Xenon hexafluoride on partial hydrolysis produces compounds ' $X$ ' and ' $Y$ '. Compounds ' $X$ ' and ' $Y$ ' and the oxidation state of $Xe$ are respectively

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${XeO}_{2} (+ 4)$ and ${XeO}_{3} (+ 6)$

${XeOF}_{4} (+ 6)$ and ${XeO}_{3} (+ 6)$

${XeO}_{2} F_{2} (+ 6)$ and ${XeO}_{2} (+ 4)$

${XeOF}_{4} (+ 6)$ and ${XeO}_{2} F_{2} (+ 6)$

answer is D.

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Detailed Solution

${XeO}_{3} + 6 HF \to {XeF}_{6} + 3 H_{2} O$

${XeOF}_{4} + H_{2} O \to {XeO}_{2} F_{2} + 2 HF$

Xenon hexafluoride on partial hydrolysis forms xenon oxy tetrafluoride ${XeOF}_{4}$ or xenon dioxide difluoride ${XeO}_{2} F_{2}$ .

The oxidation state of xenon here is +6 in both the cases.

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