JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Chemical Bonding and Molecular Structure

When $\mathrm{Xe}{\mathrm{F}}_4$ reacts with $\mathrm{Sb}{\mathrm{F}}_5$ it forms $[\mathrm{Xe}{\mathrm{F}}_3{]}^{+}[\mathrm{Sb}{\mathrm{F}}_6{]}^{-}$, and in doing so the xenon changes from $s{p}^3{d}^2$ in $\mathrm{Xe}{\mathrm{F}}_4$ to $s{p}^{3}d$ in the $[\mathrm{Xe}{\mathrm{F}}_3{]}^{+}$ cation. In the other choices the marked atom keeps its hybridisation: sulphur stays $s{p}^3$ when ${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$ reacts with $\mathrm{NaCl}$, phosphorus stays $s{p}^3$ through the disproportionation of ${\mathrm{H}}_3\mathrm{P}{\mathrm{O}}_2$, and nitrogen stays $s{p}^3$ as $\mathrm{N}{\mathrm{H}}_3$ is protonated to $\mathrm{N}{\mathrm{H}}_4^{+}$. Only the xenon reaction alters the hybridisation of the underlined atom.