Homework 2 Solution - Organometallic Chemistry 4571 HW 2...

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Organometallic Chemistry - 4571 ANSWER KEY HW # 2 Due: October 30, 2008 ( by 2 PM ) 1. (15 pts) For each of the following pairs of metal complexes, circle the one that will have the highest CO stretching frequency. Briefly and clearly discuss your reasoning for each case. The metal complex that has the least amount of electron density on the metal center (most electron deficient) will be the one with the highest IR CO stretching frequency. a) TiO 2 (PMe 3 ) 2 (CO) 2 -or- Cp 2 Ti(CO) 2 Although both complexes can be counted as 18 e- systems, the first Ti complex is d 0 and has no d electrons to -backbond to the CO ligand. It is, therefore, the most electron deficient and will have the highest IR CO stretching frequency (and the most labile CO ligand). The second Ti complex is in the +2 oxidation state and has a rather electron-rich d 2 configuration. b) RhH(CO)(PPh 3 ) 2 -or- IrCl(CO)[P(OMe) 3 ] 2 The Ir complex is more electron-deficient due to the poorly donating Cl and P(OMe) 3 ligands. The P(OMe) 3 ligands are also moderately good -backbonders. The Ir complex, therefore will have the higher CO stretching frequency. The Rh complex, on the other hand, has a strongly donating hydride and better donating PPh 3 ligands. The Ir is less electronegative than Rh and as a 3 rd row metal will bind more strongly to the CO, but these are relatively minor factors compared to the more dominate ligand effects. c) PtCl 2 (CO) 2 -or- Ni(CO) 2 (PPh 3 ) 2 The Pt complex is more electron-deficient due to the poorly donating Cl ligands, the +2 oxidation state, and d 8 electron configuration. The Ni complex is in the zero oxidation state ( d 10 ) and has moderately good donor PPh 3 ligands.

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