NAME: Inorganic Chemistry 412/512 Final Exam. Please show all work, partial credit may be awarded.

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1 NAME: Inorganic Chemistry 412/512 inal Exam 110 minutes Please show all work, partial credit may be awarded. 1. Given the following ligand field splitting parameters (in cm 1 ): Δ (e 3+ ) = Δ T (e 3+ ) = 6200 Δ (Ni 2+ ) = 8600 Δ T (Ni 2+ ) = 3600 (a) Explain in detail WHY Nie 2 4 forms the inverse instead of the normal spinel structure. [10] Consider Nie 2 4 (e 3+ ) tet (Ni 2+ e 3+ ) oct 4 with the inverse spinel structure The LSE of Ni 2+ sitting in octahedral sites is significantly greater than in tetrahedral sites. LSE = 6*0.4 o 2*0.6 o = 1.2 o LSE = 4*0.6 t 4*0.4 t = 0.8 t = cm 1 = 2880 cm 1 or e 3+, the LSE is zero for both octahedral and tetrahedral sites. This means that e 3+ has no site preference in this structure. (b) At room temperature, is Nie 2 4 a ferro, ferri, or an antiferromagnet? Explain. [10] errimagnetism. spins are antiparallel, but do not cancel. µ = 2(2+1)µ b = 2.5µ b (c) Show the general shape of a plot of magnetic susceptibility vs. temperature for Nie 2 4, and label the ordering temperature, T N, on your plot. [10 pts]

2 2. Describe in detail the structure and most abundant defects present in e 1.1, and explain why the same defects occur only at very low concentration in Ni. [15 pts] A stoichiometric e structure is a rock salt structure with the CC array of 2. The octahedral sites are fully occupied by e 2+. However, nonstoichiometric structure of e x is typically found with the presence of cationic vacancy and the ease oxidation of e 2+ to e 3+. The generated e 3+ occupies tetrahedral sites rather than octahedral sites. 2 Compared to e oxidation, Ni 2+ oxidation to Ni 3+ require much more energy due to the greater effective nuclear charge on Ni, and this defect is less abundant. 3. Why are Ce and Eu the easiest lanthanides to isolate from a mixed lanthanide ore using chemical methods? [10 pts] Ce has a relatively stable 4+ oxidation state, and Eu has a relatively stable 2+ oxidation state. All the other lanthanides have only 3+ oxidation states available. Thus, Ce can be selectively oxidized, and Eu selectively reduced during separation reactions. 4. or each of the following, provide a balanced reaction. [10 pts each] (a) Disproportionation of an acidic aqueous solution of sodium chlorite, Na 2. H H (b) Addition of Sb 5 to Xe 2 Sb 5 + Xe 2 [Xe + ][Sb 6 ]

3 3 (c) Reaction of P 3 with moist air. 2P H H 3 P 4 + 6H (d) Decomposition of 20% H 2 2 in aqueous acid. 2H 2 2 2H (e) Reaction of potassium permanganate, KMn 4, with excess e 2 (aqu) in aqueous acid. 8H + + Mn 4 + 5e 2+ 5e 3+ + Mn H 2 5. Describe two distinct methods that can be used to prepare LiTiS 2 starting from TiS 2 and any reagents/equipment of your choice. [15 pts] 1) TiS 2 (s) + Li(g) Li x TiS 2 (s) high temp. 2) Electrochemical synthesis 6. ZrS 2 reacts with cobaltocene, Co(η 5 C 5 H 5 ) 2, to form an intercalation compound. Explain why ZrS 2 does not react with ferrocene, e(η 5 C 5 H 5 ) 2, even though the two metallocenes have approximately the same dimensions.. [10 pts]

4 4 Based on the 18electron rule, ferrocene (an 18electron complex) is stable. However, cobaltocene is a 19electron complex and is a strong reducing agent. It reduces ZrS 2 according to the following: Co(C 5 H 5 ) 2 + ZrS 2 > [Co(C 5 H 5 ) 2 ] + ZrS 2 The cobaltocenium cation intercalates into the layered ZnS 2 to form the intercalation compound. 7. Which is more easily oxidized, e(m) or Cu(m)? ully explain your reasoning in terms of periodic trends. [10 pts] e(m) is more easily oxidized than Cu(m) of the increasing effective nuclear charge, Z*, on Cu. This Z* is slowly increasing left to right across the dblock. 8. Predict the appearance of the 19 NMR spectrum of Xe 4 [10 pts] The Lewis structure of Xe 4 is shown below. Xe The point group is C 4v, and all atoms are identical. Therefore, only one peak is observed in the 19 NMR spectrum. If we consider coupling to Xe, the NMR spectrum is more complex, with two NMR active nuclei (Xe129 is spin ½ and Xe131 is spin 3/2). This is not necessary for full credit on this question. 9. Draw the structure for B 5 H 9 (indicate clearly whether this is a closo, nido, or arachno structure) and explain your reasoning by calculating the number of cluster bonding electron pairs. [10 pts] # of electrons from the BH unit = 5*2 = 10 e # of electron from 4 additional H atoms = 4 e Total # of electrons involved = 14 e So, # of cluster bonding electron pairs = 7 pairs ( n + 2) The structure of B 5 H 9 is nido.

5 10. Iodine trichoride, I 3, oligomerizes, but Br 3 is a monomer. State the general trend that is being illustrated by this example. or full credit you must provide a complete statement of a general trend. Give a two more specific examples that illustrate this trend. [15 pts] 5 Heavier congeners tend to higher coordination number. This drives oligomerization in I 3 to allow 4 coordination. I I ther examples, 1) Sb 5 is oligomeric, unlike As 5 and P 5 Sb Sb 2) periodic acid is H 5 I 6 with octahedral coordination around I, rather than Td as in H 4 and HBr Ammonium persulfate, (NH 4 ) 2 S 2 8, is a common oxidant used in aqueous solutions. Explain why this compound acts as a strong oxidant while potassium sulfate, K 2 S 4, is not an oxidant. [10 pts] The reactive group in persulfate is the peroxo bond, which is not present in sulfate: S S 12. Multiple choice, 5 points each. Circle the NE best answer. i. What is the MoMo bond order for the cluster [Mo 2 (S 4 ) 4 ] 3? (a) 1 (b) 2 (c) 3 (d) 3.5 (e) 4 (f) 4.5 (g) 0 (the metal atoms are not bonding to one another) ii. What is the structure and point group symmetry of N 2? (a) linear and C v

6 6 (b) linear and D h (c) bent and C 2v (d) bent and C 3v (e) trigonal planar (f) Tshaped (g) it s not a known compound iii. Which one of the following compounds or elements is the most air stable? (a) Xe 2 (b) P 3 (c) Si 4 (d) Pb (e) Na (f) B 2 H 6 (g) B 3 (h) [AuXe 4 ](Sb 6 ) 2