NAME: 3rd (final) EXAM

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1 1 Chem 64 Winter 2003 AME: 3rd (final) EXAM THIS EXAM IS WORTH 100 POITS AD COTAIS 9 QUESTIOS THEY ARE OT EQUALLY WEIGHTED! YOU SHOULD ATTEMPT ALL QUESTIOS AD ALLOCATE YOUR TIME ACCORDIGLY IF YOU DO'T KOW THE ASWER, GUESS! ALL BOOKS AD PAPERS OTHER THA TABLES WHICH HAVE BEE HADED OUT SHOULD BE PLACED O THE FLOOR DURIG THIS EXAM IF YOU DO WORK AYWHERE OTHER THA THE SPACE PROVIDED FOR EACH QUESTIO, IDICATE CLEARLY WHERE IT IS LOCATED A PERIODIC TABLE, SPECTROCHEMICAL SERIES, MAGETISM IFO, AD TAABE-SUGAO DIAGRAMS ARE ATTACHED AT THE ED OF THE EXAM PLEASE PRIT YOUR AME BOTH O THIS PAGE AD O PAGE 2

2 2 AME: FOR GRADIG USE OLY Question 1 (13 pts)... Question 2 (8 pts)... Question 3 (8 pts)... Question 4 (10 pts)... Question 5 (12 pts)... Question 6 (12 pts)... Question 7 (12 pts)... Question 8 (12 pts)... Question 9 (13 pts)... TOTAL (100 pts)...===========

3 3 1. (13 pts) Consider data for the two ligand substitution reactions shown below. 1 L SnPh 3 SnPh 3 Pt L + L* L Pt L* + L L = PMe 2 Ph 2 SnPh 3 SnPh 3 SiMePh 2 SiMePh 2 Ph 2 MeSi Pt L + L* Ph 2 MeSi Pt L* + L L L Reaction number 1 2 H (kj/mol) S (J/mol K) a. (4 pts) Suggest mechanisms for both reactions. Briefly explain your answers. b. (3 pts) The SnPh 3 group is a π-acceptor. Why might this be important in your proposed mechanism for reaction 1?

4 4 c. (6 pts, 3 each) This table shows X-ray structural data for the complex in reaction 2 and for analogous complexes with Me and Cl ligands. [ote: L = PMe 2 Ph] Complex Pt-P bond length (Å) cis-ptl 2 (SiMe 2 Ph) cis-ptl 2 (Me) cis-ptl 2 (Cl) i. Circle your choice for the trans influence of the SiMe 2 Ph group, relative to Me, Cl, and other ligands with which you are familiar. Briefly explain your answer. The trans influence of SiMe 2 Ph is high medium low ii. Why might this be important in your proposed mechanism for reaction 2?

5 5 2. (8 pts, 4 each) a. For water exchange on complex 1, S = 34 J/mol K and V = 4.7 cm 3 /mol. Propose a mechanism for the ligand substitution and briefly explain your answer Cu H 2 H 2 O H 2 H 2 Cu Me 2 H 2 O 1 2 Me 2 Me 2 b. Water exchange in complex 2 is 10 5 times slower than in 1. Explain this observation in terms of the mechanisms of these ligand substitutions.

6 6 3. (8 pts, 4 each) The figure below shows the electronic spectra of the cis and trans isomers of approximately octahedral [Co(en) 2 F 2 ] + (en = H 2 CH 2 CH 2 H 2 ). There are 2 curves, labeled A and B. Fill in the box to match the spectrum with the isomer and explain your answer. [ote: the y-axis shows ε values.] cis trans b. Electronic spectra of [RhCl 6 ] 3 and [Rh(H 3 ) 6 ] 3+ are shown in the figure. Fill in the box to match the spectrum to the complex and explain your answer. [RhCl 6 ] 3 [Rh(H 3 ) 6 ] 3+

7 7 4. (10 pts) a. (4 pts) The complex [Cr(H 2 O) 6 ] 2+ reacts with [Co(H 3 ) 5 Cl] times faster than it does with [Co(H 3 ) 6 ] 3+. Explain this observation. b. (6 pts) The substitution of azide ( 3 ) for triflate (OSO 2 CF 3 ) in the reaction below is very slow in acidic aqueous solution. [Co(H 3 ) 5 (OSO 2 CF 3 )] 2+ a 3 [Co(H 3 ) 5 ( 3 )] 2+ very slow However, when 1.0 M 3 in the presence of 0.10 M OH is used, the ligand substitution reaction is much faster, and the products are the cobalt azide complex (10%) and the cobalt hydroxide complex (90%). [Co(H 3 ) 5 (OSO 2 CF 3 )] 2+ a 3 [Co(H 3 ) 5 ( 3 )] 2+ OH much faster [Co(H 3 ) 5 (OH)] 2+ Explain these observations in terms of mechanisms for the ligand substitutions. Don't forget to explain why the reaction in acid solution is slow.

8 8 5. (12 pts) a. (4 pts) Both [Fe(C) 6 ] 4 and [Fe(C) 6 ] 3 are toxic because they dissociate cyanide, C. Circle the one you predict will be more poisonous, and briefly explain your answer. [Fe(C) 6 ] 4 [Fe(C) 6 ] 3 b. (8 pts) The gemstone ruby consists of Cr(III) impurities in octahedral sites in Al 2 O 3. The electronic spectrum of ruby contains some very narrow (peak width ~ 10 cm 1 ), low-intensity lines. Identify (as specifically as possible) the electronic transitions responsible for the narrow peaks in the ruby spectrum, and explain why the peaks are unusually narrow in comparison to "normal" ones.

9 9 6. (12 pts) [Fe(C) 6 ] 3 and [Co(EDTA)] 2 react slowly to yield [Fe(C) 6 ] 4 and [Co(EDTA)]. In this reaction (1), the cyano-bridged intermediate [(C) 5 Fe-C- Co(EDTA)] 5 is formed within milliseconds. The rates of the analogous reactions 2 and 3 are very similar to each other and to that of 1. Suggest mechanisms for these electron-transfer reactions and explain your answers. 1 2 C C C C Fe C C Fe C C 3 C [Co(EDTA)] 2 [Co(EDTA)] C [Fe(C) 6 ] 4 C 2 [Co(EDTA)] 2 [Co(EDTA)] [Fe(C) 5 (py)] 3 C C Fe C C C C [Co(EDTA)] intermediate 5 O 2 C py = EDTA O 2 C CO 2 CO 2 3 C C C Fe C C 2 [Co(EDTA)] 2 [Co(EDTA)] [Fe(C) 5 (bipy)] 3 bipy =

10 10 7. (12 pts) Phosphines PR 3 can act as π-acceptors using the P-C σ* MO as an acceptor orbital. Using this model, explain a. why the P-C bond shortens on oxidation of the Co (1 ---> 2). b. why the Co-P bond lengthens on oxidation of the Co (1 ---> 2). Be sure to consider both σ- and π-bonding effects. number Compound Co-P bond length (Å) P-C bond length (Å) 1 [CpCo(PEt 3 ) 2 ] 2.218± ± [CpCo(PEt 3 ) 2 ][BF 4 ] 2.230± ±0.003

11 8. (12 pts) The isoelectronic ions [VO 4 ] 3, [CrO 4 ] 2, and [MnO 4 ] all show intense charge transfer peaks in their UV-vis spectra. The wavelengths of these transitions increase in this series, with [MnO 4 ] having its charge transfer absorption at longest wavelength. Explain these observations in terms of a suitable MO diagram for these complexes; be sure to identify the specific transitions responsible for the charge transfer peaks. 11

12 12 9. (13 pts) a. (4 pts) Consider a linear ML 2 complex (the z-axis is the molecular axis). According to crystal field theory, the energies of the M atom's d orbitals in such a complex split into 3 different groups. Fill in the boxes with the appropriate orbitals and explain your answers. Energy d-orbitals b. (4 pts) The complex [io 2 ] 2, which contains two O 2 ligands, is linear and paramagnetic. Its absorption spectrum shows two peaks attributed to d-d transitions at 9000 and cm 1. Use your CFT result above to explain the origin of these transitions in terms of ground state and excited state electron configurations.

13 13 c. (5 pts) The extinction coefficient, ε, for the peaks in part b is about 5 M 1 cm 1. Compare the magnitude of ε to that for peaks in the spectra of generic octahedral complexes (circle your choice) and explain qualitatively why ε is similar (or different) to that in octahedral complexes. In comparison to ε values for octahedral complexes, this ε value is much smaller about the same much bigger

14 14 PERIODIC TABLE OF THE ELEMETS 1 H Li a K Rb Cs Fr 87 (223) Be Mg Ca 21 Sc Sr Ba Ra Y La Ac Ti Zr Hf Unq (261) V b Ta (262) Unp Cr Mn Mo Tc (98) W Unh Re Uns (263) (262) Fe Ru Os Uno (265) 27 Co Rh Ir Une (266) i Pd Pt Cu Ag Au Zn Cd Hg B Al Ga In Tl C Si P Ge As Sn Sb Pb Bi O Se Te Po 84 (209) S F 9 2 He e Cl Ar Br I At (210) 85 Kr Xe Rn (222) Ce Pr d Pm Sm Eu Gd Tb Dy Ho (145) Th Pa U p Pu Am Cm Bk Cf Es Er Tm Yb Lu Fm Md o Lr (237) (244) (243) (247) (247) (251) (252) (257) (258) (259) (260) SPECTROCHEMICAL SERIES Increasing (Fixed Metal) I < Br < Cl ~ *SC ~ 3 < F < urea < OH < CH 3 CO 2 < ox < H 2 O < *CS < EDTA < H 3 ~ py < en ~ tren < o-phen < *O 2 < H ~ CH 3 < C ~ CO ote: * denotes the ligating atom Increasing (Fixed Ligand) Mn 2+ < Co 2+ ~ i 2+ ~ Fe 2+ < V 2+ < Fe 3+ < Cr 3+ < V 3+ < Co 3+ < Mn 4+ < Mo 3+ < Rh 3+ < Ru 3+ < Ir 3+ < Re 4+ < Pt 4+ Spin-only magnetic moment formula µs = 2 [S(S+1)] 0.5 B.M. = [n(n+2)]0.5 B.M. (Bohr magnetons) Ligand abbreviations Cp = η 5 -C 5 H 5

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