Xenon hexafluoride on partial hydrolysis prod | vedclass

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(C) The partial hydrolysis of $XeF_6$ occurs in steps:$1. XeF_6 + H_2O 	o XeOF_4 + 2HF$Here,$X$ is $XeOF_4$. The oxidation state of $Xe$ is $x + (-2)

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$XeOF_4 (+6)$ and $XeO_2F_2 (+6)$

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(C) The partial hydrolysis of $XeF_6$ occurs in steps:$1. XeF_6 + H_2O 	o XeOF_4 + 2HF$Here,$X$ is $XeOF_4$. The oxidation state of $Xe$ is $x + (-2) + 4(-1) = 0$,so $x = +6$.$2. XeOF_4 + H_2O 	o XeO_2F_2 + 2HF$Here,$Y$ is $XeO_2F_2$. The oxidation state of $Xe$ is $x + 2(-2) + 2(-1) = 0$,so $x = +6$.Thus,the compounds are $XeOF_4 (+6)$ and $XeO_2F_2 (+6)$.

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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