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How are XeO(3) and XeOF(4) prepared ?...

How are `XeO_(3)` and `XeOF_(4)` prepared ?

Text Solution

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`XeF_(6)` on hydrolysis produce `XeO_(3)`
`XeF_(6)+3H_(2)O toXeO_(3)+6HF`
Partial hydrolysis of `XeF_(6)` gives `XeOF_(4)`
`XeF_(6)+H_(2)O toXeOF_(4)+2HF`
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