Class JEE Chemistry ALL Q #1166
KNOWLEDGE BASED
APPLY
4 Marks 2026 JEE Main 2026 (Online) 21st January Morning Shift MCQ SINGLE
In the following reaction, $MnO4^{2-} \xrightarrow{H^{+}}$ ?. Manganate ion undergoes disproportionation to form:
(A) $MnO_2, MnO_4^-$
(B) $MnO_4^-, MnO_2$
(C) $MnO_2, MnO$
(D) $MnO_4^-, MnO$
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Step-by-Step Solution

  1. First, determine the oxidation state of Mn in $MnO_4^{2-}$. Let the oxidation state of Mn be x. Then, x + 4(-2) = -2, so x = +6.

  2. In a disproportionation reaction, the same element is both oxidized and reduced. This means Mn(+6) will be oxidized to a higher oxidation state and reduced to a lower oxidation state.

  3. Consider the options:

    • (A) $MnO_2, MnO_4^-$: Mn oxidation states are +4 and +7.
    • (B) $MnO_4^-, MnO_2$: Mn oxidation states are +7 and +4.
    • (C) $MnO_2, MnO$: Mn oxidation states are +4 and +2.
    • (D) $MnO_4^-, MnO$: Mn oxidation states are +7 and +2.

  4. The correct disproportionation products are $MnO_4^-$ (where Mn is in +7 oxidation state) and $MnO_2$ (where Mn is in +4 oxidation state). Therefore, the correct option is (B).

Correct Answer: $MnO_4^-, MnO_2$

AI Suggestion: Option B

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because the student needs to apply their knowledge of disproportionation reactions and oxidation states to predict the products of the given reaction.
Knowledge Dimension: CONCEPTUAL
Justification: The question requires understanding the concept of disproportionation and how it relates to changes in oxidation states of manganese.
Syllabus Audit: In the context of JEE, this is classified as KNOWLEDGE. The question directly tests knowledge of redox reactions and disproportionation, which are core concepts covered in the syllabus.

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