Colorless nature of carbonyl complexes

Researching the coloration of transition metal complexes one day, I came across this reference on Wikipedia 1:

I am not sure why this is so. Take $\ce{Mo(CO)_6}$, or any other carbonyl complex of the metals in group 6. If we have a strong $\pi$ acceptor like carbonyl, and hence a low-spin $d^6$ system, would we not get an orbital splitting pattern like this:

The excitation of an electron to one of the empty $e_g$ orbitals seems feasible to me. Why can't one of those paired electrons absorb radiation to exhibit colour for the complex?

Reference

• Remember that going from t2g to eg in an octahedral complex is Laporte-forbidden, too.
– Jan
Commented Sep 19, 2015 at 17:42

Why can't one of those paired electrons absorb radiation to exhibit colour for the complex?

They can, they do and they color. The problem is, the corresponding transition requires so much energye, that the adsorption is well above upper range of visible spectrum in most cases. Linked article provides us with 3 optical spectra of $\ce{Cr}$, $\ce{Mo}$ and $\ce{W}$ carbonyls. For all adsoprtion mostly happens above 300 nm, with only small bits between 300 and 400 nm. Visible spectrum is between ~400 and ~700 nm. Since only bit of upper blue part of the spectrum is adsorbed, only a pale yellow coloring is observed.

This, of course, if you want so badly to apply crystal field theory to carbonyls, which does not work well in the case.

Because the transition need too much energy. If orbital relaxation is neglected, the energy required to excite an electron from the $j^{th}$ (occupied) orbital into the $i^{th}$ (unoccupied) orbital is

$\displaystyle \Delta E_{ij}$ $\textstyle =$ $\displaystyle (\epsilon_{i} - \epsilon_{j})$
$\textstyle -$ $\displaystyle \int \rho_i({\bf r}) \rho_j({\bf r}') v_{\rm E}({\bf r}-{\bf r}') \, {\rm d}{\bf r} \, {\rm d}{\bf r}'$
$\textstyle +$ $\displaystyle \delta_{s_is_j} \int \phi_i^*({\bf r}) \phi_j^*({\bf r}') \phi_j({\bf r}) \phi_i({\bf r}') f({\bf r}-{\bf r}') \, {\rm d}{\bf r} \, {\rm d}{\bf r}'$
$\textstyle +$ $\displaystyle \frac{1}{2} \int \rho_i({\bf r}) \rho_i({\bf r}') \left[ v_{\rm E... ...bf r}-{\bf r}')-f({\bf r}-{\bf r}')\right] \, {\rm d}{\bf r} \, {\rm d}{\bf r}'$
$\textstyle +$ $\displaystyle \frac{1}{2} \int \rho_j({\bf r}) \rho_j({\bf r}') \left[ v_{\rm E... ...r}-{\bf r}')-f({\bf r}-{\bf r}')\right] \, {\rm d}{\bf r} \, {\rm d}{\bf r}'\;,$

where $\rho_k = \vert\phi_k\vert^2$ is the charge density from the $k^{th}$ orbital.

So you need a lot of energy for the transition and because $E=hc/\lambda$, the wavelength will be so low that you can't see it. Therfor no colour.