The given reaction occurs because XeO64- oxidises F- and F-
reduces XeO64- In this reaction, the oxidation number (O.N.) of Xe decreases from +8 in XeO64- to +6 in XeO3 and the O.N. of F increases from –1 in F– to O in F2.
Hence, we can conclude that Na4XeO6 is a stronger oxidising agent than F– .