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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
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Hint: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.