$N_2$ has 14 electrons.

Moleculer orbital configuration of $N_2$

= ${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}$

$\therefore\,\,\,\,$ N_b = 10

N_a = 4

$\therefore$ BO = ${1 \over 2}\left[ {10 - 4} \right]$ = 3

N₂⁺ has 13 electrons.

Moleculer orbital configuration of N₂⁺

= ${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^1}}$

$\therefore\,\,\,\,$ N_b = 9

N_a = 4

$\therefore$ BO = ${1 \over 2}\left[ {9 - 4} \right]$ = 2.5

As the bond order in N₂ is more than N₂⁺ so the dissociation energy of N₂ is higher than N₂⁺.