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JEE Main 202430 JanuaryEvening Shift
Exam
JEE Main
Year
2024
Date
30 January
Shift
Evening Shift
Subject
Chemistry
Chapter
d and f Block Elements

JEE Main 2024 · 30 January · Evening Shift · Chemistry · d and f Block Elements

A and B formed in the following reactions are: CrO _ 2 Cl _ 2 +...

AA and BB formed in the following reactions are:
CrO2Cl2+4NaOH→A+2NaCl+2H2O{CrO}_{2}{Cl}_{2} + 4NaOH \to A + 2NaCl + 2{H}_{2}O

A+2HCl+2H2O2→B+3H2OA + 2HCl + 2{H}_{2}{O}_{2} \to B + 3{H}_{2}O

Options

  1. AA=Na2CrO4,B=CrO5A = {Na}_{2}{CrO}_{4}, B = {CrO}_{5}
  2. BA=Na2Cr2O4,B=CrO4A = {Na}_{2}{Cr}_{2}{O}_{4}, B = {CrO}_{4}
  3. CA=Na2Cr2O7,B=CrO3A = {Na}_{2}{Cr}_{2}{O}_{7}, B = {CrO}_{3}
  4. DA=Na2Cr2O7,B=CrO5A = {Na}_{2}{Cr}_{2}{O}_{7}, B = {CrO}_{5}

Correct answer

A. A=Na2CrO4,B=CrO5A = {Na}_{2}{CrO}_{4}, B = {CrO}_{5}

Step-by-step solution

Chromyl chloride dissolves in NaOHNaOH solution to give a yellow solution. The chromate solution changes to blue colour chromium peroxide with hydrogen peroxide.

The reaction will be as follows,

CrO2Cl2+4NaOH→Na2CrO4+2NaCl+2H2OCrO42−+2H++2H2O2→CrO5+3H2O{CrO}_{2}{Cl}_{2} + 4NaOH \to {Na}_{2}{CrO}_{4} + 2NaCl + 2{H}_{2}O{{CrO}_{4}}^{2 -} + 2{H}^{+} + 2{H}_{2}{O}_{2} \to {CrO}_{5} + 3{H}_{2}O

Hence, option A is the answer.

 

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