A xenon compound 'A' upon partial hydrolysis gives $\mathrm{XeO}_{2} \mathrm{~F}_{2}$. The number of lone pair of electrons present in compound $A$ is \_\_\_\_. (Round off to the nearest integer).

JEE Main · 2021 · 18 Mar 2021 · Shift 2 · p-Block Elements

A xenon compound 'A' upon partial hydrolysis gives

XeO_{2} F_{2}

. The number of lone pair of electrons present in compound

A

is ____. (Round off to the nearest integer).

Answer: 19

Solution

Partial hydrolysis of a xenon fluoride replaces fluorine atoms by oxygen in stages, and the compound that yields

Xe O_{2} F_{2}

on partial hydrolysis is xenon hexafluoride, so A is

Xe F_{6}

. In

Xe F_{6}

the xenon atom forms six bonds and, of its eight valence electrons, six are engaged in bonding, leaving one lone pair on xenon. Each of the six fluorine atoms carries three lone pairs of its own, which account for

6 \times 3 = 18

lone pairs, so together with the single lone pair on xenon the molecule holds

18 + 1 = 19

lone pairs in all.

JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Integer answer

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