Question

What is the shape of Xenon difluoride?

Answer 1

Hybridization:

• It is the process of intermixing of the orbitals of slightly different energies so as to redistribute their energies, resulting in the formation of a new set of orbitals of equivalent energies and shapes.
• The hybridized orbitals are always equivalent in energy and shape.

The formula used to calculate the hybridization of a molecule is as follows:

• $\mathrm{Number}\mathrm{of}\mathrm{electron}\mathrm{pair}=\frac{1}{2}[\mathrm{V}+\mathrm{N}-\mathrm{C}+\mathrm{A}]$

where,

$\mathrm{V}$ is the number of valence electrons present in the central atom

$\mathrm{N}$ is the number of monovalent atoms bonded to the central atom

$\mathrm{C}$ is the charge of cation

$\mathrm{A}$ is the charge of anion

• The given molecule is, xenon difluoride $\mathbf{\left(}{\mathbf{XeF}}_{\mathbf{2}}\mathbf{\right)}$

$\mathrm{Number}\mathrm{of}\mathrm{electron}\mathrm{pair}=\frac{1}{2}\times [8+2-0+0]=\frac{10}{2}=5$

The number of electron pairs is $5$ which means the hybridization will be ${\mathrm{sp}}^3\mathrm{d}$ and the electronic geometry of the molecule will be trigonal bipyramidal.

But as there are $3$ atoms around the central xenon atom, the $3^{\mathrm{rd}},4^{th}\mathrm{and}{5}^{\mathrm{th}}$ position will be occupied by lone pair of electrons. Hence the molecular geometry will be linear.

Therefore, the shape of xenon difluoride is linear.