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Chemistry · d and f Block Elements

JEE Main 2025 — 23 January, Evening Shift — Question 42

Consider the following reactions:

K2Cr2O7→−H2OKOH[A]→−H2OH2SO4[ B]+K2SO4\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \xrightarrow[-\mathrm{H}_{2} \mathrm{O}]{\mathrm{KOH}}[\mathrm{A}] \xrightarrow[-\mathrm{H}_{2} \mathrm{O}]{\mathrm{H}_{2} \mathrm{SO}_{4}}[\mathrm{~B}]+\mathrm{K}_{2} \mathrm{SO}_{4}

The products [A][\mathrm{A}] and [B][\mathrm{B}], respectively are :

  1. Option A:

    K2Cr(OH)6{{\text{K}}_{2}}\text{Cr}{{(\text{OH})}_{6}} and Cr2O3\text{C}{{\text{r}}_{2}}{{\text{O}}_{3}}

  2. Option B:

    K2CrO4{{\text{K}}_{2}}\text{Cr}{{\text{O}}_{4}} and Cr2O3\text{C}{{\text{r}}_{2}}{{\text{O}}_{3}}

  3. Option C:

    K2CrO4{{\text{K}}_{2}}\text{Cr}{{\text{O}}_{4}} and K2Cr2O7{{\text{K}}_{2}}\text{C}{{\text{r}}_{2}}{{\text{O}}_{7}}

    Correct
  4. Option D:

    K2CrO4{{\text{K}}_{2}}\text{Cr}{{\text{O}}_{4}} and CrO\mathrm{CrO}

Answer: C

Step-by-step solution

In alkaline medium:

K2Cr2O7+2KOH→2K2CrO4+H2O\mathrm{K_2Cr_2O_7 + 2KOH \rightarrow 2K_2CrO_4 + H_2O}

Therefore, [A]=K2CrO4[A] = \mathrm{K_2CrO_4}.

On acidification:

2K2CrO4+H2SO4→K2Cr2O7+K2SO4+H2O\mathrm{2K_2CrO_4 + H_2SO_4 \rightarrow K_2Cr_2O_7 + K_2SO_4 + H_2O}

Therefore, [B]=K2Cr2O7[B] = \mathrm{K_2Cr_2O_7}.

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
23 January, Evening Shift
Subject
Chemistry
Chapter
d and f Block Elements
Topic
Compounds of Chromium

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