JEE Challenger
More from Coordination Compounds
ChemMedAdv 2014 • Paper 2

Identification of Metal Ion M1 and Ligand Reagents

Comprehension Passage

An aqueous solution of metal ion M1\mathbf{M1} reacts separately with reagents Q\mathbf{Q} and R\mathbf{R} in excess to give tetrahedral and square planar complexes, respectively. An aqueous solution of another metal ion M2\mathbf{M2} consistently forms tetrahedral complexes with these reagents. The reaction of an aqueous solution of M2\mathbf{M2} with reagent S\mathbf{S} yields a white precipitate which subsequently dissolves in an excess of S\mathbf{S}. These transformations are summarized in the scheme below:

Passage Diagram 1

Identify M1\mathbf{M1}, Q\mathbf{Q}, and R\mathbf{R}, respectively:

Options

A

Zn2+\text{Zn}^{2+}, KCN\text{KCN} and HCl\text{HCl}

B

Ni2+\text{Ni}^{2+}, HCl\text{HCl} and KCN\text{KCN}

Correct
C

Cd2+\text{Cd}^{2+}, KCN\text{KCN} and HCl\text{HCl}

D

Co2+\text{Co}^{2+}, HCl\text{HCl}, and KCN\text{KCN}

Topics & Concepts

Coordination CompoundsCoordination CompoundsCrystal Field Theoryd-Block Elements

Step-by-Step Solution

To determine the identities of M1\mathbf{M1}, Q\mathbf{Q}, and R\mathbf{R}, we analyze the coordination chemistry and stereochemistry of the metal complexes described in the passage:

  1. Reaction of M1\mathbf{M1} with excess Q\mathbf{Q} and R\mathbf{R}:

    • With excess reagent Q\mathbf{Q}, the metal ion M1\mathbf{M1} forms a tetrahedral complex.
    • With excess reagent R\mathbf{R}, the metal ion M1\mathbf{M1} forms a square planar complex.

  2. Analysis of Metal Ions:

    • For Ni2+\text{Ni}^{2+} (3d83d^8 configuration):

      • With a weak field ligand like Cl−\text{Cl}^- (provided by excess HCl\text{HCl}), pairing of electrons does not occur. The complex formed is [NiCl4]2−[\text{NiCl}_4]^{2-}, which undergoes sp3sp^3 hybridization and possesses a tetrahedral geometry: Ni2++4Cl−⟶[NiCl4]2−(sp3,tetrahedral)\text{Ni}^{2+} + 4\text{Cl}^- \longrightarrow [\text{NiCl}_4]^{2-} \quad (sp^3, \text{tetrahedral})
      • With a strong field ligand like CN−\text{CN}^- (provided by excess KCN\text{KCN}), pairing of the two unpaired 3d3d electrons occurs, vacating one 3d3d orbital. The complex formed is [Ni(CN)4]2−[\text{Ni(CN)}_4]^{2-}, which undergoes dsp2dsp^2 hybridization and possesses a square planar geometry: Ni2++4CN−⟶[Ni(CN)4]2−(dsp2,square planar)\text{Ni}^{2+} + 4\text{CN}^- \longrightarrow [\text{Ni(CN)}_4]^{2-} \quad (dsp^2, \text{square planar})

    • For other given ions:

      • Zn2+\text{Zn}^{2+} and Cd2+\text{Cd}^{2+} (d10d^{10} configurations) have completely filled dd-orbitals, so they universally form tetrahedral complexes (sp3sp^3) with both weak and strong field 4-coordinate ligands (e.g., both [ZnCl4]2−[\text{ZnCl}_4]^{2-} and [Zn(CN)4]2−[\text{Zn(CN)}_4]^{2-} are tetrahedral). This identifies Zn2+\text{Zn}^{2+} as M2\mathbf{M2}, not M1\mathbf{M1}.
      • Co2+\text{Co}^{2+} (3d73d^7) typically forms tetrahedral complexes with Cl−\text{Cl}^-, but with excess CN−\text{CN}^-, it oxidizes readily to form low-spin octahedral complexes such as [Co(CN)6]3−[\text{Co(CN)}_6]^{3-} or five-/six-coordinate complexes.

  3. Conclusion:

    • Metal ion M1=Ni2+\mathbf{M1} = \text{Ni}^{2+}
    • Reagent Q=HCl\mathbf{Q} = \text{HCl}
    • Reagent R=KCN\mathbf{R} = \text{KCN}

Therefore, the correct option is B.

PreviousNext