The hybridization of orbitals of N atom in $\mathrm{NO}_{3}^{-}, \mathrm{NO}_{2}^{+}$ and $\mathrm{NH}_{4}^{+}$ are respectively :

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

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

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

JEE Main · 2011 · 10 Apr 2011 · Offline · Chemical Bonding and Molecular Structure

The hybridization of orbitals of N atom in

NO_{3}^{-}, NO_{2}^{+}

and

NH_{4}^{+}

are respectively :

Answer: (A)

Count the sigma framework on nitrogen in each species. In

N O_{3}^{-}

nitrogen forms three sigma bonds and is trigonal planar, so it is

s p^{2}

. In

N O_{2}^{+}

it forms two sigma bonds and is linear, so it is

s p

. In

N H_{4}^{+}

it forms four sigma bonds and is tetrahedral, so it is

s p^{3}

. The hybridisations are therefore

s p^{2}

s p

and

s p^{3}

respectively.

JEE Main · 2011 · 10 Apr 2011 · Offline · Single correct option