JEE Main · 2026 · 23 Jan 2026 · Shift 2 · p-Block Elements

$\mathrm{Xe}{\mathrm{O}}_2{\mathrm{F}}_2$ places four bonding regions (two Xe=O, two Xe--F) and one lone pair about xenon, a trigonal-bipyramidal arrangement that gives a see-saw shape, with the lone pair and the two oxygen atoms equatorial and the two fluorine atoms axial. The F--Xe--F angle is therefore close to $180^{\circ }$, so those two descriptions are correct. The O--Xe--O angle is the equatorial angle of about $105^{\circ }$, not $180^{\circ }$; xenon brings eight valence electrons, not sixteen; and counting its $\sigma$ bonds, $\pi$ bonds and lone pair gives seven electron pairs, not five. The statements that are not true are that xenon has five electron pairs, that the O--Xe--O angle is near $180^{\circ }$, and that xenon has sixteen valence electrons.