Question

What is the hybridization of $\mathrm{Xe}$ in ${\mathrm{XeO}}_3$?

Answer 1

Step 1: 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 monovalent atoms attached to the central atom, $\mathrm{C}$ is the positive charge, and $\mathrm{A}$represents the negative charge.

Step 2: Hybridisation of $\mathrm{Xe}$ in ${\mathrm{XeO}}_3$:

• The central atom is Xenon $\mathrm{Xe}$ and its atomic number is $54$, so its electronic configuration is $\left[\mathrm{Kr}\right]4{\mathrm{d}}^{10}5{\mathrm{s}}^{2}5{\mathrm{p}}^6$.
• So, $\mathrm{Xe}$ has $8$ valence elecetrons.
• $\mathrm{Hybridization}\mathrm{of}{\mathrm{XeO}}_3=\frac{1}{2}\times (8+0-0+0)=4$
• The hybridization number of ${\mathrm{XeO}}_3$ is $4$. Thus, the hybridization of $\mathrm{Xe}$ in ${\mathrm{XeO}}_3$ is ${\mathrm{sp}}^3$.