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

Hint: $ 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 $ .