Question

What is the structure of ${\mathrm{XeOF}}_4$ ?

Answer 1

Step 1: 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:

• $\mathrm{Steric}\mathrm{number}=\frac{1}{2}\left[\mathrm{V}+\mathrm{M}-\mathrm{C}+\mathrm{A}\right]$
• Where V is the number of valence electrons of the central atom, M is the monovalent electron, C is the positive charge and A represents the negative charge.

Step 3: Calculation for Hybridisation of ${\mathrm{XeOF}}_4$:

• The total number of valence electrons in ${\mathrm{XeOF}}_4$ is $8$ and there are $4$ monovalent atoms.
• $$\begin{gathered} \mathrm{Steric}\mathrm{number}=\frac{1}{2}\left[8+4\right]=\frac{1}{2}\times 12 \\ \mathrm{Steric}\mathrm{number}=6 \end{gathered}$$
• The Steric number is $6$, thus the hybridization of ${\mathrm{XeOF}}_4$ is ${\mathrm{sp}}^3{\mathrm{d}}^2$.

Step 4: Structure and shape of ${\mathrm{XeOF}}_4$:

• The xenon $\left(\mathrm{Xe}\right)$ atom must bond with the four fluorine $\left(\mathrm{F}\right)$atoms into four single bonds and double-bonded with an oxygen atom $\left(\mathrm{O}\right)$.
• It has one lone pair of electrons.
• As per VSEPR theory, ${\mathrm{sp}}^3{\mathrm{d}}^2$ has a square pyramidal shape.
• Therefore, the Xenon oxyfluoride(${\mathrm{XeOF}}_4$) is square pyramidal in shape with ${\mathrm{sp}}^3{\mathrm{d}}^2$ hybridization.
• The structure of ${\mathrm{XeOF}}_4$ is given below: