1. Criterion for $\mathrm{sp}^3$ hybridisation

A central atom uses $\mathrm{sp}^3$ hybrid orbitals when its steric number is $4$, i.e.:

Now check each species.

2. Examine each molecule

(i) $\mathrm{NH_3}$

• Central atom: $\mathrm{N}$
• It has 3 $\sigma$ bonds and 1 lone pair.
• Steric number $= 3+1=4$
• Hence, hybridisation of $\mathrm{N}$ is:

✅ Counts

(ii) $\mathrm{SO_2}$

• Central atom: $\mathrm{S}$
• Sulfur has 2 $\sigma$ bonds with O and 1 lone pair.
• Steric number $=2+1=3$
• Hence hybridisation is:

❌ Does not count

(iii) $\mathrm{SiO_2}$

• As a discrete molecule, structure is $\mathrm{O=Si=O}$
• Central atom: $\mathrm{Si}$
• 2 $\sigma$ bonds, 0 lone pairs
• Steric number $=2$
• Hybridisation:

❌ Does not count

(iv) $\mathrm{BeCl_2}$

• Central atom: $\mathrm{Be}$
• 2 $\sigma$ bonds, 0 lone pairs
• Steric number $=2$
• Hybridisation:

❌ Does not count

(v) $\mathrm{CO_2}$

• Central atom: $\mathrm{C}$
• 2 $\sigma$ bonds, 0 lone pairs
• Steric number $=2$
• Hybridisation:

❌ Does not count

(vi) $\mathrm{H_2O}$

• Central atom: $\mathrm{O}$
• 2 $\sigma$ bonds and 2 lone pairs
• Steric number $=2+2=4$
• Hybridisation:

✅ Counts

(vii) $\mathrm{CH_4}$

• Central atom: $\mathrm{C}$
• 4 $\sigma$ bonds, 0 lone pairs
• Steric number $=4$
• Hybridisation:

✅ Counts

(viii) $\mathrm{BF_3}$

• Central atom: $\mathrm{B}$
• 3 $\sigma$ bonds, 0 lone pairs
• Steric number $=3$
• Hybridisation:

❌ Does not count

3. Total count

Species with central atom using $\mathrm{sp}^3$ hybrid orbitals:

So, the number of such species is: