JEE Main · 2021 · Shift-IImediumPB12-121

The difference in the oxidation state of Xe between the oxidised product of Xe formed on complete hydrolysis of XeF4…

p-Block Elements (Class 12) · Class 12 · JEE Main Previous Year Question

Question

The difference in the oxidation state of Xe between the oxidised product of Xe formed on complete hydrolysis of XeF4\mathrm{XeF_4} and XeF4\mathrm{XeF_4} is

Options
  1. a

    1

  2. b

    2

    ✓
  3. c

    3

  4. d

    4

Correct Answerb

2

Detailed Solution

Strategy: Compare the oxidation steps in Xenon fluoride hydrolysis.

Step 1: Complete Hydrolysis of XeF4 The complete hydrolysis reaction is a disproportionation: \ce3XeF4+6H2O→2Xe+XeO3+12HF+1.5O2\ce{3XeF4 + 6H2O \to 2Xe + XeO3 + 12HF + 1.5O2} The "oxidised product" is \ceXeO3\ce{XeO3}, where Xenon is in the +6+6 oxidation state.

Step 2: Difference Calculation

  • Oxidation state in \ceXeF4=+4\ce{XeF4} = +4.
  • Oxidation state in \ceXeO3=+6\ce{XeO3} = +6.
  • Difference =6−4=2= 6 - 4 = 2.

Answer: (Option) [b] 2\boxed{\text{Answer: (Option) [b] 2}}

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