Chemistry - Redox Reactions Question with Solution | TestHub

Given the atomic ratio of $\mathrm{Xe}:\mathrm{F}$ is $0.4:2.4$.

Simplifying the ratio by dividing by $0.4$:

$\mathrm{Xe}:\mathrm{F}=\frac{0.4}{0.4}:\frac{2.4}{0.4}=1:6$

The empirical formula of the compound is $\mathrm{Xe}{\mathrm{F}}_6$.

Let the oxidation number of $\mathrm{Xe}$ be $x$.

Since the oxidation number of fluorine is $-1$:

$x+6(-1)=0$

$x=+6$

The oxidation number of $\mathrm{Xe}$ is $+6$.

Answer: $+6$