I$\mathrm{XeO_6^{4-}{}_{(aq)}} + 2\mathrm{F^{-}_{(aq)}} + 6\mathrm{H^{+}_{(aq)}}$ → $\mathrm{XeO_{3(g)}} + \mathrm{F_{2(g)}} + 3\mathrm{H_2O_{(l)}}$ n this reaction, O.N. of Xe decreases from +8 in $\mathrm{XeO_6^{4-}}$ to +6 in $\mathrm{XeO_3}$ while that of F increases from –1 in $\mathrm{F^{-}}$ to 0 in $\mathrm{F_2}$. Therefore, $\mathrm{XeO_6^{4-}}$ is reduced while $\mathrm{F^{-}}$ is oxidised. From this reaction it is concluded that $\mathrm{Na_4XeO_6}$ is a stronger oxidising agent than $\mathrm{F_2}$.