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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? |
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Answer» `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)`. |
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