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

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(C) In $V(CO)_6$, the vanadium atom is in a neutral oxidation state. In $[V(CO)_6]^-$, the vanadium atom is in a $-1$ oxidation state.Due to the negat

Vedclass · Accepted Answer

$200, 193$

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(C) In $V(CO)_6$, the vanadium atom is in a neutral oxidation state. In $[V(CO)_6]^-$, the vanadium atom is in a $-1$ oxidation state.Due to the negative charge on the complex $[V(CO)_6]^-$, there is an increased back-bonding from the metal $d$-orbitals to the $\pi^*$ antibonding orbitals of the $CO$ ligands.This increased back-bonding strengthens the $V-C$ bond (increasing its double bond character) and weakens the $C-O$ bond.Consequently, the $V-C$ bond length in $V(CO)_6$ is approximately $200 \ pm$, whereas in $[V(CO)_6]^-$, the increased electron density on the metal facilitates stronger back-donation, resulting in a shorter $V-C$ bond length of approximately $193 \ pm$.

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

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