Redox and Equivalent ConceptHard
Question
The compound of Xe and F is found to have 53.5% Xe. What is the oxidation number of Xe in this compound? (Xe = 131, F = 19)
Options
A.−4
B.0
C.+4
D.+6
Solution
$N_{Xe}:N_{F} = \frac{53.5}{131}:\frac{46.5}{19} = 1:6 \Rightarrow XeF_{6}$
∴ Oxidation state of Xe = +6.
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