Question

Which are the species in which sulphur undergoes $s{p}^3$ hybridisation?

• A. $S{F}_4$
• B. $SC{l}_2$
• C. $S{O}_4^{2-}$
• D. $H_{2}S$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $SC{l}_2$
C $H_{2}S$
D $S{O}_4^{2-}$

Concept involved:
Hybridization of a molecule or an ion is decided by using the following formula:
Number of Hybrid orbitals $H=n=\frac{1}{2}$[group no of central atom +no of atoms attached to central metal atom (like H , Cl, F.except O) + magnitude of negative charge - magnitude of positive charge].
1) If $H=2$; Hybridization : $sp$.

2) If $H=3$ ; Hybridization : $s{p}^2$.

3) If $H=4$ ; Hybridization : $s{p}^3$.

4) If $H=5$ ; Hybridization : $s{p}^{2}d$.

5) If $H=6$ ; Hybridization: $s{p}^3{d}^2$.

A) $S{F}_4$: $H=\frac{1}{2}$(6+4)=5 $s{p}^{2}d$ hybridization.

B) $SC{l}_2$: $H=\frac{1}{2}$(6+2)=4 $s{p}^3$ hybridization.

C) $S{O}_4^{2-}$: $H=\frac{1}{2}$(6+2) = 4 $s{p}^3$ hybridization .

D) $H_{2}S$: $H=\frac{1}{2}$(6+2) = 4 $s{p}^3$ hybridization.

Hence option B,C & D are corect.