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So there was a question in coordination chemistry regarding the nature of $\ce{[Cr(CO)_6]}$

The book had the question under VBT and solved it using "strong ligands cause all electrons to pair up". $$\ce{[Ar] 3d^5 4s^1\rightarrow [Ar] 3d^6}\text{ and all electrons paired}$$ Book solution

Later we learnt about Crystal Field Theory and splitting in octahedral complexes. From that the configuration of the above complex should be $$\ce{3d_{t_{2g}}^5 \ 3d_{e_g}^0\ 4s^1}$$

My answer was paramagnetic and outer orbital complex.

So why does electron from $\ce{4s}$ orbital pair up with $\ce{4d}$ electron and make the complex diamagnetic? I only know VBT and CFT for coordinate chemistry, so how is this explained by CFT?

I didn't find much good of an explanation on the internet except that $\ce{s}$-orbitals must be emptied to take part in hybridization. But then why is it wrong to use the concept of hybridization for transition metal complexes? My personal guess is maybe the splitting energy is so high for $\ce{4s}$ that somehow lower energy gap/mixes with $\ce{3d}$ orbital.

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    $\begingroup$ Due to the strong-field nature of the CO ligand, chromium hexacarbonyl is low-spin. Cr has 6 d-electrons (the d-orbitals fill first for all transition metals), which are all 6 in t2g and paired. $\endgroup$
    – Mäßige
    Mar 16 at 14:06
  • $\begingroup$ @Mäßige According to n+l rule 4s is filled before 3d right and ground state configuration of Cr is exceptionally $3d^5\ 4s^1$ and ligands can rearrangement of electrons in d-orbitals by splitting it can't force s orbital (non-directional) to d-orbital, right? $\endgroup$
    – Aurelius
    Mar 16 at 14:15
  • $\begingroup$ The n+l rule strictly applies to main group elements. The general consensus is that the transition metals’ 3d-orbitals are completely filled before the 4s orbitals in transition metal complexes. This is due to the lowered energy as per the increased effective nuclear charge, because of the electron withdrawing nature of the ligands relative to the central atom. You would be right, if the chromium center would be in the gas phase and an isolated atom without ligands. $\endgroup$
    – Mäßige
    Mar 17 at 6:17
  • $\begingroup$ @Mäßige Sorry to bother. But "You would be right if ........gas phase...isolated atom...." So The presence of ligands and/or water increases the energy of 4s ? So how does that happen? Also if 3d is filled before 4s, then $[Cr(H_2O)_6]$ if it exists will also have an empty s orbital?( irrespective of ligand strength) $\endgroup$
    – Aurelius
    Mar 18 at 20:45
  • $\begingroup$ Yes this complex would also have an empty s orbital, only the 3d orbitals would be filled. The reason is described in this question: chemistry.stackexchange.com/questions/33297/… $\endgroup$
    – Mäßige
    Mar 19 at 6:17

1 Answer 1

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The energy gap between 4s and 3d is very low. This, primarily is the reason for transition metals exhibiting a huge number of oxidation states. In the case of $[Cr(CO)6]$, carbonyl group acts as a strong ligand. The CFSE can be calculated by

$$CFSE = Δo[0.6 n(eg)- 0.4 n(t2g)]$$

Where n(eg) = no of electrons in eg orbital

And n(t2g) = no of electrons in t2g orbital.

In this case, Carbonyl group creates a very strong field. As we know for achieving stability, the energy possessed has to be the least, pairing takes place, which considerably provides a more stable compound i.e. a low spin complex. Thus, $[Cr(CO)6]$ is a low spin where there are 5 t2g electrons.

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  • $\begingroup$ But s orbital in non-directional, so how does ligand repulsion affect it? Strong or weak , there should be no effect of ligand repulsion on a non-directional orbital, right? $\endgroup$
    – Aurelius
    Mar 18 at 20:37
  • $\begingroup$ Repulsion affects s orbital before d orbital because it is in valence shell. D is in penultimate shell, so first electrons migrate from s to d then splitting takes place because the interaction of ligands is limited to valence shell. $\endgroup$ Mar 25 at 1:29

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