Total number of species from the following with central atom utilising $$sp^2$$ hybrid orbitals for bonding is ___________. $$NH_3, SO_2, SiO_2, BeCl_2, C_2H_2, C_2H_4, BCl_3, HCHO, C_6H_6, BF_3, C_2H_4Cl_2$$

We need to count the species with the central atom using sp$$^2$$ hybrid orbitals from the given list.

sp$$^2$$ hybridisation occurs when the central atom has 3 electron domains (regions of electron density), giving a trigonal planar arrangement (or bent if lone pairs are present).

Analyze each species:

(i) $$NH_3$$: N has 3 bond pairs + 1 lone pair = 4 electron domains → sp$$^3$$. ✗

(ii) $$SO_2$$: S has 2 bond pairs + 1 lone pair = 3 electron domains → sp$$^2$$. ✓

(iii) $$SiO_2$$: Each Si is bonded to 4 O atoms in a tetrahedral arrangement → sp$$^3$$. ✗

(iv) $$BeCl_2$$: Be has 2 bond pairs + 0 lone pairs = 2 electron domains → sp. ✗

(v) $$C_2H_4$$: Each C has 3 electron domains (2 C-H + 1 C=C) → sp$$^2$$. ✓

(vi) $$C_2H_2$$: Each C has 2 electron domains (1 C-H + 1 C≡C) → sp. ✗

(vii) $$BCl_3$$: B has 3 bond pairs + 0 lone pairs = 3 electron domains → sp$$^2$$. ✓

(viii) $$HCHO$$: C has 3 electron domains (2 C-H + 1 C=O) → sp$$^2$$. ✓

(ix) $$C_6H_6$$: Each C has 3 electron domains → sp$$^2$$. ✓

(x) $$BF_3$$: B has 3 bond pairs + 0 lone pairs → sp$$^2$$. ✓

(xi) $$C_2H_4Cl_2$$: Each C has 4 electron domains → sp$$^3$$. ✗

Count: SO$$_2$$, C$$_2$$H$$_4$$, BCl$$_3$$, HCHO, C$$_6$$H$$_6$$, BF$$_3$$ = 6 species.

The answer is 6.