Question

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

Answer 1

$C{N}^{-}$ has 14 electrons

$(\sigma 1s{)}^2,({\sigma }^{\ast }1s{)}^2,(\sigma 2s{)}^2,({\sigma }^{\ast }2s{)}^2,(\pi 2p{)}^4,(\sigma 2p{)}^2$

So, Bond Order$=\frac{1}{2}(a-b)$
where,

a is the number of electrons in bonding molecular orbitals and

b is the number of electrons in antibondng molecular orbitals.

Bond Order$=\frac{1}{2}(10-4)=3$.