JEE Challenger
More from Coordination Compounds
ChemMedAdv 2011 • Paper 1

Identification of Soluble Complex Ions in Final Solution

Comprehension Passage

When a metal rod M\mathbf{M} is dipped into an aqueous colourless concentrated solution of compound N\mathbf{N}, the solution turns light blue. Addition of aqueous NaCl\text{NaCl} to the blue solution gives a white precipitate O\mathbf{O}. Addition of aqueous NH3\text{NH}_3 dissolves O\mathbf{O} and gives an intense blue solution.

The final solution contains

Options

A

[Pb(NH3)4]2+[\text{Pb(NH}_3)_4]^{2+} and [CoCl4]2−[\text{CoCl}_4]^{2-}

B

[Al(NH3)4]3+[\text{Al(NH}_3)_4]^{3+} and [Cu(NH3)4]2+[\text{Cu(NH}_3)_4]^{2+}

C

[Ag(NH3)2]+[\text{Ag(NH}_3)_2]^+ and [Cu(NH3)4]2+[\text{Cu(NH}_3)_4]^{2+}

Correct
D

[Ag(NH3)2]+[\text{Ag(NH}_3)_2]^+ and [Ni(NH3)6]2+[\text{Ni(NH}_3)_6]^{2+}

Topics & Concepts

Coordination CompoundsCoordination Compoundsd-Block Elements

Step-by-Step Solution

To determine the composition of the final solution, we analyze the sequence of chemical reactions described:

  1. Reaction between metal M\mathbf{M} and compound N\mathbf{N}:
    The metal rod M\mathbf{M} is copper (Cu\text{Cu}), and the concentrated colourless solution N\mathbf{N} is silver nitrate (AgNO3\text{AgNO}_3). When copper is placed in an aqueous solution of silver nitrate, a redox reaction occurs:
    Cu(s)+2AgNO3(aq)⟶Cu(NO3)2(aq)+2Ag(s)\text{Cu(s)} + 2\text{AgNO}_3\text{(aq)} \longrightarrow \text{Cu(NO}_3)_2\text{(aq)} + 2\text{Ag(s)}
    The formation of hydrated cupric ions, [Cu(H2O)6]2+[\text{Cu(H}_2\text{O)}_6]^{2+}, imparts a light blue colour to the solution. Since concentrated AgNO3\text{AgNO}_3 solution is used, unreacted Ag+\text{Ag}^+ ions remain in the solution alongside the newly formed Cu2+\text{Cu}^{2+} ions.

  2. Precipitation with NaCl\text{NaCl}:
    When aqueous sodium chloride (NaCl\text{NaCl}) is added to the mixture, the chloride ions (Cl−\text{Cl}^-) selectively precipitate the remaining silver ions (Ag+\text{Ag}^+) as insoluble silver chloride (AgCl\text{AgCl}), which is a white precipitate (O\mathbf{O}):
    Ag+(aq)+Cl−(aq)⟶AgCl(s)↓(white precipitate, O)\text{Ag}^+\text{(aq)} + \text{Cl}^-\text{(aq)} \longrightarrow \text{AgCl(s)}\downarrow \quad (\text{white precipitate, } \mathbf{O})
    Copper(II) chloride remains soluble under these conditions.

  3. Reaction with excess aqueous ammonia (NH3\text{NH}_3):
    When aqueous ammonia is added to the mixture containing the precipitate O\mathbf{O} and the blue solution:

    • The white precipitate of AgCl\text{AgCl} dissolves due to the formation of the soluble diamminesilver(I) complex ion:
      AgCl(s)+2NH3(aq)⟶[Ag(NH3)2]+(aq)+Cl−(aq)\text{AgCl(s)} + 2\text{NH}_3\text{(aq)} \longrightarrow [\text{Ag(NH}_3)_2]^+\text{(aq)} + \text{Cl}^-\text{(aq)}
    • The copper(II) ions in the solution react with ammonia to form the deep/intense blue tetraamminecopper(II) complex ion:
      Cu2+(aq)+4NH3(aq)⟶[Cu(NH3)4]2+(aq)\text{Cu}^{2+}\text{(aq)} + 4\text{NH}_3\text{(aq)} \longrightarrow [\text{Cu(NH}_3)_4]^{2+}\text{(aq)}

Therefore, the final solution contains the complex ions [Ag(NH3)2]+[\text{Ag(NH}_3)_2]^+ and [Cu(NH3)4]2+[\text{Cu(NH}_3)_4]^{2+}.

Correct Option: C

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