A compound of $Xe$ and $F$ is found to have $53.5$% of $Xe$. What is oxidation number of $Xe$ in this compound?

The correct option is D $+6$
A compound of $Xe$ and $F$ is found to have $53.5$% of $Xe$.
100 g of this compound will have $53.5$ g of Xe and $46.5$ g of F.
The atomic masses of Xe and F are 131.2 g/mol and 19 g/mol respectively.
The number of moles of $Xe=\frac{53.5}{131.2}=0.4$
The number of moles of $F=\frac{46.5}{19}=2.4$
The mole ratio $Xe:F=0.4:2.4=1:6$
Molecular formula of the compound is $Xe{F}_6$
Let n be the oxidation number of $Xe$ in $Xe{F}_6$
The oxidation number of F is $-1$
$n+6(-1)=0$
$n-6=0$
$n=6$
The oxidation number of $Xe$ in $Xe{F}_6$ is $+6$