Question

a)
how many CO stretching absorption are in the IR spectrum of fac-(MeCN)3Cr (CO)3

b) Which irreducible representations correspond to these absoprtions?

Answer 1

(a) There are total two CO stretching absorption frequencies in the IR spectrum of .

That variation one is due to Charge on central Cr metal ion and another is due to the effect of other ligand

In the compound the central metal ion has charge. As the electron density on the Cr metal atom increase, more pi‐back‐bonding to the CO ligands occur. This back bonding weakens the C–O bond. Then rest electron density is pushed into the empty pi* anti‐bonding carbonyl ligand orbital. This results in the Cr–C bond order more than 1 and the C‐O bond order less than 2. That is, the resonance structure becomes more dominant. More back bonding results in more lowering of the C=O bond order and simultaneously more lower CO stretching frequency. The compound show CO two stretching frequencies around 1462 and 1657 cm^-.

The structure of compound is octahedral. In octahedral symmetry the transition splits into e and t2 energy levels. In which the and components coupled to a T1 irreducible representation. ​According to CO groups octahedral symmetry the matrices will be formed as 3 x (1 +3+3).