The hybridization of N in  ${\mathrm{NO}}_2^{+},{\mathrm{NO}}_3^{-}$ and ${\mathrm{NH}}_4^{+}$ respectively are

Solution:

${\mathrm{NO}}_2^{+}$ has 16 valence electrons. There arrangement will be $:\underset{..}{\overset{..}{o}}- \mathrm{N} = \underset{..}{\overset{..}{o}}$ Two pairs of electrons around N implies linear geometry involving sp hybridization. NO3 has 24 (= 5 x 3 x 6 + 1) valence electrons. These are arranged as   

There are three pairs of electrons around N. Its geometry will be trigonal planar and will result from ${\mathrm{sp}}^2$ hybridization. ${\mathrm{NH}}_4^{+}$ has  ${\mathrm{sp}}^3$ hybridization.