JEE Main 2021 · 25 July · evening · Single correct · +4 / −1

Identify the process in which change in the oxidation state is five :

  1. (a) Cr₂O₇²⁻ → 2Cr³⁺
  2. (b) MnO₄ - → Mn²⁺
  3. (c) CrO₄²⁻ → Cr³⁺
  4. (d) C₂O₄²⁻ → 2CO₂
Answer and solution

Answer (b) MnO₄ - → Mn²⁺

Solution

Chemistry ma’am · 2 steps

Find the oxidation state of the central atom on each side, then subtract.

  1. Find each state. Take O as -2. The oxidation states must add up to the charge of the ion. For example, in MnO₄⁻, x + 4(-2) = -1, so x = +7.
    ProcessBeforeAfterChange per atom
    (a) Cr₂O₇²⁻ → 2Cr³⁺Cr: +6Cr: +33 (6 for both Cr atoms)
    (b) MnO₄⁻ → Mn²⁺Mn: +7Mn: +25
    (c) CrO₄²⁻ → Cr³⁺Cr: +6Cr: +33
    (d) C₂O₄²⁻ → 2CO₂C: +3C: +41 (2 for both C atoms)
  2. So. Only in (b) does the oxidation state change by five: Mn goes from +7 to +2. So the answer is (b).

The idea to keep. Fix oxidation states with O as -2 and the total equal to the ion's charge. Then check whether the question asks for the change per atom or the total electrons transferred.

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

Study Redox Reactions

Asked about 90 times in JEE Main since 2020. This question is from 2021.