NAME: SECOND EXAMINATION - PDF Free Download

1 Chemistry 64 Winter 1994 NAME: SECOND EXAMINATION THIS EXAMINATION IS WORTH 100 POINTS AND CONTAINS 4 (FOUR) QUESTIONS THEY ARE NOT EQUALLY WEIGHTED! YOU SHOULD ATTEMPT ALL QUESTIONS AND ALLOCATE YOUR TIME ACCORDINGLY ALL BOOKS AND PAPERS OTHER THAN TABLES WHICH HAVE BEEN HANDED OUT SHOULD BE PLACED ON THE FLOOR DURING THIS EXAMINATION IF YOU DO WORK ANYWHERE OTHER THAN THE SPACE PROVIDED FOR EACH QUESTION, INDICATE CLEARLY WHERE IT IS LOCATED The Periodic Table is provided as detachable Supplementary Material at the back of this exam. The Spectrochemical Series and Magnetic Moment information is provided below. Spectrochemical Series Increasing o (fixed metal): I < Br < Cl ~ SCN ~ N 3 < F < OH < CH 3 CO 2 < oxalate 2 < H 2 O < NCS < EDTA 4 < NH 3 ~ py < en ~ tren < o phen < NO 2 < H ~ CH 3 < CO ~ CN Spin-only formula: µ s = 2{S(S+1)} 1/2 BM = {n(n+2)} 1/2 BM Abbreviations: Me = methyl; t-bu = tert-butyl; dppe = 1,2-bis(diphenylphosphino)ethane; phen = 1,10-phenanthroline; en = ethylenediamine

2 FOR GRADING USE ONLY Question 1 (33 pts)... Question 2 (24 pts)... Question 3 (33 pts)... Question 4 (10 pts)... TOTAL (100 pts)...===========

3 Question 1 (33 pts) The reaction of BH 3 and CO gives the compound H 3 B-CO. (a) (4 pts) Identify the Lewis acid in this reaction and sketch its LUMO. (b) (4 pts) Identify the Lewis base in this reaction and sketch its HOMO. (c) (4 pts) Sketch the geometry at the boron atom in BH 3 and in the BH 3 -CO molecule. (d) (4 pts) Describe the energetic changes associated with the above changes in the geometry at boron.

4 (e) (4 pts) Explain why compound A is a stronger Brønsted base than is compound B, but is a weaker Lewis base towards BMe 3 than is compound B. Me N A N B (f) (4 pts) Explain why Me 3 B is a stronger Lewis acid than BCl 3 towards NMe 3.

5 (g) (9 pts) Predict whether the equilibrium constants for the following reactions are greater than or less than 1. Explain your answers. (i) (t-bu) 3 In-PMe 3 + Me 3 In Me 3 In-PMe 3 + (t-bu) 3 In (ii) HgF 2 + BeI 2 HgI 2 + BeF 2 (iii) GaH 3 -NMe 3 + BF 3 Me 3 N-BF 3 + GaH 3

6 Question 2 (24 pts) (a) (4 pts) The reaction of palladium(ii) with sources of chloride and ammonia ligands gives 2 compounds, whose empirical formulas are: A: PdCl 2 3NH 3 B: PdCl 2 KCl NH 3 Compounds A and B both contain ions, but only A reacts with Ag + to give a precipitate. IR spectroscopy shows that both palladium containing ions belong to the C 2v point group. Draw the structures of A and B. (b) (6 pts) Similarly, it is possible to make a series of compounds each of which has the empirical formula PdCl 2 2NH 3, as 3 isomers C, D, and E. Isomers C and D do not contain ions, but C has a molecular dipole moment and D has none. C belongs to the point group C 2v and D to the point group D 2h. Isomer E has a molecular weight twice the empirical formula weight, and contains two ions of equal but opposite charge, each of which belongs to the point group D 4h. Draw the structures of C, D, and E.

7 (c) (4 pts) The complex [Co(dppe) 2 Cl]+ can be isolated in two different crystalline forms, one red and the other green. Suggest possible structures for these 2 isomers (you do not have to explain the colors). (d) (4 pts) The compound Ni(en) 2 Cl 2 exists in two forms. Form A is yellow and diamagnetic, and gives a precipitate on treatment with Ag +. Form B is blue, paramagnetic, and does not give a precipitate on treatment with Ag +. Suggest structures for A and B, and explain your answers (again, you need not explain the colors).

(e) (6 pts) Assuming an octahedral geometry, sketch all the possible isomers (including enantiomers if appropriate) of [Co(NH 3 ) 2 (H 2 O) 2 Cl 2 ] +. Be sure to indicate how many isomers there are. 8

9 QUESTION 3 (33 pts) (a) (6 pts) Consider a tetragonal distortion (elongating the 2 M-L bonds along the z-axis) of the octahedral metal complex ML 6 shown below. Tetragonal distortion L L L M L L L z y x (i) Give the crystal field splitting of the metal d-orbitals in the original octahedral compound. Explain your answer. (ii) Show how this splitting pattern is changed by the tetragonal distortion. Explain your answer.

10 (b) (3 pts) At high temperatures, the complex Fe(phen) 2 (NCS) 2 shows a magnetic moment consistent with the presence of 4 unpaired electrons. When the compound is cooled to 175 K, however, the magnetic moment suddenly decreases and approaches zero. Explain these observations.

11 (c) (8 pts) Spinels are crystalline inorganic compounds of the formula AB 2 O 4, where A is a metal in the +2 oxidation state and B is a metal in the +3 oxidation state. Spinels adopt 2 different possible structures: In normal spinels the A(II) ions are tetrahedrally coordinated by the oxygen ligands, while the B(III) ions are octahedrally coordinated. In inverse spinels, however, the A(II) ions are octahedral, while the B(III) ions are equally divided between octahedral and tetrahedral coordination geometries. The compound NiFe 2 O 4 adopts the inverse spinel structure. Rationalize this observation. (HINT--LFSE).

12 (d) (16 pts) Give the metal oxidation state, the d n electron configuration, the LFSE, and the spin-only magnetic moment for each of the following compounds. If you assume that a particular complex is high or low spin, explain your reasoning. (i) [Co(H 2 O) 6 ] 3+ (ii) [Mn(H 2 O) 6 ] 2+ (iii) [CoCl 4 ] 2 (iv) [V(CO) 6 ]

13 QUESTION 4 (10 pts) Part of the molecular orbital diagram for an octahedral metal complex ML 6 (where L is a σ-only donor ligand) is shown below. e g t 2g o (a) Identify the MO's shown as bonding, non-bonding, or antibonding. (b) What changes in the diagram would occur if : (1) The ligands were π-acceptors (2) The ligands were π-donors Your answer should consider both (i) the nature of the MO's, as in part (a) above, and (ii) the magnitude of o, and should include an explanation for these results.

14 DETACHABLE, HANDY DANDY PERIODIC TABLE OF THE ELEMENTS 1 H 1.0079 Li 3 6.941 11 Na 22.9898 19 K 39.0983 37 Rb 85.4678 55 Cs 132.905 Fr 87 (223) Be 4 9.01218 12 Mg 24.305 20 Ca 40.08 38 Sr 87.62 56 Ba 137.33 88 Ra 226.025 21 Sc 44.9559 Y 39 88.9059 57 La 138.906 89 Ac 227.028 Ti 22 47.88 40 Zr 91.224 72 Hf 178.49 104 Unq (261) V 23 50.9415 Nb 41 92.9064 Ta (262) 73 180.948 105 Unp Cr 24 51.996 95.94 42 Mo W 74 183.85 106 Unh (263) 25 Mn 54.9380 Tc (98) Re 43 75 186.207 107 Uns (262) Fe 26 55.847 Ru 44 101.07 Os 190.2 76 108 Uno (265) 27 Co 58.9332 45 Rh 102.906 Ir 77 192.22 109 Une (266) 28 Ni 58.69 Pd 46 106.42 Pt 78 195.08 Cu 29 63.546 Ag 47 107.868 Au 79 196.967 Zn 65.39 Cd 30 48 112.41 Hg 80 200.59 B 5 10.81 13 Al 26.9815 31 Ga 69.72 In 49 114.82 Tl 81 204.383 C 6 N 7 12.011 14.0067 14 15 Si P 28.0855 30.9738 32 33 Ge As 72.59 74.9216 Sn 50 51 Sb 118.71 121.75 Pb 82 83 Bi 207.2 208.980 O 32.06 34 Se 78.96 Te 52 127.60 Po 84 (209) 8 15.9994 16 S F 9 2 He 4.00260 10 Ne 18.9984 20.179 17 18 Cl Ar 35.453 39.948 35 36 Br 79.904 I 53 126.905 At (210) 85 Kr 83.80 54 Xe 131.29 86 Rn (222) 58 59 60 61 62 63 64 65 66 67 68 69 70 71 Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho 140.12 140.908 144.24 (145) 150.36 151.96 157.25 158.925 162.50 164.930 167.26 168.934 173.04 174.967 90 91 92 93 94 95 96 97 98 99 100 101 102 103 Th Pa U Np Pu Am Cm Bk Cf Es Er Tm Yb Lu Fm Md No Lr 232.038 231.036 238.029 (237) (244) (243) (247) (247) (251) (252) (257) (258) (259) (260)