Answer:
Option D
Explanation:
Plan This problem can be solved by using concept involved in chemical properties of xenon oxide and xenon fluoride
XeF4 on complete hydrolysis produce XeO3
XeO3 on reaction with OH- produces HXeO4- which on further treatment with OH- undergo slow disproportionation reaction and produces XeO64- along with Xe(g) , H2O (l) and O2 (g) as a by-product
OXidation half cell in basic aqueous solution
$HXeO_4^-+5OH^{-}\rightarrow XeO_{6}^{4-}+3H_{2}O+2e$
Reduction half-cell in basic aqueous solution
$HXeO_4^-+3H_{2}O+6e^{-}\rightarrow Xe+7OH^{-}$
Balanced overall disproportionation reaction is
$4HXeO_4^-+8OH^{-}\rightarrow 3XeO_6^{4-}+Xe+6H_{2}O$
2 products
Complete sequence of reaction can be shown as
$XeF_{6}+3H_{2}O\rightarrow XeO_{3}+3H_{2}F_{2}$ $\underrightarrow{OH^{-}}$ $HXeO_4^-$
$\underrightarrow{OH^{-}/H_{2}O (disproportionation)}XeO^{4-}_{6}(s)+Xe(g)+H_{2}O(l)+O_{2}(g)$
