Answer the question on basis of passage given below:

The noble gases have closed shell electronic configuration and are monatomic gases under normal conditions. The low boiling points of lighter noble gases are due to the weak dispersion forces between the atoms and the absence of other inter atomic interaction. The direct reaction of xenon with fluorine leads to a series of compounds with oxidation numbers +2, +4 and +6.

\(XeF_4\) reacts violently with water to give \(XeO_3\). The compound of Xenon exhibit stereoisomerism and their geometries can be deduced considering the total number of electrons in the valence shell.

Structure of \(XeO_3\) is :

Pyramidal

The correct answer is option 3. Pyramidal.

Xenon belongs to the group of inert gases and it has the atomic number \(54\). Its electronic configuration is,

\(Xe = 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6\)

Oxygen has the atomic number \(8\). Its electronic configuration is,

\(O = 1s^2 2s^2 2p^4\)

For Xenon trioxide, \(XeO_3\), let’s calculate the number of electron pairs.

Xenon has \(8\) valence electrons and oxygen has \(6\) valence electrons.

Therefore, \(8 + (3 × 6) = \frac{26}{2} = 13\) electron pairs. So, \(XeO_3\) has 13 electron pairs present in it.

Let’s draw the Lewis structure for \(XeO_3\)

In this structure, we can see that all \(13\) electron pairs are present either as bond-pairs or as lone-pairs. Let’s take a look at its hybridization now. Xenon has vacant \(5d\) orbitals so, when xenon gets excited its fifth orbital configuration will become \(5s^2 5p^3 5d^3\). \(5s\) and \(5p\) orbitals will undergo \(sp^3\) hybridization.

Therefore, Xenon in xenon trioxide will have three \(sp^3\) hybrid orbitals, containing three unpaired electrons and one lone pair of electrons, and unhybridized \(5d\) orbitals having three unpaired electrons.

Oxygen will undergo \(sp^2\) hybridization and will have two \(sp^2\) hybridized orbitals having one lone pair of electrons each and one \(sp^2\) hybridized orbital containing one unpaired electron. It will have one unhybridized \(2p\) orbital having one electron.

The three \(sp^3\) orbitals of xenon having three unpaired electrons will axially overlap with three \(sp^2\) orbitals of three individual oxygen atoms to form three \(\sigma \)-bonds. The three \(5d\) orbitals will laterally overlap with three \(2p\) unhybridized orbitals of three oxygen atoms to form three \(\pi \)−bonds.

Now, each oxygen atom has two lone pairs of electrons and a xenon atom has one lone pair of electrons. Xenon and oxygen are doubly bonded to each other.

Due to the presence of lone pair on central atom Xenon, there will be lone pair-bond pair repulsion in the molecule. So, the trigonal planar geometry will be distorted to form pyramidal geometry.

Therefore, the geometry of \(XeO_3\) is pyramidal.