In reaction (1), XeF_6 undergoes complete hyd | vedclass

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(A) Complete hydrolysis of $XeF_6$ is given by the reaction:$XeF_6 + 3 H_2O \longrightarrow XeO_3 + 6 HF$Thus, $X = XeO_3$.Partial hydrolysis of $XeF_

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$XeO_3, XeOF_4, XeO_2 F_2$

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(A) Complete hydrolysis of $XeF_6$ is given by the reaction:$XeF_6 + 3 H_2O \longrightarrow XeO_3 + 6 HF$Thus, $X = XeO_3$.Partial hydrolysis of $XeF_6$ occurs in two steps:$XeF_6 + H_2O \longrightarrow XeOF_4 + 2 HF$$XeF_6 + 2 H_2O \longrightarrow XeO_2 F_2 + 4 HF$Thus, $Y$ and $Z$ are $XeOF_4$ and $XeO_2 F_2$ respectively.Therefore, the products $X, Y, Z$ are $XeO_3, XeOF_4, XeO_2 F_2$.

In reaction $(1)$, $XeF_6$ undergoes complete hydrolysis to form $HF$ and $X$. In reaction $(2)$, $XeF_6$ undergoes partial hydrolysis to form $HF$, $Y$, and $Z$. The products $X, Y, Z$ respectively, are

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