Question
Why does the following reaction occur?
$\text{XeO}^{4-}_6(\text{aq})+2\text{F}^{-}(\text{aq})+6\text{H}^+(\text{aq})\rightarrow\\\text{XeO}_3(\text{g})+\text{F}_2(\text{g})+3\text{H}_2\text{O(l)}$
What conclusion about the compound $Na_4XeO_6$ (of which $\text{XeO}^{4-}_6$ is a part) can be drawn from the reaction.

Answer

The given reaction occurs because $\text{XeO}^{4-}_6$ oxidises $F^-$ and $F^-$ reduces $\text{XeO}^{4-}_6.$ $\stackrel{{+8}}{\ \ \ \ \ \hbox{XeO}}^{4-}_{6(\text{aq})}+\stackrel{{-1}}{2\ \ \hbox{F}^{-}}_{(\text{aq})}+6\text{H}^{+}_{(\text{aq})}\rightarrow\\\stackrel{{+6}}{\ \ \ \ \ \ \hbox{XeO}}_{3(\text{g})}+\stackrel{{0}}{\ \ \hbox{F}_2}_{(\text{g})}+3\text{H}_2\text{O}_{(\text{l})}$In this reaction, the oxidation number (O.N.) of Xe decreases from +8 in $\text{XeO}^{4-}_6$ to +6 in $XeO_3$ and the O.N. of F increases from -1 in $F^-$ to O in $F_2$​​​​​​​​​​​​​​.
Hence, we can conclude that $Na_4XeO_6​​​​​​​$​​​​​​​ is a stronger oxidising agent than $F^-$​​​​​​​.

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