Bond angle is maximum in NH₄⁺ tetrahedral molecule with bond angle 109°.

As X and Y both are electronegative elements thus both attracts the electron density from H thus, electron density on H decreases and on X it increases.

Compounds with same shape and same hybridisation are known as isostructural.

XeF₂, IF₂^- $$ \to $$ both are sp³ hybridised linear molecules.

$$N_2$$ 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⁺.