JEE Mains · Chemistry · STD 12 - p -Block elements - ll
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
- A \(4\)
- B \(6\)
- C \(2\)
- D \(8\)
Answer & Solution
Correct Answer
(C) \(2\)
Step-by-step Solution
Detailed explanation
\(6 XeF _4+12 H _2 O \longrightarrow 2 XeO _3+4 Xe +24 HF +3 O _2\) in \(XeO _3\), Oxidation state of \(Xe =+6\) in \(XeF _4\), Oxidation state of \(Xe =+4\) So difference in oxidation state \(=2\)
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