Question

What is the hybridization of xenon tetrafluoride?

Answer 1

Step 1: Formula used for calculation:

• 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 monovalent atoms attached to the central atom, $\mathrm{C}$ is the positive charge, and $\mathrm{A}$represents the negative charge.

Step 2: Hybridization of Xenon tetrafluoride ${\mathrm{XeF}}_4$:

• In ${\mathrm{XeF}}_4$, the central atom is $\mathrm{Xe}$ and it has $8$ electrons in its valence shell and $4$ monovalent $\mathrm{F}$ atoms are attached to it.
• By using the above formula, the hybridization of ${\mathrm{XeF}}_4=\frac{1}{2}\left[8+4-0+0\right]=\frac{1}{2}\times 12=6$
• The hybridization number is $6$. So, the hybridization of ${\mathrm{XeF}}_4$ is ${\mathrm{sp}}^3{\mathrm{d}}^2$.

Hence, the hybridization of Xenon tetrafluoride is ${\mathrm{sp}}^3{\mathrm{d}}^2$.