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

Explanation

The given reaction occurs because `XeO_(6)^(4-)` oxidised `F^(-)` and `F^(-)` reduces `XeO_(6)^(4-)`
`overset(-8) (X)eO_(6 (aq))^(4-) + 2 overset(-1)(F)_((aq))^(-) + 6 H_((aq))^(+) to overset(+6)(X)e O_(3 (g)) + overset(0)(F)_(2 (g)) + 3H_(2)O _((l))`
In this reaction , the oxidation number (O.N.) of Xe decreases from + 8 in `XeO_(6)^(4-)` to +6 in `XeO_(3)` and the O.N. of F increases from -1 in `F^(-)` to O in `F_(2)` .
Hence , we can conclude that `Na_(4)XeO_(6)` is a stronger oxidising agent than `F^(-)`