
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
467.1k+ views
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.
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.
Recently Updated Pages
The correct geometry and hybridization for XeF4 are class 11 chemistry CBSE

Water softening by Clarks process uses ACalcium bicarbonate class 11 chemistry CBSE

With reference to graphite and diamond which of the class 11 chemistry CBSE

A certain household has consumed 250 units of energy class 11 physics CBSE

The lightest metal known is A beryllium B lithium C class 11 chemistry CBSE

What is the formula mass of the iodine molecule class 11 chemistry CBSE

Trending doubts
Is Cellular respiration an Oxidation or Reduction class 11 chemistry CBSE

In electron dot structure the valence shell electrons class 11 chemistry CBSE

What is the Pitti Island famous for ABird Sanctuary class 11 social science CBSE

State the laws of reflection of light

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells
