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Q.

When ${XeF}_{4}$ is completely hydrolyzed. Find the difference between the oxidation state of xenon in the oxidized product and ${XeF}_{4}$ .

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answer is 2.

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Detailed Solution

$6 X e F_{4} + 12 H_{2} O \to 2 X e O_{3} + 4 X e + 24 H F + 3 O_{2}$

$X e O_{3},$ oxidation state of Xe = + 6

$X e F_{4},$ oxidation state of Xe = + 4

Difference in oxidation state = +2

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