Calculate the oxidation number of $Xe$ in $Xe{F}_4$ and $Xe{O}_2$.

1) In $Xe{F}_4$

$Xe{F}_4=0$

$Xe+4\times (-1)=0$

$Xe=+4$

2) In $Xe{O}_2$

$Xe{O}_2=0$

$Xe+(-2)\times 2=0$

$Xe=+4$

3) Thus, Oxidation state of Xe in both the compound is $+4$.