Question

Find out difference in oxidation state of Xe in completely Hydrolysed form of XeF4 and XeF6

Answer 1

$6XeF_4+12H_2O\rightarrow 2XeO_3+4Xe+24HF+3O_2$
in $XeO_3, Oxidation \;state \; of Xe=+6$
in $XeF_4, Oxidation\;state\;of \;Xe=+4$
$\text{So difference in oxidation state = 2}$

$\text{Hence, the correct answer is : 2}$