Question

Which of the following has a regular tetrahedral geometry?

• A. $S{F}_4$
• B. ${B{F}_4}^{-}$
• C. $Xe{F}_4$
• D. $Cl{F}_3$

Answer 1

The correct option is B ${B{F}_4}^{-}$

$H=\frac{1}{2}[V+M-C+A]$
where,
H= Number of orbitals involved in hybridization.
V=Valence electrons of central atom.
M- Number of monovalent atoms linked to central atom.
C= Charge of cation.
A= Charge of anion.
Consider the hybridization and shape of the options:-
A) $S{F}_4$
$H=\frac{1}{2}[6+4]=5$ $\Rightarrow s{p}^{3}d$ hybridized state.
It has a see saw shape.
B) $B{F}_4^{-}$
$H=\frac{1}{2}[3+4+1]=4$ $\Rightarrow s{p}^3$ hybridised state
It is tetrahedral in shape.
C) $Xe{F}_4$
$H=\frac{1}{2}[8+4]=6$ $\Rightarrow s{p}^3{d}^2$ hybridized state with 2 lone pair of electrons.
It is square planar in shape.
D)$Cl{F}_3$
$H=\frac{1}{2}[7+3]=5$ $\Rightarrow s{p}^{3}d$ hybridized state.
It is T- shaped.
Hence, option B is the right answer.