The hybridisations of the atomic orbitals of nitrogen in $\mathrm{NO}_{2}^{-}, \mathrm{NO}_{2}^{+}$ and $\mathrm{NH}_{4}^{+}$ respectively are:

(A) $\mathrm{sp}^{3}, \mathrm{sp}$ and $\mathrm{sp}^{2}$

JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Chemical Bonding and Molecular Structure

The hybridisations of the atomic orbitals of nitrogen in

NO_{2}^{-}, NO_{2}^{+}

and

NH_{4}^{+}

respectively are:

Answer: (C)

The hybridisation follows from the number of

σ

bonds plus lone pairs on nitrogen. In

N O_{2}^{-}

nitrogen has two

σ

bonds and one lone pair, so three electron domains and

s p^{2}

hybridisation with a bent shape. In

N O_{2}^{+}

that lone pair has been removed, leaving two

σ

bonds and no lone pair, so two domains,

s p

hybridisation and a linear ion. In

N H_{4}^{+}

nitrogen has four

σ

bonds and no lone pair, so four domains,

s p^{3}

hybridisation and a tetrahedral ion. The hybridisations are

s p^{2}

s p

and

s p^{3}

respectively.

JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Single correct option