At 143 K. the reaction of $\mathrm{XeF}_{4}$ with $\mathrm{O}_{2} \mathrm{~F}_{2}$ produces a xenon compound $\mathbf{Y}$. The total number of lone pair(s) of electrons present on the whole molecule of $\mathbf{Y}$ is ____

JEE Advanced · 2019 · 27 May 2019 · Paper 1 · p-Block Elements

At 143 K. the reaction of

XeF_{4}

with

O_{2} F_{2}

produces a xenon compound

Y

. The total number of lone pair(s) of electrons present on the whole molecule of

Y

is ____

Answer: 19

The reaction is:

XeF_{4} + O_{2} F_{2} \to XeF_{6} + O_{2}

Compound Y is

XeF_{6}

.

In

XeF_{6}

Xe has 8 valence electrons. With 6 bonding pairs (to 6 F atoms), there is 1 lone pair on Xe.
Each F atom has 3 lone pairs (7 valence electrons, 1 used in bonding).

Total lone pairs = 1 (on Xe) + 6 × 3 (on F atoms) = 1 + 18 = 19.

Answer: 19

JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Numerical answer