Xenon trioxide, $X e O_{3}$ reacts with aqueous base to form the Xenon containing anion. This ion reacts further with ${OH}^{-}$ to form the perxenate anion $X e O_{6}^{4 -}$ in the following reaction: $2 H X e O_{4}^{-} (a q) + 2 O H^{-} (a q) \to X e O_{6}^{4 -} (a q) + X e + O_{2} (g) + 2 H_{2} O$
In this reaction:

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Xenon is both oxidized and reduced

Oxygen in $O H^{-}$ is oxidized.

The electrons required for the reduction of one xenon atom are supplied by both, another xenon atom and OH^- ions.

In the conversion of 2 moles of $H X e O_{4}^{-}$ to $X e$ and $X e O_{6}^{4 -}$ , the net electrons required are 4

answer is A, B, C, D.

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Detailed Solution

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In the first step xenate ion $H X e O_{4}^{-}$ is formed. In the second step it undergoes both oxidation and reduction. In the second step the perxenate ion $X e O_{6}^{4 -}$ is formed in which Xe oxidation number is +8 and the second Xe is reduced to Xe. In addition O2- is oxidized to O2.
One Xe atom in $H X e O_{4}^{-}$ loses 2 electrons to convert into $H X e O_{6}^{4 -}$ . Second Xe atom in $H X e O_{4}^{-}$ gains 6 electrons to give Xe. Thus 2 $H X e O_{4}^{-}$ ions require a net 4 electrons. These electrons can be supplied by the oxidation of $2 O^{2 -}$ ions to give $O_{2} .$ Structure of perxenate ion is