0
$\begingroup$

I am not sure what is the electron configuration in benzene chromium tricarbonyl?

From what I can work out, Cr is in 0 oxidation state (since all ligands are neutral). The complex is tetrahedral so likely high spin. But the CO ligand is strong field and so is more likely to cause low spin. There are three of these ligands and one weak field benzene ligand. I'm confused by the conflicting information.

$\endgroup$

1 Answer 1

2
$\begingroup$

When there are strong pi interactions, a tetrahedral geometry no longer corresponds to weak splitting. With tetrahedral geometry the splitting is small between sigma orbitals but large between pi orbitals. So with the strong pi interactions from carbon monoxide and benzene that tetrahedral geometry can indeed impart strong splitting and a low-spin configuration. Add the fact that we have eighteen net valence electrons around the metal (six from the core atom, six from the benzene ring, two apiece from the three carbon monoxide ligands), which most strongly stabilizes the low-spin configuration, and low-spin becomes the natural choice. And the real one.

$\endgroup$
4
  • $\begingroup$ So this will leave one unpaired electron for the metal (two in $d_{x^2-y^2}$, two in $d_{z^2}$ and one in one of the $t_2$ orbitals), making it ever so slightly paramagnetic? $\endgroup$
    – b20
    Dec 29, 2023 at 0:15
  • $\begingroup$ No, there are no unpaired electrons. When you meet the 18-electron rule as I describe,you are typically low-spin and the low-spin cilonfiguration has no unpaired electrons. $\endgroup$ Dec 29, 2023 at 0:26
  • $\begingroup$ Chromium is $3d^5 4s^1$ so I'm confused how the electrons pair? Does the 4s electron pair in a $t_2$ orbital with the fifth $d$ electron? $\endgroup$
    – b20
    Dec 29, 2023 at 0:30
  • $\begingroup$ The s electron mixes in with the d. Then you have three pairs if electrobs in the mixed orbitals and six pairs from the ligands. $\endgroup$ Dec 29, 2023 at 0:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.