Q. A compound of $Xe$ and $F$ is found to have $53.5 \%\, Xe$. What is the oxidation number of $Xe$ in this compound?
Redox Reactions
Solution:
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 $XeF _{6}$
$\therefore $ Oxidation state of $Xe$ is +6
| Element | % | At. mass | Relative number of atoms | Simplest ratio |
|---|---|---|---|---|
| $Xe$ | 53.5 | 131 | 0.408 | 1 |
| $F$ | 46.5 | 19 | 2.44 | 6 |