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In reaction (1), $\mathrm{XeF}_6$ hydrolysis to form $\mathrm{HF}$ and $X$. In reaction (2), $\mathrm{XeF}_6$ on partial hydrolysis form HF, $Y$ and $Z$.
The products $X, Y, Z$ respectively, are
Options:
The products $X, Y, Z$ respectively, are
Solution:
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Verified Answer
The correct answer is:
$\mathrm{XeO}_3, \mathrm{XeOF}_4, \mathrm{XeO}_2 \mathrm{~F}_2$
Complete hydrolysis of $\mathrm{XeF}_6$ produces xenon trioxide.
$$
\mathrm{XeF}_6+3 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{KeO}_3+6 \mathrm{HF}
$$
Whereas partial hydrolysis of $\mathrm{XeF}_6$ form different products.
$$
\begin{gathered}
\mathrm{XeF}_6+2 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{XeO}_2 \mathrm{~F}_2+4 \mathrm{HF} \\
\mathrm{XeF}_6+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{XeOF}_4+2 \mathrm{HF}
\end{gathered}
$$
$$
\mathrm{XeF}_6+3 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{KeO}_3+6 \mathrm{HF}
$$
Whereas partial hydrolysis of $\mathrm{XeF}_6$ form different products.
$$
\begin{gathered}
\mathrm{XeF}_6+2 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{XeO}_2 \mathrm{~F}_2+4 \mathrm{HF} \\
\mathrm{XeF}_6+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{XeOF}_4+2 \mathrm{HF}
\end{gathered}
$$
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