH-, in that hydroxyl groups are uncharged. In the illustration below the "R" group bonded tothe hydroxyl groups can stand for any organic group and is unimportant in our consideration of

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the bonding that takes place. The O-H bonds in hydroxyl groups are extremely polar due to thedifferences in electronegativity in oxygen and hydrogen.

When two alcohol molecules approach each other with the correct orientation, thehydrogen atom of one alcohol, which has a relatively large partial positive charge, begins toshare one of the two non-bonding pairs of electrons on the oxygen atom (which has a largepartial negative charge) of the hydroxyl group of the other alcohol molecule. This results in theformation of a partial coordinate covalent bond and a very strong intermolecular interactionbetween the two alcohol molecules. The dashed lines in the image below represent hydrogenbonds between the pairs of alcohol molecules.

In case you don’t recall, we discussed coordinate covalent bonds in Chapter 5. Still, while hydrogenbonds are a very strong intermolecular force, they are on average only about 10-25% as strong asan average covalent bond. This is because the charges involved in hydrogen bonds are onlypartial charges.

Remember that hydrogen bonds are intermolecular bonds. There are exceptions to this,especially in very large biological molecules such as proteins, but we will not consider thoseexceptions in this class. The bond between the hydrogen atom and the nitrogen, oxygen, orfluorine atom within the molecule is simply a polar covalent bond.

Ion-dipole interactions

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An ion-dipole interaction is the intermolecular bonding force that occurs between apermanently polar molecule and an ion. As one might expect, these are on average strongerintermolecular forces than hydrogen bonds, as they involve interactions between a polarmolecule, which has its partial charges, with an ion, which has a full charge.

Predicting the intermolecular forces in compounds

Intermolecular forces play an important role in many important physical properties. Wewill expand on this topic below. To predict the role of intermolecular forces in the physicalproperties of a substance, we need to be able to accurately state the intermolecular forces ofwhich the substance is capable. Remember the general order of intermolecular bond strengths:on average, ion-dipole bonds are the strongest, followed by hydrogen bonds, and then bydipole-dipole interactions. London forces are generally the weakest intermolecular force, but allatoms and molecules are capable of London forces.

To correctly identify the intermolecular forces of which a substance is capable we mustbe able to correctly identify the substance as being polar or nonpolar. To do this we need to beable to draw its Lewis structure and to identify whether or not any of its bonds are polar bonds,and also whether or not the compound has symmetry.

Let's look at two examples of problems in which we need to be able to identify theintermolecular forces of which compounds are capable.

• Rank the following in order of the strength of their intermolecular forces: CH4 (methane),CH2F2 (difluoromethane), H2 (hydrogen), and CH3OH (methanol).

Methane.

Methane is a nonpolar molecule. It therefore has London forces, because allsubstances are capable of London forces. As a nonpolar molecule it cannot havedipole-dipole interactions or hydrogen bonds. It is not capable of forming ion-dipole interactions. Note: while we can and will find London forces, dipole-dipoleinteractions, and hydrogen bonds in pure substances, in this class we will only find ion-dipole interactions occurring in aqueous solutions of dissolved ionic compounds andnever - in this class - in samples of pure substances, such as pure water, or puremethane, or pure - well, I hope you get the idea.

Difluoromethane.

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Note that the nonbonding pairs of electrons do not appear on the fluorine atoms in thisLewis structure - sorry! I need a new software package that permits me to do betterLewis structures. Difluoromethane is capable of London forces because allcompounds, polar and nonpolar alike, are capable of London forces. The C-Fbonds in the compound are polar bonds. This molecule has two polar bonds anddoes not have symmetry and is therefore a polar molecule. It is also capable ofdipole-dipole interactions. It is not capable of forming hydrogen bonds.

Hydrogen.

Hydrogen is capable of London forces, as all substances are capable of Londonforces. It is a nonpolar molecule so it is not capable of dipole-dipole interactions,neither is it capable of forming hydrogen bonds.

Methanol.

Methanol is capable of London forces, as all substances are capable of Londonforces. The C-O and O-H bonds are polar, and the molecule does not havesymmetry. It is therefore a polar molecule and is capable of forming dipole-dipoleinteractions. Further, as there is a covalent O-H bond within the molecule, this isa compound that can form hydrogen bonds with other similar molecules.

Ok, we’re now ready to rank these four compounds in order of the strength of theirintermolecular forces.

Methanol is capable of hydrogen bonding, which are the strongest intermolecular forceswe see in this collection of compounds, so it sits at the top of the list for these four compounds.

Difluoromethane cannot form hydrogen bonds but it can form dipole-dipole interactions,which are stronger than London forces, so we rank it second in this group of four compounds.

Methane and hydrogen both form London forces and nothing else, but we would expectthe London forces to be stronger in methane than in hydrogen because methane is largermolecule. So we’d rank methane third and hydrogen last (or weakest) in ranking the strengths ofthe intermolecular forces for these four compounds. No, I do not expect you to remember thenames or molecule formulas of the compounds used in this example, or in the next example either.

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• Rank the following in order of the strength of their intermolecular forces:CH2OHCHOHCH2OH (glycerin), Ar, Ne, CH3CH2F (fluoroethane), and CH3NH2

(methylamine).

We must again begin with the Lewis structures for the compounds in which weare interested and use these to determine whether the compounds are polar ornonpolar.

Glycerin.

Again, I need to remind you that these Lewis structures do not show the nonbondingpairs of electrons on the oxygen atoms in glycerine, the fluorine atom in fluoroethane,or the nitrogen atom in methylamine. And again, I apologize. Also: yes, glycerine doeshave a Lewis structure a bit more complicated than those we have seen in this class.Don’t panic. I’m not going to ask you to do a Lewis structure this complicated on yourown in this class. We’re using it for the purposes of this example only. Glycerine iscapable of London forces, as all substances are capable of London forces. The C-O and O-H bonds are polar, and the molecule does not have symmetry. It istherefore a polar molecule and capable of forming dipole-dipole interactions.Further, as there is a covalent O-H bond within the molecule, this is a compoundthat can form hydrogen bonds with other similar molecules. Moreover, as thereare three O-H groups on this one molecule, we would expect this molecule to beable to form as many as three hydrogen bonds per molecule.

Argon. The Lewis structure for this noble gas atom consists of a single atom (Ar)surrounded by four nonbonding pairs of electrons. Argon is capable of Londonforces, as all substances are capable of London forces. It is nonpolar (how couldan individual atom be otherwise than nonpolar?) so it is not capable of dipole-dipole interactions, neither is it capable of forming hydrogen bonds.

Neon. The Lewis structure for this noble gas atom consists of a single atom (Ne)surrounded by four nonbonding pairs of electrons. Neon is capable of Londonforces, as all substances are capable of London forces. It, as argon, is nonpolar soit is not capable of dipole-dipole interactions, neither is it capable of forminghydrogen bonds.

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Fluoroethane.

Fluoroethane is capable of London forces, as all substances are capable of Londonforces. The C-F bond is polar and the molecule does not have symmetry. It istherefore a polar molecule and is capable of forming dipole-dipole interactions.But at this point take care. Even though this a molecule which contains bothfluorine and hydrogen, as there is no hydrogen covalently bonded to the fluorineatom, it is *not* capable of forming hydrogen bonds with other molecules.

Methylamine.

Methylamine is capable of London forces, as all substances are capable ofLondon forces. The C-N and N-H bonds are polar (remember what I said earlierabout the C-N bond), and the molecule does not have symmetry. It is therefore apolar molecule and is capable of forming dipole-dipole interactions. Further, asthere is a covalent N-H bond within the molecule, this is a compound that canform hydrogen bonds with other similar molecules.

We can now rank these five compounds in order of the strength of their intermolecularforces.

Both glycerine and methylamine are both capable of hydrogen bonding, the strongest ofthe intermolecular forces we see in pure substances. But as glycerine can form up to threehydrogen bonds per molecule while methylamine can only form one, we would predict that theintermolecular forces are stronger in glycerine than in methylamine.

Next we come to the dipole-dipole interactions in fluoroethane.

Argon and neon both form London forces and nothing else. We expect the London forcesto be stronger in argon because it is larger molecule. So we’d say that glycerine has the strongestintermolecular forces in this collection of substances, followed in order by methylamine,fluoroethane, argon, and neon.

The relative strengths of intermolecular forces can be estimated by comparing the boilingpoints of compounds. For boiling to occur in any substance enough energy must be added to the

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system to overcome the intermolecular forces between the liquid phase molecules. The strongerthe intermolecular forces, the more energy must be added in order to overcome those forces. Thisis reflected as a higher boiling point.

As an example, three substances have boiling points of 175°C, -30°C, and 50°Crespectively. Rank these three in order from strongest to weakest intermolecular forces:

(strongest) 175°C > 50°C > -30°C (weakest)

Did you notice that it doesn't even matter if we know what the compounds are. As long as weknow their boiling points we can make correct assumptions about the strengths of theintermolecular forces in the substances. We cannot, however, come to any correct conclusionsabout the types of intermolecular forces in these three substances. For this we need molecularformulas and Lewis structures. A comparison of melting points and other physical properties canalso provide insight into the relative strengths of the intermolecular forces between compounds.We will discuss this below.

Intermolecular forces and solutions

We first learned of solutions in Chapter 1 and we will study them in greater detail inChapter 9. Intermolecular forces are an essential feature of all solutions, but let's become familiarwith the vocabulary of solutions before we begin discussing them. You'll be seeing this first bitagain in Chapter 9, so save yourself some time and pay attention now.

• solutions are homogeneous mixtures that are equally dispersed at the molecular level andthat are uniform throughout in their physical and chemical properties

• solvent: does the dissolving in a solution and is generally the substance present ingreatest abundance. There is only ever one solvent per solution.

• solute: gets dissolved and is the substance (or substances) that is less abundant. Asolution may have more than one solute.

• solvation is the molecular process by which solutes are dissolved by solvents

A knowledge of the intermolecular forces of which a substance is capable can help uspredict the substance's solubility behavior. The phrase "like attracts like," or "like dissolves like"is a very good general rule that can help us predict whether or not one substance will dissolve inanother. "Like dissolves like" means that substances with similar intermolecular forces will mixand form solutions, while substances with different intermolecular forces will not form solutionswell, if at all. Let's work a few examples of this.

Will the following pairs of substances form solutions?

• Table salt and water. Table salt, sodium chloride, is an ionic compound. Water is a polarliquid. Since the hydrogen bonds that hold water together in the liquid phase are similar

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to the ionic bonds that hold sodium chloride together in the solid state, we predict thatsodium chloride and water should dissolve in each other.

• Water and alcohol. Water is a polar liquid capable of forming hydrogen bonds. This isalso true of alcohols. Alcohols and water dissolve in each other.

• Water and oil. Water is a polar liquid. Oil is a homogeneous mixture of various nonpolarcompounds. Since the intermolecular forces are dissimilar, oil and water do not dissolvein each other.

• Oil and vinegar. Oil is a homogeneous mixture of various nonpolar compounds. Vinegaris an aqueous solution of acetic acid, a polar compound. Since the intermolecular forcesare dissimilar, oil and vinegar should not dissolve in each other.

• Oil and hexane. Oil is a homogeneous mixture of various nonpolar compounds. Hexaneis a nonpolar organic compound. Since the intermolecular forces are similar, oil andhexane should dissolve in each other.

• Oil and carbon tetrachloride. Oil is a homogeneous mixture of various nonpolarcompounds. Carbon tetrachloride is a nonpolar organic compound. Since theintermolecular forces are similar, oil and carbon tetrachloride should dissolve in eachother.

Intermolecular forces and physical properties

Many of the physical properties of substances are determined by the intermolecularforces between the substance’s particles. Freezing point, boiling point, vapor pressure and theease with which a substance evaporates, density, and the solubility - or lack of solubility - of onecompound in another are all examples of properties directly determined by intermolecular forces.Let’s take a look at the boiling points and aqueous solubilities of some organic compounds.

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alkanes (hydrocarbons) alcohols

substance BP (oC) solubility substance BP (oC) solubility

CH4 -162 no CH3OH 65 yes

C2H6 -89 no C2H5OH 78 yes

C3H8 -42 no C3H7OH 97 yes

C4H10 -1 no C4H9OH 117 moderately

C5H12 36 no C5H11OH 138 slightly

C6H14 69 no C6H13OH 158 no

Although it may not be immediately apparent to you, there is a wealth of information containedin this table. Relax! You don’t have to memorize this table or the names and molecular formulas ofthe compounds we discuss in this table. But you do need to understand what we’re looking at in thistable and I’m going to explain it to you.

Let’s begin with the alkanes, methane (CH4), ethane (C2H6), propane (C3H8), butane(C4H10), pentane (C5H12), and hexane (C6H14). Alkanes are all nonpolar compounds. The onlytypes of bonding forces that can occur between hydrocarbon molecules are London forces. Theirstructures may be seen in the following image.

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Note that the boiling points of these hydrocarbon compounds increase as the size of themolecules increases. This is consistent with the behavior of London forces: the larger themolecules, the stronger the London forces between the molecules. It is important to rememberthat boiling point of a substance, which is a physical property, is determined by the strength ofthe intermolecular forces between the particles of the substance (as we mentioned earlier). Andas we said, in a general sense, if we compare the boiling points of any two substances, the onewith the higher boiling point will have the stronger intermolecular forces between its particles,while the one with the lower boiling point will have the weaker intermolecular forces between itsparticles.

If we examine the solubilities of the six hydrocarbons we see that none of them is solublein water. But as water is polar and the hydrocarbons are nonpolar, this should not surprise us.

Now let’s look at the six alcohols, methanol (CH3OH), ethanol (C2H5OH), propanol(C3H7OH), butanol (C4H9OH), pentanol (C5H11OH), and hexanol (C6H13OH). All alcohols arepolar compounds which are also capable of forming hydrogen bonds. Their structures may beseen in the following image.

If we compare the boiling points of the alkane and the alcohol with the same number ofcarbons i.e., methane and methanol, ethane and ethanol, propane and propanol, etc., we see thatthe alcohols have much higher boiling points than the corresponding alkanes. The alcohols areslightly larger, it’s true, but the principal reason for the difference in boiling points is a reflectionof the difference in strength of the stronger hydrogen bonds of the alcohols and the weakerLondon forces of the alkanes.

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Notice that the boiling points of the alcohols increase as the size of the moleculesincreases. This is similar to the behavior demonstrated by the alkanes, and for exactly the samereason: the larger the molecules, the stronger the London forces between the molecules. It mayseem odd to you to think that London forces play a role in the behavior demonstrated by alcoholsbut remember that all substances experience London forces. And, the larger the molecules thestronger the London forces between them.

We need to remember this to explain the aqueous solubility of the alcohols. The firstthree alcohols are soluble in water and this should come as no great surprise, as both water andalcohols are polar and also capable of forming hydrogen bonds. But it is curious to note thatbutanol is less soluble than the three smaller alcohols, pentanol even less so, and hexanol isinsoluble water. How can it be that these three polar hydrogen-bond forming compounds havedifficulty dissolving in water? Look at the structures of the alcohols. As they become larger, theybegin to resemble the alkanes as the carbon backbone of the molecules becomes longer. Andremember: the chains of carbon atoms bonded to hydrogen atoms are not polar. Although thealcohols retain their polar hydroxy group and the capability to form hydrogen bonds, the longernonpolar hydrocarbon chains become more strongly attracted to each other than the polarhydroxy groups are to each other or to water. In other words, at around 5-6 carbon atoms inalcohols, we reach the point at which polar compounds are more stable behaving as though theywere nonpolar.

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Chapter 9: Solutions

General solution properties and definitions

Let's begin our study of solutions with some definitions.

Solutions are homogeneous mixtures that are evenly dispersed at the molecular level andthat are uniform throughout in their physical and chemical properties. They consist of a singlephase. Most of us are familiar with solutions made by dissolving a solid in a liquid, but asolution may be made by dissolving:

• a solid in another solid (all alloys)• a solid in a liquid (all aqueous solutions)• a liquid in a solid (mercury in gold)• a liquid in another liquid (rubbing alcohol, for example)• a gas in a solid (hydrogen in palladium is a good example although most of us in this

class don't have any real world experience with this mixture)• a gas in a liquid (carbon dioxide in soda, beer, and champagne)• a gas in another gas (oxygen and nitrogen in air)

There are always at least two components in every solution. The solvent is the materialpresent in greater abundance in the solution. The solute is present in lesser abundance. Ifsolvation occurs, the solvent does the dissolving and the solute is the substance that is dissolved.A solution can only have one solvent, but it may have more than one solute. If you doubt this,read the label for a can of your favorite soda, which, regardless of the product you select, is anaqueous solution of carbon dioxide, sugar, and various other substances. Unless it's diet soda -then it has an artificial sweetener in place of the sugar.

Dissolution (the processing of dissolving ) is often a spontaneous process. At a molecularlevel, the dispersion of solute molecules throughout the solvent results in an increase in theentropy of the solution system. For many solutes dissolution in aqueous solution is anendothermic process, although it is not uncommon for dissolution to be an exothermic process.The enthalpy of solvation is dependent both on the solvent and on the solute and thus varies fromone solution system to another.

Solvation is a molecular process by which a solute may become dispersed throughout asolvent. This is an especially important process when discussing the dissolution of solids inliquids. For solvation to occur, solvent particles (atoms, ions, or molecules) must form attractiveintermolecular interactions with solute particles at the surface of the bulk solute. The bulk solute isthe big mass of solute particles as it is placed in the solvent, before the individual particles aredispersed throughout the solvent. These interactions take place most easily at the edges and cornersof the bulk solute. As each particle leaves the bulk solute, other solvent particles crowd aroundit, forming a number of solvent-solute intermolecular bonds. This phenomenon is called

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clustering. The number of solvent particles that cluster around solute particles depends onvarious attributes of the solution, such as the particular solvent and solute, solute concentration,and temperature. For an aqueous 1M sodium chloride solution at room temperature and pressure,it is common for four to seven water molecules to cluster around each of the ions as they arepulled away from the solute surface. As solvated solute particles are randomly dispersed throughthe solvent they are accompanied by their cluster of solvent particles at all times. Bare ionsalmost never exist in nature and especially in solution. They are nearly always in the company ofsolvent clusters. This dispersion can occur due to collisions with other molecules, or through physicalmixing of the solution, or both.

For solvation to occur, the solvent particles must successfully attract individual soluteparticles away from the bulk solute by forming intermolecular interactions with them. If theinteractions between individual particles in the bulk solute are too strong, or if the soluteparticles are too small, it will not be possible for a sufficient number of solvent particles to formintermolecular bonds with individual solute particles to induce them away from the bulk solutesurface. When this occurs, a solute will not dissolve in a solvent. A solute that will not dissolvein a solvent is said to be insoluble.

Hydration is the specific name for the process of solvation when water is the solvent. Anillustration of a hydrated (solvated) ion may be found at http://en.wikipedia.org/wiki/Solvation.In this example of the hydration of sodium ion, water molecules form ion-dipole interactionswith the sodium ion. The water molecules orient themselves such that their oxygen atoms, whichhave a partially negative charge due to the polarity of the water molecules, are attracted to apositively charged sodium cation.

As we stated in Chapter 8, "like attracts like" or the equivalent "like dissolves like" is avery good general rule that can help us predict whether or not one substance will dissolve inanother. It qualitatively describes the probability of the formation of the favorable intermolecularinteractions between solvent-solute particles required for dissolution to occur. "Like dissolveslike" means that generally speaking, substances with similar intermolecular forces will mix andform solutions, while substances with different intermolecular forces will not form solutionswell, if at all.

The solubility of a substance is the maximum amount of that substance that will dissolvein a stated amount of a specific solvent to give a thermodynamically stable solution. Solubility isusually expressed in grams of solute for a given amount of solvent, at specific temperature andpressure conditions. As examples:

• the solubility of NaCl is 35.7 g per 100 mL of water at 0°C and 1 atm of pressure• the solubility of CO2 is 171.3 g per 100 mL of water at 0°C and 1 atm of pressure• the solubility of barium sulfate is 0.000222 g per 100 mL of water at 18°C and 1 atm of

pressure. This small number tells us that barium sulfate does not dissolve very well in water atthis temperature.

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Solubility may also be expressed as the number of moles, rather than grams, that will dissolve ina given amount of solvent. This is called the molar solubility of a substance.

A saturated solution is a solution in which as much solute as is physically possible hasbeen dissolved in the solvent, with some solute remaining, undissolved in the container holdingthe solution. The point at which a solution becomes saturated is dependent on the particularsolvent-solute system. It also depends on the temperature of the solution. An unsaturatedsolution is one in which more solute may be dissolved. Occasionally, supersaturated solutionsmay be formed. These are solutions in which more solute has been dissolved than isthermodynamically stable. This does not necessarily mean that they are dangerous, but it doesmean that they can be a challenge to prepare and store. Portable chemical hand warmers andliquid infant heel warmers contain a supersaturated solution of sodium acetate dissolved inwater.

Substances that are miscible can form solutions in all proportions. In other words, if twosubstance are miscible, they are infinitely soluble in each other.

A colloid consists of small aggregates of particles (clusters of particles) of one materialdispersed throughout another. Small aggregates are defined as those that are about 500 nm orsmaller in size. ("Physical Chemistry," Peter W. Atkins; W. H. Freeman and Co. 5th Ed., p. 971)These collections of particles often behave differently than they do as individual particlesdispersed throughout a solvent. There are different types of colloids, depending on the types ofmaterials involved. A dispersion of solid particles in a solid, or of solid particles in liquid, iscalled a sol. A dispersions of solid particles in a gas, or of liquid particles in gas, is called anaerosol. A dispersion of liquid particles in a liquid is called an emulsion. Colloids are oftenreferred to as suspensions, which contain molecular-level aggregations of particles physicallysuspended in liquid that are often so small that gravity has no effect on the aggregates.

Calculating concentration

One of the most important attributes of a solution is its concentration. Concentration is ameasure of the amount of solute dissolved in either a specific amount of solvent or solution.

Concentration can be calculated in a variety of ways. Concentration can be stated interms of percent composition which is, in general, the ratio of amount of solute to some amountof solution multiplied by 100. There are three types of percent composition:

• Weight/weight (w/w) = (grams solute/grams solution) x 100 • Weight/volume (w/v) = (grams solute/volume solution) x 100• Volume/volume (v/v) = (volume solute/volume solution) x 100

Some examples of the calculation of percent composition are as follow.

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• What is the (w/w) concentration of 100. g of solution containing 1.22 g of sodiumchloride?

(w/w) = (1.22 g NaCl/100 g solution) x 100 = 1.22% w/w

• What is the (w/v) concentration of 50.0 mL of solution containing 1.22 g of sodiumchloride?

(w/v) = (1.22 g NaCl/50.0 mL solution) x 100 = 2.44% w/v

• What is the (v/v) concentration of a solution containing 100 mL of isopropyl alcohol thatis diluted with water to 150 mL?

(v/v) = (100 mL isopropyl alcohol/150.0 mL solution) x 100 = 66.7% v/v

The concentrations of dilute solutions can be expressed in terms of mg%, parts permillion (ppm), and parts per billion (ppb).

• mg % = (mg solute/100 mL solution) x 100• ppm: = (mg solute/L solution)• ppb = (ug solute/L solution)

When concentration is expressed in parts per million it is equivalent to one part solute forevery one million parts of solution. Another way of looking at this is to think of it in terms ofpopulations of people. If the state of Utah has a population of three million people, then 1 ppm isequal to three individuals out of the state's population. Parts per billion can also be thought ofalong these lines. If the country of China has a population of one billion people, then 1 ppb isequivalent to 1 person out of the total population of the country.

While ppm and ppb may seem like very small concentrations, they still are oftensignificant. For example, many substances are toxic at a ppm level, some lethally so, and thereare numerous chemicals that can induce physiological effects at a ppb level. As another example,our noses are extremely sensitive to many chemicals. Many of the chemicals we smell in theworld around us are only present in ppm concentrations. We can smell the isopropyl alcohol inrubbing alcohol beginning at concentrations of about 22 ppm and the acetone in nail polishremover at concentrations of about 5 ppm and higher. Of course we're talking about the airborneconcentrations of these compounds as they evaporate; this ability to smell these compounds in notdirectly linked to the concentrations of the compounds in the products which I have mentioned. Wehave a particularly acute sense of smell when it comes to a class of compounds called sulfides,all of which contain a sulfur atom. The smallest member of the sulfide family is hydrogensulfide, H2S. Note: this is the common name of compound, not the IUPAC name. But this particularcompound is known almost exclusively by its common name and so we’ll use it here. This compoundis also known as rotten egg gas, and it is what one smells when one drives by oil refineries or

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visits geysers, hot springs, and mud pots as are found in Yellowstone National Park. The averageperson can smell H2S gas beginning at a concentration of about 0.5 ppb, or 500 parts per trillion!The levels at which we can begin to smell chemicals are called odor thresholds. There are a some goodreferences available online, if you find this topic interesting. I'd suggest you begin with the most currentedition of 3M's "Respirator Selection Guide," which can be found at their corporate website and at nocharge.

Molality is also known as molal concentration. It is the ratio of moles of solute per kg ofsolvent. Molal concentration is indicated with a lower case "m".

• What is the molal concentration of 125 grams of glucose that is dissolved in 750 grams ofwater?

m = mol solute/kg solvent

convert 125 grams of glucose to moles of glucoseconvert the mass of water to kg of water

(125 g glucose) x (1 mole glucose/180.2 g glucose) x (1/750 g water) x (1000 g water/1 kg water) = 0.925 m

Molarity is also known as molar concentration. It is the ratio of moles of solute per literof solution. Molar concentration is indicated with an upper case "M".

• What is the molar concentration of 125 grams of glucose that is dissolved in 750 mLs ofwater?

M = mol solute/L solution

convert 125 grams of glucose to moles of glucoseconvert the volume of water to liters of water

(125 g glucose) x (mole glucose/180.2 g glucose) x (1/0.750 L solution) = 0.925 M

It is not uncommon for the molar and the molal concentrations of a solutions to be thesame, but it does not always work out that way. Molality is not as commonly used as it once was.Molarity is much more commonly encountered. Some examples of typical molarity calculationsfollow.

• If 0.100 moles of H2SO4 is dissolved in 450 mL of water, what is the molarity of theresulting solution?

convert 450 mL of water to liters

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0.100 mol/0.450 L = 0.222 M

• If 37.6 grams of copper (II) nitrate is dissolved in 500 mL of water, what is the molarityof the resulting concentration?

convert 37.6 g of copper (II) nitrate to moles; the Mm = 187.56 g/molconvert 500 mL of water to L

(37.6 g copper (II) nitrate) x (1 mol copper (II) nitrate/187.56 g copper (II) nitrate) x (1/0.500 L) = 0.401 M

• How many moles of copper (II) nitrate are contained in 25.0 mL of 2.5 M solution?

if M = mol/L then mol = M x Lconvert volume to L

2.5 mol/L x 0.025 L = 0.0625 mol copper (II) nitrate

• How many grams of ammonium hydroxide are contained in 75 mL of concentrated(15M) solution?

To do this problem we need to calculate the number of moles of ammonium hydroxidefound in 75 mL of concentrated solution and then convert it to a mass in grams using themolar mass of the compound (Mm = 35.05 g/mol)

again, if M = mol/L then mol = M x Lconvert volume to L

15 mol/L x 0.075 L x (35.05 g/mol) = 39.4 g ammonium hydroxide

Dilutions

It is often necessary to take concentrated solutions and to lessen or lower the concentrations.This is typically done by adding the concentrated acid or base to water. In doing so, the number ofmoles of acid or base remains constant but as the volume of the solution increases, the concentrationof the solution decreases.

compound solution strength

acetic acid 17.5 M

hydrochloric acid 12 M

sulfuric acid 18 M

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nitric acid 16 M

ammonium hydroxide 15 M

No, you don't need to remember these concentrations, they are simply provided to help illustrate a point.Too, we should point out explicitly what is being implied: all of the substances listed in the table are actuallyaqueous solutions of the various compounds for which they are named.

Assume we begin with exactly 1.00 L of 12 M hydrochloric acid. This means that we haveexactly 12.0 moles of hydrochloric acid in each liter of solution.

12 M HCl = 12 mol HCl/1.00 L solution

Now let’s assume we begin to add water to the solution. The number of moles of HCl is unaffectedby the addition of water but the solution volume begins to increase. In this solution, water is the solventand HCl is the solute. We’re making an assumption here: that the solvent does not react with the soluteand therefore does not consume it as more solvent is added. In terms of concentration, this means thatthe numerator (the number of moles of acid) remains constant while the denominator (the solutionvolume) becomes larger. As a consequence, the overall solution concentration decreases.

moles of HCl solution volume solution concentration

12.0 1.00 L 12.0 M

12.0 2.00 L 6.0 M

12.0 3.00 L 4.0 M

12.0 4.00 L 3.0 M

12.0 6.00 L 2.0 M

12.0 12.00 L 1.0 M

We can dilute any solution by adding more solvent to the solution. This does not change thetotal number of moles of solute in the solution, again, operating under the constraint that the solventand solute do not react with each other but it does lessen the ratio of the amount of solute per amountof solvent. In other words, when a solution is diluted, the solute concentration becomes less.

A useful equation when studying dilutions is

M1V1 = M2V2

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in which M1 is the initial concentration of the solution, V1 is the initial volume of the concentratedsolution, and M2 andV2 are the final concentration and final volume of the diluted solution. If wethink of this equation in terms of the units involved

(mol1/L) x (L) = (mol2/L) x (L)

remembering that the units of molarity are mol/L, it can be simplified to

mol1 = mol2

since the units of volume cancel. This corroborates what was stated above: during the course of adilution the total number of moles of solute does not change. It is constant.

Here are a few examples of how this dilution equation can be used.

• How many milliliters of concentrated NH4OH are required to form 100. mL of 1.0 Msolution?

make a table of the given values to help clarify what you have and what you are solving for

M1 15 M

V1 ?

M2 1.0 M

V2 100 mL

if M1V1 = M2V2

thenV1 = (M2V2)/M1

(1.0 mol/L x 100. mL)/15 mol/L = 6.67 mL of concentrated ammonium hydroxide

This means that if 6.67 mL of 15 M ammonium hydroxide are diluted to 100 mL, theconcentration of the diluted solution will be 1.0 M

Note that it was not necessary to convert the volume from milliliters to liters. As long as oneis consistent, either unit may be safely used.

• If 10.0 mL of 12 M HCl is diluted to 600 mL, what is the new concentration of the acid?

make a table of the given values to help clarify what you have and what you are solving for

M1 12 M

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V1 10.0 mL

M2 ?

V2 600 mL

if M1V1 = M2V2

then (M1V1)/V2 = M2

(12.0 mol/L x 10. mL)/600 mL = 0.200 M HCl

In plain English, by taking 10 mL of 12 M HCl and adding another 590 mL of water to thesolution (so that the total solution volume is 590 + 10 = 600. mL), we change the concentration ofHCl to 12 M to 0.200 M.

• The dilution of 50.0 mL of a solution of phosphoric acid results in 1.00 L of a 0.75 Msolution. What was the initial concentration of the acid?

make a table of the given values to help clarify what you have and what you are solving for

M1 ?

V1 50.0 mL

M2 0.75 M

V2 1.00 L

if M1V1 = M2V2

then M1 = (M2V2)/V1

(0.75 mol/L x 1.00 L)/0.050 L = 15 M phosphoric acid

Note that we have to be consistent in our choice of units of volume. We chose to solve thisproblem using L, but we could also have correctly solved it using mL.

Some properties of liquids and solutions: surface tension and capillary action

For any substance, in the vapor phase (gaseous state) its particles are, relatively speaking,far apart, possess high kinetic energy, and have few if any interaction with each other. For this samesubstance in the liquid phase, the particles will be lower in kinetic energy than in the vapor phase,closer together than in the vapor phase, and there will be attractive interactions between theparticles.

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The particles at the boundary of a liquid and it's vapor behave differently than particles faraway (5-10 particle diameters) from the phase boundary. In the vapor state, away from the interfacewith the liquid, the spacing between the particles is so great that, in relative terms, they seldom havethe opportunity to interact. In the bulk liquid, away from the interface with the gas, liquid particlesattract each other equally and, in turn, are equally attracted in all directions. But at the phaseboundary the attraction in the liquid is one-sided. The liquid particles are attracted inwardly, towardother molecules in the liquid phase. The liquid particles at the surface can engage in intermolecularinteractions with other particles in the liquid phase but they can form few or no such interactionswith particles in the gas phase, due to the low density of the particles in the vapor phase and alsobecause of the relatively high kinetic energy of the vapor phase particles. This one-sided attractionmakes the liquid resist expansion. It also drives the bulk liquid to attempt to minimize its surfacearea. The particles at the interface pack together more closely than in the bulk liquid. The moleculesclosest to the interface wind up acting like a kind of "skin" on the surface of the liquid. This explainsthe behavior of water droplets on the waxed surface of a car. Like attracts like. The molecules at thesurface of a water droplet are attracted to the hydrogen-bonding interior of the droplet and not to thewaxy surface of the car, which can only engage in dispersion forces. As a consequence, the dropletsbead up. This is also the reason that rain drops falling through the air assume a more or less sphericalshape. And this attractive tendency is the reason that liquids in containers form a meniscus at theirsurface.

Surface tension is related to this attribute of liquids attempting to make their surface area assmall as possible. Surface tension is the energy required to overcome the tendency of liquids tominimize their surface area.

There are exceptions to this behavior. If the intermolecular forces in the substances on bothsides of phase interface have similar bonding interactions, the meniscus may shrink to the point ofbeing barely observable. As an example, the meniscus in a container of a nonpolar liquid is oftenless pronounced than that seen with polar liquids. And as nonpolar liquids have similarintermolecular forces as those observed in air, this is not surprising.

There may also be attractive intermolecular interactions between particles in the liquid phaseand those on the surface of the container holding the liquid, if they are similar. The surface of glassvessels is polar. If the diameter of the glass container is very small, as is the case with a thin glasstube, attractive interactions between a polar liquid - such as water - and the tube's interior surfacecan cause the liquid to creep up the interior walls of the tube. This phenomenon is called capillaryaction.

Some properties of liquids and solutions: vaporization and vapor pressure

Let's discuss the behavior of a substance in the liquid state in a sealed container. Thecontainer will hold not only the substance in its liquid form, but it will also hold some of thesubstance in its vapor (gas) phase.

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An important difference between molecules in the liquid and vapor phases is the kineticenergy of the particles. Particles in the vapor phase of a substance generally have higher kineticenergies than particles in the liquid phase of the same substance.

If we examine the kinetic energy of all of the particles in any phase of any substance, be itsolid, liquid or gas, we find that the particles do not all have a single kinetic energy. Rather, thereis a (more or less) Gaussian distribution of energies amongst the particles. A Gaussian distribution isa bell-shaped curve. In other words, in a collection of the particles in any phase, some of the particleshave high energies, some have low energies, and most have a kinetic energy intermediate betweenhigh energy and low energy. This kinetic energy is affected by a number of factors but most notablyby temperature. We talk about this relationship in the “General gas properties” section of Chapter 8.

Particles in the liquid phase of a substance that have (a.) a high kinetic energy and (b.) thatare near the liquid-gas phase boundary can actually have enough kinetic energy to "escape" fromthe liquid phase and join the vapor phase. This is why heating a liquid facilitates the transition ofthe liquid to the vapor phase. As the liquid is heated, the kinetic energy of the particles increases andthe fraction of particles with sufficient kinetic energy to escape the liquid phase also increases.

Molecules in the vapor phase travel randomly. They collide with the walls of their containerand with each other. They also collide with the liquid surface during their random travels. If thesecolliding vapor phase particles have high kinetic energy they generally bounce off liquid surfacewhen they bump into it. But vapor phase particles with low kinetic energy may "stick" and becomepart of liquid phase when they collide with it. By cooling a vapor we lower the average kineticenergy of the molecules in the vapor phase and make it easier for them to " stick" to the liquidsurface if they collide with it.

At any given temperature, an equilibrium exists between the rates of particles leaving theliquid phase and joining the vapor phase and the reverse process, particles leaving the vapor phaseand rejoining the liquid phase. As per our discussion of equilibrium in the Chapter 7 notes under theheading “Chemical equilibrium.” The pressure exerted by the vapor when this equilibrium exists iscalled the vapor pressure.

A liquid will boil when its vapor pressure is equal to atmospheric pressure. We call thetemperature at which this occurs the normal boiling point of the liquid. In other words, the normalboiling point of a liquid is the temperature at which the vapor pressure of the liquid is equal toexactly 1 atm of pressure. That water boils at 100°C tells us that at this temperature, the vaporpressure of water is equal to exactly 1 atm. The boiling point of isopropyl alcohol, a compoundfound in rubbing alcohol, is 82.4°C. This means that in pure isopropyl alcohol at this temperature,the vapor pressure of the liquid will equal exactly 1 atm of pressure.

There is an interesting implication to this. As atmospheric pressure decreases, any liquid willboil at an increasingly lower temperature. Water boils at 100oC when atmospheric pressure is exactly1.00 atm. If one tries to boil water in the mountains, where atmospheric pressure is significantly less

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than 1.00 atm, the water will boil at a temperature less than 100oC.This is why it takes longer to boilan egg or to cook pasta when one is camping in the mountains. Conversely, if one were vacationingin a place where atmospheric pressure was greater than 1.00 atm, water would then boil at atemperature greater than 100oC and your egg or pasta would cook more quickly. There is a onlinevideo clip at YouTube that demonstrates the boiling of water at room temperature. The demonstratorlowers the pressure by using a vacuum pump. Take a look if you’re interested at the link found athttp://www.youtube.com/watch?v=LxtAeGtL9SE.

In a comparative respect vapor pressure is rather like boiling point. It gives a general senseof the relative strengths of intermolecular forces. If we compare the vapor pressure of twosubstances at the same temperature, the substance with the highest vapor pressure is the substancewith the weakest intermolecular forces. These weaker intermolecular forces make it easier forparticles to escape the liquid phase and enter the vapor phase. The vapor pressure of such asubstance will be higher than that of a substance with stronger intermolecular forces, which makeit more difficult for particles to escape the liquid phase.

Some properties of liquids and solutions: diffusion

In the solid state the particles of a substance are more or less locked in place, but in pureliquids and gases the particles of a substance tend to move randomly. In solutions this randommovement results in solute particles moving from areas of high concentration to areas of lowconcentration until they are uniformly distributed throughout the solvent particles in the solution.The solvent particles in like manner tend to move from areas of high concentration to areas of lowconcentration until they too are uniformly distributed throughout the solute particles. It is statisticallypossible for the solvent and solute molecules in a solution to behave in some other manner, butmathematically it is extremely unlikely. This behavior is called diffusion. This is how a solute achievesuniform distribution throughout a solvent in the process of forming a solution, whether the solutionis a gas/gas solution, a gas/liquid solution, a liquid/solid solution, or etc.

If we add a single crystal of sodium chloride to a container of water, the salt will dissolveand ionize. The ions will eventually become uniformly distributed throughout the container of water.At a molecular level, upon addition of the crystal, the concentration of sodium chloride within thecrystal is much higher than the concentration of sodium chloride in the pure water. Diffusiondisperses the sodium ions and chloride ions throughout the water until they are evenly dispersedthroughout the water. Conversely, the concentration of pure water outside the crystal is much greaterthan the concentration of water within the crystal. Diffusion disperses the water moleculesthroughout the sodium chloride until they are evenly dispersed throughout the dissolving solid.

Diffusion makes the system in which it occurs more random, i.e., the entropy of the systemincreases as a consequence of diffusion. This means that diffusion should usually be a spontaneousprocess. It is, in fact, one of the fundamental laws of thermodynamics that the overall entropy of a systemwill increase during a spontaneous process. This randomizing effect will occur independent of anyexternal forces (e.g. mixing) and is due to random collisions between molecules. These random

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collisions are called Brownian motion (see Einstein's Explanation of Brownian Motion athttp://galileo.phys.virginia.edu/classes/109N/more_stuff/Applets/brownian/brownian.html). The rate of diffusion depends on those factors that affect the rates atwhich these random collisions may occur and include temperature, viscosity, pressure, the molecularweight of the solute, and solute concentration.

Some properties of liquids and solutions: osmosis

Let's begin with a few definitions.

• A membrane is a sheet-like structure, often porous, that can regulate the passage ofsubstances from one side of the membrane to the other based on molecular size or charge.

• Membranes that allow some substances to pass but not others are called semi-permeable.• Osmosis is the flow of water through a semi-permeable membrane from areas of low to high

solute concentration.

I would strongly advise you to view the short video entitled “Osmosis Experiment” athttp://www.youtube.com/watch?v=7WX8zz_RlnE. It will take you less than a minute. In the videoa column of blue liquid moves up a tube over a period of several hours. Why does this occur?

The blue liquid is probably a concentrated solution of table sugar, dissolved in water, witha few drops of blue food coloring added to make it easier to see the behavior of the solution. Thesugar solution sits in a beaker of distilled water. A semi-permeable membrane separates the sugarsolution from the distilled water. It is the natural tendency of water to pass from areas of highconcentration to areas of low concentration. As water is at a higher concentration in the beaker, andat a lower concentration within the sugar solution, water molecules pass through the semi-permeablemembrane into the solution. As the semi-permeable membrane does not stretch, the addition of waterto the solution increases its volume, which forces it slowly but surely up the tube.

Diffusion is a two-way street. Just as it is the natural tendency of water to diffuse from anarea of high concentration (pure water) to an area of low concentration (sugar solution), it is alsothe natural tendency of the sugar in the solution to diffuse from areas of high concentration (sugarsolution) to areas of low concentration (pure water). But in this example it cannot occur. Watermolecules are small enough to pass freely through the pores of the semi-permeable membrane. Butthe sugar molecules are too large. Hence the sugar remains trapped in the solution, while water canflow freely into the solution.

Osmotic pressure is a force that is generated by osmosis. It is the pressure that would berequired to stop the net flow of water (i.e., the force required to stop osmosis) from one side of thesemi-permeable membrane to the other.

Reverse osmosis is a process which forces water across a semi-permeable membrane fromareas of high to low solute concentration. Reverse osmosis is used to purify sea water. Sea water,which is salty, this comes as a real revelation to you, doesn't it? is placed on one side of a

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semi-permeable membrane and then forced through a membrane with small enough pores to preventions from passing through it. The natural tendency of the purified water is to diffuse back throughthe membrane, to areas of high salt concentration, but means are devised to prevent this fromoccurring. While reverse osmosis is a very effective way to purify water, it takes a great deal ofenergy. This can make the cost of purifying water using reverse osmosis very expensive. And yetin areas like the Persian gulf and Saudi Arabia, there are entire cities that obtain their drinking andhousehold water from large reverse osmosis plants. Reverse osmosis is also used to purify water ona limited scale in many American cities, like Los Angeles.

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Chapter 10: Acids and Bases

General acid-base information: a review

We have already spent a bit of time discussing acids and bases in Chapter 4. Let’s summarizewhat was said.

• There are three common acid-base theories.• The Arrhenius theory defines acids as substances that can donate a hydrogen ion (H+) and

bases as compounds that can donate a hydroxide ion (OH-). • The molecular formulas of Arrhenius acids are usually written with hydrogen listed first,

e.g., HCl, HBr, HC2H3O2 and so on.• The names of many Arrhenius acids are common names due to their extensive historical use,

but they are still accepted by IUPAC. • The names of Arrhenius acids always include the word “acid.”• The common bases are the Group 1A and 2A hydroxides, ammonia, and ammonium

hydroxide.• Not every substance that contains hydroxide can behave as a base.

The Arrhenius acid-base theory

There are a number of different acid-base theories. We will concern ourselves with three thatare more commonly used, the Arrhenius theory, the Brønsted-Lowry theory, and the Lewis theory.

The Arrhenius theory is named for the Swedish chemist Svante Arrhenius, who not onlycontributed to our knowledge of acids and bases but also to our understanding of strong and weakelectrolytes in aqueous solution. He received a Nobel Prize for his work in Chemistry in 1903, the thirdawarded in this branch of science. A brief biography and description of his work may be found at a website maintained by the Nobel Foundation which awards the annual prizes. If you’ve never visited this site,even if you don’t like chemistry, it’s rather interesting all the same. Arrhenius’s page can be found athttp://nobelprize.org/nobel_prizes/chemistry/laureates/1903/arrhenius-bio.html. As we have said,according to the Arrhenius theory an acid is a substance that can donate a hydrogen ion in aqueoussolution, and a base is a substance that can donate a hydroxide ion in aqueous solution. This is anextremely useful working definition and in fact, the one most commonly used in elementary andgeneral chemistry. But as it only describes things as acids and bases in aqueous solution, it is oftenrather limited in its ability to be useful when studying subjects outside the purview of elementarychemistry.

The Brønsted-Lowry acid-base theory

In the 1920s, several decades after Arrhenius’s work was published, Johannes Brønsted andMartin Lowry, working independent of each other and hundreds of miles apart, arrived at the sameconclusion at more or less the same time. According to their theory an acid is a substance which can

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donate a proton in chemical reactions and a base is a substance which can accept a proton inchemical reactions. Proton donation is accomplished through the donation of a hydrogen ion and notthrough the donation of protons from the nuclei of other atoms, which would be nuclear (non-chemical) events requiring huge amounts of energy.

Let’s be sure we’re clear on this. A hydrogen atom consists of its nucleus, in which we finda single proton, and a single electron which is found in an orbital in the 1s shell and subshell outsidethe nucleus. If we remove the electron from a hydrogen atom all that remains is its nucleus - inwhich we find one proton. Knowing this single fact can make a particular aspect of physiology a bit moreclear. An important class of medicinal compounds are the proton pump inhibitors, which include suchmedicines as Nexium and Prilosec. What does it mean to say that they inhibit the production of protons?It’s not that they interfere with nuclear reactions taking places in human stomachs. Rather, they simplymediate the production of hydrochloric acid which is produced in cells lining the stomach and which isessential to digestion.

The principal advantage of the Brønsted-Lowry theory is that it is a broader, moreencompassing theory than that of Arrhenius. The Brønsted-Lowry theory can be used to explainacid-base behavior in aqueous solution, but also in non-aqueous solution, i.e., solutions in which thesolvent is not water. Everything that is an Arrhenius acid is also a Brønsted-Lowry acid. Everythingthat is an Arrhenius base is also a Brønsted-Lowry base. While it is theoretically possible for asubstance to be a Brønsted-Lowry acid and not an Arrhenius acid, the exceptions will not bediscussed or worried about in this class. But there are many thousands of substances that can accepta proton in chemical reactions besides hydroxide ion and in doing so, behave as Brønsted-Lowrybases.

In this image we see that ammonia and sulfide ion can accept a proton (hydrogen ion), just ashydroxide ion does. In doing so they behave as Brønsted-Lowry bases. We also note that hydrogensulfide ion, HS-, also possesses the ability to accept a proton and to form hydrogen sulfide as aresult. Any anion can act as a Brønsted-Lowry base. All of the monatomic and polyatomic anionswe’ve discussed in this class can behave as Brønsted-Lowry bases, as can all of the thousands ofpolyatomic anions we’ve not discussed in this course. There are also many neutral molecules whichcan accept protons, such as ammonia and water. We’re going to be talking a little more about the acid-base behavior of water later in this chapter.

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The Lewis acid-base theory

In 1923, the same year that Brønsted and Lowry proposed their acid-base theory, GilbertLewis proposed his own acid-base theory in which he described acids as electron-pair acceptors andbases as electron pair donors. This is the broadest and most useful of the acid-base definitions.Although it is one we will little use in this course (and in fact, it is only mentioned once very brieflyin your text), we want to spend a few minutes trying to better understand this theory.

As is the case with Brønsted-Lowry acid-base reactions, Lewis acid-base reactions may takeplace either in aqueous or non-aqueous solution.

As the broadest acid-base definition, we can say that anything that is an Arrhenius acid isalso a Brønsted-Lowry acid and therefore also a Lewis acid. But, there are many substances that can act as Lewis acids but which do not donate protons (hydrogen ions) and which are therefore notBrønsted-Lowry or Arrhenius acids. As an example, all transition metal cations have the capacityto act as Lewis acids (i.e., accept an electron pair from a donor), but as they do not have hydrogenions (or protons) to donate, they cannot act as Brønsted-Lowry or Arrhenius acids.

And, as the broadest definition, we can also say that anything that is a Arrhenius base anda Brønsted-Lowry base is also a Lewis base. But, there are substances which can act as Lewis bases(i.e. donate an electron pair) but which do not accept protons (hydrogen ions) in the process andwhich are therefore not Brønsted-Lowry bases.

The reaction of silver (I) ion and ammonia (NH3) is an example of a Lewis acid-basereaction. In this reaction

Ag+(aq) + 2 NH3 (aq) => Ag(NH3)2

+(aq)

This is a rather odd and completely unexpected reaction, to us in this class at least. And yet ithappens easily and spontaneously. In this reaction silver(I) ion acts as a Lewis acid as it accepts anelectron pair from the nitrogen atom in each of ammonia molecules (remember the Lewis structureof ammonia!). Each ammonia molecule acts as a Lewis base, in that the nitrogen atom in eachammonia molecule has a nonbonding pair of electrons it shares with the silver(I) ion. The resultingsilver product is a called a complex ion. Complex ion chemistry is a topic on which we could easilyspend an entire semester, if we had the time and if you as students had the interest. There arehundreds of these complex ions that occur naturally in the world around us. They formspontaneously. Their physical and chemical properties typically differ significantly from that of the uncomplexed ions (e.g., the physical and chemical properties of silver(I) ion - such as melting point, boilingpoint, density, color, chemical reactivity, toxicity, and etc. - should be different from those of the Ag(NH3)2

+(aq) ion). But it will suffice to point out that the formation of each of a complex ions is

usually the result of a Lewis acid-base reaction.

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The typical attribute of a Lewis acid is one or more empty orbitals. In order to accept anelectron pair there must be a place for the accepting atom (molecule) to place it and empty orbitalsare typically used. The typical attribute of a Lewis base is the presence of one or more atoms withat least one pair of electrons to donate. Most often this will be a nonbonding pair of electrons on oneof the atoms in a compound, or the electron pairs found in pi bonds. The pi bond electrons found indouble and triple covalent bonds make it possible for substances with these types of bonds to act as Lewisbases as well. But as we have not discussed pi electrons, now that I’ve mentioned it, you can promptlyforget about it - unless you study organic chemistry outside of this class. The sharing of a nonbonding pairby the Lewis base with the empty orbital of the Lewis acid results in the formation of a coordinatecovalent bond, a covalent bond in which one atom donates both of the bonding electrons. Soundfamiliar? This is a topic we discussed in Chapter 5. The definition of a coordinate covalent bond is one youreally should be sure you remember. And as we have seen in the reaction of silver(I) ion and ammonia,it is possible for a single Lewis acid to form coordinate covalent bonds with more than one Lewisbase, depending on the number of empty orbitals in the Lewis acid.

Given that all Arrhenius acid-base reactions are also Brønsted-Lowry acid-base reactions,and that all Brønsted-Lowry acid-base reactions are also Lewis acid-base reactions, then it is wellworth our time to take a look at a few reactions to prove this concept is correct. We need to useLewis structures as illustrations to make this possible.

Let’s begin with the reaction of hydrogen ion and hydroxide ion, an Arrhenius acid-basereaction.

In this reaction we see that the oxygen atom of the hydroxide ion possesses two nonbonding pairsof electrons. The 1s orbital of the hydrogen ion is empty. Remember: the full electron configuration ofhydrogen is 1s1. If it loses its electron in becoming an ion, it now has a configuration of 1s0. One ofoxygen’s two nonbonding pairs is used to form a coordinate covalent bond and water is formed asa product. In this illustration the curved arrow running from the oxygen atom in hydroxide ion to thehydrogen ion indicates that it is the atom which is sharing a nonbonding pair with the hydrogen ion. Andas we said in Chapter 5, the coordinate covalent bond formed in this reaction is identical in length andstrength to regular covalent bond that existed in the hydroxide ion. In this reaction hydrogen ion is actingas a Lewis acid and hydroxide ion is acting as a Lewis base. To be clear: hydrogen ion is acting as anArrhenius acid as well as a Brønsted-Lowry acid and a Lewis acid, and hydroxide ion is acting as anArrhenius base, as a Brønsted-Lowry base, and also as a Lewis base.

In the reaction of hydrogen ion and ammonia

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the nitrogen atom of the ammonia molecule possesses one nonbonding pair of electrons. As in theprevious example, the 1s orbital of the hydrogen ion is empty. Nitrogen’s nonbonding pair is usedto form a coordinate covalent bond with hydrogen ion. Again, the curved arrow running from thenitrogen atom in ammonia to the hydrogen ion indicates that it is the atom sharing a nonbonding pair withthe hydrogen ion. As a result, ammonium ion is formed. In this reaction hydrogen ion is acting as aLewis acid and ammonia is acting as a Lewis base. Hydrogen ion is acting as an Arrhenius acid as wellas a Brønsted-Lowry acid and a Lewis acid. Ammonia is acting as a Brønsted-Lowry base and also as aLewis base.

Finally, as an example, let’s look at the reaction between silver (I) ion and ammonia.

In this reaction the silver(I) ion has two empty orbitals. The nitrogen atom in each ammoniamolecule uses it’s nonbonding pair of electrons to form a coordinate covalent bond with one ofsilver(I) ion’s empty orbitals. As a result, the silver diammine complex ion is formed. (Note: theIUPAC name of the Ag(NH3)2

+ complex ion is silver(I) diammine ion. We’re not going to worry about howto name complex ions. Silver(I) diammine ion is not a name you’re required to remember. I only mentionit in case you’re curious. If you’re not, feel free to forget this bit.) In this reaction silver(I) ion is actingas a Lewis acid and ammonia is acting as a Lewis base. Silver(I) ion is acting only as a Lewis acid,not as an Arrhenius or Brønsted-Lowry acid. Ammonia is acting only as a Lewis base and not alsoas an Arrhenius or Brønsted-Lowry base in this particular reaction. Why? How can you tell? Youneed to remember your definitions. Silver(I) ion is not donating a hydrogen ion (Arrheniusdefinition) or a proton (Brønsted-Lowry definition) in this reaction. Ammonia is not hydroxide ion;in this reaction it is not accepting either a hydrogen ion (Arrhenius definition) or a proton(Brønsted-Lowry definition).

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Strong and weak acids and bases

As we covered this material back in Chapter 6, we will review a few things you havehopefully learned and remember which are relevant for the remainder of this chapter. If you don’tremember this stuff, turn back to Chapter 6 and spend a few minutes reviewing it. You’re going to needit. It’s also information you’ll probably want to include on your 3" x 5" card for the final exam.

• All compounds are classified as strong electrolytes, weak electrolytes, or non-electrolytes.• This is true of all ionic compounds and also of all covalent compounds.• Electrolytes are substances that form aqueous solutions capable of conducting electricity. A

substance does this by dissociating when in aqueous solution. The ions serve as chargecarriers.

• Substances must dissolve before they can dissociate. Dissolution, or, the process ofdissolving, and dissociation are two different things. Non-electrolytes are substances that candissolve without dissociating. But - and this is important to remember! - no substance candissociate unless it first dissolves.

• Strong electrolytes are substances that dissociate completely in aqueous solution. This meansthat roughly100% of all of the molecules in a sample of strong electrolyte will dissociate inaqueous solution if they dissolve.

• All ionic compounds are strong electrolytes, as are all strong acids and bases.• Not all ionic compounds are soluble in aqueous solution. But we still classify them as strong

electrolytes based on their potential for complete ionization if they did dissolve.• Weak electrolytes are substances that dissociate incompletely in aqueous solution. This

means that far less than 100% of all of the molecules in a sample of weak electrolyte willdissociate in aqueous solution if they dissolve. Typical values usually range from 1-10%.This means that in a sample of weak electrolyte, only about 1-10% of the particles willionize if they dissolve in water.

• The most common weak electrolytes are the weak acids and bases. Note: which are the strongacids? The strong bases? Which are the weak acids and weak bases? How do we know? If youdon’t recall, review!!!

• Non-electrolytes do not ionize at all in aqueous solution, even though they may dissolve.Note: which compounds are typically nonelectolytes? Again, if you don’t remember, be sure toreview!

• Nearly all organic compounds are non-electrolytes, except for the organic acids and organicbases.

• In chemistry the words “strong” and “weak” are used almost exclusively to indicate theextent to which a compound ionizes in water. There is no apparent correlation betweenstrong and weak and how corrosive or dangerous a compound may be.

Hopefully this is all familiar and well understood to you. And hopefully you retain the abilityto categorize substances as strong acids, weak acids, strong bases, and weak bases, based on the listyou were given in Chapter 6.

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The behavior of acids and conjugate bases in water

The generic behavior of any Arrhenius acid in aqueous solution, whether strong or weak, isdescribed by the following equation:

HA(aq) + H2O(l) => H3O+

(aq) + A-(aq)

In this equation HA represents any (and all) Arrhenius acids. The anion of the acid HA is A- and iscalled the conjugate base of the acid HA. We will discuss conjugate bases further in just a moment. Thecompound H3O

+is called hydronium ion. The reaction by which hydronium ion is formed is asfollows:

The oxygen atom in water shares one of its two nonbonding pairs of electrons with hydrogen ion,forming a coordinate covalent bond. It should be noted that bare ions of any sort are extremelyunstable and almost never occur in aqueous solution without being surrounded by a cluster ofsolvent (water) molecules. As was the case for sodium ion in Chapter 9, it is also true for hydrogenion in that it seems to move about in aqueous solution surrounded by four to seven water molecules.Studying clustering in solutions is one of the more difficult problems chemists attempt. Hence the lack ofprecision in relating the number of water molecules per cluster. And it does not help when one learns thatcluster size, i.e., the number of solvent molecules clustering around the solute molecule, depends not onlyon the given solvent and the solute but also on temperature, solute concentration, and other factors.Hydronium ion does exist, but it’s lifetime is extremely short. Different research papers offerdifferent values for this lifetime, but it’s thought to be fractions of a picosecond before thecoordinate covalent bond between a hydrogen ion and one water molecule in the solvent clusterbreaks, the hydronium molecule decomposes to water and hydrogen ion, and then a new coordinatecovalent bond is formed with the hydrogen ion and a different water molecule in the ion’s solvationcluster, forming a new hydronium ion as a part of the solvation cluster. As far as we’re concernedin this class, it’s far easier to use the molecular formula of hydronium when balancing equations thanthe molecular formula for a small, four water molecule cluster, which would be expressed as H9O4

+.

The anion of any acid is called its conjugate base, as mentioned above. This is not a randomchoice of words. The anion of any acid, by itself, will behave in the following way in aqueoussolution:

A-(aq) + H2O(l) => HA(aq) + OH-

(aq)

As we know, any substance that forms hydroxide ion in aqueous solution can be correctly classifiedas a base. By labeling the anions of acids as conjugate bases we are simply acknowledging theircapacity to do just that: to behave as bases.

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This generic equation showing the behavior of any and all conjugate bases in water is aspecial type of reaction called a hydrolysis reaction. Hydrolysis reactions are water-splittingreactions, i.e., reactions in which water molecules are split into hydrogen ions and hydroxide ions.

In this reaction the conjugate base shares one of its nonbonding pairs of electrons with one of thehydrogen atoms of a water molecule and forms a coordinate covalent bond with it. The hydrogenatom in water loses its electron. The O-H covalent bond in water is broken. Both electrons in thebroken bond are transferred to the oxygen atom which now belongs to a hydroxide ion. There arenumerous reactions in which hydrolysis reactions occur besides the reactions of conjugate bases. Inphysiology hydrolysis plays an important role in many common and important reactions. For example, whenproteins, fats, or complex carbohydrates are ingested, one of the first steps in the process of digestion ishydrolysis of the bonds between the various amino acids in the protein (or hydrolysis of the componentsof the fat molecules, or hydrolysis of the complex carbohydrate into mono and disaccharides). There aremany other examples. You should know, by definition, what a hydrolysis reaction is.

You need to be able to identify the conjugate base of any acid. This is far easier than youmight believe. We simply remove the acidic “H” at the beginning of the molecular formula for anacid. The remainder will have a net negative charge as the acidic hydrogen atom leaves its electronbehind when the acid dissociates. Here are a few examples:

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acid conjugate base

hydrofluoric acid, HF fluoride ion, F-

hydrochloric acid, HCl chloride ion, Cl-

hydrobromic acid, HBr bromide ion, Br-

hydroiodic acid, HI iodide ion, I-

nitric acid, HNO3 nitrate ion, NO3-

perchloric acid, HClO4 perchlorate ion, ClO4-

acetic acid, HC2H3O2 acetate ion, C2H3O2-

sulfuric acid, H2SO4 hydrogen sulfate ion, HSO4-

hydrogen sulfate ion, HSO4- sulfate ion, SO4

2-

phosphoric acid, H3PO4 dihydrogen phosphate ion, H2PO4-

dihydrogen phosphate ion, H2PO4- hydrogen phosphate ion, HPO4

2-

hydrogen phosphate ion, HPO42- phosphate ion, PO4

3-

There are several important things to note in this table.

The relationship between an acid and it conjugate base is the same for all acids, regardlessof whether they are strong acids or weak acids. Simply remove the acidic “H” from the molecularformula and add a negative charge to the acid’s anion to determine its conjugate base.

Note that the molecular formula for sulfuric acid is H2SO4, something you’ve known forsome time now. The number “2" following the initial “H” in the molecular formula tells us that asingle molecule of sulfuric acid can donate two protons in a chemical reaction. In a broader sense,this is true of all acids. The subscript following the initial “H” in the molecular formula of any acidtells us how many protons one molecule of that acid can donate in a reaction. So the first seven acidsin the above table (hydrofluoric acid, hydrochloric acid, hydrobromic acid, hydroiodic acid, nitricacid, perchloric acid, and acetic acid) can donate one hydrogen ion per molecule of acid. The number“1" is not written expressly but is implied. This means that phosphoric acid, H3PO4, can donate threehydrogen ions per molecule. And an imaginary acid, H4A, could donate how many hydrogen ionsper molecule of acid? (4)

Acids that can donate one hydrogen ion per molecule of acid are called monoprotic acids.Or if you’d rather, we can also think in terms of moles and say that a monoprotic acid is an acid that candonate one mole of hydrogen ions per mole of acid molecules. Acids that donate two hydrogen ions permolecule of acid are classified as diprotic acids, while those that can donate three hydrogen ions per

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acid molecule are referred to as triprotic acids. Most of the common acids in the world around usare monoprotic or diprotic acids. There are a few common triprotic acids, such as phosphoric acidand citric acid. And there are one or two tetraprotic acids known, which can donate four hydrogenions per acid molecule. In general, acids that are not monoprotic may be referred to as polyproticacids, acids that can donate more than one hydrogen ion per molecule.

The seven monoprotic acids in the above table behave exactly as described when predictingtheir conjugate bases. But when we reach sulfuric acid, we find that it’s conjugate base is thehydrogen sulfate ion and not simply sulfate as we might have predicted. This is because a diproticacid does not lose both of its acidic protons at once. Instead, they dissociate in a series of steps. Thenumber of dissociative steps is equal to the number of protons a molecule of acid can donate in areaction. Sulfuric acid can donate two protons in a chemical reaction. It will therefore lose theseprotons in a series of two steps.

In the first step sulfuric acid reacts with water to form hydronium ion and sulfuric acid’s conjugatebase, hydrogen sulfate ion. In the second step hydrogen sulfate ion, which is itself an acid, reactswith water to form hydronium ion and sulfate ion, the conjugate base of hydrogen sulfate ion. Howdo we know hydrogen sulfate ion is an acid? It’s molecular formula begins with “H.” If we add these twosteps together we obtain the observed reaction, which tells us that one molecule of sulfuric acid willreact with two molecules of water to form two molecules of hydronium ion and one molecule ofsulfate ion. The hydrogen sulfate ion does not cancel out in this case because it is a spectator ion. It issimply a function of the arithmetic and nothing more.

We have observed that hydrogen sulfate ion can behave both as an acid and also as a base.Substances with this ability are classed as amphoteric, possessing the capacity to act either as an acidor as a base. We can also classify hydrogen sulfate are amphiprotic, possessing the capacity to eitherdonate or accept a proton. Amphoteric and amphiprotic do not mean exactly the same thing, but thetwo definitions are sufficiently close that we, for our purposes in this class, will consider them tobe the same. Amphiprotic substances are actually quite common in the world around us.

We have observed that phosphoric acid is a triprotic acid. We might therefore reasonablydeduce that it will not lose all three acid protons at once but rather, in a series of three dissociativesteps which are as follows:

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In the first step phosphoric acid and water react to form hydronium ion and dihydrogen phosphateion, the conjugate base of phosphoric acid. In the second step dihydrogen phosphate ion, which isamphoteric, reacts with water to form hydronium ion and hydrogen phosphate ion, the conjugatebase of dihydrogen phosphate ion. And in the third step hydrogen phosphate ion, which is alsoamphoteric, reacts with water to form hydronium ion and phosphate ion, the conjugate base ofhydrogen phosphate ion. In the overall reaction one molecule of phosphoric acid reacts with threewater molecules to form three hydronium molecules and a molecule of phosphate ion. As was thecase above, hydrogen phosphate ion and dihydrogen phosphate ion do not cancel because they arespectator ions, but rather, simply as a function of the arithmetic.

To re-iterate what we’ve just learned: diprotic acids lose those two H+ ions in two steps.Triprotic acids lose their 3 H+ ions in three steps. For any diprotic acid, strong or weak:

step 1: acid - H2A; conjugate base - HA-

step 2: acid - HA-; conjugate base A2-

Over all, H2A forms two H+ ions and an A2- anion, but it's important to note that this occurs in a pairof steps. If you're asked for the conjugate base to a diprotic acid, the correct answer depends onwhether you're talking about the acid H2A in step 1, or the acid HA- in step 2.

The behavior of bases and conjugate acids in water

The Arrhenius bases, which consist of the Group 1 and 2 hydroxides and ammoniumhydroxide, behave as ionic compounds in aqueous solution. They dissolve and they also dissociate.The Group 1 and group 2 hydroxides are strong bases, and as such they ionize completely.Ammonium hydroxide is a weak base, and while it does dissolve readily, only a fraction of itssolvated molecules also ionize. As we’ve noted previously, this is extremely unusual behavior foran ionic compound because, as a rule, nearly all ionic compounds are strong electrolytes. Usingsodium hydroxide to serve as a model for all Arrhenius bases, we see they behave as follows:

NaOH(aq) => Na+(aq) + OH-

(aq)

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The generic behavior of any Brønsted-Lowry base in aqueous solution (except for theArrehenius bases), whether strong or weak, is described by the following generic equation:

B(aq) + H2O(l) => HB+(aq) + OH-

(aq)

I hope you noticed this is a hydrolysis reaction! In this equation B represents all Brønsted-Lowry bases(except for the Arrhenius bases). The protonated form of the base HB+ is called the conjugate acidof the base B. The conjugate acid of any base, by itself, will behave in aqueous solution in a mannerdescribed by this generic equation:

HB+(aq) + H2O(l) => B(aq) + H3O

+(aq)

You need to be able to identify the conjugate acid of any base. We simply add a hydrogenatom to the molecular formula of the base and change it’s charge by the amount of “+1". Here area few examples:

base conjugate acid

ammonia, NH3 ammonium ion, NH4+

hydroxide ion, OH- water, H2O

chloride ion, Cl- hydrochloric acid, HCl

bromide ion, Br- hydrobromic acid, HBr

nitrate ion, NO3- nitric acid, HNO3

acetate ion, C2H3O2- acetic acid, HC2H3O2

sulfate ion, SO42- hydrogen sulfate ion, HSO4

-

hydrogen sulfate ion, HSO4- sulfuric acid, H2SO4

phosphate ion, PO43- hydrogen phosphate ion, HPO4

2-

hydrogen phosphate ion, HPO42- dihydrogen phosphate ion, H2PO4

-

dihydrogen phosphate ion, H2PO4- phosphoric acid, H3PO4

As with the table of acids and conjugate bases, there are several things to take note of in thistable as well.

Many of these conjugate acids are amphiprotic.

Conjugate acids do not always have a net positive charge. Regardless of the base’s charge,the charge on its conjugate acid is always “1+" more than that of the base. Ammonia is a base withno net charge. It’s conjugate acid ammonium ion has a net charge of 1+. Chloride ion has a net

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charge of 1-. It’s conjugate acid hydrochloric acid has no net charge as it is neutral. Phosphate ionhas a net charge of 3-. It’s conjugate acid, hydrogen phosphate, has a net charge of 2-. Hydrogenphosphate ion has a net charge of 2- while its conjugate acid, dihydrogen phosphate, has a net chargeof 1-. And so on.

You have been given a total of four generic equations in this section and the previous section.These are the equations showing the behavior of all acids in water, the behavior of conjugate basesin water, the behavior of Brønsted-Lowry bases in water, and the behavior of the conjugate acidsof Brønsted-Lowry bases in water. I do not expect you to remember the specific equationsdescribing the behavior of sulfuric acid and phosphoric acid in aqueous solution, but you shouldremember and be able to apply all four of these generic equations to any acid or base.

The acid-base behavior of water

Water is amphiprotic. It can behave as an acid or as a base, depending on circumstances. Weknow that water is an acid as its molecular formula, H2O, begins with “H.” As we see in the tablein the previous section water is the conjugate acid of hydroxide ion. Water behaves as a base in theformation of hydronium ion, which we discussed above, and also in many other reactions we havenot discussed.

The equation describing the equilibrium behavior of pure water is

2 H2O(l) W H3O+

(aq) + OH-(aq)

This equation describes behavior referred to as the auto-ionization of water (it is also called theauto-dissociation of water, the self ionization of water, and etc.).

The equilibrium constant expression for this reaction is

K H O OHw [ ][ ]3

in which the equilibrium constant Kw is called the auto-ionization constant of water (or the auto-dissociation constant of water, the self-ionization constant of water, etc.). The numerical value ofthis auto-ionization constant has been measured and is known to be 1.0 x 10-14 at roughly 0 oC.

Knowing this number permits us to make several very important observations. That Kw isless than 1 tells us that this is a reactant-favored reaction. The forward reaction in which hydroniumion and hydroxide ion are formed only happens to a very slight extent. The reverse reaction, inwhich water remains in its undissociated state, is energetically favored and therefore spontaneous.We discussed the meaning of equilibrium constants in the Chapter 7 section entitled “Equilibrium constantsand Gibbs free energy.”

Knowing the value of the equilibrium constant permits us to calculate the concentrations ofhydronium ion and hydroxide ion. In pure water

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[H3O+] = [OH-]

as there is a one to one relationship between the two compounds in the balanced equation thatdescribes the auto-ionization of water. And by using the equilibrium constant expression and a bitof arithmetic, we can calculate that

[H3O+] = [OH-] = 1.0 x 10-7

The pH scale

In chemistry we often find it useful to employ logarithms when we work with numbers thatare either very large or very small. This is because in the days before calculators and computers, sliderules and logarithm tables were commonly used to facilitate work with very large and very small numbers.If we take the log (which is short for “logarithm”) of the hydrogen ion concentration in pure waterand change the sign of the resulting calculation

-log (1.0 x 10-7) = 7.00

we have calculated the pH of pure water. Note: throughout the remainder of this chapter when we speakof taking the log of something, and from this point on we’ll mention it frequently, we mean the base 10log. To do this on your calculator you use the “log” button, not the “ln” button which is a different type oflogarithmic function. This value serves as a reference point for determining whether things in theworld around us are acidic or basic. In an acidic solution the concentration of hydronium ion isalways greater than the concentration of hydroxide ion

[H3O+] > [OH-]

and the pH of an acidic solution is always less than 7. In a basic solution the concentration ofhydroxide ion is always greater than the concentration of hydronium ion

[H3O+] < [OH-]

and the pH of basic solutions is always greater than 7. In a neutral solution, the hydronium ionconcentration and hydroxide ion concentrations are equal

[H3O+] = [OH-]

and the pH is exactly equal to 7, as is the case in pure water.

Most of the things in the world around us are either acidic or basic, not neutral. There arevarious places you can turn to get a better idea of which things are acidic and which things are basic.Your text has a number of examples, and the Wikipedia entry for pH has several illustrations worth

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examination. (http://en.wikipedia.org/wiki/PH note -should you look at this reference I would encourageyou to confine yourself to looking at the illustrations, as the accompanying text is something that will berather confusing and frustrating to most of you)

One important thing to note is that the pH scale is a logarithmic scale. This means that littledifferences in pH correspond to big differences in acidity or basicity. One can calculate the relativedifference in the acidities of two solutions using the following equation

difference = 10ΔpH

where ΔpH is the difference between the two pH values being compared.

If we want to know how much more acidic a solution with a pH of 6 is than a solution witha pH of 7,

difference = 10(7-6) = 101

This means that the solution with a pH of 6 is ten times more acidic than the solution with a pH of 7.If we compare two solutions, one with a pH of 7 and the other with a pH of 5, the difference in pHis

difference = 10(7-5) = 102

or, the solution with the pH of 5 is 100 (102) times more acidic than the solution with a pH of 7. Ifwe compare the pH of pure water, with it a value of 7, with the pH of 1.0 M hydrochloric acid, witha pH of 0, the difference in acidities is

difference = 10(7-0) = 107

or, the solution of hydrochloric acid is 10,000,000 (107) times more acidic than the pure water.

Calculating [H3O+], [OH-], and pH

Before we begin, you need to have a calculator that can do logarithms. A TI-30, a CasioFX-260, or something equivalent is necessary. And you need to know how to use it. This issomething I can’t help you with. You need to sit down and take a look at the manual or to havesomeone help you. You should also be sure you know how to use the exponential notation featureof your calculator. For many calculators there is an “EE” button or an “EXP” button which shouldbe used to enter a number using scientific notation. Students are often surprised to learn that if theysimply type, for example, “1 x 10-14" into their calculator that the instrument understands the inputdifferently than if they used “1 EE ± 14" which is something the calculator will correctly interpret.

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We will only discuss how to calculate the pH for strong acids and bases. It is not difficultto calculate the pH of weak acids and bases but it requires the use of equilibrium theory in a way wedo not discuss in this class.

To calculate the pH of a strong acid solution we must know the equilibrium hydronium ionconcentration of that solution. Fortunately there is always a direct correlation between the startingconcentration of the acid and the equilibrium hydronium ion concentration which we can use tomake these calculations.

Let’s take hydrochloric acid as an example. The equation describing its behavior in wateris

HCl(aq) + H2O(l) => H3O+

(aq) + Cl-(aq)

This balanced equation shows us that there is a 1:1 correlation between the starting acidconcentration and the equilibrium hydronium ion concentration. So if we begin with 0.25 M HCl,the concentration of hydronium ion at equilibrium is also 0.25 M. If we are given 1.50 M HCl, theconcentration of hydronium ion at equilibrium is also 1.50 M. And so on.

More generally speaking, what we have just seen for hydrochloric acid is also true for anystrong monoprotic acid. There is always a 1:1 relationship between the starting acid concentrationand the equilibrium hydronium ion concentration.

Diprotic acids are a bit different in that every mole of strong diprotic acid will produce twomoles of hydronium ion at equilibrium. Using sulfuric acid as an example the balanced equationwhich shows its overall behavior in water is

H2SO4 (aq) + 2 H2O(l) => 2 H3O+

(aq) + SO42-

(aq)

We can see that there is a 1:2 correlation between the starting acid concentration and the equilibriumhydronium ion concentration. So if we begin with 0.25 M H2SO4 , the concentration of hydroniumion at equilibrium will be 0.50 M (2 x 0.25 M). If we are given 1.50 M H2SO4 , the concentration ofhydronium ion at equilibrium will be 3.00 M. And so on.

There are no strong triprotic or tetraprotic acids. But if there were, the relationship betweenstarting acid concentration and equilibrium hydronium ion concentration would be 1:3 and 1:4respectively. In other words, if we had a 0.25 M concentration of a strong triprotic acid, theequilibrium hydronium ion concentration would be 0.75 M (3 x 0.25). And if we had a 0.25 Msolution of a strong tetraprotic acid solution, the equilibrium hydronium ion concentration wouldbe 1.00 M (4 x 0.25).

Remember, this is only true for strong acids. As weak acids do not dissociate completely wenever see the direct relationships between starting weak acid concentration and equilibrium

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hydronium ion concentration that we see with strong acids. If we write the equation that describesthe behavior of acetic acid in water

HC2H3O2 (aq) + H2O(l) => H3O+

(aq) + C2H3O2-(aq)

we see the 1:1 relationship between the starting acid concentration and the equilibrium hydroniumion concentration. But as a weak acid, acetic acid does not dissociate completely. Typically onlyabout 1% of the acetic acid molecules ionize. This means that if we start with 1.00 M of acetic acid,our equilibrium hydronium ion concentration will only be around 0.01 mol. And we see similarbehavior with all of the other weak acids as well.

We take a similar approach to that we used with strong acids when calculating the hydroxideion concentration of solutions of strong bases. We can establish the relationship between the startingbase concentration and the equilibrium hydroxide ion concentration by examining the equation thatdescribes the base’s behavior in water. If we consider sodium hydroxide as an example we see that

NaOH(aq) => Na+(aq) + OH-

(aq)

This balanced equation shows us that there is a 1:1 correlation between the starting baseconcentration and the equilibrium hydroxide ion concentration. So if we begin with 0.25 M NaOH,the equilibrium hydroxide ion concentration is also 0.25 M. If we are given 1.50 M NaOH, theequilibrium hydroxide ion concentration is also 1.50 M. And so on.

The behavior shown by sodium hydroxide is the same for any other Group I hydroxide.There is a 1:1 relationship between the starting base concentration and the equilibrium hydroxideion concentration. This is correct because these compounds behave as strong bases.

The Group II hydroxides differ from those in Group I in that one mole of base will producetwo moles of hydroxide ion at equilibrium. Using calcium hydroxide as an example, the balancedequation which shows its behavior in water is

Ca(OH)2 (aq)=> Ca2+(aq) + 2 OH-

(aq)

We can see that there is a 1:2 correlation between the starting base concentration and the equilibriumhydroxide ion concentration. So if we begin with 0.25 M Ca(OH)2 , the concentration of hydroxideion at equilibrium will be 0.50 M (2 x 0.25 M). If we are given 1.50 M Ca(OH)2 , the equilibriumconcentration of hydroxide ion will be 3.00 M. And so on.

As was the case for acids, what you have just read is only true for strong bases. Weak basesdo not dissociate completely. We therefore never see the direct relationships between starting weakbase concentration and equilibrium hydroxide ion concentration that we see with strong bases. If weare given a 1.00 M solution of ammonium hydroxide, we cannot say that the equilibrium hydroxideion concentration will also be 1.00 M. We only know that, as a weak base, the equilibrium hydroxideion concentration will be considerably less than 1.00 M.

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When we calculate pH we need a hydronium ion concentration. Is it possible to determinethe hydronium ion concentration for a strong base solution? The answer is yes. We have twodifferent techniques, both of which will produce exactly the same results when used correctly.

We saw above that

As Kw is a constant and always equal to 1 x 10-14 (at least, this is always the value we will assumefor our calculations in this class), this means that any time we know the hydroxide ion concentrationof a solution we can rearrange this equation to find the hydronium ion concentration

Let’s work a few examples. We mentioned above that the equilibrium hydroxide ion concentrationof a 0.25 M solution of sodium hydroxide will also be 0.25 M. What is the hydronium ionconcentration of this solution? We can simply plug numbers into this equation to find the answer.

If we begin with 0.25 M Ca(OH)2 , the concentration of hydroxide ion at equilibrium will be 0.50 M,as we said above. What is the hydronium ion concentration of this solution? Again, we can simplyplug numbers into the equation used in the previous example to calculate our answer.

Once we know the hydronium ion concentration of a solution we can calculate the pH of thatsolution. Again, let’s work through a few examples. At this point you need to get out your calculatorand actually try to get the same answers as I do in the following table.

solution [H3O+] pH = -log [H3O

+]

0.25 M HCl 0.25 M 0.60

1.50 M HCl 1.50 M -0.18

0.25 M H2SO4 0.50 M 0.30

1.50 M H2SO4 3.00 M -0.48

0.25 M NaOH 4.00 x 10-14 M 13.4

0.25 M Ca(OH)2 2.00 x 10-14 M 13.7

K H O OHw [ ][ ]3

K

OHH Ow

[ ][ ]

3

K

OHH O

xxw

[ ][ ]

..

3

14141 10

0 254 00 10

K

OHH O

xxw

[ ][ ]

..

3

14141 10

0502 00 10

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How did you do with your calculations? A point of interest may be seen in this table. You observe thatthe 1.50 M solutions of hydrochloric acid and of sulfuric acid both have negative pH values. Is thatpossible? Yes, it is. The pH scale is often depicted as having endpoints at pH values of 0 and 14 butthis is not correct. Any acid solution with a hydronium ion concentration of 1.0 M or greater willhave a pH less than zero, i.e. a negative pH. Any base solution with a hydroxide ion concentrationof 1.0 M or greater will have a ph above 14. Concentrated strong acid solutions may have pH valuesranging as low as -2 to -3. Concentrated strong base solutions may have pH values as high as 15 to16.

Buffer solutions

If you’ve studied human physiology you know that our bodies operate at a very slightly basicpH of around 7.4. If our physiological pH climbs above 7.45 we enter respiratory alkalosis. If ourphysiological pH drops below 7.35 it causes respiratory acidosis. And if our pH changes muchbeyond either of these values our bodies stop working properly and we die. The difference inhydronium ion content between these values is very small, so in theory it should not take too muchacid or base to cause a dramatic and very unhealthy change in pH for the average person.

pH [H3O+]

7.35 4.46 x 10-8 M

7.40 3.98 x 10-8 M

7.45 3.54 x 10-8 M

The pH of pure water is exactly 7.0. If we take 1.00 L of pure water and add just 1.00 mLof 1.0 M hydrochloric acid to it, the pH of the resulting solution will drop from 7.0 to 3.0. As thepH scale is a logarithmic scale this means that adding this small amount of a rather dilute acidsolution makes the resulting solution 10,000 times more acidic than pure water.

We see something similar occur if we add just 1.00 mL of 1.0 M sodium hydroxide to 1.00 Lof pure water. The solution pH will jump from 7.0 to 11.0, 10,000 times more basic than pure water.

Think about the implications of this. Every day most of what we eat and drink is either acidicor basic. Take a look at the following web page kept by the U.S. Food and Drug Administrationwhich lists the pH values of some of the foods we commonly eat (it can be found athttp://www.webpal.org/SAFE/aaarecovery/2_food_storage/Processing/lacf-phs.htm). We need tomaintain a pH of about 7.4 but most of what we eat has a pH far different from this. If we could onlyeat foods with a pH of about 7.4, according to this web page our diets would consist mostly ofCamembert cheese, cooked frozen corn, graham crackers, chocolate cake, cooked lobster, and tofu.We would risk death were we to eat those things that most of us prefer in our diets. We couldn’teven drink pure water, let alone tap water or bottled water which are usually much more acidic than

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pure water. And no soft drinks, milk, coffee, beer, wine, or hard liquors either. Life would get ratherthirsty for most of us very quickly.

So how is it that we can eat and drink these things we love without running the risk ofimminent death? Buffers (or, buffer solutions) are solutions with the ability to resist changes in pHwhen acids or bases are added to them. And our body contains a buffer that makes it possible to livein a world where a neutral pH is seldom ever found.

A buffer solution usually consists of a weak acid and its conjugate base. Although it is alsopossible to have a buffer solution made with a weak base and its conjugate acid. The buffer solution wefind in our bodies consists of carbonic acid, H2CO3, and its conjugate base bicarbonate ion, HCO3

-.Remember that bicarbonate ion is also known as hydrogen carbonate ion which is one of the polyatomicions you hopefully learned and remember from Chapter 3. The carbonic acid in our blood is formedfrom the combination reaction of the carbon dioxide produced as a waste product in cellularmetabolism and the water in our plasma. Bicarbonate ion is formed by the ionization of carbonicacid. While this is certainly the most interesting and important buffer system to us, there are manyother buffer systems that play a role in other living systems, biochemistry, and other studies inchemistry.

How does a buffer resist a change in pH when an acid or base is added? When an acid isadded to a buffer solution it reacts with the conjugate base of the buffer. And when a base is addedto a buffer solution it reacts with the acid of the buffer.

Let’s take a closer look at this. We can imagine the way in which an acid is neutralized bya buffer solution as occurring in three steps. Assume we add some generic acid HX to a carbonicacid/bicarbonate ion buffer solution. In water the acid will dissociate to some extent, depending onwhether it is a strong or a weak acid

HX(aq) + H2O(l) W H3O+

(aq) + X-(aq)

The hydronium ion formed in this step will then react with bicarbonate ion found in the buffersolution and will be completely consumed, regardless of whether the added acid is a strong acid ora weak acid

H3O+

(aq) + HCO3-(aq) W H2CO3(aq) + H2O(l)

The carbonic acid formed in this reaction will then behave as all acids behave in water

H2CO3(aq) + H2O(l) W H3O+

(aq) + HCO3-(aq)

but as carbonic acid is a weak acid it will only ionize slightly, which means the solution hydroniumion concentration will increase only slightly, and the solution pH will also only change to a slightextent.

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We can also imagine the way in which a carbonic acid-bicarbonate ion buffer solution actson a base as taking place in a series of three steps. In the first step the base “B” either formshydroxide ion directly or forms hydroxide ion through the hydrolysis of water

B(aq) + H2O(l) W HB+(aq) + OH-(aq)

In the second step carbonic acid reacts with the hydroxide ion formed in the first step and consumesit completely

H2CO3(aq) + OH-(aq) W H2O(l) + HCO3

-(aq)

In the third step the bicarbonate ion hydrolyzes water and forms carbonic acid and hydroxide ionas products

HCO3-(aq) + H2O(l) W H2CO3(aq) + OH-

(aq)

but as bicarbonate ion is a weak base, the increase in the hydroxide ion concentration of the solutionwill be small, which means the resulting change in pH will also be small.

A common buffer solution in chemistry is made with acetic acid and acetate ion. If we have 1.00 L of a buffer solution that is 0.70 M in acetic acid and 0.60 M in acetate ion, the pH of thisbuffer will be 4.68. We saw above that if we added 1.00 mL of 1.0 M HCl to a liter of water that thepH changed from 7.0 to 3.0. But if we add the same volume of this same hydrochloric acid to ourliter of acetic acid-acetate buffer solution, there is no measurable change in the buffer solution pH. I’m not going to show you how we calculate this as it is beyond the scope of this course. But if you’reinterested you can see how we do this at my Chem 1220 web site. Ask me and I’ll tell you where to findit.

Every buffer solution has its limits. This limit is called the buffer capacity and it depends onthe amounts of weak acid and conjugate base found in the buffer solution. In the acetic acid-acetatebuffer solution we just discussed, if we have exactly 1.00 liter of this solution then we have 0.70moles of acetic acid and 0.60 moles of acetate ion. When we add an acid to this solution it reactswith the acetate ion on a 1:1 basis. If, then, we add any amount of acid beyond 0.600 moles, say forexample, 0.601 moles of acid to a liter of this solution, we will consume all of the acetate ion in thesolution and we now will see a large change in pH as a consequence. The same is true for the aceticacid. If we add any amount of base beyond 0.70 moles, say, for example, 0.701 moles of a base toa liter of this solution, the acetic acid - which reacts with the base on a 1:1 ratio, will all beconsumed and a large change in pH will result.

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