Question

Using VBT theory describe shape of $Xe{O}_4$.

Answer 1

Xenon undergoes $s{p}^3$ hybridization in the fourth excited state in the formation of $Xe{O}_4$. In fourth excited state, $Xe$ atom has $8$ unpaired electrons. One $s$ and three $p$ orbitals undergo $s{p}^3$ hybridization. Since it has no lone pair of electrons, the shape of $Xe{O}_4$ is tetrahedral with $4\sigma$ and $\pi$ bonds each with bond angle of $109°$.