Question

The hybridization of the central atom in $Xe{F_4}$?
A) $s{p^2}$
B) $s{p^3}{d^2}$
C) $s{p^3}$
D) $s{p^3}d$

Answer 1

As we know that the mixing of two orbitals results in a hybridized orbital formation and the xenon in xenon tetrafluoride consists of a total of six valence electrons in its outermost subshell and the electronic configuration of xenon is given as $[Kr]4{d^{10}}5{s^2}5{p^6}$.

Complete solution:
As we know that there are a total of six electrons pair in Xenon and out of these six electrons two are lone pair of electrons. There are a total of six electrons in the $5p$ orbital and two electrons in the $5s$ orbitals, and there are no electrons in the d-orbital and the f-orbital. Therefore, in the ground state the two electrons in the $5p$ orbitals will be transferred to the vacant $5d$orbitals in the excited state and thus there are four unpaired electrons now two of them coming from $5p$ and remaining two coming from the $5d$orbital. Now the four fluorine atoms will bind with these four unpaired electrons. Therefore, the hybridisation of xenon tetrafluoride becomes $s{p^3}{d^2}$.
Also, we know that the other hybridization involves one s-orbital and three p-orbitals only which made it impossible for the accommodation of electrons of the central atom in the subshells. Therefore all the other options are incorrect.
We know the $s{p^3}d$ hybridization is formed when one s-orbital combines with three p- and one d-orbital. For the hybridization of $Xe{F_4}$ we need two d-orbitals instead of one orbital for pairing the electrons in $s{p^3}d$. Thus it is also incorrect.

Therefore,the correct option is (B).

Note: Always remember that the xenon tetrafluoride possesses two lone pairs of electrons and according to VSEPR theory the net electronic repulsion should be minimum and thus the compound acquires a stable state. The lone pairs of electrons are in a perpendicular plane to the central atoms resulting in a square planar geometry to the molecule and a $s{p^3}{d^2}$ hybridization.