IR spectrum of an octahedral complex

[mycotheology]

I read that bond vibrations are only IR active if they produce a change in dipole moment. I'm trying to visualise how this works with chromium hexacarbonyl:

If I'm not mistaken, this compound has one significant IR absorption at around 1900 cm-1, which is a C-O stretching absorption. Is this produced by an asymmetrical stretch? For example, if the ligand at the top is stretched, but the ligand at the bottom is compressed (as in the bond length is temporarily shorter)? Thats the only way I can see the dipole changing. Then again, what about bending? If one of the C-O bonds bends so that it is no longer colinear with the M-C bond axis, that would change the dipole moment, wouldn't it?

Also, what about Jahn-Taller distortions? Would they cause a change in dipole moment? Because all the ligands are identical and the molecule is completely, I'm guessing Jahn-Taller distortions don't change the dipole moment of the molecule. Am I right?

 

[JohnRC]

The carbonyl stretch mode that you describe will indeed give a strong signal. It is a triply degenerate T_1u mode in the O_h symmetry of the complex.

As a low-spin Cr(0) complex, Cr(CO)₆ will have full octahedral symmetry, and there will be no question of Jahn-Teller distortion.

In full octahedral symmetry, only the T_1u modes will be infrared active. There are only 4 such modes for this complex; the others are both at much lower frequency and intensity. Two are bending modes, roughly equivalent to an umbrella type motion of the four equatorial ligands (out-of-plane co-ordinated L-M-L motion), and the other equivalent to an arm-waving motion of these ligands (out-of-plane co-ordinated O-C-M bending). The other is an asymmetric metal-ligand stretching mode for the pair of axial ligands. My guess (and it is a very rough guess) is that you will find them represented by very weak bands around 150, 550, and 250 cm^–1. Remember that CO is hardly polar at all, neither as a free molecule nor as a ligand.