Question

End product of the complete hydrolysis of $Xe{F_6}$ is:
a) $XeO{F_4}$
b) $Xe{O_2}{F_2}$
c) $Xe{O_3}$
d) $Xe{O_3}^ -$

Answer 1

$Xenon$ unlike any other noble gas element from group- $18$ , is proven to be reactive. It reacts with most electronegative elements like $Fluorine$ , $Oxygen$ and from different compounds.

Complete step by step solution:
$Xenon$ Belongs to the noble gas family thus it was considered as inert. But later it was discovered that it is reactive among all other members in its group. $Xenon$ Is used in mostly arc lamps to produce ultraviolet light. Also it is used in various instruments to detect the kind of radiation like-ray.
$Xenon$ Reacts directly with only $Fluorine$ as it is the most electronegative atom. The compounds of $Xenon$ with $Oxygen$ are obtained from the $fluorides.$
$Xenon$ Reacts with $Fluorine$ when their gases are heated up to $400^\circ C$ in a sealed Nickel vessel.
Different ratios of the mixture of reactants give different products.
Like in $2:1$ ratio-
$Xe + {F_2} \to Xe{F_2}$
In $1:5$ ratio-
$Xe + 2{F_2} \to Xe{F_4}$
In $1:20$ ratio-
$Xe + 3{F_2} \to Xe{F_6}$
These compounds are solids. They can sublime at room temperature, thus stored in nickel vessels.
Compounds like $Xe{F_2}$ react with water but it undergoes very slow hydrolysis.
$2Xe{F_2} + 2{H_2}O \to 2Xe + 4HF + {O_2}$
$Xe{F_6}$ Reacts very violently with water. It also undergoes slow hydrolysis by atmospheric moisture forming a very highly explosive solid $Xe{O_3}$ .
$Xe{F_6} + 6{H_2}O \to Xe{O_3} + 6HF$
Also it undergoes partial hydrolysis forming a colourless liquid xenon oxyfluoride, ( $XeO{F_4}$ ).
$Xe{F_6} + {H_2}O \to XeO{F_4} + 2HF$ .
Thus on complete hydrolysis $Xe{O_3}$ is formed and it is also hygroscopic in nature. Further it reacts with $Xe{F_6}$ and $XeO{F_4}$ .
$Xe{O_3} + 2Xe{F_6} \to 3XeO{F_4}$
Though it is soluble in water, it doesn't ionise but in basic medium it exists as xenate ion ${[HXe{O_4}]^ - }$ .
Hence in complete hydrolysis the end product will be $Xe{O_3}$ .
In $Xe{F_6}$ , Xenon is an $+ 6$ oxidation state and the structure it acquires will be a distorted octahedron.
In $Xe{O_3}$ , Xenon is also in $+ 6$ oxidation state and it acquires Pyramidal structure.
$\therefore$ Correct option is (c).

Note:
The formation of xenate ion particularly occurs in an alkaline medium with a $pH = 10.5$ . $Xe{O_3} + NaOH \to N{a^ + }{[HXe{O_4}]^ - }$
$XeO{F_4}$ also has $+ 6$ oxidation state with seven electron pairs and one lone pair acquiring square pyramidal structure. The hybridisation of square pyramidal is $s{p^3}d$ .