Which of the following molecules have linear shape?

\(XeF_2\)

The correct answer is option 1: \(XeF_2\).

1. \(XeF_2\):

\(XeF_2\) has a linear molecular geometry. Molecule \(Xe\) has eight valence electrons, while fluorine has seven, totaling \(22\) valence electrons. This indicates that both fluorines must be bound to the Xe molecule, resulting in three unshared pairs and two bonded pairs on the Xe molecule. Although the lone pairs are at equatorial positions (bonds perpendicular to the axis), the molecule is a trigonal bipyramid. Due to the \(180^o\) bond angles, the structure must be linear.

It is left with three lone pairs placed in the equatorial plane and two fluorines arranged perpendicularly to the lone pairs in the axial plane after sharing 1–1 electron with fluorine.

2.\(XeF_4\):

In xenon tetrafluoride, the hybridization takes place in the central atom which is Xenon (Xe). If we look at the valence shell of Xe there are a total of six electrons in the \(5p\) orbital and two electrons in the 5s orbital. If we observe the \(5^{th}\) shell then there are the d orbital and f orbital in which no electrons are present. In the formation of \(XeF_4\), two of the \(5p\) orbital electrons which, in the excited state move to fill the vacant \(5d\) orbitals. As a result, there are 4 unpaired electrons which include 2 in \(5p\) and 2 in \(5d\) orbitals. This results in \(sp^3d^2\) hybridization.

In the case of fluorine, four F atoms bond with these four half filled orbitals. These fluorine atoms will then be placed on either side of the central atom.

3. \(XeF_6\):

Xenon is a noble gas element which belongs to the fifth period.

Let’s write the electronic configuration of Xenon.

\(Xe = [Kr]4d^{10}5s^25p^6\)

There are \(7\) valence shell electrons in each fluorine and eight valence shell electrons in xenon atom. So, the number of electrons present is \(8+(6 × 7) = 50\) and the total number of electron pairs present is \(25\).

The Lewis structure of Xenon hexafluoride is,

In xenon, there are eight electrons in its valence shell. It also has an empty \(5d\) orbital. There are six fluorine atoms in xenon hexafluoride, and six covalent bonds are formed. Xenon undergoes \(sp^3d^3\) hybridization to form seven \(sp^3d^3\) hybrid orbitals. Six \(sp^3d^3\) orbitals have one electron each and one \(sp^3d^3\) contains a lone pair of electrons. The six unpaired electrons get paired with six electrons from six fluorine atoms to form xenon hexafluoride, \(XeF_6\). Due to the presence of one lone pair of electrons, they will be lone–pair–bond–pair repulsion in the molecule and the geometry will be distorted octahedral.

4. \(XeO_3\):

The central atom is a xenon atom. Xenon is a noble gas element or inert gas element.

The atomic number of xenon is 54 and its electronic configuration is \([Kr]4d^{10}5s^25p^6\).

The atomic number of oxygen is 8 and its electronic configuration is \([He]2s^22p^4\).

Thus, an oxygen atom has six valence electrons.

In the xenon trioxide molecule, the central xenon atom forms three double bonds with three oxygen atoms. Thus, there are three \(Xe = O\) double bonds in one molecule of xenon trioxide.

Thus, the number of bonding domains is 3. The central xenon atom has one lone pair of electrons. Thus, the steric number for the central xenon atom in xenon trioxide is 4. The steric number of four is associated with tetrahedral electron pair geometry due to \(sp^3\) hybridization. Since one lone pair of electrons is present, the molecular geometry is pyramidal.

The diagram of xenon trioxide is shown below: