Lowest C-O Bond Order in Metal Carbonyls

In metal carbonyls, bonding involves synergic bonding:

1. $\sigma$-donation: electron density from the carbonyl carbon lone pair is donated into an empty $d$-orbital of the metal.
2. $\pi$-backbonding: filled $d$-orbitals of the metal donate electron density into the empty $\pi^*$ antibonding molecular orbitals of $\text{CO}$.

Greater electron density on the central metal ion increases $\pi$-backdonation into the $\pi^*$ antibonding orbital of $\text{CO}$, weakening the $\text{C}-\text{O}$ bond and lowering its bond order.

Among $[\text{Mn(CO)}_6]^+$, $[\text{Fe(CO)}_5]$, $[\text{Cr(CO)}_6]$, and $[\text{V(CO)}_6]^-$, vanadium has a negative oxidation state ($-1$), giving it the highest electron density on the metal center. Consequently, $[\text{V(CO)}_6]^-$ exhibits maximum $M \rightarrow \text{CO}$ $\pi$-backdonation, making the $\text{C}-\text{O}$ bond the weakest and yielding the lowest $\text{C}-\text{O}$ bond order.

Correct Answer: D