QJEEM24Cheb55: JEE Main 2024 Chemistry - Class 11 JEEM24Cheb55 Solution: The number of species from the following in which the central…

To determine the number of species in which the central atom uses $\mathrm{sp}^3$ hybrid orbitals in its bonding, we need to analyze the hybridization of each central atom in the given species.

1. $\mathrm{NH}_3$ (Ammonia):

The central atom is nitrogen. Ammonia has 3 sigma bonds and 1 lone pair. Thus, the steric number is 4, which corresponds to $\mathrm{sp}^3$ hybridization.

2. $\mathrm{SO}_2$ (Sulfur dioxide):

The central atom is sulfur. Sulfur dioxide has 2 sigma bonds and 1 lone pair. Thus, the steric number is 3, which corresponds to $\mathrm{sp}^2$ hybridization.

3. $\mathrm{SiO}_2$ (Silicon dioxide):

The central atom is silicon. Silicon dioxide has a linear structure with double bonds, leading to $\mathrm{sp}$ hybridization.

4. $\mathrm{BeCl}_2$ (Beryllium chloride):

The central atom is beryllium. Beryllium chloride has 2 sigma bonds and no lone pair. Thus, the steric number is 2, which corresponds to $\mathrm{sp}$ hybridization.

5. $\mathrm{CO}_2$ (Carbon dioxide):

The central atom is carbon. Carbon dioxide has a linear structure with double bonds, leading to $\mathrm{sp}$ hybridization.

6. $\mathrm{H}_2\mathrm{O}$ (Water):

The central atom is oxygen. Water has 2 sigma bonds and 2 lone pairs. Thus, the steric number is 4, which corresponds to $\mathrm{sp}^3$ hybridization.

7. $\mathrm{CH}_4$ (Methane):

The central atom is carbon. Methane has 4 sigma bonds and no lone pair. Thus, the steric number is 4, which corresponds to $\mathrm{sp}^3$ hybridization.

8. $\mathrm{BF}_3$ (Boron trifluoride):

The central atom is boron. Boron trifluoride has 3 sigma bonds and no lone pair. Thus, the steric number is 3, which corresponds to $\mathrm{sp}^2$ hybridization.

From the above analysis, the species in which the central atom uses $\mathrm{sp}^3$ hybrid orbitals are:

• $\mathrm{NH}_3$
• $\mathrm{H}_2\mathrm{O}$
• $\mathrm{CH}_4$

Therefore, the number of species where the central atom uses $\mathrm{sp}^3$ hybrid orbitals is 3.