QJEEM24Chd3b8: JEE Main 2024 Chemistry - Class 11 JEEM24Chd3b8 Solution: Number of molecules/ions from the following in which the…

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.