JEE Main 2024 · 30 January · evening · Single correct · +4 / −1

Reduction potential of ions are given below:
ClO₄⁻, IO₄⁻, BrO₄⁻; E°=1.19 V, E°=1.65 V, E°=1.74 V
The correct order of their oxidising power is :

  1. (a) IO₄⁻>BrO₄⁻>ClO₄⁻
  2. (b) BrO₄⁻>ClO₄⁻>IO₄⁻
  3. (c) ClO₄⁻>IO₄⁻>BrO₄⁻
  4. (d) BrO₄⁻>IO₄⁻>ClO₄⁻
Answer and solution

Answer (d) BrO₄⁻>IO₄⁻>ClO₄⁻

Solution

Chemistry ma’am · 3 steps

Higher reduction potential means the ion grabs electrons more eagerly, so just sort by E°.

  1. What E° tells you. A reduction potential measures how strongly a species pulls electrons towards itself. An oxidising agent is the one that gets reduced, so the larger its E°, the stronger it oxidises.
    IonE° (V)Rank as oxidant
    BrO₄⁻1.741st, strongest
    IO₄⁻1.652nd
    ClO₄⁻1.193rd, weakest
  2. Why bromine is odd. Perbromate does not follow the group trend. Bromine sits just after the 3d series, so its 4s and 4p electrons are held tightly and the +7 state is hard to keep. That makes BrO₄⁻ especially eager to be reduced.
  3. So. The order is BrO₄⁻ > IO₄⁻ > ClO₄⁻, which is option (d).

The idea to keep. Oxidising power follows reduction potential: the higher the E°, the stronger the oxidant. Trust the given numbers over a remembered group trend.

Answer from the NTA final answer key. Solution by PrepWiser.

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Asked about 170 times in JEE Main since 2020. This question is from 2024.