To determine the number of molecules/ions in which the central atom is involved in $$\mathrm{sp}^3$$ hybridization, we must analyze the hybridization state for each central atom. Hybridization is typically determined by the number of sigma bonds and lone pairs on the central atom.

Let's evaluate each molecule/ion:

1. $$\mathrm{NO}_3^{-}$$:

The central atom is nitrogen (N). In $$\mathrm{NO}_3^{-}$$, nitrogen forms three sigma bonds with oxygen atoms and has no lone pairs. Using the formula for hybridization:

Number of hybrid orbitals = number of sigma bonds + number of lone pairs

For $$\mathrm{NO}_3^{-}$$: $$3 + 0 = 3$$ hybrid orbitals. Therefore, nitrogen is $$\mathrm{sp}^2$$ hybridized.

2. $$\mathrm{BCl}_3$$:

The central atom is boron (B). In $$\mathrm{BCl}_3$$, boron forms three sigma bonds with chlorine atoms and has no lone pairs. Using the same formula:

For $$\mathrm{BCl}_3$$: $$3 + 0 = 3$$ hybrid orbitals. Therefore, boron is $$\mathrm{sp}^2$$ hybridized.

3. $$\mathrm{ClO}_2^{-}$$:

The central atom is chlorine (Cl). In $$\mathrm{ClO}_2^{-}$$, chlorine forms two sigma bonds with oxygen atoms and has two lone pairs. Therefore,:

For $$\mathrm{ClO}_2^{-}$$: $$2 + 2 = 4$$ hybrid orbitals. Thus, chlorine is $$\mathrm{sp}^3$$ hybridized.

4. $$\mathrm{ClO}_3^{-}$$:

The central atom is chlorine (Cl). In $$\mathrm{ClO}_3^{-}$$, chlorine forms three sigma bonds with oxygen atoms and has one lone pair. Therefore,:

For $$\mathrm{ClO}_3^{-}$$: $$3 + 1 = 4$$ hybrid orbitals. Thus, chlorine is $$\mathrm{sp}^3$$ hybridized.

Based on the analysis, the molecules/ions $$\mathrm{ClO}_2^{-}$$ and $$\mathrm{ClO}_3^{-}$$ have their central atoms involved in $$\mathrm{sp}^3$$ hybridization.

Therefore, the total number is 2.

Thus, the correct option is Option A.