Q. Xenon hexafluoride reacts with silica to form a xenon compound X. The oxidation state of xenon in X is
A) +2
B) +4
C) +6
D) 0
Explanation
(c)
Xenon hexafluoride reacts with silica to form XeOF_{4} as
2 XeF_{6}+SiO_{2} ⟶XeOF_{4}+SiF_{4}
The oxidations state of xenon in XeOF_{4} is calculated as
$x$ ^{-2-1}
XeOF_{4}
$x$+(-2)+4$\times ( - 1) = 0$
$x$-2-4=0
$x$=+6
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.