A compound of Xe and F is found to have an at | vedclass

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(D) The atomic ratio of $Xe : F$ is given as $0.4 : 2.4$. Dividing both by $0.4$,we get the molar ratio as $1 : 6$. Thus,the empirical formula of the

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$+6$

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(D) The atomic ratio of $Xe : F$ is given as $0.4 : 2.4$. Dividing both by $0.4$,we get the molar ratio as $1 : 6$. Thus,the empirical formula of the compound is $XeF_6$. In $XeF_6$,let the oxidation number of $Xe$ be $x$. Since the oxidation number of $F$ is $-1$,we have: $x + 6(-1) = 0$ $x - 6 = 0$ $x = +6$. Therefore,the oxidation number of $Xe$ is $+6$.

$A$ compound of $Xe$ and $F$ is found to have an atomic ratio $Xe : F$ as $0.4 : 2.4$. Find the oxidation number of $Xe$.

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