JEE Challenger
More from d and f Block Elements
ChemMedAdv 2023 • Paper 1

Identify Species X and Y in Inorganic Qualitative Analysis Scheme

In the scheme given below, X\mathbf{X} and Y\mathbf{Y}, respectively, are

Metal halide→aq. NaOHWhite precipitate (P)  +  Filtrate (Q)P→heataq. H2SO4,PbO2 (excess)X (a coloured species in solution)Q→warmMnO(OH)2,Conc. H2SO4Y (gives blue-coloration with KI-starch paper)\begin{aligned} \text{Metal halide} &\xrightarrow{\text{aq. }\mathrm{NaOH}} \text{White precipitate } (\mathbf{P}) \;+\; \text{Filtrate } (\mathbf{Q}) \\[1.5em] \mathbf{P} &\xrightarrow[\text{heat}]{\substack{\text{aq. }\mathrm{H_2SO_4}, \\ \mathrm{PbO_2}\text{ (excess)}}} \mathbf{X} \text{ (a coloured species in solution)} \\[1.5em] \mathbf{Q} &\xrightarrow[\text{warm}]{\substack{\mathrm{MnO(OH)_2}, \\ \text{Conc. }\mathrm{H_2SO_4}}} \mathbf{Y} \text{ (gives blue-coloration with KI-starch paper)} \end{aligned}

Options

A

CrO42−\text{CrO}_4^{2-} and Br2\text{Br}_2

B

MnO42−\text{MnO}_4^{2-} and Cl2\text{Cl}_2

C

MnO4−\text{MnO}_4^- and Cl2\text{Cl}_2

Correct
D

MnSO4\text{MnSO}_4 and HOCl\text{HOCl}

Topics & Concepts

d and f Block Elementsd-Block Elementsp-Block Elements

Step-by-Step Solution

To identify the species X\mathbf{X} and Y\mathbf{Y}, we analyze the reactions step-by-step:

  1. Reaction of Metal Halide with Aqueous NaOH\text{NaOH}: When the metal halide (manganese(II) chloride, MnCl2\text{MnCl}_2) reacts with aqueous NaOH\text{NaOH}, a white precipitate of manganese(II) hydroxide, Mn(OH)2\text{Mn(OH)}_2, is formed, while sodium chloride (NaCl\text{NaCl}) remains in the filtrate: MnCl2(aq)+2NaOH(aq)→Mn(OH)2(s)⏟White precipitate (P)+NaCl(aq)⏟Filtrate (Q)\text{MnCl}_2\text{(aq)} + 2\text{NaOH(aq)} \rightarrow \underbrace{\text{Mn(OH)}_2\text{(s)}}_{\text{White precipitate }(\mathbf{P})} + \underbrace{\text{NaCl(aq)}}_{\text{Filtrate }(\mathbf{Q})}

    • Precipitate P=Mn(OH)2\mathbf{P} = \text{Mn(OH)}_2
    • Filtrate Q=NaCl(aq)\mathbf{Q} = \text{NaCl(aq)} (containing Cl−\text{Cl}^- ions)

  2. Formation of Species X\mathbf{X}: When the white precipitate P\mathbf{P} (Mn(OH)2\text{Mn(OH)}_2) is heated with aqueous H2SO4\text{H}_2\text{SO}_4 and excess lead dioxide (PbO2\text{PbO}_2), the manganese(II) ions are oxidized to permanganate ions (MnO4−\text{MnO}_4^-), giving a characteristic pink/purple coloured solution: 2Mn(OH)2(s)+5PbO2(s)+3H2SO4(aq)→Δ2HMnO4(aq)+5PbSO4(s)↓+4H2O(l)2\text{Mn(OH)}_2\text{(s)} + 5\text{PbO}_2\text{(s)} + 3\text{H}_2\text{SO}_4\text{(aq)} \xrightarrow{\Delta} 2\text{HMnO}_4\text{(aq)} + 5\text{PbSO}_4\text{(s)}\downarrow + 4\text{H}_2\text{O(l)} The coloured species X\mathbf{X} in the solution is the permanganate ion, MnO4−\mathbf{MnO_4^-}.

  3. Formation of Species Y\mathbf{Y}: When filtrate Q\mathbf{Q} (containing Cl−\text{Cl}^- ions) is warmed with MnO(OH)2\text{MnO(OH)}_2 (hydrated manganese dioxide) and concentrated H2SO4\text{H}_2\text{SO}_4, chloride ions are oxidized to chlorine gas (Cl2\text{Cl}_2): MnO(OH)2(s)+2Cl−(aq)+4H+(aq)→ΔMn2+(aq)+Cl2(g)↑+3H2O(l)\text{MnO(OH)}_2\text{(s)} + 2\text{Cl}^-\text{(aq)} + 4\text{H}^+\text{(aq)} \xrightarrow{\Delta} \text{Mn}^{2+}\text{(aq)} + \mathbf{Cl_2(g)}\uparrow + 3\text{H}_2\text{O(l)} The evolved chlorine gas (Cl2\text{Cl}_2) oxidizes iodide ions (I−\text{I}^-) in the KI-starch paper to iodine (I2\text{I}_2), which forms an intense blue complex with starch: Cl2(g)+2I−(aq)→2Cl−(aq)+I2(aq)\text{Cl}_2\text{(g)} + 2\text{I}^-\text{(aq)} \rightarrow 2\text{Cl}^-\text{(aq)} + \text{I}_2\text{(aq)} I2+Starch→Blue-colored complex\text{I}_2 + \text{Starch} \rightarrow \text{Blue-colored complex} Thus, species Y\mathbf{Y} is Cl2\mathbf{Cl_2}.

Conclusion:

  • X=MnO4−\mathbf{X} = \text{MnO}_4^-
  • Y=Cl2\mathbf{Y} = \text{Cl}_2

Hence, the correct option is (C).

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