Question

Among the following metal carbonyls, CO bond order is lowest in
A. ${\left[ {Mn{{\left( {CO} \right)}_6}} \right]^{ + \,\,}}$
B. $\left[ {\,Fe{{\left( {CO} \right)}_5}} \right]\,$
C. $\left[ {\,Cr{{\left( {CO} \right)}_6}} \right]$
D. $\left[ {V{{\left( {CO} \right)}_6}} \right]$

Answer 1

Hint: Whenever there is a metal and electrons in ligands, then there occurs a back bonding ,which strengths the $M - CO$ bond but it weakens the $C - O$ carbonyl by pumping more electron into the empty ${\pi _{{y^ * }}}\,or\,{\pi _{{z^ * }}}$molecular orbital.

Complete step by step answer:
The structure of mononuclear carbonyls has been studied by X-ray diffraction, electron diffraction, and electron diffraction. All the mononuclear carbonyls have linear $M - CO$ bonds in which CO group is linked to the metal atom through the carbon atom, since oxygen atom is more electronegative than carbon.
All the mononuclear carbonyls $M{\left( {CO} \right)_6}\,\,\,\,\left( {M - V,Cr,Mo,W} \right)$ with M having coordination number (CN) equal to 6 have octahedral shape ,pentacarbonyl $M{\left( {CO} \right)_5}\,\,\,\,\left( {M - Fe,Ru,Os} \right)$with coordination number (CN) equal to 5 have trigonal bipyramidal shape and $Ni{\left( {CO} \right)_4}$(CN=4) has tetrahedral shape.
In case of $M{\left( {CO} \right)_5}\,\,\,\,and\,\,M{\left( {CO} \right)_6}$type carbonyls $M \leftarrow CO\,\,\sigma $bond results by the overlap between vacant $ds{p^3}\,and\,{d^2}s{p^3}$hybrid orbital on the metal atom and filled $sp$hybrid orbital on C-atom of CO molecule the formation of $M \leftarrow CO\,\,\sigma $-bond ,metal atom acts as an acceptor while $CO$molecule acts as a donor.
In the formation of $M \leftarrow CO\,\,\pi $-bonds in $M{\left( {CO} \right)_6}$carbonyls metal atom acts as a donor while $CO$molecule acts as an acceptor. Since $CO$molecule receives back donated electron from the metal atoms in low or zero oxidation state into its empty ${\pi _{{y^ * }}}\,or\,{\pi _{{z^ * }}}$molecular orbital. This $M \leftarrow CO\,\,\pi $-bond is also called back bonding.
As the electron density increases on the metal centre then there will be more -backconding to $CO$will take place. This weakens the $C - O$ bond strength in carbonyl by pumping more electron into the empty ${\pi _{{y^ * }}}\,or\,{\pi _{{z^ * }}}$molecular orbital. This increases the $M - CO$ strength making it more double bond like.
Electronic configuration of central atom-
$\begin{array}{l}
M{n^ + } = 3{d^5}4{s^1}\\
F{e^0} = 3{d^5}4{s^2}\\
C{r^0} = 3{d^5}4{s^1}\\
{V^ - } = 3{d^4}4{s^2}
\end{array}$
Due to the highest electron density in Case of $Fe$,so highest back bonding will be there therefore CO bond order is lowest here.

Therefore, the correct option is option (B).

Note:
As the electron density increases on the metal centre then there will be more -backconding to $CO$ will take place. This weakens the $C - O$ bond strength in carbonyl by pumping more electron into the empty ${\pi _{{y^ * }}}\,or\,{\pi _{{z^ * }}}$molecular orbital. This increases the $M - CO$ strength making it more double bond like.