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

Options

A. `+2`
B. 0
C. `+4`
D. `+6`

Answer

D

Explanation

Xe : F `= (53.3)/(133): (46.7)/(19)`
`= 0.398: 2.45 =1:6`
Thus the compound is Xenon fluoride `XeF_6`
`:.` O. N. of Xe in `XeF_6` is + 6.