Question

In which of the following molecules, the central atom does not have $s{p}^3$ hybridization?

• A. $C{H}_4$
• B. $S{F}_4$
• C. ${B{F}_4}^{-}$
• D. $N{H}_4^{+}$

Answer 1

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

For neutral molecules,
Number of electron pairs $=$ Number of atoms bonded to it $+\frac{1}{2}$ [Gp. no of central atom - Valency of central atom]
$\therefore$ For $C{H}_4$, no. of e${}^{-}$ pairs $=4+$ $\frac{1}{2}[4-4]=4$ (sp${}^3$ hybridisation)
For $S{F}_4$, no. of e${}^{-}$ pairs $=4+\frac{1}{2}[6-4]=$ $5$ ($s{p}^{3}d$ hybridisation)
For ions,
No. of electron pairs = No. of atoms bonded to it $+\frac{1}{2}$ [Gp. no. of central atom - Valency of central atom $\pm$ No. of electrons]
$\therefore$ For $B{F}_4^{-}$, no. e${}^{-}$ pairs $=4+$ $\frac{1}{2}[3-4+1]=4$ ($s{p}^3$ hybridisation)
For $N{H}_4^{+}$, no. of e${}^{-}$ pairs $=4+$ $\frac{1}{2}[5-4-1]=4$ ($s{p}^3$ hybridisation)
Thus, in $S{F}_4$, the central atom does not have $s{p}^3$ hybridization.

Therefore, the correct option is B.