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

Answer 1

Answer: (D)

Element	%	At. mass	Relative number of atoms	Simplest ratio
$Xe$	53.5	131	0.408	1
$F$	46.5	19	2.44	6

$\therefore$ The empirical formula is $Xe{F}_6$

$\therefore$ Oxidation state of $Xe$ is +6