Q. The oxidation number of Xe in ${XeF}_{4}$ and ${XeO}_{2}$ is
A) +6
B) +4
C) +1
D) +3
Explanation
(b)
The oxidation state of Xe in both $XeO_{2}\ and\ {XeF}_{4}\ is\ 4.\$
${XeO}_{2}$ ${XeF}_{4}$
$x + 2( - 2) = 0$ $x + 4( - 1) = 0$
$x = 4$ $x = 4$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.