The types of hybrid orbitals of nitrogen in \(\text{NO}_2^+, \text{NO}_3^-\) and \(\text{NH}_4^+\) respectively are

The hybridization of the central atom can be determined using the formula for the steric number \(H = \dfrac{1}{2}(V + M - C + A)\), where \(V\) is the number of valence electrons on the central atom, \(M\) is the number of monovalent surrounding atoms, \(C\) is the cationic charge, and \(A\) is the anionic charge.

For \(\text{NO}_2^+\):
\(V = 5\), \(M = 0\) (oxygen is divalent), \(C = 1\), \(A = 0\)
\(H = \dfrac{1}{2}(5 + 0 - 1 + 0) = 2\)
A steric number of \(2\) corresponds to \(sp\) hybridization.

For \(\text{NO}_3^-\):
\(V = 5\), \(M = 0\), \(C = 0\), \(A = 1\)
\(H = \dfrac{1}{2}(5 + 0 - 0 + 1) = 3\)
A steric number of \(3\) corresponds to \(sp^2\) hybridization.

For \(\text{NH}_4^+\):
\(V = 5\), \(M = 4\) (hydrogen is monovalent), \(C = 1\), \(A = 0\)
\(H = \dfrac{1}{2}(5 + 4 - 1 + 0) = 4\)
A steric number of \(4\) corresponds to \(sp^3\) hybridization.

The types of hybrid orbitals are \(sp\), \(sp^2\), and \(sp^3\) respectively.

Answer: \(sp, sp^2\) and \(sp^3\)