Download - Electronic Structures of Atoms / Periodic Trends / Ionic Bonding / Solids / Phase Changes

Electronic Structures of Atoms / Periodic Trends / Ionic Bonding / Solids / Phase Changes

H Advanced ChemistryUnit 3

Objectives #1-3 Atomic Theory

*review of electromagnetic radiation characteristics:

(diagrams)

Examples of Electromagnetic Radiation


frequency, wavelength, energyfrequency vs. wavelength (inverse

relationship)frequency vs. energy (direct relationship)wavelength vs. energy (inverse

relationship)c=fl (c = speed of light in m/s, f =

frequency in Hz (1/s), l = wavelength in m)

Max Planck (1858-1947)


E = hf or hc/l h = Planck’s Constant (energy for waves)

6.626 X 10-34 Js

Albert Einstein (1879-1955)


E = mc2 (energy for particles)*Wave particle-dualityMatter has wave and particle

characteristics; acts as particle when interacting with matter; acts as wave when traveling through space (clip)

Louis de Broglie (1892-1987)

Derivation of de Broglie’s Equation:Ewaves = Eparticles hc / l = mc2

l (hc / l) = (mc2)lh = mcll = h / mv

(examples)

#1 Calculate the frequency of light having a wavelength of 6.50 X 102 nm.

c = fl (convert nm to meters)6.50 X 102 nm X 1m / 1 X 109 nm = 6.50 X 10-7 mf = (3.00 X 108 m/s / 6.5 X 10-7 m) = 4.62 X 1014 Hz

#2 Calculate the energy of the blue color of wavelength 4.50 X 102 nm emitted by an atom of copper.

E = hc / l = (6.626 X 10-34 Js) (3.00 X 108 m/s) /

4.5 X 10-7 m = 4.42 X 10-19 J

#3 Calculate the wavelength for an electron having a mass of 9.11 X 10-31 kg and traveling at a speed of 1.0 X 107 m/s. Calculate the wavelength for a ball having a mass of .10 kg and traveling at a speed of 35 m/s.

l = h/mv = (6.626 X 10-34 Js) / (9.11 X 10-31 kg) (1.0 X 107 m/s) 7.27 X 10-11 m

l = h / mv = (6.626 X 10-34 Js) / (.10 kg) (35 m/s) = 1.9 X 10-34 m


*Work Function (Photoelectric Effect)Φ = hfo

Φ = work functionminimum energy required to remove

electron from surface of metalfo = threshold frequencyminimum frequency required to remove

electrons from surface of metal(clip, diagram and examples to follow)

Photoelectric Effect (Albert Einstein Nobel Prize 1921)

#1 A gold strip is irradiated with radiation of a frequency of

6.9 X 1012 Hz. Calculate the energy of this radiation and determine if it is sufficient to cause electrons to be released from the metal. The work function of gold is 7.7 X 10-19 J.

E = hf = 6.626 X 10-34 Js X 6.9 X 1012 Hz = 4.57 X 10-21 J; no electron ejection

#2 The ionization energy of gold is 890.1 kJ/mole of electrons. Calculate the threshold

frequency required to cause the photoelectric effect and eject an electron.

Φ = hfo

(convert I.E. into energy per electron)890.1 kJ/mole X 1000 J / 1 kJ X 1 mole electrons / 6.02 X 1023 electrons= 1.479 X 10-18 Jfo = 1.47 X 10-18 J / 6.626 X 10 -34 Js = 2.232 X 1015 Hz

Niels Bohr (1892-1962)


*Bohr’s Equation:E = -2.178 X 10-18 J (z2/n2) OR ∆E = -

2.178 X 10-18 J (z2) X (1/n2final –

1/n2initial)

used for: determining energy changes when electrons change energy levels

for hydrogen; z = 1(clip and examples)

#1 Calculate the energy required to excite an electron from energy level 1 to energy level 3.

ΔE = -2.178 X 10-18 J (1/9 – 1/1) = -2.178 X 10-18 J (-.8889) = 1.936 X 10-18 J

#2 Calculate the energy required to remove an electron from a hydrogen atom.

*this energy represents the ionization energy for hydrogen

ΔE = -2.178 X 10-18 J (-1) =2.178 X 10-18 J

Johannes Rydberg (1854-1919)


*Rydberg Equation:1/λ = 1/91 nm (1/nL

2 – 1/nH2)

*used for: determining wavelength of photons released when electrons change energy levels

(Examples)

#1 Calculate the wavelength of light released when an electron drops from the following states:

2 to 1:1 / l = 1 /91 nm (1/1 – 1/4) = .75/91 nm l = 91nm/.75 = 121.33 nm

4 to 1:1 / l = 1 /91 nm (1/1 – 1/16) = .9375/91 nm l = 91nm/.9375 = 97.07 nm

6 to 1:1 / l = 1 /91 nm (1/1 – 1/36) = .9722/91 nm l = 91nm/.9722 = 93.60 nm

*relationships of answers: the greater the energy difference, the smaller the wavelength

Erwin Schrodinger (1887-1961)(clip)

Objectives #4-5 The Quantum Numbers and Quantum States

*Review of Quantum Theory:1. Quantum NumbersA. Principle (n)*energy level of shell of electron*n = 1,2,3…..*(old system) n = K, L, M, ….*indicates the number of sublevels in

energy level

Illustration of Principle Quantum Number


B. Orbital (l)*indicates orbital shape*l = 0, n-1*s, p, d, f

Illustration of Orbital Quantum Number / Orbital Shapes


C. Magnetic (ml)*indicates orientation of orbital in

space*ml = 0, +/-l *the number of ml values indicate the

number of orbitals within sublevel

Illustration of Magnetic Quantum Number


D. Spin (ms)*indicates spin of electron*+1/2 or -1/2*allows for up to 2 electrons per orbital*s 2 electrons p 6 electrons d 10 electrons f 14 electrons

Illustration of Spin Quantum Number


*Quantum Number Sets for Electrons in Atoms:

Illustration of Quantum States


(examples of quantum number states problems)

Objectives #7-9 Electron Configurations of Ions / Orbital Filling and Periodic Trends

*valence electrons and occasionally the electrons contained within the d sublevel are involved in chemical bonding

*atoms tend to lose or gain electrons in such a way to complete octets (s2p6) or to form similarly stable arrangements called pseudo noble-gas configurations

(clip and examples)

Objectives #7-9 Electron Configurations of Ions / Orbital Filling and Periodic Trends

*Orbital Filling and Periodic Trends1. Ionization Energy (from period 2)Group 1 Li 5.4 evGroup 2 Be 9.3 ev (spike)Group 15 N 14.5 ev (spike)Group 17 F 17.4 ev (spike)Group 18 Ne 21.6 ev (spike)

Trends in Ionization Energy

The effective nuclear charge on a particular electron in an atom is less than the actual nuclear charge of the atom due to the screening or shielding effect of the inner core electrons. For example, the effective nuclear charge on the 3s electrons in magnesium is less than the 2p electrons because there is less nuclear charge acting on the 3s electron. Since the 2p electrons are attracted more strongly, they are more stable, and have less energy than the 3s electron

Objective #10-12 Formation of the Ionic Bond / Born-Haber Cycle and Lattice Energy

*ionic bonds involve the transfer of valence electrons from a metal to a nonmetal

*the tendency for a metal to lose electrons depends on its ionization energy and the tendency of a nonmetal to gain electrons depends on its electron affinity

*the loss of an electron requires a gain of energy and is therefore an endothermic process

example: Na + energy › Na+1 + e-

*the gain of an electron releases energy and is therefore an exothermic process

example: Cl + e- › Cl-1 + energy

Formation of Sodium Chloride

Formation of Crystal Lattice


*combinations of elements with low ionization energies and high electron affinities will cause an extremely exothermic reaction and generally be the most stable*example: Na(s) + Cl2(g) › NaCl(s) + energy

*the energy produced when the ionic bond forms is referred to as the lattice energy; this energy is also equal to the energy required to break apart the ionic bond

*chemical bonding not only involves a rearrangement of electrons but it also involves changes in energy (clip)

Illustration of Born-Haber Cycle


*the formation of an ionic compound; such as the following reaction:

Na(s) + 1/2Cl(2)(g) › NaCl(s) + ∆Hof =

-410.9 kJwhere ∆Ho

f refers to the standard heat of formation which is the energy change involved when a compound is formed from its elements, involves a series of energy changing steps known as the Born-Haber cycle

*these steps are as follows:


1. Sublimation or Vaporization of nongaseous reaction components:

Here: Na(s) › Na(g) 108 kJ which represents the energy of sublimation or vaporization (an endothermic process)

2. Breaking the bonds of any gaseous components:Here 1/2Cl2(g) › Cl(g) 122 kJ which represents the

dissociation energy (an endothermic process)(now that all reactants are gaseous, ions must be

formed)


3. Formation of the positive ion:Here: Na(g) › Na+1

(g) + e- 496 kJ which represents the ionization energy (an endothermic process)

4. Formation of the negative ion:Here: Cl(g) + e- › Cl(g)

-1 -349 kJ which represents the electron affinity energy (an exothermic process)

5. Formation of the ionic compound by combining the two ions formed together:

Here: Na+1(g) + Cl(g)

-1 › NaCl(s) -788 kJwhich represents the lattice energy (an exothermic process)*the overall energy change, ∆Ho

f, is equal to the sum of all these changes:

∆Hof = ∆Ho

fNa + ∆HofCl + IE Na - EA Cl - ∆Hlattice

= -411 kJ

Example II Show all of the energy changing steps and determine the heat of formation for LiF given the following:

Sublimation energy for Li = 161 kJIonization energy for Li = 520 kJBond dissociation energy for F2 = 77 kJ

per ½ moleElectron affinity of fluorine = -328 kJLattice energy for LiF = -1047 kJ

Li(s) > Li(g) +161 kJLi(g) > Li+1 +520 kJ1/2F2(g) > F(g) +77 kJF(g) + e-1 > F-1

(g) -328 kJLi+1

(g) + F-1 (g)

> LiF(s) -1047 kJ--------------------------------------------

-617 kJ

Example III

Mg(s) > Mg(g) +147 kJMg(g) > Mg+1 (g) +738 kJMg(g)

+1 > Mg+2 (g) +1451 kJ

Cl2(g) > 2Cl(g) +242 kJ2Cl(g) + 2e-1 > 2Cl-1(g) -698 kJMg+2

(g) + 2Cl-1 > MgCl2(s) -2326 kJ--------------------------------------------

-446 kJ


*Relationship of lattice energy and ionic chargeConsider the following lattice energy data from the above

example problems:NaCl 788 kJLiF 1030 kJMgCl2 2326 kJKCl 701 kJ**strength of ionic bonds:KCl ‹ NaCl ‹ LiF ‹ MgCl2**size of ions:LiF ‹ MgCl2 ‹ NaCl ‹ KCl**charge of ions:Na +1, Cl -1 Li +1, F -1 K +1, Cl -1 Mg +2, Cl-1


**formula: Eel. = KQ1Q2/d where “K” is a constant of electrical charge, where “Q1” and “Q2” are the charges of the ions involved, where “d” distance separating the ions

As the magnitude of the charges in an ionic compound increases, the lattice energy increases (affects lattice energy the most)

As the size of the ions involved decrease, the lattice energy increases

(examples)

Examples:

Arrange the following ionic compounds in order of increasing lattice energy:

NaF CsI CaOCsI, NaF, CaO

Arrange the following ionic compounds in order of increasing lattice energy:

AgCl, CuO, CrN

Objectives #13-14 Phase Changes and Phase Diagrams


*Important Parts of a Heating/Cooling Curve(see curve in lecture guide)A. Specific heat of solid added (endothermic)B. Specific heat of liquid added (endothermic)C. Specific heat of gas added (endothermic)D. Melting (heat of fusion added) (endothermic)E. Freezing (heat of solidification released) (exothermic)F. Boiling (heat of vaporization added) (endothermic)G. Condensing (heat of condensation released)

(exothermic)(graph interpretation practice)


*where the graph is increasing or decreasing, specific heat is being added or subtracted (which results in the temperature changing)

*where the graph is not changing, a phase change is occurring and these is no change in the temperature of the substance

*key equations:to change temperature: Q =mc∆tto change phase:∆H = moles of material X molar heat of phase

change(examples)

Example I:Calculate the energy change involved

to convert 1 mole of ice at -25oC to steam at 125oC.

(analysis of steps involved)

Example I

Step I: -25C > 0CQ = (18.0 g) (2.09J/gC) (25C) = 941 J = .941 kJStep II: meltΔH = (1 mole) (6.01 kJ/mol) = 6.01 kJStep III: 0C > 100C Q = (18.0 g) (4.18 J/gC) (100C) = 7524 J = 7.52 kJ

Example I

Step IV: boil ΔH = (1 mole) (40.67 kJ/mole) = 40.67 kJStep V: 100C > 125C Q = (18.0 g) (1.84 J/gC) (25C) = 828 J = .828 kJ Total = 56.0 kJ

Example II

Step I: 140C > 100CQ = (100.0 g) (1.84J/gC) (40C) = -7360 J = -7.360 kJStep II: condensationΔH = (100.0 g/18.0 g) (40.67 kJ/mol) = -225.9 kJStep III: 100C > 0C Q = (100.0 g) (4.18 J/gC) (100C) = -41800 J = -41.8 kJ

Example II

Step IV: freeze ΔH = (100.0 g/18.0 g) (6.01 kJ/mole) = -33.4 kJStep V: 0C > -30C Q = (100.0 g) (2.09 J/gC) (30C) = -6270 J = -6.27 kJ Total = -315 kJ


*Interpreting Phase Diagrams*a phase diagram allows one to

determine the phase that a substance is in at a given temperature and pressure

*the phase diagram only shows one substance in its various phases

*a typical phase diagram:(see diagram in lecture guide and next

slide)


*Interpreting Phase Diagrams*a phase diagram allows one to

determine the phase that a substance is in at a given temperature and pressure

*the phase diagram only shows one substance in its various phases

*a typical phase diagram:(see diagram in lecture guide)


*the boundaries between different phase regions represent areas of equilibrium in which the two phase changes are occurring at the same rate; for example at the liquid – gas boundary, molecules of gaseous vapor are moving into the liquid phase while molecules of liquid are moving into the gaseous phase

*if a point on the diagram does not fall on any line, only one phase is present


*the following lines on the graph represent phase change boundaries:Line Segment Phase Change

BoundaryA-B Liquid – gas

(vaporization ↔ condensation

A-C Solid – gas (sublimation ↔ deposition)

A-D Solid – liquid (melting ↔ freezing)


Critical Point the endpoint of the vapor-pressure curve; beyond this point of critical temperature and critical pressure, the liquid and gas phases can not be distinguished from each other

Normal Boiling Point the location on the vapor-pressure curve where the vapor pressure is 1 atm

Normal Melting Point the location on the solid-liquid curve where the melting (freezing) point is at 1 atm

Triple Point the point where all 3 curves intersect and all three phases are in equilibrium

Objectives #13-14 Phase Changes and Phase Diagram

(practice worksheets)

*some general relationships and observations to note:

*if the solid-liquid line curves to the right with increasing pressure, then the melting point is also increasing (this is the norm)

*if the solid-liquid line curves to the left with increasing pressure, then the melting point is decreasing (this is not the norm; water follows this pattern)

Illustration of Phase Diagram for Water

Objectives #15-17 Structure of Solids, Properties and Applications

*solids come in two general types: crystalline or amorphous*crystalline solids contain particles arranged in a well-defined

pattern called a crystal lattice with flat faces and definite angles; examples include NaCl or diamonds

*amorphous solids lack any well defined structure; examples include wax, rubber, or glass

*the crystal lattice of a crystalline solid, which is a three dimensional array showing the location of individual particles, is actually made up of many repeating individual parts called the unit cell; for example the repeating pattern on wall paper

*the simplest common type of unit cell is the cubic unit cell where all sides are equal in length and consist of all 90o angles


*the three types of cubic unit cells are:primitive or simple cubic – lattice points

only occur at the corners of unit cellbody-centered cubic – lattice points

occur at the corners and in the centerface-centered cubic – lattice points

occur at the corners and faces

Types of Cubic Unit Cells


*except for the atom in the center of the body-centered cubic unit cell, all of the atoms located at the lattice points are actually shared to various degrees by other unit cells; in order to determine the net number of atoms in a unit cell and thus its chemical formula, one must know the fraction of an atom that occurs in each position of the unit cell as follows:


Position in Unit Cell Fraction in Unit CellCenter 1Face ½Edge ¼Corner 1/8

Example of Face Centered Unit Cell in Sodium Chloride


(examples)Example #1: Determine the number of sodium

and chloride ions in the NaCl unit cell using the diagram on the screen as a guide.

For Na: 12 edges X 1/4 each = 3 Na ions 1 ion NaFor Cl: 6 faces X ½ each = 3 Cl ions 8 corners X 1/8 each = 1 Cl ion4 Na ions to 4 Cl ions; NaCl

Example of Body Centered Unit Cell in Cesium Chloride

Example #2 The element iron crystallizes in a form called alpha iron, which has a body centered unit cell. How many iron atoms are in the unit cell?

1 Fe atom at center8 corners X 1/8 each = 1 Fe atomresults in 2 iron atoms

X-Ray Crystallography

*the layers of atoms in the crystal lattice acts as an effective diffraction grating that can be used to scatter a beam of x-rays

*the diffraction, or scattering, of the x-rays produces a characteristic pattern of light and dark areas on a x-ray detector

*by examining the areas of light and dark and measuring the angles of deflection, the original crystal structure of the material can be deduced

*this analytical technique has been used to determine the structure of DNA and other molecular crystals

Instrumentation(clip)


*Types of Bonding in Solids and their Influence on Properties

Type of Solid

Form of Unit Particle

Forces Between Particles

Properties Examples

Molecular Atoms or molecules

London, DD, HB

Fairly soft, low melting points, poor conductors

Gases, sugars, dry ice

Covalent-Network

Atoms connected in a network of covalent bonds

Covalent bonds

Very hard, high melting points, poor conduction

Diamond, quartz, SiO2

Ionic Positive and negative ions

Electrostatic attractions

Hard and brittle, high melting point, poor conductor

Salts such as NaCl


*Types of Bonding in Solids and their Influence on Properties

Metallic Atoms Metallic bonds

Soft to very hard, low to high melting point, malleable and ductile, excellent conductor

Metallic elements such as copper

Dry Ice – Example of Molecular Solid

Diamond – Example of Covalent Network Solid

Sodium Chloride – Example of Ionic Solid

Copper – Example of Metallic Solid

Download - Electronic Structures of Atoms / Periodic Trends / Ionic Bonding / Solids / Phase Changes

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