Perxenate ion is:

X e O_{6}^{4 -}

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H X e O_{4}^{-}

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X e O_{4}^{2 -}

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X e O_{4}^{-}

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Solution

The correct option is A $X e O_{6}^{4 -}$
$X e O_{3}$ reacts with aqueous alkali to form the hydrogen xenate ion $H X e O_{4}^{-}$ which slowly disproportionates to give xenon and perxenate ion in which $X e$ is present in $+ 8$ oxidation state.

$X e O_{3} + {O H}^{-} ⟶ H X e O_{4}^{-}$

$2 H X e O_{4}^{-} + 2 {O H}^{-} ⟶ X e O_{6}^{4 -} + X e + 2 H_{2} O + O_{2}$

${X e O_{6}}^{4 -}$ is perxenate ion.

Hence, the correct answer is option $A$ .

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