# 8.16   Why does the following reaction occur?$XeO_{6}^{4-}(aq)+2F^{-}(aq)+6H^{+}(aq)\rightarrow XeO_{3}(g)+F_{2}(g)+3H_{2}O(l)$ What conclusion about the compound $Na_{4}XeO_{6}$   (of which$XeO_{6}^{4-}$ is a part) can be drawn from the reaction ?

G Gautam harsolia

we conclude that the oxidation state of Xenon changes from  +8 to +6

$XeO_{6}^{4-} (+8)\rightarrow XeO_{3}(+6)$

and oxidation state of F changes from -1 to 0

$F^{-}(-1)\rightarrow F_{2}(0)$

$Na_{4}XeO_{6}$  is reduced by accepting an electron.

It is a strong oxidizing agent than F

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