Question

The $\pi$ acid ligands donate their lone pairs to the metal to form a normal $\sigma$ bond with the latter in addition to it, the vacant orbitals accept $e^{-}$ from the filled metal orbitals to form a type of $\pi -$bond which supplements the $\sigma$ bond.

Which of the following has lowest M-C bond length ?

• A. $\left[Ni\right(CO{)}_4]$
• B. $\left[Co\right(CO{)}_4{]}^{-}$
• C. $\left[Fe\right(CO{)}_4{]}^{2-}$
• D. $\left[Mn\right(CO{)}_6{]}^{+}$

Answer 1

The correct option is C $\left[Fe\right(CO{)}_4{]}^{2-}$

In metal carbonyls, CO acts a $\pi$ -ligand ie it transfers electrons to LUMO (Lowest Unoccupied Molecular Orbital) of the metal. Since LUMO is Anti-bonding orbital, more the electron density in it, less will be the Bond Order (BO).

$BondOrder=\frac{Bondingelectrons+Anti-bondingElectrons}{2}$

So the complex with most effective electron donation, will have the least Bond Order. The most effective electron donation will be in the metal which is most electron deficit.

In short,

$BondOrder\propto \frac{1}{PositiveChargeonthemetal}\propto NegativeCharge$

$BondLength\propto PositiveChargeonthemetal\propto \frac{1}{NegativeChargeonthemetal}$

In the given options, $\left[Fe\right(CO{)}_4{]}^{2-}$ has least postive charge on the metal ie -2 , hence least bond length.