The difference in the oxidation state of Xe between the oxidised product of Xe formed on complete hydrolysis of XeF$$_4$$ and XeF$$_4$$ is _______

We need to find the difference in oxidation state of Xe between the oxidized product formed on complete hydrolysis of $$XeF_4$$ and $$XeF_4$$ itself.

The complete hydrolysis of $$XeF_4$$ is a disproportionation reaction:

In this reaction, $$XeF_4$$ undergoes disproportionation - some Xe is reduced to elemental form (Xe) and some is oxidized to $$XeO_3$$.

In $$XeF_4$$: F has oxidation state $$-1$$. So $$x + 4(-1) = 0$$, giving $$x = +4$$.

In $$XeO_3$$ (oxidized product): O has oxidation state $$-2$$. So $$x + 3(-2) = 0$$, giving $$x = +6$$.

In Xe (reduced product): Oxidation state = 0.

The oxidized product is $$XeO_3$$ (Xe in +6 state). The oxidation state of Xe in $$XeF_4$$ is +4.

The difference in the oxidation state is 2.