Question

At $143\mathrm{K}$, the reaction of ${\mathrm{XeF}}_4$ with ${\mathrm{O}}_2{\mathrm{F}}_2$ produces a Xenon compound $\mathrm{Y}$. The total number of lone pairs of electrons present on the whole molecule of $\mathrm{Y}$ is -

Answer 1

At $143\mathrm{K}$, the reaction of ${\mathrm{XeF}}_4$ with ${\mathrm{O}}_2{\mathrm{F}}_2$ which produces a Xenon compound $\mathrm{Y}$is given below-

$$\begin{gathered} {\mathrm{XeF}}_4+\mathrm{O}{}_2\mathrm{F}_2\stackrel{143\mathrm{K}}{\to }{\mathrm{XeF}}_6+{\mathrm{O}}_2 \\ \left(\mathrm{Y}\right) \end{gathered}$$

$\mathrm{Y}$ is Xenon hexafluoride $\left({\mathrm{XeF}}_6\right)$.

Xenon hexafluoride has a Xenon atom and six Fluorine atom. The xenon atom has a lone pair of electrons. Each Fluorine atom has three lone pairs of electrons.

So, the total lone pairs on the compound $=1+3\times 6$

$=19$

Structure of Xenon hexafluoride is given below-

So, the compound $\mathrm{Y}$ has $19$ lone pairs of electrons.