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Question
Why does the following reaction occur?
\[\ce{XeO^{4-}_6 (aq) + 2F- (aq) + 6H+ (aq) -> XeO3(g) + F_2(g) + 3H_2O(l)}\]
What conclusion about the compound Na4XeO6 (of which `"XeO"_6^(4+)` is a part) can be drawn from the reaction.
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Solution
The given reaction occurs because `"XeO"_6^(4-)` oxidises `"F"^(-)` and `"F"^(-)` reduces `"XeO"_6^(4-)`
\[\ce{^{+8}XeO^{4-}_6 (aq) + 2^{-1}F- (aq) + 6H+ (aq) -> ^{+6}XeO3(g) + ^{0}F_2(g) + 3H_2O(l)}\]
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"_4"XeO"_6` is a stronger oxidising agent than F–.
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