$\mathrm{XeF}_{6}$ on hydrolysis gives
- $\mathrm{XeO}_{3}$
- $\mathrm{XeO}$
- $\mathrm{XeO}_{2}$
- $\mathrm{Xe}$
Solution
\mathrm{XeF}_{6}+3 \mathrm{H}_{2} \mathrm{O} ightarrow \mathrm{XeO}_{3}+6 \mathrm{HF}
$$ ~
Asked in: JEE-TOPICTESTS-CHEMISTRY
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