# electronic structures of atoms / periodic trends / ionic bonding / solids / phase changes

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Electronic Structures of Atoms / Periodic Trends / Ionic Bonding / Solids / Phase Changes. H Advanced Chemistry Unit 3. Objectives #1-3 Atomic Theory. *review of electromagnetic radiation characteristics: (diagrams). Examples of Electromagnetic Radiation. Objectives #1-3 Atomic Theory. - PowerPoint PPT Presentation

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

Electronic Structures of Atoms / Periodic Trends / Ionic Bonding / Solids / Phase ChangesH Advanced ChemistryUnit 3

3Objectives #1-3 Atomic Theoryfrequency, 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)

4Max Planck (1858-1947)

5Objectives #1-3 Atomic TheoryE = hf or hc/l h = Plancks Constant (energy for waves)6.626 X 10-34 Js

6Albert Einstein (1879-1955)

7Objectives #1-3 Atomic TheoryE = 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)

8Louis de Broglie (1892-1987)

9Derivation of de Broglies Equation:Ewaves = Eparticles hc / l = mc2l (hc / l) = (mc2)lh = mcll = h / mv(examples)

10#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.= 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

Objectives #1-3 Atomic Theory*Work Function (Photoelectric Effect) = hfo = 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)

15Photoelectric Effect (Albert Einstein Nobel Prize 1921)

16#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. = 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)

19Objectives #1-3 Atomic Theory*Bohrs 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 levelsfor hydrogen; z = 1(clip and examples)

20#1 Calculate the energy required to excite an electron from energy level 1 to energy level 3.E = -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 hydrogenE = -2.178 X 10-18 J (-1) =2.178 X 10-18 J

Johannes Rydberg (1854-1919)

23Objectives #1-3 Atomic Theory*Rydberg Equation:1/ = 1/91 nm (1/nL2 1/nH2)*used for: determining wavelength of photons released when electrons change energy levels(Examples)

24#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 nm4 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)

29Objectives #4-5 The Quantum Numbers and Quantum States*Review of Quantum Theory:Quantum NumbersPrinciple (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

30Illustration of Principle Quantum Number

31Objectives #4-5 The Quantum Numbers and Quantum StatesOrbital (l)*indicates orbital shape*l = 0, n-1*s, p, d, f32Illustration of Orbital Quantum Number / Orbital Shapes

33Objectives #4-5 The Quantum Numbers and Quantum StatesMagnetic (ml)*indicates orientation of orbital in space*ml = 0, +/-l *the number of ml values indicate the number of orbitals within sublevel

34Illustration of Magnetic Quantum Number

35Objectives #4-5 The Quantum Numbers and Quantum StatesSpin (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

36Illustration of Spin Quantum Number

37Objectives #4-5 The Quantum Numbers and Quantum States*Quantum Number Sets for Electrons in Atoms:38Illustration of Quantum States

39Objectives #4-5 The Quantum Numbers and 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)

41Objectives #7-9 Electron Configurations of Ions / Orbital Filling and Periodic Trends*Orbital Filling and Periodic TrendsIonization Energy (from period 2)Group 1Li5.4 evGroup 2Be9.3 ev (spike)Group 15N14.5 ev (spike)Group 17F17.4 ev (spike)Group 18Ne21.6 ev (spike)42Trends in Ionization Energy

43The 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 processexample: Na + energy Na+1 + e-*the gain of an electron releases energy and is therefore an exothermic processexample: Cl + e- Cl-1 + energy 45Formation of Sodium Chloride

46Formation of Crystal Lattice

47Objective #10-12 Formation of the Ionic Bond / Born-Haber Cycle and Lattice Energy*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)

48Illustration of Born-Haber Cycle

49Objective #10-12 Formation of the Ionic Bond / Born-Haber Cycle and Lattice Energy*the formation of an ionic compound; such as the following reaction:Na(s) + 1/2Cl(2)(g) NaCl(s) + Hof = -410.9 kJwhere Hof 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: 50Objective #10-12 Formation of the Ionic Bond / Born-Haber Cycle and Lattice EnergySublimation or Vaporization of nongaseous reaction components:Here: Na(s) Na(g) 108 kJ which represents the energy of sublimation or vaporization (an endothermic process)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)51Objective #10-12 Formation of the Ionic Bond / Born-Haber Cycle and Lattice EnergyFormation of the positive ion:Here: Na(g) Na+1(g) + e- 496 kJ which represents the ionization energy (an endothermic process)Formation of the negative ion:Here: Cl(g) + e- Cl(g)-1 -349 kJ which represents the electron affinity energy (an exothermic process)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, Hof, is equal to the sum of all these changes:Hof = HofNa + HofCl + IE Na - EA Cl - Hlattice = -411 kJ52Example 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 k

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