Question

Which of the following molecule has a planar structure ?

• A. $O_{2}S{F}_2$
• B. $OS{F}_2$
• C. $Xe{F}_4$
• D. $Cl{O}_4^{-}$

Answer 1

Answer: (C)

$Xe{F}_4$

$H=\frac{1}{2}[V+M-C+A]$
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) $O_{2}S{F}_2$
$H=\frac{1}{2}[6+2]=4$.
$\Rightarrow s{p}^3$ hybridized state.
Tetrahedral in shape.
B) $OS{F}_2$
$H=\frac{1}{2}[6+2]=4$.
$\Rightarrow s{p}^3$ hybridized state with 1 lone pair.
Trigonal Pyramidal 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.
Square Planar in shape.
D)$Cl{O}^{4-}$
$H=\frac{1}{2}[7+1]=4$.
$\Rightarrow s{p}^3$ hybridized state.
Tetrahedral in shape.
Hence option C is the right answer.