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Final Exam Spring 2017 Descriptive Inorganic Chemistry CHE 3464 Solution Key There are 10 questions on this examination covering 10 pages for 100 total points. Read each question carefully and answer only what is asked. Use the space provided to record the final answer for each question. Notes, hand...

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Descriptive Inorganic Chemistry CHE 3464

Final Exam Spring 2017

Solution Key There are 10 questions on this examination covering 10 pages for 100 total points. Read each question carefully and answer only what is asked. Use the space provided to record the final answer for each question. Notes, handouts, or other materials are not permitted. An appendix of useful information and a periodic table are attached. You have 150 minutes. Good luck!

1. This question concerns aspects of atomic structure. (a) How many radial nodes are present in a 5p orbital? (2 points) 3 radial nodes

(b) Assuming there are no radial nodes, what is the identity of the atomic orbital shown below? Your answer should be in the form n!m! (e.g. 2pz). (2 points)

z x

3dxz

(c) Order the following atoms in terms of increasing ionization energy. (2 points)

Rb, Cl, Cs, Ca, K Cs < Rb < K < Ca < Cl (d) What is the ground-state electron configuration for a free Cu+ ion? (2 points)

[Ar]3d10 or 1s22s22p63s23p63d10 (e) What atomic property is most responsible for the difference in atomic radius between magnesium (Mg) and barium (Ba)? (2 points)

The principal quantum number, n, is primarily responsible for the difference

in atomic radius.

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2. Draw Lewis structures for each of the following molecules or molecular ions showing the correct VSEPR geometry (using wedges and dashes or by writing the name of the geometry in words). Show all stereochemically-active lone pairs and all major resonance contributors. (10 points)

(a) CO32−

O

O

C O

O

C O

C

O

O

O

O

trigonal planar

(b) TeF4

Te

Cl Cl

Cl

Cl

disphenoidal (seesaw)

(c) H3O+

O

H H trigonal pyramidal H

(d) N3− 2-

N

N

N

2-

N

N

linear

(e) XeF2

F Xe F linear

N

N

N

N

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3. The cyanide ion (CN−) is an aggressive ligand that binds irreversibly to the Fe centers of heme proteins including those responsible for mitochondrial respiration. As a result, cells are unable to utilize oxygen from the bloodstream accounting for the high toxicity of cyanide. (a) Draw the molecular orbital diagram for CN−. As a basis set, use only the valence orbitals on carbon and nitrogen. A table of valence orbital ionization energies (VOIEs) can be found in the appendix. (8 points)

CN− +4

σ*

C

N

0 π* −4 −8

~E (eV)

−12

2p

σnb 2p π

−16 −20

2s σnb

−24 2s −28

σ

−32

(b) Based on your diagram and recognizing that the site of Lewis basicity for cyanide is localized on carbon, draw a picture of the HOMO. (2 points)

C

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4. The unit cell of a perovskite material composed of Na, W, and O is displayed below along with its density of states diagram. Three additional oxygen atoms from adjacent unit cells are displayed to show the coordination geometry about tungsten. The unit cell is cubic with an edge length of 385 pm. O O

W O

W

W

O O

O Na

O W

O O

O W

Fermi Level

Energy

O

W O

O

W

total DOS

O O

W

Density of States

(a) What is the density of this material in units of g/cm3? (6 points) Density = mass of atoms in unit cell / volume of unit cell Mass of unit cell contents: NaWO3 = 4.23 × 10-22 g Volume of unit cell = (385 pm)3 = 5.71 × 10-23 cm3 Density = (4.23 × 10-22 g)/(5.71 × 10-23 cm3) = 7.41 g/cm3

(b) Based on the DOS diagram, is this perovskite material best regarded as an insulator, metallic conductor, or semiconductor. (2 points) The Fermi level passes directly through a band so it is a metallic conductor.

(c) Draw the d orbital splitting diagram for the tungsten ions. (2 points)

eg

t

t 2g

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5. Ruthenium(IV) oxide (RuO2) is an important catalytic material for a variety of processes from chlorine production to hydrocarbon refinement. The material adopts a rutile structure. (a) Using the following thermochemical data, calculate the lattice energy of ruthenium(IV) oxide. (8 points) ΔatH º(Ru) = +652 kJ/mol IE1(Ru) = +710.2 kJ/mol IE2(Ru) = +1617 kJ/mol IE3(Ru) = +2747 kJ/mol IE4(Ru) = +4340 kJ/mol

ΔfH º(RuO2) = −314.2 kJ/mol BDE(O2) = +497.0 kJ/mol EA1(O) = +141.0 kJ/mol EA2(O) = −844.0 kJ/mol

ΔLH º(RuO2) = −ΔfH º(RuO2) + ΔatH º(Ru) + BDE(O2) + IE1-4(Ru) − 2EA1-2(O) ΔLH º(kJ/mol) = −(−314.2) + 652 + 497.0 + 9414 –2(–703) ΔLH º(RuO2) = +12283 kJ/mol

(b) Under acidic conditions, RuO2 exists as Ru(OH)22+ and can be oxidized with periodate (IO4−) to afford ruthenium tetroxide (RuO4). RuO4 is a potent reagent capable of oxidizing hydrocarbons. Balance the reaction. (2 points)

Ru(OH)22+ (aq) + IO4− (aq) → RuO4 (l) + I− (aq) 2 Ru(OH)22+ (aq) + IO4− (aq) → 2 RuO4 (l) + I− (aq) + 4 H+ (aq)

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