Question

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

• A. -4
• B. 0
• C. +4
• D. +6

Answer 1

Answer: (D)

+6

Answer 2

Xe = 53.5 % $\therefore$ F = 46.5%
Relative number of atoms Xe
$= \frac{53.5}{131.2} = 0.4 F = \frac{46.5}{19} = 2.4$
Simple ratio Xe = 1 and F = 6 ; Molecular
formula is ${XeF_{6}}$
O.N.of Xe is +6