JEE Main 2026 — Chemical Bonding and Molecular Structure Question with Solution

The central atom in $Xe{O}_6^{4-}$ is Xenon (Xe).

The number of valence electrons in Xe is $8$.

Let the oxidation state of Xe in $Xe{O}_6^{4-}$ be $x$.

$x+6(-2)=-4$

$x=+8$

Since the oxidation state of Xe is $+8$, all $8$ of its valence electrons are involved in bonding with the $6$ oxygen atoms. There are no non-bonding electrons left on the central atom.

Number of lone pairs on Xe $=0$

Xe forms $6$ $\sigma$ bonds with the $6$ oxygen atoms (and $2$ $\pi$ bonds to satisfy its $+8$ oxidation state and minimize formal charge).

Number of $\sigma$ bond-pairs $=6$

Total number of $\sigma$ bond-pairs and lone pairs around the central atom $=6+0=6$.

Answer: $6$