Chem 400. Inorganic Chemistry. Practice Exam 1. 1 of 5. 2 of 5. 3 of of of of of of of of 10.

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1 Chem 400 Inorganic Chemistr Practice Eam 1 1 of 5 2 of 5 3 of 10 4 of 10 5 of 10 6 of 10 7 of 10 8 of 10 9 of of of 10 Σ of 100 Name KEY (please print)

2 1. a. Predict the structure of 4 using the VEPR model. (2 points) Aial Equatorial b. Eplain the fact that at 298 K and in solution the 19 NMR spectrum of 4 ehibits a singlet but that at 175 K, two equal intensit triplets are observed. (3 points) Temperature dependent 19 NMR spectrum indicates that 4 is stereochemicall non-rigid (fluional). There is more thermal energ at 298 K than at 175 K. At 298 K, a singlet is observed, therefore all atoms must be equivalent, and the aial and equatorial positions are echanging over the Berr pseudo-rotation. At 175 K, two equal intensit triplets are observed. This arises as follows. There are two environments (aial and equatorial) as shown above. Each aial couples to 2 equatorial s to give a binominal triplet (and vice versa). The two triplets occur at different chemical shifts. 2. Does VB (valence bond) theor indicate that the diatomic molecule He 2 is a viable species? Rationalize our answer using arguments derived from VB theor onl. (5 points) Ground state electron configuration of He (Z = 2) is 1s 2. Within VB theor, the resonance structures that could be drawn (remembering that electrons are paired so far as possible) are: He=He He 2+ He 2- He 2- He 2+ The double bond formation is not possible with onl the 1s orbital per He atom, and the ionic form is unreasonable (look at ionization energies for He). It is concluded that He 2 is not a viable species. (The real question is What is the stabilit of He 2 with respect to 2 He? and VB theor does not give an answer to this.) 1

3 3. Draw Lewis structures for each of the following molecules. (2 points each) a. e 4 e b. Xe 3 Xe c. V 3 V d. 3 - e. 4 2

4 4. Draw three-dimensional structures for each of the compounds in problem 3 using VEPR theor. (2 points each) a. e disphenoidal molecular shape b. Xe Trigonal pramidal shape, similar to NH 3, but with Xe- double bonds. c. V Distorted tetrahedral shape, with -V- angles of 111º, and -V- angles of 108º. d has a trigonal pramidal shape, with one lone pair and three principal resonance structures. e. 4 has a distorted trigonal bipramidal shape. The aial fluorine atoms are nearl linear with the atom. 3

5 5. Determine the appropriate point group for each of the compounds in problem 4. (2 points each) a) e 4 has C 2v smmetr b) Xe 3 has C 3v smmetr c) V 3 has C 3v smmetr d) 3 - has C 3v smmetr e) 4 has C 2v smmetr 4

6 6. Zinc sulfide has two crstalline forms (shown below). Describe these two structures in terms of closest packed spheres, interstitial occupanc, and coordination number. (10 points) zinc-blende wurtzite A cubic closest packed (ccp) arra of sulfide anions with zinc cations occuping ½ of the tetrahedral holes. CN ( 2+ ) = 4 CN ( 2- ) = 4 A heagonal closest packed (hcp) arra of sulfide anions with zinc cations occuping ½ of the tetrahedral holes. CN ( 2+ ) = 4 CN ( 2- ) = 4 5

7 7. a. Derive a set of diagrams using 3s, 3p, 3p and 3d 2-2 atomic orbitals to describe the formation of sp 2 d hbrid orbitals. (8 points) Take the shaded lobes of the p and d 2-2 orbital to point along the + ais, and the shaded lob of the p orbital to point along the + ais. In the plane, the orbital combinations to give 4 sp 2 d hbrid orbital are: b. What is the percentage character of each sp 2 d hbrid orbital in terms of constituent atomic orbitals? (2 points) Available for hbridization are one s, two p and one d orbital. Each hbrid orbital must contain the same amount of s character; since there are 4 hbrid orbitals, each contains 25% s character. Each hbrid orbital also must contain the same amount of p character, i.e. 50% p character. Each hbrid orbital contains 25% d character. 1

8 8. ketch out the molecular orbital energ level diagram for NH 3 appling the LG approach. Define the spatial orientation of NH 3 using,,z directions and identif all atomic and ligand group orbitals involved in bonding. (10 points) ψ 7 node z 2p z 2p 2p ψ 5 ψ 6 LG(3) LG(2) LG(1) LG(2) LG(3) Energ ψ 4 ψ 2 ψ 3 LG(1) 2s ψ 1 H N NH 3 H H 2

9 9. Draw schematic representations for all of the bonding molecular orbitals (M s) of NH 3 (use the information derived in Problem 8). (10 points) ψ 1 ψ 2 ψ 4 ψ 3 ψ 5 ψ 6 ψ 7 3

10 10. How man normal modes of vibration are IR active for (2 points each): a. H 2 H 2 is polar and possesses a bent molecular shape. All three modes of vibration are IR active. b. i 4 i 4 is tetrahedral and non-polar; T d smmetr. 4 modes of vibrational freedom (2 degenerate pairs) are IR active and give rise to 2 bands in the IR spectrum. c. HCN HCN is linear and polar. Thus, both smmetric and asmmetric stretches as well as the doubl degenerate deformation are IR active. Three fundamental absorptions are seen in the IR spectrum. d. Al 3 Al 3 has a trigonal planar structure and is non-polar. Therefore, the smmetric stretch is IR inactive. The IR active modes are the smmetric deformation, doubl degenerate stretch and doubl degenerate deformation 5 modes of vibrational freedom giving rise to 3 bands in the IR spectrum. e. PBr 3 PBr 3 has a trigonal pramidal structure and is polar. Both smmetric stretch and smmetric deformation are IR active, the doubl degenerate (asmmetric stretch and the doubl degenerate deformation are also IR active. In summar, there are 6 modes of vibrational freedom giving rise to 4 bands in the IR spectrum. 11. B 2 3 is acidic, Al 2 3 is amphoteric, and c 2 3 is basic. Wh? Give a detailed eplanation writing down the equations describing the reaction of these compounds with water. (10 points) In general, oide ion reacts with water to form hdroide: 2- + H 2 2 H - unless other factors prevent it. In B 2 3, the small, hard B 3+ holds on the oide ions strongl. As a result: B H 2 2 B(H) 3 H + + H 2 B 3 - and the solution is ver weakl acidic (pk a = 9.25). In Al 2 3, the Al 3+ ion is larger and softer. It can form either [Al(H) 4 ] - (acting as an acid) or [Al(H 2 ) 6 ] 3+ (acing as a base), depending on the other species in solution. γ-al 2 3 (s) + 3H 2 (l) + 6[H 3 ] + (aq) 2[Al(H 2 ) 6 ] 3+ (aq) γ-al 2 3 (s) + 3H 2 (l) + 3[H] - (aq) 2[Al(H) 4 ] - (aq) c 3+ is still larger and softer, so it combines better with water than with hdroide ion. As a result c H 2 2 [c(h 2 ) 6 ] H - is possible. 4