Question

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

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

Hint:

Fluorine always has oxidation number -1.

Answer 1

The Correct Option is C

Step 1: Let the compound be XeFₓ.
Step 2: Percentage of Xe: (131)/(131+19x)×100=53.5
Step 3: Solving: 131=0.535(131+19x)⟹ x≈4
Step 4: Oxidation state of Xe in XeF₄: Xe=+4