Why does the following reaction occur ?
`XeO_(6)^(4-) (aq) + 2 F^(-) (aq) + 6 H^(+) (aq) to XeO_(3) (g) + F_(2)(g) + 3H_(2)O(1)`
What conclusion about the compound `Na_(4)XeO_(6)` (of which `XeO_(6)^(4-)` is a part ) can be drawn from the reaction.

Explanation

`overset(+8)(Xe)O_(6)^(4-)(aq)+2F^(Θ)(aq)+6H^(o+)(aq)rarr overset(+6)(Xe)O_(3)(s)+overset(0)(F_(2))(g)+3H_(2)O(l)`
Here, oxidation number of `Xe` decreases from `+8` in `XeO_(6)^(4-)` to `+6` in `XeO_(3)`, while that of F increases from `-1` in `F^(Θ)` to `0` in `F_(2)`. Therefore, `XeO_(6)^(4-)` is reduced while `F^(Θ)` is oxidised. This reaction occure because `Na_(2)XeO_(6) (or XeO_(6)^(4-))` is a stronger oxidising agent than `F_(2)`.