Inorganic Chemistry 412 Final Exam 110 minutes. (a) Disproportionation of an acidic aqueous solution of sodium chlorite, NaClO 2.
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1 NAME: KEY Inorganic Chemistry 412 inal Exam 110 minutes Please show all work, partial credit may be awarded. 1. or each of the following, provide a balanced reaction. [8 pts each] (a) Disproportionation of an acidic aqueous solution of sodium chlorite, NaO 2. H + + 2O 2 - HO + O 3 - (b) Reaction of P 3 with moist air. 2P H 2 O + O 2 2H 3 PO 4 + 6H 3 O (c) Decomposition of 20% H 2 O 2 in aqueous acid. 2H 2 O 2 2H 2 O + O 2 2. Which is more easily oxidized, e(m) or Cu(m)? ully explain your reasoning in terms of periodic trends. [8 pts] e(m) is more easily oxidized than Cu(m) because of the increasing effective nuclear charge, Z*, on Cu. This Z* is slowly increasing left to right across the d-block. 3. Predict the number of peaks and their multiplicities in the 19 NMR spectrum of XeO 4 [8 pts] The Lewis structure of XeO 4 is shown below. O 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 (Xe-129 is spin ½ and Xe-131 is spin 3/2). This is not necessary for full credit on this question. 4. 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. [8 pts]
2 2 # of electrons from the B-H 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. Iodine trichoride, I 3, oligomerizes, but Br 3 is a monomer. Give the general trend that is illustrated by this example. or full credit you must provide a complete statement of a general trend. [8 pts] Heavier congeners tend to higher coordination number. This drives oligomerization in I 3 to allow 4 coordination. I I Other examples, 1) Sb 5 is oligomeric, unlike As 5 and P 5 Sb Sb 2) periodic acid is H 5 IO 6 with octahedral coordination around I, rather than Td as in HO 4 and HBrO 4 6. BieO 3 is a perovskite with a Neel temperature of 653K. (a) Sketch a unit cell of BieO 3. [8 pts]
3 3 Either is fine (b) Draw the Curie plot for BieO 3. [5 pts] Curie-Weiss plot (straight-line, negative intercept) is OK, too (c) Sketch the antiferromagnetic coupling interactions in BieO 3. Label the orbitals involved, show their overlap, and indicate the spins in these orbitals. [8 pts] (d) How many unpaired spins does iron have in BieO 3? [6 pts] e(iii) is d 5, high spin case has 5 unpaired spins 7. (a) How is charge neutrality maintained in TiO 1.25? What types of defects are most abundant? [8 pts] Some Ti(II) is oxidized to Ti(IV). This creates cation vacancies and also Ti(IV) in interstitial Td sites. (b) Describe the origin of the t 2g conduction band in TiO 1.25 and sketch the orbitals involved. [8 pts]
4 4 The orbitals are relatively large for early TM, so more overlap possible. The t2g are oriented towards other metal cations in the structure. (c) TiO 1.25 is properly described as a metallic conductor. How does its conductivity change with increasing temperature? [5 pts] Metallic conduction means that electronic conductivity decreases with increasing temp. 8. What is the mobile ionic species in the Ag 2 HgI 4 ionic conductor? Explain briefly. [6 pts] The mobile species is Ag+. Hg 2+ is more highly charged and therefore much less mobile (higher activation energy for ion hopping). However, the presence of Hg 2+ creates cation vacancies that increase the mobility of Ag + 9. A Pourbaix diagram for Eu is shown to the right. (a) At what ph is Eu 2+ the most stable in water? [3 pts] 7 (neutral) (b) Is Eu 2+ an oxidant or reductant? [3 pts] reductant (c) Why is Eu 2+ relatively stable compared to other lanthanide(ii) cations? [6 pts] Because it has a more symmetric valence shell (it is an f7 ion). 10. (a) Give a balanced chemical reaction for the preparation of [Mo 6 O 19 ] 2- (aq) from K 2 MoO 4 (s). [8 pts] 6 K 2 MoO H + [Mo 6 O 19 ] H 2 O + 12K +
5 (b) Is the reaction in part (a) an acid/base reaction, a redox reaction, or both? Explain. [5 pts] This is acid/base only, Mo is VI oxidation state in reactant and product (c) Explain the number of shared vertices in each of the linked octahedral in the structure of [Mo 6 O 19 ] 2- (shown to the right). [6 pts] There are 6 edge-sharing Oh, Each Oh has 1 unique O 1 4 shared O 4 x ½ 1 center O 1 x 1/6 total O / M 3 1/6 = M 6 O Give two different reactions where Au(m) is oxidized and dissolved into a solution. [8 pts] Many ligands, such as CN -, dramatically lower the reduction potential of Au + and many other TM cations. So: 2 Au(m) + 4 CN - (aq) + ½ O 2 (g) + 2 H + (aq) 2 [Au(CN) 2 ] - (aq) + H 2 O Or, use aqua regia: 5 Au(m) + NO 3 - (aq) + 2 H + (aq) Au + (aq) + NO 2 + H 2 O 12. Barium titanate is a classic ferroelectric material. (a) Explain how Ti atoms displace off their ideal position in the ferroelectric phase. Include a sketch with two or more connected octahedra. It may help to put an arrow on the Ti atoms to indicate the displacement directions. [8 pts] Looking for something like the E phase on the right where the Ti is displace toward an oxygen atom. And show the same displacement for adjacent Oh.
6 6 (b) Sketch a hysteresis curve for BaTiO 3 and label the spontaneous polarization, the saturation point, and the coercive field. [6 pts] Ps = spontaneous polarization. C is the saturation point. Ec is the coercive field. Ec can be at +E=0 or E=0 (c) How can the Curie temperature of a perovskite be changed in a controlled manner? [4 pts] By full or partial substitution of ions with different radii into the A site. (Other methods possible, too) 13. What are the electronic configurations for the following metal-metal complexes? [4 pts each] (a) [Mo 2 (O 2 CCH 3 ) 4 ] In this case, we have 2 Mo(II) ions, each is d 4. The electron configuration is therefore σ 2 π 4 δ 2 (b) [Cu 2 (O 2 CCH 3 ) 4 ] 2 Cu(II) ions, each is d 9. The electron configuration is therefore σ 2 π 4 δ 2 δ* 2 π* 4 σ* 2 plus an additional e- in each Cu d x2-y2 orbital 14. Why do the late transition metals have very few stable oxidation states? [5 pts] The oxidation potential (or IE or Z*) is increasing across the row, so late TM s can only be oxidized to low oxidation numbers. Also, there are few unfilled orbitals available for favorable overlap to form bonds. 15. Multiple choice, 4 points each. Circle the ONE best answer. i. What is the structure and point group symmetry of N 2 O? a. linear and C v
7 7 b. linear and D h c. bent and C 2v (OOPS should be bent and C s this one is not scored) d. bent and C 3v e. trigonal planar f. T-shaped g. it s not a known compound ii. Which one of the following compounds or elements is the most air stable? a. Xe 2 b. P 3 c. Si 4 d. PbO e. Na f. B 2 H 6 g. B 3 h. [AuXe 4 ](Sb 6 ) 2 iii. The best way to tune the color of gadolinium oxide is: a. Controlled heating b. Substituting the oxo ligands for fluoro ligands c. Controlling the oxygen partial pressure d. Doping another cation into the system e. Changing the ph f. Oxidizing the gadolinium iv. Which of the following oxides does not form a glass? a. BeO b. Al 2 O 3 c. SiO 2 d. GeO 2 e. P 2 O 5 v. Zeolites are excellent absorbers for which of the following molecules? a. H 2 O b. NH 3 c. NO 2 d. CO 2 e. All of the above
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