Why does the following reaction occur?
XeO_6^(4-) (aq) + 2F^(-) (aq) + 6H^(+) (aq) -> XeO_3(g) + F_2(g) + 3H_2O(l)
What conclusion about the compound Na4XeO6 (of which `XeO_6^(4+)` is a part) can be drawn from the reaction.
Here O.N of Xe decreases from +8 in XeO_6^(4-) to +6 in `XeO_3` while that of F increase from-I in `F^(-)` to 0 in `F_2`. Therefore `XeO_6^(4-)` is reduced while `F^(-)` is oxidised. This reaction occur because `Na_2XeO_6^(4-)` (or `XeO_6^(4-))` is stronger oxidising agent than `F_2`
The given reaction occurs because `XeO_6^(4-)` oxidises `F^(-)` and `F^(-)` reduces `XeO_6^(4-)`
In this reaction, the oxidation number (O.N.) of Xe decreases from +8 in `XeO_6^(4-)`
to +6 in XeO3 and the O.N. of F increases from –1 in F– to O in F2.
Hence, we can conclude that `Na_4XeO_6` is a stronger oxidising agent than F–.