Question

Complete the following reaction:
$Xe{F_6} + 2{H_2}O \to $

Answer 1

Hint: Partial and controlled hydrolysis of $Xe{F_6}$ (xenon hexafluoride) gives xenon oxyflourides. $Xe{F_6}$ undergoes partial hydrolysis with $2$ molecules of ${H_2}O$ to produce xenon dioxy difluoride $(Xe{O_2}{F_2})$.

Complete step by step answer:
-$Xe{F_6}$ (xenon hexafluoride) is a fluoride of xenon , a noble gas. It is $s{p^3}{d^3}$ hybridised molecule and has a distorted octahedral shape. $Xe{F_6}$ reacts vigorously with water. The products of partial hydrolysis of $Xe{F_6}$ depends on the amount of water.
-If $Xe{F_6}$ is hydrolysed by 2 molecules of water, then the product will be xenon dioxydiflouride $(Xe{O_2}{F_2})$.
$Xe{F_6}\,(s) + 2{H_2}O\,(l) \to Xe{O_2}{F_2}\,(s) + 4HF\,(aq)$
-$Xe{O_2}{F_2}$( xenon dioxydiflouride) are metastable colourless crystals. Its melting point is ${31^ \circ }C$. It decomposes into $Xe{F_2}$ (xenon difluoride) and ${O_2}$ on standing, at room temperature. It is $s{p^3}d$ hybridised. In xenon dioxydifluoride the central atom will contain 4 bond pairs and a lone pair, due to the presence of the lone pair it has a see-saw shape. Due to the presence of a lone pair, $Xe{O_2}{F_2}$ is polar in nature.

Note:
If the xenon hexafluoride is partially hydrolysed by one mole of water then $XeO{F_4}$ (xenon oxytetrafluoride) will be formed.
$Xe{F_6}\,(s) + {H_2}O\,(l) \to XeO{F_4}\,(s) + 2HF\,(aq)$
Complete hydrolysis of xenon hexafluoride yields xenon trioxide.
$Xe{F_6}\,(s) + 3{H_2}O\,(l) \to Xe{O_3}\,(s) + 6HF\,(aq)$