JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Chemical Bonding and Molecular Structure

The shape of each xenon compound is fixed by the number of bond pairs and lone pairs on xenon, and is paired with the species of identical geometry. $\mathrm{Xe}{\mathrm{O}}_3$ has three bond pairs and one lone pair, a pyramidal shape, the same geometry as $\mathrm{N}{\mathrm{H}}_3$. $\mathrm{Xe}{\mathrm{F}}_2$ has two bond pairs and three lone pairs, a linear shape, the same geometry as ${\mathrm{I}}_3^{-}$. $\mathrm{Xe}{\mathrm{O}}_2{\mathrm{F}}_2$ has four bonding regions and one lone pair (AX${}_4$E), a see-saw shape, the same geometry as $\mathrm{S}{\mathrm{F}}_4$. $\mathrm{XeO}{\mathrm{F}}_4$ has five bonding regions and one lone pair (AX${}_5$E), a square pyramidal shape, the same geometry as $\mathrm{Br}{\mathrm{F}}_5$. The correct matching is $\mathrm{Xe}{\mathrm{O}}_3$ with $\mathrm{N}{\mathrm{H}}_3$, $\mathrm{Xe}{\mathrm{F}}_2$ with ${\mathrm{I}}_3^{-}$, $\mathrm{Xe}{\mathrm{O}}_2{\mathrm{F}}_2$ with $\mathrm{S}{\mathrm{F}}_4$, and $\mathrm{XeO}{\mathrm{F}}_4$ with $\mathrm{Br}{\mathrm{F}}_5$.