The hybridizations of atomic orbitals of nitrogen in $N{O}_2^{+}$,$N{O}_3^{-}$ and ${\mathrm{NH}}_4^{+}$ respectively are

Number of hybrid orbitals = Number of ${\mathrm{Bpe}}^{-}+\frac{1}{2}$ [Group number- Valency of central atom(Q)]
$\begin{array}{l}{\mathrm{NO}}_2^{+}\\ =2+\frac{1}{2}[5-4-1]=2=\mathrm{sp}\\ {\mathrm{NO}}_3^{-}\\ =3+\frac{1}{2}[5-6(-1\left)\right]=3={\mathrm{sp}}^2\\ {\mathrm{NH}}_4^{+}\\ =4+\frac{1}{2}[5-4-1]=4={\mathrm{sp}}^3\end{array}$