In $$SO_2$$, $$NO_2^-$$ and $$N_3^-$$ the hybridizations at the central atom are respectively :

The hybridization of a given atom can be predicted from its steric number.
Steric number $$=\,$$ (number of $$\sigma$$-bonds around the atom) $$+\,$$ (number of lone pairs on that atom).

• Central atom: $$S$$.
• Lewis structure: $$O = S - O$$ with one lone pair on $$S$$.
• $$\sigma$$-bonds on $$S = 2$$ (one to each $$O$$).
• Lone pairs on $$S = 1$$.
Therefore, steric number $$= 2 + 1 = 3$$.

Steric number 3 corresponds to $$sp^2$$ hybridization, giving a bent (V-shaped) molecule.

• Central atom: $$N$$.
• Lewis structure: $$O = N - O^-$$ (resonance forms) with one lone pair on $$N$$.
• $$\sigma$$-bonds on $$N = 2$$.
• Lone pairs on $$N = 1$$.
Steric number $$= 2 + 1 = 3$$.

Steric number 3 again implies $$sp^2$$ hybridization, giving a bent ion.

• Central atom: the middle $$N$$ of $$N - N - N$$.
• Resonance structures involve $$N \equiv N^+ - N^-$$ and $$N^- - N^+ \equiv N$$, but the central $$N$$ always has:
  $$\sigma$$-bonds = 2$$\,$$(to the terminal nitrogens).
  Lone pairs = 0 (all its electrons are in bonds or as formal charges on terminals).
Steric number $$= 2 + 0 = 2$$.

Steric number 2 corresponds to $$sp$$ hybridization, giving a linear ion.

Thus, the hybridizations are $$sp^2$$ in $$SO_2$$, $$sp^2$$ in $$NO_2^-$$ and $$sp$$ in $$N_3^-$$.

Matching with the options, we select Option A.

Answer: Option A