The number of species from the following in which the central atom uses $$sp^3$$ hybrid orbitals in its bonding is _________. $$NH_3, SO_2, SiO_2, BeCl_2, CO_2, H_2O, CH_4, BF_3$$

The hybridization of the central atom can be determined using the steric number.

$$\text{Steric Number}=\text{Number of bonded atoms}+\text{Number of lone pairs}$$

A steric number of $$4$$ corresponds to $$sp^3$$ hybridization.

For $$NH_3$$, nitrogen is bonded to three hydrogen atoms and possesses one lone pair.

$$SN=3+1=4$$

Therefore, nitrogen is $$sp^3$$ hybridized.

For $$SO_2$$, sulfur is bonded to two oxygen atoms and possesses one lone pair.

$$SN=2+1=3$$

Therefore, sulfur is $$sp^2$$ hybridized.

For $$SiO_2$$, each silicon atom in the three-dimensional covalent network is bonded to four oxygen atoms.

$$SN=4+0=4$$

Therefore, silicon is $$sp^3$$ hybridized.

For $$BeCl_2$$, beryllium is bonded to two chlorine atoms.

$$SN=2+0=2$$

Therefore, beryllium is $$sp$$ hybridized.

For $$CO_2$$, carbon is bonded to two oxygen atoms.

$$SN=2+0=2$$

Therefore, carbon is $$sp$$ hybridized.

For $$H_2O$$, oxygen is bonded to two hydrogen atoms and possesses two lone pairs.

$$SN=2+2=4$$

Therefore, oxygen is $$sp^3$$ hybridized.

For $$CH_4$$, carbon is bonded to four hydrogen atoms.

$$SN=4+0=4$$

Therefore, carbon is $$sp^3$$ hybridized.

For $$BF_3$$, boron is bonded to three fluorine atoms.

$$SN=3+0=3$$

Therefore, boron is $$sp^2$$ hybridized.

Thus, the species containing a central atom with $$sp^3$$ hybridization are $$NH_3$$, $$SiO_2$$, $$H_2O$$, and $$CH_4$$.

Hence, the total number of $$sp^3$$ hybridized species is

$$4$$