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A compound of Xe and $\mathrm{F}$ is found to have $53.5 \%$ of Xe. What is oxidation number of Xe in this compound?
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1991 Upvotes
Verified Answer
The correct answer is:
6
$\mathrm{Xe}=53.5 \% \therefore \mathrm{F}=46.5 \%$
Relative number of atoms Xe
$$
=\frac{53.5}{131.2}=0.4 \text { and } \mathrm{F}=\frac{46.5}{19}=2.4
$$
Simple ratio $\mathrm{Xe}=1$ and $\mathrm{F}=6 ;$ Molecular formula is $\mathrm{XeF}_{6}$
O. $\mathrm{N}$.of $\mathrm{Xe}$ is +6
Relative number of atoms Xe
$$
=\frac{53.5}{131.2}=0.4 \text { and } \mathrm{F}=\frac{46.5}{19}=2.4
$$
Simple ratio $\mathrm{Xe}=1$ and $\mathrm{F}=6 ;$ Molecular formula is $\mathrm{XeF}_{6}$
O. $\mathrm{N}$.of $\mathrm{Xe}$ is +6
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