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Question
Complete the following equations : XeF4 + O2F2→
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Solution
XeF6 can be prepared by the interaction of XeF4 and O2F2 at 143 K.
XeF4 + O2F2 → XeF6 + O2
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\[\ce{XeF6 + H2O ->[Partial][Hydrolysis] \underline{}\underline{}\underline{}\underline{} + \underline{}\underline{}\underline{}\underline{}}\]
\[\ce{XeF4 + H2O - \underline{}\underline{}\underline{}\underline{} + \underline{}\underline{}\underline{}\underline{}}\]
