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

`overset(+3)(Xe)O_6^(4-)(aq)+overset(-1)(2F^(-))(aq)+6^(+)(aq)rarroverset(+6)(Xe)O_3(s)+overset(0)F_2(g)+3H_2O(l)`
O.N. of Xe decreases form +8 (in `XeO_6^(4-)`) to +6(in `XeO_3`) and that of F increases from -1 (in `F^-`) to 0 (in `F_2`) . Therefore , `XeO_6^(4-)` is reduced while `F^(-)` is oxidised . The reaction occurs because `Na_4XeO_6` (or `XeO_6^(4-)` ) is a stronger oxidising agent than `F_2` .