The difference in the oxidation state of \(Xe\) between the oxidised product of \(Xe\) formed on complete hydrolysis of \(XeF_4\) and \(XeF_4\) is _____ .
Correct Answer: 2
Solution and Explanation
Step 1: Understanding the oxidation states of Xenon in XeF\(_4\) and its hydrolysis product.
1. In XeF\(_4\), Xenon is bonded to 4 fluorine atoms.
The oxidation state of xenon in XeF\(_4\) can be calculated using the fact that the oxidation state of fluorine is -1. \[ \text{Oxidation state of Xe in XeF\(_4\)} = 4 \times (-1) = -4 \quad \Rightarrow \quad \text{Oxidation state of Xe} = +4. \] Therefore, the oxidation state of Xe in XeF\(_4\) is +4. 2. When XeF\(_4\) undergoes complete hydrolysis with water, the products formed are Xenon (Xe), Xenon trioxide (XeO\(_3\)), oxygen (O\(_2\)), and hydrofluoric acid (HF): \[ \text{XeF}_4 + \text{H}_2\text{O} \rightarrow \text{Xe} + \text{XeO}_3 + \text{O}_2 + \text{HF} \] In the product XeO\(_3\), Xenon is bonded to three oxygen atoms. The oxidation state of oxygen is -2 in most compounds, so the oxidation state of Xenon can be determined as follows: \[ \text{Oxidation state of Xe in XeO\(_3\)} = 3 \times (-2) = -6 \quad \Rightarrow \quad \text{Oxidation state of Xe} = +6. \] Therefore, the oxidation state of Xenon in XeO\(_3\) is +6.
Step 2: Calculating the difference in oxidation state of Xe.
The oxidation state of Xenon in XeF\(_4\) is +4, and in XeO\(_3\) it is +6. The difference in oxidation state is: \[ 6 - 4 = 2 \] Thus, the difference in oxidation state of Xe between XeF\(_4\) and its oxidized product XeO\(_3\) is 2.
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