Question

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 _____ .

Answer 1

Correct Answer: 2

To determine the oxidation state difference for \(Xe\) between \(XeF_4\) and its hydrolysis product, we follow these steps:

1. Identify the oxidation state of \(Xe\) in \(XeF_4\):
   • In \(XeF_4\), the oxidation state of \(F\) is \(-1\). With four fluorine atoms, the total is \(-4\). Since \(XeF_4\) is a neutral molecule, \(Xe\) must balance this out.
   • Let the oxidation state of \(Xe\) be \(x\). Therefore, \(x + 4(-1) = 0 \implies x = +4\).
2. Determine the oxidation state of \(Xe\) in the hydrolysis product:
   • When \(XeF_4\) undergoes complete hydrolysis, the products are \(XeO_3\), \(HF\), and water as a by-product.
   • In \(XeO_3\), the oxidation state of \(O\) is \(-2\). With three oxygen atoms, the total is \(-6\).
   • Let \(x\) be the oxidation state of \(Xe\) in \(XeO_3\). Thus, \(x + 3(-2) = 0 \implies x = +6\).
3. Calculate the difference in oxidation state:
   • The difference = \(+6 - (+4) = +2\).
4. Verify the solution falls within the provided range \(2,2\):
   • The calculated difference is exactly \(2\), which matches the given range, confirming our solution is correct.

Thus, the difference in oxidation states is 2.