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

The correct option is D $+6$
Let the total weight of the compound is 100 g.

So the amount of $Xe$ present is 53.5 g and of $F$ is 46.7 g.

Now we find moles of $Xe$ is $\frac{53.5}{133}$=0.4 and of $F$ is $\frac{46.7}{19}$ is 2.45.

Divide both by 0.4 so that we get a simple whole-number ratio, we get

Moles of $Xe=\frac{0.4}{0.4}$=1

Moles of $F=\frac{2.45}{0.4}$=6

So the formula is $Xe{F}_6$.

The oxidation state of $Xe$ is +6.

Hence, the correct option is $D$