Question

The strongest ${\text{ - CO}}$ bond is present in :
A) ${{\text{[Cr(CO}}{{\text{)}}_{\text{6}}}{\text{]}}^{\text{ + }}}$
B) $[{\text{Fe(CO}}{{\text{)}}_{\text{5}}}{\text{]}}$
C) ${{\text{[V(CO}}{{\text{)}}_{\text{6}}}{\text{]}}^ - }$
D) All have equal strength

Answer 1

Carbonyl is $\pi$ acceptor ligand. It has a tendency to accept the electrons from the metal. The $\pi$ back bonding tendency of metal depends on the electron cloud around the metal. Complex having lower metal to ligand $\pi$ back bonding has the strongest ${\text{ - CO}}$bond.

Complete solution:
Carbonyl has a tendency to accept electrons from metal so it is known as $\pi$ acceptor ligand. So in complexes having carbonyl ligands we observe ${\text{M}} \to {\text{L}}$ charge transfer.
This electron transfer tendency of metal depends on the electron cloud around the metal. The greater the electron cloud around the metal the greater is ${\text{M}} \to {\text{L}}$ charge transfer.
Metals having a higher oxidation state have lower electrons around it so it has a lower tendency of ${\text{M}} \to {\text{L}}$ charge transfer. The strength of ${\text{ - CO}}$the bond increases with a decrease in $\pi$ back bonding.
To determine which complex has the strongest ${\text{ - CO}}$bond we will determine the oxidation state of metal in each complex.
In all complexes, the only carbonyl is the ligand. We know that carbonyl is a neutral ligand so the oxidation state of the metal is nothing but the charge on the complex.
${{\text{[Cr(CO}}{{\text{)}}_{\text{6}}}{\text{]}}^{\text{ + }}}$
Here, the charge on the complex is +1 so the oxidation state of ${\text{Cr}}$ is +1.
$[{\text{Fe(CO}}{{\text{)}}_{\text{5}}}{\text{]}}$
Here, the charge on the complex is 0 so the oxidation state of ${\text{Fe}}$ is 0.
${{\text{[V(CO}}{{\text{)}}_{\text{6}}}{\text{]}}^ - }$
Here, the charge on the complex is -1 so the oxidation state of ${\text{V}}$ is -1.
As ${\text{Cr}}$ has a positive oxidation state we can say that it has lower electron density so it has a lower ${\text{Cr}} \to {\text{ CO}}$ charge transfer tendency than ${\text{Fe}} \to {\text{ CO}}$ and ${\text{V}} \to {\text{ CO}}$.
The increased order of electron transfer tendency of given metal is as follows:
${\text{Cr}} \to {\text{ CO}} < {\text{Fe}} \to {\text{ CO}} < {\text{V}} \to {\text{ CO}}$
So, the strongest ${\text{ - CO}}$bond is present in${{\text{[Cr(CO}}{{\text{)}}_{\text{6}}}{\text{]}}^{\text{ + }}}$ the complex.

Thus, the correct options are (A).

Note: ${\text{M - CO}}$ bond strength increases with a decrease in $\pi$ back bonding tendency. A $\pi$ back bonding tendency of the metal decreases with an increase in the oxidation state. So carbonyl complexes having greater positive charge show stronger ${\text{M - CO}}$ bond.