JEE Main 2024 — Chemical Bonding And Molecular Structure Question with Solution

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.