The V-C distance in [V(CO)_6] and [V(CO)_6]^- | vedclass

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(C) In the neutral complex $[V(CO)_6]$, the vanadium atom is in a lower oxidation state compared to the anionic complex $[V(CO)_6]^-$. However, the ex

Vedclass · Accepted Answer

$200, 193$

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(C) In the neutral complex $[V(CO)_6]$, the vanadium atom is in a lower oxidation state compared to the anionic complex $[V(CO)_6]^-$. However, the extent of back-bonding from the metal $d$-orbitals to the $\pi^*$ orbitals of the $CO$ ligand is significant in both. Experimental data shows that the $V-C$ bond distance in $[V(CO)_6]$ is approximately $200 \ pm$. In the anionic complex $[V(CO)_6]^-$, the increased electron density on the metal enhances the back-bonding, which strengthens the $V-C$ bond and shortens the bond length to approximately $193 \ pm$. Therefore, the distances are $200 \ pm$ and $193 \ pm$ respectively.

The $V-C$ distance in $[V(CO)_6]$ and $[V(CO)_6]^-$ are respectively (in $pm$):

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