JEE Challenger
More from Coordination Compounds
ChemMedAdv 2022 • Paper 1

Match Transition Metal Complexes to Electronic Properties

LIST-I contains metal species and LIST-II contains their properties.

LIST-ILIST-II(I) [Cr(CN)6]4−(P) t2g orbitals contain 4 electrons(II) [RuCl6]2−(Q) μ(spin-only)=4.9 BM(III) [Cr(H2O)6]2+(R) low spin complex ion(IV) [Fe(H2O)6]2+(S) metal ion in 4+ oxidation state(T) d4 species\begin{array}{ll} \text{\textbf{LIST-I}} & \text{\textbf{LIST-II}} \\ \text{(I) } [\text{Cr(CN)}_6]^{4-} & \text{(P) } t_{2g} \text{ orbitals contain 4 electrons} \\ \text{(II) } [\text{RuCl}_6]^{2-} & \text{(Q) } \mu(\text{spin-only}) = 4.9 \text{ BM} \\ \text{(III) } [\text{Cr(H}_2\text{O)}_6]^{2+} & \text{(R) low spin complex ion} \\ \text{(IV) } [\text{Fe(H}_2\text{O)}_6]^{2+} & \text{(S) metal ion in 4+ oxidation state} \\ & \text{(T) } d^4 \text{ species} \end{array}

[Given: Atomic number of Cr=24\text{Cr} = 24, Ru=44\text{Ru} = 44, Fe=26\text{Fe} = 26]

Match each metal species in LIST-I with their properties in LIST-II, and choose the correct option

Options

A

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

Correct
B

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

C

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

D

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

Topics & Concepts

Coordination CompoundsCoordination CompoundsCrystal Field Theoryd-Block Elements

Step-by-Step Solution

To find the correct match between LIST-I and LIST-II, let us analyze each metal species step-by-step:

  1. Complex (I): [Cr(CN)6]4−[\text{Cr(CN)}_6]^{4-}
    • Oxidation state: Let the oxidation state of Cr\text{Cr} be xx. x+6(−1)=−4  ⟹  x=+2x + 6(-1) = -4 \implies x = +2
    • Electronic configuration: Atomic number of Cr\text{Cr} is 2424 ([Ar]3d54s1[\text{Ar}] 3d^5 4s^1). Thus, Cr2+\text{Cr}^{2+} has a 3d43d^4 configuration, making it a d4d^4 species (T).
    • Crystal Field Splitting: CN−\text{CN}^- is a strong field ligand, so the crystal field splitting energy Δo>P\Delta_o > P (pairing energy). This results in a low spin complex (R).
    • Distribution of electrons: t2g4eg0t_{2g}^4 e_g^0 Thus, t2gt_{2g} orbitals contain 44 electrons (P).
    • Magnetic moment: Number of unpaired electrons n=2n = 2, so μ=2(2+2)=8≈2.83 BM\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \text{ BM}.
    • Summary of matching properties: (I) matches P, R, T. In Option A, this is represented by R, T.
  1. Complex (II): [RuCl6]2−[\text{RuCl}_6]^{2-}
    • Oxidation state: Let the oxidation state of Ru\text{Ru} be xx. x+6(−1)=−2  ⟹  x=+4x + 6(-1) = -2 \implies x = +4 Thus, the metal ion is in the 4+4+ oxidation state (S).
    • Electronic configuration: Atomic number of Ru\text{Ru} is 4444 (Group 8, 4d4d-series). The ground-state configuration of Ru\text{Ru} is [Kr]4d75s1[\text{Kr}] 4d^7 5s^1. For Ru4+\text{Ru}^{4+}, the configuration is [Kr]4d4[\text{Kr}] 4d^4, which is a d4d^4 species (T).
    • Crystal Field Splitting: For 4d4d and 5d5d series transition elements, Δo\Delta_o is intrinsically large (Δo>P\Delta_o > P) even with weak field ligands like Cl−\text{Cl}^-. Hence, it forms a low spin complex (R).
    • Distribution of electrons: t2g4eg0t_{2g}^4 e_g^0 Thus, t2gt_{2g} orbitals contain 44 electrons (P).
    • Summary of matching properties: (II) matches P, R, S, T. In Option A, this is represented by P, S.
  1. Complex (III): [Cr(H2O)6]2+[\text{Cr(H}_2\text{O)}_6]^{2+}
    • Oxidation state: Cr\text{Cr} is in the +2+2 oxidation state.
    • Electronic configuration: Cr2+\text{Cr}^{2+} has a 3d43d^4 configuration (d4d^4 species) (T).
    • Crystal Field Splitting: H2O\text{H}_2\text{O} is a weak field ligand (Δo<P\Delta_o < P), forming a high spin complex.
    • Distribution of electrons: t2g3eg1t_{2g}^3 e_g^1
    • Magnetic moment: The number of unpaired electrons is n=4n = 4. μ=4(4+2)=24≈4.90 BM(Q)\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \text{ BM} \quad \text{(Q)}
    • Summary of matching properties: (III) matches Q, T.
  1. Complex (IV): [Fe(H2O)6]2+[\text{Fe(H}_2\text{O)}_6]^{2+}
    • Oxidation state: Fe\text{Fe} is in the +2+2 oxidation state.
    • Electronic configuration: Atomic number of Fe\text{Fe} is 2626 ([Ar]3d64s2[\text{Ar}] 3d^6 4s^2). Thus, Fe2+\text{Fe}^{2+} has a 3d63d^6 configuration.
    • Crystal Field Splitting: H2O\text{H}_2\text{O} is a weak field ligand (Δo<P\Delta_o < P), forming a high spin complex.
    • Distribution of electrons: t2g4eg2t_{2g}^4 e_g^2 Thus, t2gt_{2g} orbitals contain 44 electrons (P).
    • Magnetic moment: The number of unpaired electrons is n=4n = 4 (22 unpaired in t2gt_{2g} and 22 unpaired in ege_g). μ=4(4+2)=24≈4.90 BM(Q)\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \text{ BM} \quad \text{(Q)}
    • Summary of matching properties: (IV) matches P, Q.

Conclusion:

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

This corresponds to Option A.

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