The hybridization of atomic orbitals of nitrogen in  ${\mathrm{NO}}_2^{+},{\mathrm{NO}}_3^{-}$ and ${\mathrm{NH}}_4^{+}$ ions are respectively.

To predict the hybridization,

$$\begin{gathered} \mathrm{no}. \mathrm{of} \mathrm{H}.\mathrm{O}=\frac{1}{2}\mathrm{n}\left(\mathrm{o}. \mathrm{of} \mathrm{valance} {\mathrm{e}}^{-}\mathrm{s} \mathrm{on} \mathrm{central} \mathrm{atom} \\ +\mathrm{no}. \mathrm{of} \mathrm{monovalent} \mathrm{atoms}-\mathrm{charge}\right) \\ \mathrm{For} {\mathrm{NO}}_2^{+}, \mathrm{no}. \mathrm{of} \mathrm{H}.\mathrm{O}=\frac{1}{2}\left(5-1\right)=2\to \mathrm{sp} \\ \mathrm{For} {\mathrm{NO}}_3^{-}, \mathrm{no}. \mathrm{of} \mathrm{H}.\mathrm{O}=\frac{1}{2}\left(5+1\right)=3\to {\mathrm{sp}}^2 \\ \mathrm{For} {\mathrm{NH}}_4^{+}, \mathrm{no}. \mathrm{of} \mathrm{H}.\mathrm{O}=\frac{1}{2}\left(5+4-1\right)=4\to {\mathrm{sp}}^3 \end{gathered}$$