Question

What is the bond-order of CN?
[Note: Write the answer after multiplying it with 4]

Answer 1

Hint:Bond order gives the information about the number of bonds which are present in between two atoms in molecules. There is a formula to calculate the bond order and it is as follows:
\[\text{Bond order=}\frac{\text{1}}{\text{2}}\text{ }\!\![\!\!\text{ B-A }\!\!]\!\!\text{ }\]
Here B = number of electrons in bonding orbital
A = number of electrons in antibonding orbital

Complete step-by-step answer:- In the question it is asked about the bond order of CN.
- We should know the molecular electronic configuration of the CN molecule to calculate the bond order.
- The number of electrons in carbon is 6 and the number of electrons in nitrogen is 7.
- Therefore the CN molecule contains 13 electrons.
- The molecular electronic configuration of CN molecules is as follows.
\[\sigma 1{{s}^{2}}{{\sigma }^{*}}1{{s}^{2}}\sigma 2{{s}^{2}}{{\sigma }^{*}}2{{s}^{2}}\pi 2p_{y}^{2}=\pi 2p_{z}^{2}\sigma 2p_{x}^{1}\]
- From the above electronic configuration we can say that the number electrons in bonding orbitals are 9 and the number of electrons in antibonding orbital are 4.
- Substitute the above known values in the below formula to get the bond order of the CN molecule.
\[\text{Bond order=}\frac{\text{1}}{\text{2}}\text{ }\!\![\!\!\text{ B-A }\!\!]\!\!\text{ }\]
Here B = number of electrons in bonding orbital = 9
A = number of electrons in antibonding orbital = 4
\[
 \text{Bond order=}\frac{\text{1}}{\text{2}}\text{ }\!\![\!\!\text{ B-A }\!\!]\!\!\text{ } \\
\text{=}\frac{1}{2}[9-4] \\
 =2.5 \\
\]
- In the question it is given that to multiply the bond order with 4.
- Therefore bond order = (4) (2.5) = 10.

Note:By using bond order we can easily find the number of bonds which are present in between two atoms. But we should know information about the number of bonding and antibonding electrons present in the given molecule.