8.16 Why does the following reaction occur ?

XeO64– (aq) + 2F (aq) + 6H+(aq) XeO3(g)+ F2(g) + 3H2O(l)

What conclusion about the compound Na4XeO6 (of which XeO64– is a part) can be drawn from the reaction.

The given reaction occurs because ${\mathrm{XeO}}_{6}^{4-}$ oxidises
$\stackrel{+8}{\mathrm{X}}e{O}_{6\left(aq\right)}^{4-}+2{{F}^{-1}}_{\left(aq\right)}+6{{H}^{+}}_{\left(aq\right)}\to \stackrel{+6}{\mathrm{X}}e{O}_{\mathit{3}\left(g\right)}\mathit{+}{\stackrel{\mathit{0}}{\mathit{F}}}_{\mathit{2}\left(g\right)}\mathit{+}\mathit{3}{H}_{\mathit{2}}{O}_{\left(l\right)}$ 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 O.N. of F increases from -1 in ${\mathrm{F}}^{-}$ to O in ${\mathrm{F}}_{2}$.
Hence, we can conclude that ${\mathrm{Na}}_{4}{\mathrm{XeO}}_{4}$ is a stronger oxidizing agent that ${\mathrm{F}}^{-}$.