Question

Which of the following ion is not tetrahedral in shape?

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

Answer 1

Answer: (C)

$S{F}_4$

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

where,
H= Number of orbitals involved in hybridization
V=Valence electrons of the central atom
M- Number of monovalent atoms linked to the central atom
C= Charge of cation
A= Charge of anion

Consider the hybridization and shape of the following given molecules:

A) $B{F}_4^{-}$ - $s{p}^3$, Tetrahedral

B) $N{H}_4^{+}$ - $s{p}^3$, Tetrahedral

C) $S{F}_4$- $s{p}^{3}d$, See-saw

D) $C{F}_4$ - $s{p}^3$, Tetrahedral

Hence, the correct option is $C$