Question

In which of the following molecule, the central atom is used $s{p}^{3}d$ hybridized orbitals?

• A. $SF$${}_4$
• B. $NH$${}_3$
• C. $XeF$${}_2$
• D. $XeF$${}_4$

Answer 1

The correct option is A $SF$${}_4$

A) The hybridization that is involved in $S{F}_4$ is $s{p}^{3}d$ type.

When the central metal atom is surrounded with five pairs of electrons the electronic geometry is trigonal bi pyramidal. If all the electron pairs are shared, the molecular geometry will also be trigonal bi pyramidal. In this molecule $S{F}_4$, there are five electron pairs which should lead to trigonal bi pyramidal structure.

B) In $N{H}_3$ , the central atom $N{H}_3$ is $s{p}^3$ hybridized.

$\to$ The geometry of such a molecule is pyramidal.

$\to$ The bond angle is ${107}^0$ .

C)The hybridization of $Xe{F}_2$ (Xenon Difluoride) is an $s{p}^{3}d$ type.

$\to$ It involves $S{p}^3$ hybridization of xenon atoms.

$\to$ The molecule has trigonal bipyramidial structure with 3 position occupied by lone pair. So, $Xe{F}_2$ is linear.

D)$Xe{F}_4$

$\to$ The central atom has 6 electron pairs out of which two are lone pair electrons. The hybridization in Xenon is $s{p}^3{d}^2$ because there is a migration of two electrons of p to d orbital which results in the formation of sigma bond with F.

$\to$ $Xe{F}_4$ is $A{B}_4{L}_2$ in which the Xenon is bonded with four fluorine atoms with 4 sigma bonds and xenon has two lone pairs (4 sigma bonds and 2 lone pairs).

So, the hybridisation is $s{p}^3{d}^2$.

Hence, The correct option is A .