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

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Question

(A) In metal carbonyls,the metal-carbon bond involves both $\sigma$-donation from $CO$ to the metal and $\pi$-back-bonding from the metal to the $CO$

Vedclass · Accepted Answer

$[V(CO)_6] < [V(CO)_6]^-$

Vedclass · Answer

(A) In metal carbonyls,the metal-carbon bond involves both $\sigma$-donation from $CO$ to the metal and $\pi$-back-bonding from the metal to the $CO$ $\pi^*$ antibonding orbital.As the negative charge on the complex increases,the electron density on the metal increases.This leads to greater $\pi$-back-bonding from the metal to the $CO$ ligand.Increased $\pi$-back-bonding populates the antibonding $\pi^*$ orbital of $CO$,which weakens the $C-O$ bond and increases its bond length.Since $[V(CO)_6]^-$ has a higher negative charge than $[V(CO)_6]$,it exhibits more $\pi$-back-bonding.Therefore,the $C-O$ bond length in $[V(CO)_6]^-$ is greater than in $[V(CO)_6]$,which means $[V(CO)_6] < [V(CO)_6]^-$.

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

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