
Among the following metal carbonyls the $C - O$ bond order is lowest in
A. $[Mn{\left( {CO{)_6}} \right]^ + }$
B. $\left[ {{\text{Fe}}{{\left( {{\text{CO}}} \right)}_5}} \right]$
C. $\left[ {Cr{{\left( {CO} \right)}_6}} \right]$
D. ${\left[ {V\left( {CO} \right)6} \right]^ - }$
Answer
565.8k+ views
Hint: We know that higher the strength of metal carbonyl, weaker the $C - O$ bond will be. We also know that $CO$ is a strong field ligand and this gives a lot of stability to the complexes.
Complete step by step answer:
(a) For $[Mn{\left( {CO{)_6}} \right]^ + }$ we know that the charge for Carbon monoxide $(CO)$ is zero.
Therefore, the charge for $Mn$ will be +1.
The electron configuration of $M{n^ + }$ will be $3{d^5}4{s^1}$
But in the presence of $CO$ the effective configuration will be $3{d^6}4{s^0}$. This gives us three lone pairs for back bonding with a vacant orbital of $C$ In $CO$.
(b) For $\left[ {{\text{Fe}}{{\left( {{\text{CO}}} \right)}_5}} \right]$ the charge of $Fe$ is zero.
The electron configuration of $Fe$ will be $3{d^6}4{s^2}$
But in presence of $CO$ the effective configuration will be $3{d^8}$. This gives us 4 lone pairs for backing bonding with a vacant orbital of $C$ In $CO$.
(c) For $\left[ {Cr{{\left( {CO} \right)}_6}} \right]$ the charge of $Cr$ is zero.
The electron configuration of $Cr$ will be $3{d^5}4{s^1}$
But in presence of $CO$ the effective configuration will be $3{d^6}$. This gives us 3 lone pair for backing bonding with a vacant orbital of $C$ In $CO$.
(d) For ${\left[ {V\left( {CO} \right)6} \right]^ - }$ we know that the charge for Carbon monoxide $(CO)$ is zero.
Therefore, the charge for $V$ will be -1.
The electron configuration of ${V^ - }$ will be $3{d^4}4{s^2}$
But in presence of $CO$ the effective configuration will be $3{d^6}$. This gives us 3 lone pair for backing bonding with a vacant orbital of $C$ In $CO$.
From this we can observe that the $\left[ {{\text{Fe}}{{\left( {{\text{CO}}} \right)}_5}} \right]$ is having the maximum back bonding, hence the bond order will also be lowest for $\left[ {{\text{Fe}}{{\left( {{\text{CO}}} \right)}_5}} \right]$.
Note:
For solving this problem, we can use a shortcut also.
As we said that higher the strength of metal carbonyl, weaker the $C - O$ bond will be. From the options we can see that all of them has a coordination number of 6 except $\left[ {{\text{Fe}}{{\left( {{\text{CO}}} \right)}_5}} \right]$. In $\left[ {{\text{Fe}}{{\left( {{\text{CO}}} \right)}_5}} \right]$ we can see that the bond between $Fe$ and carbonyl group is distributed among 5 carbonyl group whereas the rest of them Is distributed among 6 carbonyl group. So the per bond strength of the $Fe$ will be greatest, which implies that $C - O$ bond will be weaker
Complete step by step answer:
(a) For $[Mn{\left( {CO{)_6}} \right]^ + }$ we know that the charge for Carbon monoxide $(CO)$ is zero.
Therefore, the charge for $Mn$ will be +1.
The electron configuration of $M{n^ + }$ will be $3{d^5}4{s^1}$
But in the presence of $CO$ the effective configuration will be $3{d^6}4{s^0}$. This gives us three lone pairs for back bonding with a vacant orbital of $C$ In $CO$.
(b) For $\left[ {{\text{Fe}}{{\left( {{\text{CO}}} \right)}_5}} \right]$ the charge of $Fe$ is zero.
The electron configuration of $Fe$ will be $3{d^6}4{s^2}$
But in presence of $CO$ the effective configuration will be $3{d^8}$. This gives us 4 lone pairs for backing bonding with a vacant orbital of $C$ In $CO$.
(c) For $\left[ {Cr{{\left( {CO} \right)}_6}} \right]$ the charge of $Cr$ is zero.
The electron configuration of $Cr$ will be $3{d^5}4{s^1}$
But in presence of $CO$ the effective configuration will be $3{d^6}$. This gives us 3 lone pair for backing bonding with a vacant orbital of $C$ In $CO$.
(d) For ${\left[ {V\left( {CO} \right)6} \right]^ - }$ we know that the charge for Carbon monoxide $(CO)$ is zero.
Therefore, the charge for $V$ will be -1.
The electron configuration of ${V^ - }$ will be $3{d^4}4{s^2}$
But in presence of $CO$ the effective configuration will be $3{d^6}$. This gives us 3 lone pair for backing bonding with a vacant orbital of $C$ In $CO$.
From this we can observe that the $\left[ {{\text{Fe}}{{\left( {{\text{CO}}} \right)}_5}} \right]$ is having the maximum back bonding, hence the bond order will also be lowest for $\left[ {{\text{Fe}}{{\left( {{\text{CO}}} \right)}_5}} \right]$.
Note:
For solving this problem, we can use a shortcut also.
As we said that higher the strength of metal carbonyl, weaker the $C - O$ bond will be. From the options we can see that all of them has a coordination number of 6 except $\left[ {{\text{Fe}}{{\left( {{\text{CO}}} \right)}_5}} \right]$. In $\left[ {{\text{Fe}}{{\left( {{\text{CO}}} \right)}_5}} \right]$ we can see that the bond between $Fe$ and carbonyl group is distributed among 5 carbonyl group whereas the rest of them Is distributed among 6 carbonyl group. So the per bond strength of the $Fe$ will be greatest, which implies that $C - O$ bond will be weaker
Recently Updated Pages
The number of solutions in x in 02pi for which sqrt class 12 maths CBSE

Write any two methods of preparation of phenol Give class 12 chemistry CBSE

Differentiate between action potential and resting class 12 biology CBSE

Two plane mirrors arranged at right angles to each class 12 physics CBSE

Which of the following molecules is are chiral A I class 12 chemistry CBSE

Name different types of neurons and give one function class 12 biology CBSE

Trending doubts
Which are the Top 10 Largest Countries of the World?

What are the major means of transport Explain each class 12 social science CBSE

Draw a labelled sketch of the human eye class 12 physics CBSE

Differentiate between insitu conservation and exsitu class 12 biology CBSE

The computer jargonwwww stands for Aworld wide web class 12 physics CBSE

State the principle of an ac generator and explain class 12 physics CBSE

