The difference in the oxidation state of $\mathrm{Xe}$ between the oxidised product of $\mathrm{Xe}$ formed on complete hydrolysis of $\mathrm{XeF}_{4}$ is ___________
Answer (integer)
2
Solution
$$
6 \mathrm{XeF}_4+12 \mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{XeO}_3+4 \mathrm{Xe}+24 \mathrm{HF}+3 \mathrm{O}_2
$$<br/><br/>
in $\mathrm{XeO}_3$, Oxidation state of $\mathrm{Xe}=+6$<br/><br/>
in $\mathrm{XeF}_4$, Oxidation state of $\mathrm{Xe}=+4$<br/><br/>
So difference in oxidation state $=2$
About this question
Subject: Chemistry · Chapter: p-Block Elements · Topic: Group 13: Boron Family
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