$X e O_{4}$ shows maximum oxidation state ( $+ 8$ ). If $X e$ reacts with fluorine atom, the maximum oxidation state shown by $X e$ is :

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Solution

${X e O}_{4}$ shows maximum oxidation state (+8)
x(-2)
${X e O}_{4}$ = x-8=0
$⟹$ x=+8
Xe can react directly with fluorine atom
$X e + F_{2} \to X e F_{2}$ = oxygen state of Xe=+2
$X e + {2 F}_{2} \to X e F_{4}$ - oxygen state of Xe=+4
$X e + 3 F_{2} \to X e F_{6}$ - oxygen state of Xe=+6
Therefore Highest oxygen state of Xe is +6

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