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← JEE Advanced 2014 · Paper 2
ChemistrySingle correct+3 / -1Question 26 of 60

JEE Advanced 2014 · Paper 2, Chemistry question 26

Under ambient conditions, the total number of gases released as products in the final step of the reaction scheme shown below is

XeF₆ undergoes complete hydrolysis to P + other product; P with OH⁻/H₂O gives Q; Q undergoes slow disproportionation in OH⁻/H₂O to products.
Choose one option
Answer and worked solution

Official answer: C

The scheme.

  • Complete hydrolysis: XeFX6+3 HX2O→XeOX3+6 HF\ce{XeF6 + 3H2O -> XeO3 + 6HF}, so P is XeOX3\ce{XeO3}.
  • In alkali: XeOX3+OHX−→HXeOX4X−\ce{XeO3 + OH- -> HXeO4-}, so Q is the hydrogen xenate ion.
  • Slow disproportionation: 2 HXeOX4X−+2 OHX−→XeOX6X4−+Xe+OX2+2 HX2O\ce{2HXeO4- + 2OH- -> XeO6^{4-} + Xe + O2 + 2H2O}.

Gases in the last step. Xe and OX2\ce{O2}: two.

(C)

Topics:Group 18
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