Why does the following reaction occur?
$XeO_6^{4-}{(aq)} + 2F^{-}{(aq)} + 6H^{+}{(aq)} \rightarrow XeO_{3_{(g)}} + F_{2_{(g)}} + 3H_2O_{(l)}$
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 given reaction occurs because $XeO_6^{4-}$ acts as an oxidizing agent and $F^{-}$ acts as a reducing agent.
In the reaction: $\mathop {Xe}\limits^{+8} O_6^{4-}{(aq)} + 2\mathop {F^{-}}\limits^{-1}{(aq)} + 6H^{+}{(aq)} \to \mathop {Xe}\limits^{+6}O_{3_{(g)}} + \mathop {F_2}\limits^{0}_{(g)} + 3H_2O_{(l)}$
The oxidation number $(O.N.)$ of $Xe$ decreases from $+8$ in $XeO_6^{4-}$ to $+6$ in $XeO_3$ (reduction).
The $O.N.$ of $F$ increases from $-1$ in $F^{-}$ to $0$ in $F_2$ (oxidation).
Since $XeO_6^{4-}$ oxidizes $F^{-}$ to $F_2$,we can conclude that $Na_4XeO_6$ is a very strong oxidizing agent.