Question

Why does the following reaction occur?
$XeO_6^{4-}(aq) + 2F^ - (aq) + 6H^ + (aq) \rightarrow XeO_3(g) + F_2(g) + 3H_2O(l)$ What conclusion about the compound $Na_4XeO_6$(of which $XeO_6^{4-}$ is a part) can be drawn from the reaction.

Answer 1

The given reaction occurs because $XeO_6^{4-}$ oxidises $F^-$ and $F^-$ reduces $XeO_6^{4-}$.

$\overset{+8}{X}eO_6^{4-}(aq)+2F^-+6H^+\rightarrow XeO_3(g)+F_2(g)+3H_2O(l)$

In this reaction, the oxidation number (O.N.) of $Xe$ decreases from $+8$ in $XeO_6^{4- }$to $+6$ in $XeO_3$ and the O.N. of $F$ increases from $-1$ in $F ^-$ to $0$ in $F_2$.
Hence, we can conclude that $Na_4XeO_6$ is a stronger oxidising agent than $F ^-$ .