Question

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3. For each of the following pairs of complexes, which will have the highest average co infrared stretching frequency? Circle your choice and briefly and clearly explain your reasoning a. TiO2(PMes)2(CO)2 or Cp2TI(CO)2 b. RhH(CO)(PPhs) or IrCl(o)(POMe)slz c. PtCl2(CO), or Ni(CO)2(PPha)2 d. [Co(CO)4] or [Fe(CO).]

Answer 1

> Ticz (rmes) (Co), Ti (+V) High oxidation state less back bonding to CO7* molecular oshital hence high ico. 2 Ti(CO), Ti(II) Low 0.5 more back bonding to со х¥ мә hence Low ico Io UI (60) [prome)]2 > Rh (41) Rh(+1) b) RhH(Co) (pph3) 4d series metal les back bonding due to good acceptor Pphs ligand. High ūco sd series metal more back bonding due to move cio / plome), ligand Heuce low oco > Nico c) ptcy (Co), Ni(CO), (PPhg)2 Pt (+2) less e-' density on metal more e density on hence low back bonding. metal hence more back So high o co bonding. So Low ū co d) [co (CO)A] > [fe(CO)A] ² - dlo (-1) do(-2) low oxidatim state High -ve óxidatim state less e density less back bonding, so more é density more ñ co high back bonding. So ł co law Scanned with CamScanner