A compound containing Xe and F has 63.8% of X | vedclass

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(B) Given: Mass percentage of $Xe = 63.8\%$,so mass percentage of $F = 100 - 63.8 = 36.2\%$. Atomic mass of $Xe = 133 	ext{ g/mol}$ and $F = 19 	ext

Vedclass · Accepted Answer

$+4$

Vedclass · Answer

(B) Given: Mass percentage of $Xe = 63.8\%$,so mass percentage of $F = 100 - 63.8 = 36.2\%$. Atomic mass of $Xe = 133 	ext{ g/mol}$ and $F = 19 	ext{ g/mol}$. Number of moles of $Xe = \frac{63.8}{133} \approx 0.48$. Number of moles of $F = \frac{36.2}{19} \approx 1.905$. Ratio of $Xe : F = 0.48 : 1.905 \approx 1 : 4$. The empirical formula is $XeF_4$. In $XeF_4$,let the oxidation number of $Xe$ be $x$. $x + 4(-1) = 0 \implies x = +4$. Therefore,the oxidation number of $Xe$ is $+4$.

$A$ compound containing $Xe$ and $F$ has $63.8\%$ of $Xe$ by mass. Calculate the oxidation number of $Xe$ in this compound.

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