Question

Which out of $S{O}_4^{2-}$, $S{F}_4$, and $S{F}_2$ does not undergo $s{p}^3$-hybridization?

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

Answer 1

The correct option is D $S{F}_4$

Hybridization of molecule is given by the formula,

$H=\frac{1}{2}[V+M-C+A]$

H = number of orbitals involved in hybridization

V = valence electrons of central atom

M = number of monovalent atoms linked to central atom

C = charge on the cation

A = charge on the anion

Value of H and hybridization, $[H=2(sp),3(s{p}^2),4(s{p}^3),5(s{p}^{3}d\left)\right]$

Hybridization of $S{O}_4^{2-}$,

$H=\frac{1}{2}[6+2]=4$ $\left(s{p}^3\right)$

Hybridization of $S{F}_2$,

$H=\frac{1}{2}[6+2]=4$ $\left(s{p}^3\right)$

Hybridization of $S{F}_4$,

$H=\frac{1}{2}[6+4]=5$ $\left(s{p}^{3}d\right)$

$S{O}_4^{2-}$ and $S{F}_2$ undergoes $s{p}^3$ hybridization and $S{F}_4$ is $s{p}^{3}d$ hybridization.

Hence, the correct option is (D).