Question

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

Answer 1

$C{N}^{+}$ has 12 electrons (6 from carbon, 7 for nitrogen, -1 for positive charge). So, we have configuration
$(\sigma 1s{)}^2,({\sigma }^{\ast }1s{)}^2,(\sigma 2s{)}^2,({\sigma }^{\ast }2s{)}^2,(\pi {)}^4$
So, bond order$=\frac{1}{2}(a-b)$
a is the number of electrons in bonding molecular orbitals and b is the number of electrons in antibondng molecular orbitals.
Bond Order$=\frac{1}{2}(8-4)$ $=2$.