Why does the following reaction occur?
$\mathrm{XeO}_{6}^{4-}(\mathrm{aq})+2 \mathrm{~F}^{-}(\mathrm{aq})+6 \mathrm{H}^{+}(\mathrm{aq}) \rightarrow \mathrm{XeO}_{3}(\mathrm{~g})+\mathrm{F}_{2}(\mathrm{~g})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})$
What conclusion about the compound $\mathrm{Na}_{4} \mathrm{XeO}_{6}$ (of which $\mathrm{XeO}_{6}^{4-}$ is a part) can be drawn from the reaction.
The given reaction occurs because $\mathrm{XeO}_{6}^{4}$ oxidises $\mathrm{F}^{-}$and $\mathrm{F}^{-}$reduces $\mathrm{XeO}_{6}^{4}$.
In this reaction, the oxidation number (O.N.) of Xe decreases from $+8$ in $\mathrm{XeO}_{6}^{4}$ to $+6$ in $\mathrm{XeO}_{3}$ and the $\mathrm{O} . \mathrm{N}$. of $\mathrm{F}$ increases from $-1$ in $\mathrm{F}^{-}$to $\mathrm{O}$ in $\mathrm{F}_{2}$.
Hence, we can conclude that $\mathrm{Na}_{4} \mathrm{XeO}_{6}$ is a stronger oxidising agent than $\mathrm{F}^{-}$.
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