Hybridisation

Find the steric number (σ bonds + lone pairs) of the central atom in each ion.
$NO_2^-$: N has 2 σ bonds + 1 lone pair = 3, so $sp^2$.
$NO_3^-$: N has 3 σ bonds + 0 lone pairs = 3, so $sp^2$.
$NH_2^-$: N has 2 σ bonds + 2 lone pairs = 4, so $sp^3$.
$NH_4^+$: N has 4 σ bonds + 0 lone pairs = 4, so $sp^3$.
$SCN^-$: C has 2 σ bonds + 0 lone pairs = 2, so sp.
Among the given pairs, only $NO_2^-$ and $NO_3^-$ have the same hybridisation ($sp^2$).

Count σ bonds + lone pairs on the central atom.
$CH_4$: 4 bond pairs + 0 lone pairs = 4, so $sp^3$.
$BF_4^-$: 4 bond pairs + 0 lone pairs = 4, so $sp^3$.
$NH_4^+$: 4 bond pairs + 0 lone pairs = 4, so $sp^3$.
$SF_4$: sulphur has 6 valence electrons; 4 are used in bonds and 2 remain as one lone pair. 4 bond pairs + 1 lone pair = 5, so $sp^3d$ (see-saw shape).
Hence the central atom is not $sp^3$ hybridised in $SF_4$.

Steric number = number of σ bonds + number of lone pairs on the central atom; steric number 3 means $sp^2$ and 4 means $sp^3$.
$BF_3$: 3 σ bonds + 0 lone pairs = 3, so $sp^2$.
$NO_2^-$: 2 σ bonds + 1 lone pair = 3, so $sp^2$.
$NH_2^-$: 2 σ bonds + 2 lone pairs = 4, so $sp^3$.
$NH_3$: 3 σ bonds + 1 lone pair = 4, so $sp^3$.
$H_2O$: 2 σ bonds + 2 lone pairs = 4, so $sp^3$.
Hence both central atoms are $sp^2$ hybridised in $BF_3$ and $NO_2^-$.

Count σ bonds + lone pairs on the central atom.
$PCl_5$: 5 bond pairs + 0 lone pairs = 5, so $sp^3d$.
$SF_4$: 4 bond pairs + 1 lone pair = 5, so $sp^3d$.
$I_3^-$: the central I has 2 bond pairs + 3 lone pairs = 5, so $sp^3d$.
$SbCl_5^{2-}$: Sb has 5 valence electrons + 2 from the charge = 7; five are used in bonds, leaving one lone pair. 5 bond pairs + 1 lone pair = 6, so $sp^3d^2$.
Hence $SbCl_5^{2-}$ has a different hybridisation from the other three.