Which of the following has the longest C–O bond length? (Free C–O bond length in $\mathrm{CO}$ is $1.128$ Å.)

Solution & Explanation

\textbf{Answer:} (C) $[\mathrm{Fe}(\mathrm{CO}){}_4]{}^{2-}$ has the longest C–O bond. \textbf{Solution:} In metal carbonyls the M–CO bond is synergic: the ligand donates a σ lone pair to the metal, and the metal donates electron density back from its filled d-orbitals into the empty antibonding π* orbitals of CO (π back-bonding). The more electron density pushed into the C–O π* orbital, the lower the C–O bond order and the longer the C–O bond. The greater the negative charge (lower the oxidation state) on the metal centre, the higher its electron density and the stronger the M→CO back-donation. Ranking the metal charge: $[\mathrm{Fe}(\mathrm{CO}){}_4]{}^{2-}$ (Fe in $-2$) < $[\mathrm{Co}(\mathrm{CO}){}_4]{}^{-}$ ($-1$) < $\mathrm{Ni}(\mathrm{CO}){}_4$ ($0$) < $[\mathrm{Mn}(\mathrm{CO}){}_6]{}^{+}$ ($+1$). Thus $[\mathrm{Fe}(\mathrm{CO}){}_4]{}^{2-}$, with the most negative metal centre, has the maximum back-bonding, the lowest C–O bond order, and therefore the longest C–O bond length.