$$N_2$$ 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⁺.

Those species which have unpaired electrons are called paramagnetic species.

And those species which have no unpaired electrons are called diamagnetic species.

(A) N₂⁺ has 13 electrons.

Moleculer orbital configuration of N₂⁺

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

Here 1 unpaired electrons present, so it is paramagnetic.

(B) Molecular orbital configuration of $$O_2^ - $$ (17 electrons) is

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

Here 1 unpaired electrons present, so it is paramagnetic.

(A) CO has 14 electrons.

Moleculer orbital configuration of CO

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

Here 0 unpaired electrons present, so it is diamagnetic.

(D) Molecular orbital configuration of NO (15 electrons) is

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

Here 1 unpaired electrons present, so it is paramagnetic.

In PO₄^3–, P atom has vacant d-orbitals, thus it can form p$$\pi $$ - d$$\pi $$ bond. ‘N’ and ‘C’ have no vacant ‘d’ orbital in their valence shell, so they cannot form such bond.