The difference in the oxidation state of Xe between the oxidised...

Hydrolysis of ${XeF}_{4}$ will give ${XeO}_{3}$. This hydrolysis reaction is a redox reaction. This redox reaction is an example of disproportionation reaction.

$6{XeF}_{4} + 12{H}_{2}O\overset{}{\to}4Xe + 2{XeO}_{3} + 24HF + 3{O}_{2}$

The oxidation states of Xenon in the reactant and product is

${XeO}_{3}: Xe ( + 6 )$

${XeF}_{4}: Xe ( + 4 )$

Hence, the difference in oxidation state is $2$.