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:

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

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.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Question Image

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