The hybridisations of the atomic orbitals of nitrogen in NO$$_2^-$$, NO$$_2^+$$ and NH$$_4^+$$ respectively are:

For $$NO_2^-$$ (nitrite ion): Nitrogen has 5 valence electrons, and with the negative charge, nitrogen forms 2 bonds with oxygen and has one lone pair. The electron geometry around nitrogen has 3 regions of electron density (2 bonding + 1 lone pair), giving $$sp^2$$ hybridisation. The bond angle is about 115°.

For $$NO_2^+$$ (nitronium ion): Nitrogen forms two double bonds with two oxygen atoms, giving 2 electron domains around nitrogen with no lone pair. This corresponds to $$sp$$ hybridisation with a linear geometry (180° bond angle).

For $$NH_4^+$$ (ammonium ion): Nitrogen is bonded to 4 hydrogen atoms with no lone pairs, giving a tetrahedral geometry with $$sp^3$$ hybridisation.

Therefore, the hybridisations of nitrogen in $$NO_2^-$$, $$NO_2^+$$ and $$NH_4^+$$ are $$sp^2$$, $$sp$$ and $$sp^3$$ respectively.