We need species in which the central atom uses $sp^2$ hybrid orbitals for bonding.

Recall:

• $sp^2$ hybridisation $ightarrow$ steric number $3$
• Geometry around central atom is usually trigonal planar.

Check each species one by one

(i) $\mathrm{NH_3}$

• Central atom: N
• Bonds = 3, lone pair = 1
• Steric number $=4$
• Hybridisation $= sp^3$
• Not counted

(ii) $\mathrm{SO_2}$

• Central atom: S
• Around S: two $\sigma$ bonds + one lone pair
• Steric number $=3$
• Hybridisation $= sp^2$
• Counted

(iii) $\mathrm{SiO_2}$

• Molecular form is $\mathrm{O=Si=O}$
• Central atom: Si
• Two regions of electron density
• Hybridisation $= sp$
• Not counted

(iv) $\mathrm{BeCl_2}$

• Central atom: Be
• Two $\sigma$ bonds
• Steric number $=2$
• Hybridisation $= sp$
• Not counted

(v) $\mathrm{C_2H_2}$

• Each C has two regions of electron density
• Hybridisation of C $= sp$
• Not counted

(vi) $\mathrm{C_2H_4}$

• Each C forms three $\sigma$ bonds
• Hybridisation of each C $= sp^2$
• Counted

(vii) $\mathrm{BCl_3}$

• Central atom: B
• Three $\sigma$ bonds, no lone pair
• Steric number $=3$
• Hybridisation $= sp^2$
• Counted

(viii) $\mathrm{HCHO}$ (formaldehyde)

• Central atom: C
• Around C: two C-H $\sigma$ bonds + one C=O sigma bond
• Steric number $=3$
• Hybridisation $= sp^2$
• Counted

(ix) $\mathrm{C_6H_6}$

• In benzene, each carbon is trigonal planar
• Each C is $sp^2$ hybridised
• Counted

(x) $\mathrm{BF_3}$

• Central atom: B
• Three $\sigma$ bonds
• Hybridisation $= sp^2$
• Counted

(xi) $\mathrm{C_2H_4Cl_2}$

• This is saturated (e.g. $\mathrm{ClCH_2-CH_2Cl}$)
• Each carbon forms four single bonds
• Hybridisation of each C $= sp^3$
• Not counted

Total counted species

Counted species are: $\mathrm{SO_2,\, C_2H_4,\, BCl_3,\, HCHO,\, C_6H_6,\, BF_3}$

Therefore, total number of species is $6$

Comparison with stored answer

Stored correct answer = $6$.

Our derived answer also = $6$.