Q. The oxidation number of xenon in ${XeOF}_{2}$ is
A) Zero
B) 2
C) 4
D) 3
Explanation
(c)
Let the oxidation number of Xe is $x$ in $XeOF_{2}.$
$x + ( - 2) + 2( - 1) = 0$
$x - 2 - 2 = 0$
$x = + 4$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.