JEE Challenger
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ChemMedAdv 2022 • Paper 1

Match Compounds to Their Formation Reactions

LIST-I contains compounds and LIST-II contains reactions

LIST-ILIST-II(I) H2O2(P) Mg(HCO3)2+Ca(OH)2→(II) Mg(OH)2(Q) BaO2+H2SO4→(III) BaCl2(R) Ca(OH)2+MgCl2→(IV) CaCO3(S) BaO2+HCl→(T) Ca(HCO3)2+Ca(OH)2→\begin{array}{ll} \text{\textbf{LIST-I}} & \text{\textbf{LIST-II}} \\ \text{(I) } \text{H}_2\text{O}_2 & \text{(P) } \text{Mg(HCO}_3\text{)}_2 + \text{Ca(OH)}_2 \rightarrow \\ \text{(II) } \text{Mg(OH)}_2 & \text{(Q) } \text{BaO}_2 + \text{H}_2\text{SO}_4 \rightarrow \\ \text{(III) } \text{BaCl}_2 & \text{(R) } \text{Ca(OH)}_2 + \text{MgCl}_2 \rightarrow \\ \text{(IV) } \text{CaCO}_3 & \text{(S) } \text{BaO}_2 + \text{HCl} \rightarrow \\ & \text{(T) } \text{Ca(HCO}_3\text{)}_2 + \text{Ca(OH)}_2 \rightarrow \end{array}

Match each compound in LIST-I with its formation reaction(s) in LIST-II, and choose the correct option

Options

A

I→Q\text{I} \rightarrow \text{Q}; II→P\text{II} \rightarrow \text{P}; III→S\text{III} \rightarrow \text{S}; IV→R\text{IV} \rightarrow \text{R}

B

I→T\text{I} \rightarrow \text{T}; II→P\text{II} \rightarrow \text{P}; III→Q\text{III} \rightarrow \text{Q}; IV→R\text{IV} \rightarrow \text{R}

C

I→T\text{I} \rightarrow \text{T}; II→R\text{II} \rightarrow \text{R}; III→Q\text{III} \rightarrow \text{Q}; IV→P\text{IV} \rightarrow \text{P}

D

I→Q\text{I} \rightarrow \text{Q}; II→R\text{II} \rightarrow \text{R}; III→S\text{III} \rightarrow \text{S}; IV→P\text{IV} \rightarrow \text{P}

Correct

Topics & Concepts

s-Block Elementss-Block ElementsHydrogen

Step-by-Step Solution

To determine the correct matching between the compounds in LIST-I and their corresponding formation reactions in LIST-II, let's analyze each reaction step-by-step:

  1. Compound (I): H2O2\text{H}_2\text{O}_2

    • Reaction (Q): Acidifying hydrated barium peroxide with dilute sulfuric acid yields hydrogen peroxide along with a precipitate of barium sulfate: BaO2⋅8H2O+H2SO4→BaSO4↓+H2O2+8H2O\text{BaO}_2 \cdot 8\text{H}_2\text{O} + \text{H}_2\text{SO}_4 \rightarrow \text{BaSO}_4\downarrow + \text{H}_2\text{O}_2 + 8\text{H}_2\text{O}
    • Thus, I→Q\text{I} \rightarrow \text{Q}.

  2. Compound (II): Mg(OH)2\text{Mg(OH)}_2

    • Reaction (R): Treatment of magnesium chloride with calcium hydroxide results in the precipitation of magnesium hydroxide due to its low solubility: Ca(OH)2+MgCl2→Mg(OH)2↓+CaCl2\text{Ca(OH)}_2 + \text{MgCl}_2 \rightarrow \text{Mg(OH)}_2\downarrow + \text{CaCl}_2
    • Thus, II→R\text{II} \rightarrow \text{R}.

  3. Compound (III): BaCl2\text{BaCl}_2

    • Reaction (S): Barium peroxide reacts with hydrochloric acid to yield barium chloride and hydrogen peroxide: BaO2+2HCl→BaCl2+H2O2\text{BaO}_2 + 2\text{HCl} \rightarrow \text{BaCl}_2 + \text{H}_2\text{O}_2
    • Thus, III→S\text{III} \rightarrow \text{S}.

  4. Compound (IV): CaCO3\text{CaCO}_3

    • Reaction (P): Magnesium hydrogen carbonate reacts with calcium hydroxide (slaked lime) to form a precipitate of calcium carbonate: Mg(HCO3)2+Ca(OH)2→CaCO3↓+MgCO3+2H2O\text{Mg(HCO}_3\text{)}_2 + \text{Ca(OH)}_2 \rightarrow \text{CaCO}_3\downarrow + \text{MgCO}_3 + 2\text{H}_2\text{O} (or further precipitation to form CaCO3\text{CaCO}_3 and Mg(OH)2\text{Mg(OH)}_2 depending on stoichiometry).
    • Thus, IV→P\text{IV} \rightarrow \text{P}.

Comparing the derived matches with the given options:

  • I→Q\text{I} \rightarrow \text{Q}
  • II→R\text{II} \rightarrow \text{R}
  • III→S\text{III} \rightarrow \text{S}
  • IV→P\text{IV} \rightarrow \text{P}

This corresponds to Option D.

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