Verified Solution • 2025–26 Session moderate

QJEEM24Ch4378: JEE Main 2024 Chemistry - Class 12 JEEM24Ch4378 Solution: A and B formed in the following reactions are:

A and B formed in the following reactions are:

amp;CrO2Cl2+4NaOH→A+2NaCl+2H2O,amp;A+2HCl+2H2O2→B+3H2O\begin{aligned}& \mathrm{CrO}_2 \mathrm{Cl}_2+4 \mathrm{NaOH} \rightarrow \mathrm{A}+2 \mathrm{NaCl}+2 \mathrm{H}_2 \mathrm{O}, \\& \mathrm{A}+2 \mathrm{HCl}+2 \mathrm{H}_2 \mathrm{O}_2 \rightarrow \mathrm{B}+3 \mathrm{H}_2 \mathrm{O}\end{aligned}

Multiple Choice Options

  • A A=Na2Cr2O7, B=CrO5\mathrm{A}=\mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7, \mathrm{~B}=\mathrm{CrO}_5
  • B A=Na2CrO4, B=CrO5\mathrm{A}=\mathrm{Na}_2 \mathrm{CrO}_4, \mathrm{~B}=\mathrm{CrO}_5
  • C A=Na2Cr2O4, B=CrO4\mathrm{A}=\mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_4, \mathrm{~B}=\mathrm{CrO}_4
  • D A=Na2Cr2O7, B=CrO3\mathrm{A}=\mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7, \mathrm{~B}=\mathrm{CrO}_3

Step-by-Step Verified Solution

CrO2+4NaOH→Na2CrO4(A)+2NaCl+2H2O\mathrm{Cr{O_2} + 4NaOH \to \mathop {N{a_2}Cr{O_4}}\limits_{(A)} + 2NaCl + 2{H_2}O}

Na2CrO4+2H2O2+2HCl→CrO5(B)+2NaCl(Missin⁡g from balanced equation)+3H2O\mathrm{N{a_2}Cr{O_4} + 2{H_2}{O_2} + 2HCl \to \mathop {Cr{O_5}}\limits_{(B)} + \mathop {2NaCl}\limits_{(Mis\sin g\,from\,balanced\,equation)} + 3{H_2}O}

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