Question

What is the bond order of $C{N^ + }$?

Answer 1

Bond order is defined as half the difference between the number of electrons in the bonding molecular orbitals and the number of electrons in the antibonding molecular orbitals. $C{N^ + }$ has a total of $12$ electrons and on considering the bonding and anti bonding orbitals from the configuration we can calculate the bond order.

Complete step by step answer:
Bond order of a compound is defined as half the difference between the number of electrons in the bonding molecular orbitals of the compound which is represented by $a$ and the number of electrons in the antibonding molecular orbitals of the compound which is represented by $b$.
Bond order = $\dfrac{1}{2}(a - b)$
Bond order of a compound also represents the number covalent bonds in a compound. It represents the strength of the bonds and is positive for stable bonds. It finds its applications in the valence bond theory. The electrons in the bonding orbitals stabilise the compound and the electrons in the antibonding orbitals make the compound less stable.
For the above question,
$C{N^ + }$ has a total of $12$ electrons which include $6$ of carbon atoms and $7$ of nitrogen atoms and it has a $- 1$electron because of the positive charge on the compound.
To calculate the bond order we require the configuration which is
$(\sigma 1s),{(\sigma^* 1s)^2},{(\sigma 2s)^2},{(\sigma^* 2s)^2},{(\pi )^4}$
Thus, the electrons in bonding orbitals = $8$ and in antibonding orbitals = $4$
We know that,
Bond order = $\dfrac{1}{2}(a - b)$
$\Rightarrow \dfrac{1}{2}(8 - 4)$

Therefore, bond order of $C{N^ + }$= $2$.

Note:
Cyanide is a chemical compound with a triple bond between carbon and nitrogen atoms. It forms salts with potassium and sodium which are highly toxic. It is extremely harmful for the cells of the human body as they prevent them from absorbing oxygen and it also affects the heart, brain and other organs.