Q40 / 125 · BITSAT 2016 · Chemistry

A compound of Xe and $\mathrm{F}$ is found to have $53.5 \%$ of Xe. What is oxidation number of Xe in this compound?

  • A
    -4
  • B
    0
  • C
    4
  • Correct answer
    D
    6

Solution

$\mathrm{Xe}=53.5 \% \therefore \mathrm{F}=46.5 \%$ 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

Hence, the correct answer is option: 'D'.

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