The difference in the oxidation state of Xe between the oxidised… — Solution | JEE Chemistry

NVAEasyJEE 2023Oxidation Number & Redox Reactions

The difference in the oxidation state of Xe between the oxidised product of Xe formed on complete hydrolysis of $\mathrm{XeF}_4$ and $\mathrm{XeF}_4$ is _____

Answer

Correct answer:2

Step-by-step solution

Standard Method

Given: The oxidised product of $\mathrm{Xe}$ formed on complete hydrolysis of $\mathrm{XeF}_4$ is to be compared with $\mathrm{XeF}_4$ .

Find: The difference in oxidation state of $\mathrm{Xe}$ .

In $\mathrm{XeF}_4$ , fluorine has oxidation state $-1$ . Therefore,

\text{Oxidation state of Xe} + 4(-1) = 0

So,

\text{Oxidation state of Xe} = +4

On complete hydrolysis, the oxidised product is $\mathrm{XeO}_3$ .

In $\mathrm{XeO}_3$ , oxygen has oxidation state $-2$ . Therefore,

\text{Oxidation state of Xe} + 3(-2) = 0

So,

\text{Oxidation state of Xe} = +6

Hence, the difference in oxidation state is

6 - 4 = 2

Therefore, the required answer is $2$ .

Common mistakes

Taking the hydrolysis product as elemental $\mathrm{Xe}$ only is incorrect, because the question asks for the oxidised product of xenon formed on complete hydrolysis. Use $\mathrm{XeO}_3$ for the comparison.
Assigning the oxidation state of fluorine incorrectly is wrong. In compounds, fluorine is always $-1$ , so xenon in $\mathrm{XeF}_4$ must be calculated as $+4$ .
Using oxygen as $-1$ in $\mathrm{XeO}_3$ is incorrect. Oxygen is $-2$ here, so xenon becomes $+6$ , not any other value.

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