Question

In which complex is the $C-O$ bond length maximum?

• A. $V(CO{)}_6^{-}$
• B. $Cr(CO{)}_6$
• C. $Mn(CO{)}_6^{+}$
• D. $Fe(CO{)}_6^{2+}$

Answer 1

The correct option is A $V(CO{)}_6^{-}$

Carbonyl is a neutral ligand which means the charge on the ion is the charge on the central metal atom.
Hence, we can say that the charges are as:
$V(CO{)}_6^{-}:V^{-}$ = -1
$Cr(CO{)}_6:C{r}^0$ = 0
$Mn(CO{)}_6^{+}:M{n}^{+}$ = +1
$Fe(CO{)}_6^{2+}:F{e}^{2+}$ = +2
Carbonyl has a tendency to form back bonding with the central atom and the extent of back bonding depends on the charge on the central atom. If the electron density or negative charge density on the central atom is higher, the degree of back bonding will be higher. As the strength of back bonding increases, the $M-C$ bond strength increases. This means the $M-C$ bond length decreases. As a result of this, the $C-O$ bond strength decreases and hence the $C-O$ bond length increases.

Amongst our options, we can see $V$ is the only central atom which has negative charge and hence it has the highest negative charge density. This means that its back bonding with carbonyl group will be strongest and hence the $C-O$ bond strength will be least hence the bond length will be highest.