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

`XeO_(6)^(4-)-(aq)+2 F^(-) (aq)rarroverset(+6)XeO_(3)(s)+F_(2)+3H_(2)O(l)`
here O.N of Xe decreases form +8 in `XeO_(6)^(4-)` to +6 in `XeO_(3)` while that of F increases from -I in `F^(-)` to 0 in `F_(2)` therefore `XeO_(6)^(4-)` is reduced while `F^(-)` is oxiised this reaction occurs because `Na_(2)XeO_(6)^(4-)` is a stronger oxidising agent than `F_(2)`