The hybridisations of atomic orbitals of nitrogen in $\text{NO}_2^+$, $\text{NO}_3^-$ and $\text{NH}_4^+$ are:

To determine the hybridisation of the central atom (N) in each species, we can calculate its steric number (number of $\sigma$ bond pairs + number of lone pairs):

• $\text{NO}_2^+$: Nitrogen has 5 valence electrons. A positive charge means it loses 1 electron, leaving 4 valence electrons. It forms 2 $\sigma$ bonds (and 2 $\pi$ bonds) with two oxygen atoms, leaving 0 lone pairs. Steric number = 2. Hence, its hybridisation is $sp$.
• $\text{NO}_3^-$: Nitrogen has 5 valence electrons. A negative charge gives it 1 extra electron, making 6. It forms 3 $\sigma$ bonds with three oxygen atoms and has 0 lone pairs. Steric number = 3. Hence, its hybridisation is $sp^2$.
• $\text{NH}_4^+$: Nitrogen has 5 valence electrons. A positive charge leaves it with 4 electrons. It forms 4 $\sigma$ bonds with four hydrogen atoms and has 0 lone pairs. Steric number = 4. Hence, its hybridisation is $sp^3$.

The correct order of hybridisations is $sp$, $sp^2$ and $sp^3$.