JEE Challenger
More from d and f Block Elements
ChemHardAdv 2020 • Paper 1

Identification of Metal Cations Based on Precipitation and Solubility Reactions

A clear aqueous mixture contains nitrate salts of two distinct metals, X\mathbf{X} and Y\mathbf{Y}. Addition of aqueous NaCl\text{NaCl} solution produces a white precipitate. This precipitate is only partially soluble in hot water, leaving behind a solid residue P\mathbf{P} and a filtrate solution Q\mathbf{Q}. The solid residue P\mathbf{P} dissolves in aqueous ammonia as well as in excess sodium thiosulfate solution. Treatment of the hot solution Q\mathbf{Q} with KI\text{KI} yields a bright yellow precipitate. Identify the metal elements X\mathbf{X} and Y\mathbf{Y}, respectively.

Options

A

Ag\text{Ag} and Pb\text{Pb}

Correct
B

Ag\text{Ag} and Cd\text{Cd}

C

Cd\text{Cd} and Pb\text{Pb}

D

Cd\text{Cd} and Zn\text{Zn}

Topics & Concepts

d and f Block Elementsd-Block ElementsCoordination Compounds

Step-by-Step Solution

To determine the identity of the metal cations X\mathbf{X} and Y\mathbf{Y}, we analyze each chemical reaction step-by-step:

  1. Precipitation with NaCl\text{NaCl}: When aqueous NaCl\text{NaCl} solution is added to the mixture of nitrate salts of metals X\mathbf{X} and Y\mathbf{Y}, a white precipitate is formed. Among common metal cations, Ag+\text{Ag}^+ and Pb2+\text{Pb}^{2+} form insoluble white chloride precipitates (AgCl\text{AgCl} and PbCl2\text{PbCl}_2 respectively).

  2. Solubility in Hot Water:

    • PbCl2\text{PbCl}_2 is soluble in hot water.
    • AgCl\text{AgCl} is insoluble in hot water.

    Therefore, treating the combined white precipitate with hot water partially dissolves it:

    • The insoluble residue P\mathbf{P} is silver chloride, AgCl (s)\text{AgCl (s)}.
    • The filtrate solution Q\mathbf{Q} contains dissolved lead chloride, Pb2+(aq)\text{Pb}^{2+}\text{(aq)} and Cl−(aq)\text{Cl}^-\text{(aq)}.

  3. Reactions of Solid Residue P\mathbf{P} (AgCl\text{AgCl}):

    • Dissolution in aqueous ammonia (NH3\text{NH}_3): AgCl\text{AgCl} dissolves by forming a soluble ammine complex, diammine-silver(I): AgCl (s)+2NH3(aq)→[Ag(NH3)2]+(aq)+Cl−(aq)\text{AgCl (s)} + 2\text{NH}_3\text{(aq)} \rightarrow [\text{Ag(NH}_3)_2]^+\text{(aq)} + \text{Cl}^-\text{(aq)}

    • Dissolution in excess sodium thiosulfate (Na2S2O3\text{Na}_2\text{S}_2\text{O}_3): AgCl\text{AgCl} dissolves by forming a soluble dithiosulfatoargentate(I) complex: AgCl (s)+2S2O32−(aq)→[Ag(S2O3)2]3−(aq)+Cl−(aq)\text{AgCl (s)} + 2\text{S}_2\text{O}_3^{2-}\text{(aq)} \rightarrow [\text{Ag(S}_2\text{O}_3)_2]^{3-}\text{(aq)} + \text{Cl}^-\text{(aq)}

    These reactions confirm that metal X\mathbf{X} is Silver (Ag\text{Ag}).

  4. Reactions of Filtrate Solution Q\mathbf{Q} (Pb2+\text{Pb}^{2+}):

    • Treatment of the hot solution Q\mathbf{Q} containing Pb2+\text{Pb}^{2+} ions with potassium iodide (KI\text{KI}) produces a characteristic bright yellow precipitate of lead(II) iodide (PbI2\text{PbI}_2): Pb2+(aq)+2I−(aq)→PbI2 (s)(bright yellow precipitate)\text{Pb}^{2+}\text{(aq)} + 2\text{I}^-\text{(aq)} \rightarrow \text{PbI}_2\text{ (s)} \quad (\text{bright yellow precipitate})

    This confirms that metal Y\mathbf{Y} is Lead (Pb\text{Pb}).

Hence, the metal elements X\mathbf{X} and Y\mathbf{Y} are Ag\text{Ag} and Pb\text{Pb}, respectively.

Correct Option: (A)

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