Question

What are hybridization and geometry of ${\mathrm{XeF}}_6$?

Answer 1

Xenon hexafluoride $\left({\mathrm{XeF}}_6\right)$ :

• ${\mathrm{XeF}}_6$ has distorted octahedral geometry.
• The geometry can be represented as:

Step 1: Knowing Hybridization:

• “Hybridization is defined as 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 shape.”

Step 2: Formula used for calculation:

• $\mathrm{Hybridization}=\frac{1}{2}\left[\mathrm{V}+\mathrm{M}-\mathrm{C}+\mathrm{A}\right]$
• Where $\mathrm{V}$ is the number of valence electrons of the central atom, $\mathrm{M}$ is the no. of monovalent elements atom/ions attached to the central atom, $\mathrm{C}$ is the positive charge, and $\mathrm{A}$ represents the negative charge.

Step 3: Calculation for Hybridisation of ${\mathrm{XeF}}_6$:

• Here central atom is Xenon.
• The total number of valence electrons in ${\mathrm{XeF}}_6$ is $8$ and there are $6$ monovalent Fluorine atoms.
• $$\begin{gathered} \mathrm{Hybridization}=\frac{1}{2}\left[8+6-0+0\right] \\ \mathrm{Hybridization}=\frac{1}{2}\times 14=7 \end{gathered}$$
• The hybridization number is $7$, thus the hybridization of ${\mathrm{XeF}}_6$ is ${\mathrm{sp}}^3{\mathrm{d}}^3$.

Step 4: Geometry of ${\mathrm{XeF}}_6$:

• As per VSEPR, ${\mathrm{XeF}}_6$ has a distorted octahedral geometry, this distortion is due to the presence of lone pair of electrons on the Xenon atom. L.P-L.P repulsion is more than L.P-B.P which is in turn greater than B.P-B.P repulsion.