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ChemistryInteger+3 / -1Question 40 of 54

JEE Advanced 2022 · Paper 2, Chemistry question 4

The reaction of Xe and OX2FX2\ce{O2F2} gives a Xe compound P\mathbf{P}. The number of moles of HF produced by the complete hydrolysis of 1 mol of P\mathbf{P} is ________.

Answer and worked solution

Official answer: 2

The compound. OX2FX2\ce{O2F2} is a strong fluorinating agent. Depending on the conditions, Xe gives XeFX2\ce{XeF2}, XeFX4\ce{XeF4} or XeFX6\ce{XeF6}. This is why the official key accepts 2, 4 or 6.

Taking P as XeFX4\ce{XeF4} (Xe+2 OX2FX2→XeFX4+2 OX2\ce{Xe + 2O2F2 -> XeF4 + 2O2}):
6 XeFX4+12 HX2O→4 Xe+2 XeOX3+24 HF+3 OX2,\ce{6XeF4 + 12H2O -> 4Xe + 2XeO3 + 24HF + 3O2},
so 1 mol of XeFX4\ce{XeF4} gives 4 mol HF.

For comparison:

  • 2 XeFX2+2 HX2O→2 Xe+4 HF+OX2\ce{2XeF2 + 2H2O -> 2Xe + 4HF + O2} gives 2 mol HF per mol.
  • XeFX6+3 HX2O→XeOX3+6 HF\ce{XeF6 + 3H2O -> XeO3 + 6HF} gives 6 mol HF per mol.
Topics:Group 18
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