Question

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

Answer 1

$N{O}^{+}$ has $14$ electrons.
$(\sigma 1s{)}^2,({\sigma }^{\ast }1s{)}^2,(\sigma 2s{)}^2,({\sigma }^{\ast }2s{)}^2,(\pi {)}^4,(2p_z{)}^2$
So, Bond Order$=\frac{1}{2}(a-b)$
Here, 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$