A compound of Xe and F is found to have 53.4% Xe (atomic mass 133). Oxidation number of Xe in this compound is

The correct option is D + 6

Explanation;
Compound of Xe and F:
Percentage of Xe = 53.3%
Percentage of F = 100 - 53.3 = 46.7%

Find relative moles:

Xe $\to$ Percentage of compound/Atomic mass=$\frac{53.3}{133}$ = 0.4

F $\to$ Percentage of compound/Atomic mass=$\frac{46.7}{19}$ = 2.45

Mole ratio:
For Xe = 0.40
For F = 2.45

Ratio of Xe : F = $\frac{0.40}{2.45}$ = $\frac{1}{6}$


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

Since F is present in -1 state therefore Xe exists in +6 state.