Question

The shape of $Xe{O}_4$ molecule is tetrahedral due to:

• A. $sp$
• B. $s{p}^2$
• C. $s{p}^3$
• D. none of these

Answer 1

The correct option is C $s{p}^3$

Hybridisation $=\frac{1}{2}\left[\text{No. of e in valence sheel + No. of mono valent atom + change on anion - charge on cation}\right]$
For $Xe{O}_4$
Hybridisation $=\frac{1}{2}(8+0+0-0)=4$
$\therefore$ $s{p}^3$ hybridisation