To determine the number of species in which the central atom uses

hybrid orbitals in its bonding, we need to analyze the hybridization of each central atom in the given species.

1.

(Ammonia):

The central atom is nitrogen. Ammonia has 3 sigma bonds and 1 lone pair. Thus, the steric number is 4, which corresponds to

hybridization.

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

hybridization.

3.

(Silicon dioxide):

The central atom is silicon. Silicon dioxide has a linear structure with double bonds, leading to

hybridization.

4.

(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

hybridization.

5.

(Carbon dioxide):

The central atom is carbon. Carbon dioxide has a linear structure with double bonds, leading to

hybridization.

6.

(Water):

The central atom is oxygen. Water has 2 sigma bonds and 2 lone pairs. Thus, the steric number is 4, which corresponds to

hybridization.

7.

(Methane):

The central atom is carbon. Methane has 4 sigma bonds and no lone pair. Thus, the steric number is 4, which corresponds to

hybridization.

8.

(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

hybridization.

From the above analysis, the species in which the central atom uses

hybrid orbitals are:

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

Therefore, the number of species where the central atom uses

hybrid orbitals is 3.