Question

Which of the following has the lowest ${\text{C}} - {\text{O}}$ bond order?
A. $\left[ {{\text{Ni}}{{\left( {{\text{CO}}} \right)}_{\text{4}}}} \right]$
B. ${\left[ {{\text{Co}}{{\left( {{\text{CO}}} \right)}_{\text{4}}}} \right]^ - }$
C. ${\left[ {{\text{Fe}}{{\left( {{\text{CO}}} \right)}_{\text{4}}}} \right]^{2 - }}$
D. ${\left[ {{\text{Mn}}{{\left( {{\text{CO}}} \right)}_{\text{6}}}} \right]^ + }$

Answer 1

Hint: The ${\text{CO}}$ bond order in metal carbonyls is inversely proportional to the $\pi $-back bonding. Thus, higher the $\pi $-back bonding lower is the ${\text{CO}}$ bond order.

Complete step by step answer:
In the compounds $\left[ {{\text{Ni}}{{\left( {{\text{CO}}} \right)}_{\text{4}}}} \right]$, ${\left[ {{\text{Co}}{{\left( {{\text{CO}}} \right)}_{\text{4}}}} \right]^ - }$, ${\left[ {{\text{Fe}}{{\left( {{\text{CO}}} \right)}_{\text{4}}}} \right]^{2 - }}$ and ${\left[ {{\text{Mn}}{{\left( {{\text{CO}}} \right)}_{\text{6}}}} \right]^ + }$, a metal atom lies at the center of and the ${\text{CO}}$ ligands surround the central metal atom. These compounds are known as metal carbonyls. In these compounds, the ${\text{CO}}$ ligand is a neutral ligand which has no charge.
The carbonyl carbon donates a lone pair of electrons to the vacant d-orbital of the metal. Thus, a sigma $\sigma $ bond is formed.
As the carbonyl carbon donates a lone pair of electrons to the metal, the metal becomes rich in electrons. To compensate for the high electron density, the filled d-orbital of the metal donates electrons to the vacant antibonding ${\pi ^ * }$ orbital of ${\text{CO}}$.
The $\pi $-back bonding increases as the $\sigma $ donation of the carbonyl carbon to the central metal atom increases. Thus, the strength of the bond between the carbonyl ligand and the central metal atom increases.
The most electron rich metal amongst $\left[ {{\text{Ni}}{{\left( {{\text{CO}}} \right)}_{\text{4}}}} \right]$, ${\left[ {{\text{Co}}{{\left( {{\text{CO}}} \right)}_{\text{4}}}} \right]^ - }$, ${\left[ {{\text{Fe}}{{\left( {{\text{CO}}} \right)}_{\text{4}}}} \right]^{2 - }}$ and ${\left[ {{\text{Mn}}{{\left( {{\text{CO}}} \right)}_{\text{6}}}} \right]^ + }$ is ${\text{Fe}}$ in $( - 2)$ as it has the highest negative charge. Thus, the electron rich ${\text{Fe}}$ metal donates more electrons to the vacant antibonding ${\pi ^ * }$ orbital of ${\text{CO}}$. Thus, the $\pi $-back bonding is highest in ${\left[ {{\text{Fe}}{{\left( {{\text{CO}}} \right)}_{\text{4}}}} \right]^{2 - }}$.
The ${\text{CO}}$ bond order in metal carbonyls is inversely proportional to the $\pi $-back bonding. Thus, higher the $\pi $-back bonding lower is the ${\text{CO}}$ bond order.
Thus, ${\left[ {{\text{Fe}}{{\left( {{\text{CO}}} \right)}_{\text{4}}}} \right]^{2 - }}$ has the lowest ${\text{C}} - {\text{O}}$ bond order.
Thus, the correct option is option (C).

Note: The metal carbonyl which has the highest negative charge is rich in electron density. Thus, the $\pi $-back bonding is higher.