MCQEasyJEE 2024Potassium Dichromate & Permanganate

JEE Chemistry 2024 Question with Solution

A and B formed in the following reactions are:

Cr2O72+4NaOHNa2Cr2O4+2NaCl+2H2O\text{Cr}_2\text{O}_7^{2-} + 4\text{NaOH} \rightarrow \text{Na}_2\text{Cr}_2\text{O}_4 + 2\text{NaCl} + 2\text{H}_2\text{O} A+2Cl2+2H2OB+3H2OA + 2\text{Cl}_2 + 2\text{H}_2\text{O} \rightarrow B + 3\text{H}_2\text{O}
  • A

    A = Na2Cr2O4\text{Na}_2\text{Cr}_2\text{O}_4, B = CrO3\text{CrO}_3

  • B

    A = Na2Cr2O4\text{Na}_2\text{Cr}_2\text{O}_4, B = Cr2O7\text{Cr}_2\text{O}_7

  • C

    A = Na2Cr2O4\text{Na}_2\text{Cr}_2\text{O}_4, B = NaCrO4\text{NaCrO}_4

  • D

    A = Na2Cr2O4\text{Na}_2\text{Cr}_2\text{O}_4, B = Cr3O8\text{Cr}_3\text{O}_8

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The products AA and BB are to be identified from the two reactions.

Find: The correct pair corresponding to AA and BB.

From the given reaction, AA is identified as Na2Cr2O4\text{Na}_2\text{Cr}_2\text{O}_4.

The solution statement further gives that when AA reacts with Cl2\text{Cl}_2 and H2O\text{H}_2\text{O}, it forms CrO3\text{CrO}_3.

Therefore, A=Na2Cr2O4A = \text{Na}_2\text{Cr}_2\text{O}_4 and B=CrO3B = \text{CrO}_3.

The correct option is A.

Common mistakes

  • Assuming BB is a dichromate species because the starting ion contains chromium is incorrect. The product must be taken from the given reaction outcome in the solution, which identifies BB as CrO3\text{CrO}_3.

  • Confusing option numbering with option labels can lead to the wrong response. Here, source option (1) maps to label A, so the final answer should be A, not 1.

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