Chem 400. Inorganic Chemistry. Practice Exam 2. 1 of of of of of of 5. 7 of 5. 8 of of 10.

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1 Chem 400 Inorganic Chemistry Practice Exam 2 1 of 10 2 of 10 3 of 20 4 of 10 5 of 20 6 of 5 7 of 5 8 of 10 9 of 10 Σ of 100 KEY Name (please print)

2 1. Using balanced chemical equations, give an explanation for the following observations: AlF 3 has only a low solubility in liquid F, but a combination of NaF and AlF 3 leads to dissolution of the reagents (1 st equation); when F 3 is added to this solution, a precipitate forms (2 nd equation). (10 points) AlF 3 is a Lewis acid and, in the presence of F -, forms the adduct [AlF 4 ] - : AlF 3 + NaF Na[AlF 4 ] (Equation 1) and the ionic salt Na[AlF 4 ] is soluble in liquid F. F 3 is a stronger fluoride acceptor than AlF 3, and accepts F - from [AlF 4 ] - : Na[AlF 4 ] + F 3 AlF 3 + Na[F 4 ] (Equation 2) In the absence of excess F -, AlF 3 precipitates. For acid-base chemistry, the self-ionization you need to consider is: 3F [ 2 F] + + [F 2 ] - 2. The σ bonding in the linear molecule XeF 2 may be described as a 3-center, 4- electron bond. If the z axis is assigned as the internuclear axis, use the p z orbitals on each of the atoms to prepare a molecular orbital description of the sigma bonding in XeF 2. (10 points) Two electrons are in the bonding orbital and two are in the non-bonding orbital. Energy Xe XeF 2 F- - - F 1

3 3. uggest likely products for the following reactions (which are balanced on the lefthand side): (2 points each) a. 2 XeF 2 + bf 5 [Xe 2 F 3 ] + [bf 6 ] - b. Me 3 i + Na[C 5 5 ] Na + Me 3 i[η 1 -C 5 5 ] c C 2 5 (C 2 5 ) d. n u + C 6 6 C 6 5 (Ph) + n u (or n-c 4 10 ) e. 2 (C 3 ) 3 n + Mg (C 3 ) 3 nn(c 3 ) 3 + Mg 2 f C 2 =C 2 1. TF () C 3 C 2 g. 2P 5 + N P=N-N=P h. 2XeF 6 + i 2 if 4 + 2XeF 4 i. Al C 3 C 2 Mg Al(C 2 C 3 ) + 3 Mg 2 j. 2 Na + 2F 2 F NaF + 2 2

4 4. 2 and [ 2 ] - are bent, each with equivalent - bond lengths: 147 pm in 2 and 157 pm in [ 2 ] -. Give a description of bonding in 2 and [ 2 ] - and rationalize the difference in - bond lengths. Use both resonance structures and M theory arguments. Draw a representation of the π-molecular orbitals. (10 points) Resonance structures can be drawn to rationalize equivalence of bonds: r, use M theory to show delocalized bonding schemes analogous to π-allyltype bonding. There are 12 atomic orbitals in the basis set of 2 (4 valence orbitals per atom) and therefore 12 Ms; these are occupied by 19 valence electrons. In [ 2 ] -, 10 Ms are fully occupied. The π-ms are shown below. In 2, the M is the π* M and is singly occupied. In [ 2 ] -, this M is fully occupied, giving more antibonding character to the - interactions. The M scheme illustrates delocalization of electrons over the - framework, and rationalizes the lengthening of bonds on going from 2 to [ 2 ] -. Energy π* non-bonding π π-molecular orbitals in 2. 3

5 5. Draw reasonable three-dimensional structures for each of the following inorganic molecules (2 points each) a. (C 3 ) 4 b. [ 3 6 ] 3-3 C 3 C C 3 - C c. (N) 3 N N N d. [i 4 12 ] 8- e. i i - i i i i i 4

6 f. [ 9 4 ] + g. 8 h. [ 4 ] i. (η 5 -Cp)e e or e j. Al 2 (C 3 ) 6 3 C 3 C Al Al C 3 3 C C 3 C 3 5

7 6. The triiodide ion, I 3 -, is linear, but I 3 + is bent. Explain. (5 points): I 3 - is linear because there are three lone pairs in a trigonal geometry on the central I. I 3 + is bent because there are only two lone pairs on the central I, leaving a structure similar to The sulfur-sulfur distance in 2, the major component of sulfur vapor above ~720 ºC, is 189 pm, significantly shorter than the sulfur-sulfur distance of 206 pm in 8. uggest an explanation for the shorter distance in 2. (5 points) 2 is similar to 2 with a double bond. As a result, the bond order is shorter than single bonds in uggest a balanced chemical equation to describe the autodissociation of I (1 st equation). Na behaves as a base in this solvent, while Al 3 behaves as an acid. Write balanced chemical equations for the dissolution of Na in I (2 nd equation) and Al 3 in I (3 rd equation). (10 points) Equation 1: 3 I I I 2 - oth solutes increase the concentration of ions: Na + I Na + + I 2 - Na acts as a base. Al 3 + 2I I Al 4 - Al 3 acts as an acid. 9. Describe the bonding in Ga 2 6 and Ga 2 6. (10 points) Ga 2 6 is isostructural with 2 6 ; analogous bonding pictures are appropriate. In Ga 2 6, 2c-2e localized terminal Ga- bonds and 3c-2e Ga--Ga bridge bonds: verlap of 1s orbital with two sp 3 orbitals. Ga Ga In Ga 2 6 there are sufficient valence electrons to invoke all 2c-2e Ga- interactions; analogous to Al 2 6 : Ga Ga 6