Why does the following reaction occur?
$XeO_{6(aq)}^{4-} + 2F_{(aq)}^{-} + 6H_{(aq)}^{+} \to XeO_{3(g)} + F_{2(g)} + 3H_2O_{(l)}$
What conclusion about the compound $Na_4XeO_6$ (of which $XeO_6^{4-}$ is a part) can be drawn from the reaction?

(N/A) The oxidation states of the species are as follows:
$XeO_{6(aq)}^{4-} (Xe: +8) + 2F_{(aq)}^{-} (F: -1) + 6H_{(aq)}^{+} \to XeO_{3(g)} (Xe: +6) + F_{2(g)} (F: 0) + 3H_2O_{(l)}$
In this reaction,the oxidation number of $Xe$ decreases from $+8$ to $+6$ (reduction),and the oxidation number of $F$ increases from $-1$ to $0$ (oxidation).
Since $XeO_6^{4-}$ acts as an oxidizing agent by oxidizing $F^{-}$ to $F_2$,we can conclude that $Na_4XeO_6$ is a very strong oxidizing agent.