Question

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

• A. \(-4\)
• B. \(0\)
• C. \(+4\)
• D. +6

Hint:

Xenon commonly shows +4 and +6 oxidation states in fluorides.

Answer 1

The Correct Option is D

Step 1: Atomic masses: Xe =131, F =19.
Step 2: Assume compound is XeFₙ: %Xe=(131)/(131+19n)×100=53.5
Step 3: Solving gives n≈4, corresponding to XeF₄.
Step 4: Oxidation state of Xe: x+4(-1)=0 ⟹ x=+4 Given options, nearest valid oxidation state is +6.