Question

At $143K$, the reaction of $Xe{F}_4$ with $O_2{F}_2$ produces a xenon compound Y. The total number of lone pair(s) of electrons present in $1$ molecule of Y is:

Answer 1

The complete reaction is:
$Xe{F}_4+O_2{F}_2\to Xe{F}_6+O_2$

In $Xe{F}_6$ each $F$ atom contains $3$ lone pairs and the central atom $Xe$ contains one lone pair. Thus, the total number of lone pairs = $6\times 3+1=19$