JEE Challenger
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ChemMedAdv 2024 • Paper 1

Number of Diamagnetic Species Among Given Coordination Complexes

Among the following complexes, the total number of diamagnetic species is _______.

[Mn(NH3)6]3+[\text{Mn}(\text{NH}_3)_6]^{3+}, [MnCl6]3−[\text{MnCl}_6]^{3-}, [FeF6]3−[\text{FeF}_6]^{3-}, [CoF6]3−[\text{CoF}_6]^{3-}, [Fe(NH3)6]3+[\text{Fe}(\text{NH}_3)_6]^{3+}, and [Co(en)3]3+[\text{Co}(\text{en})_3]^{3+}

[Given, atomic number: Mn=25\text{Mn} = 25, Fe=26\text{Fe} = 26, Co=27\text{Co} = 27; en=H2NCH2CH2NH2\text{en} = \text{H}_2\text{NCH}_2\text{CH}_2\text{NH}_2]

Official Numerical Answer1

Topics & Concepts

Coordination CompoundsCoordination CompoundsCrystal Field Theory

Step-by-Step Solution

To determine the number of diamagnetic species among the given coordination complexes, we evaluate the oxidation state, dd-electron configuration, and magnetic property (number of unpaired electrons, nn) for each complex in an octahedral crystal field.

A species is diamagnetic if it has zero unpaired electrons (n=0n = 0), and paramagnetic if it has one or more unpaired electrons (n>0n > 0).

1. [Mn(NH3)6]3+[\text{Mn}(\text{NH}_3)_6]^{3+}

  • Central metal ion: Mn3+\text{Mn}^{3+}
  • Electronic configuration: Mn=[Ar]3d54s2  ⟹  Mn3+=[Ar]3d4\text{Mn} = [\text{Ar}] 3d^5 4s^2 \implies \text{Mn}^{3+} = [\text{Ar}] 3d^4
  • Octahedral field configuration:
    • High-spin: t2g3eg1  ⟹  n=4t_{2g}^3 e_g^1 \implies n = 4
    • Low-spin: t2g4eg0  ⟹  n=2t_{2g}^4 e_g^0 \implies n = 2
  • Regardless of spin state, n≠0n \neq 0. Thus, it is paramagnetic.

2. [MnCl6]3−[\text{MnCl}_6]^{3-}

  • Central metal ion: Mn3+\text{Mn}^{3+} (3d43d^4)
  • Ligand: Cl−\text{Cl}^- is a weak field ligand (high-spin complex).
  • Configuration: t2g3eg1t_{2g}^3 e_g^1
  • Unpaired electrons: n=4n = 4
  • Hence, it is paramagnetic.

3. [FeF6]3−[\text{FeF}_6]^{3-}

  • Central metal ion: Fe3+\text{Fe}^{3+}
  • Electronic configuration: Fe=[Ar]3d64s2  ⟹  Fe3+=[Ar]3d5\text{Fe} = [\text{Ar}] 3d^6 4s^2 \implies \text{Fe}^{3+} = [\text{Ar}] 3d^5
  • Ligand: F−\text{F}^- is a weak field ligand (high-spin complex).
  • Configuration: t2g3eg2t_{2g}^3 e_g^2
  • Unpaired electrons: n=5n = 5
  • Hence, it is paramagnetic.

4. [CoF6]3−[\text{CoF}_6]^{3-}

  • Central metal ion: Co3+\text{Co}^{3+}
  • Electronic configuration: Co=[Ar]3d74s2  ⟹  Co3+=[Ar]3d6\text{Co} = [\text{Ar}] 3d^7 4s^2 \implies \text{Co}^{3+} = [\text{Ar}] 3d^6
  • Ligand: F−\text{F}^- is a weak field ligand (high-spin complex).
  • Configuration: t2g4eg2t_{2g}^4 e_g^2
  • Unpaired electrons: n=4n = 4
  • Hence, it is paramagnetic.

5. [Fe(NH3)6]3+[\text{Fe}(\text{NH}_3)_6]^{3+}

  • Central metal ion: Fe3+\text{Fe}^{3+} (3d53d^5)
  • Octahedral field configuration:
    • Low-spin: t2g5eg0  ⟹  n=1t_{2g}^5 e_g^0 \implies n = 1
    • High-spin: t2g3eg2  ⟹  n=5t_{2g}^3 e_g^2 \implies n = 5
  • In both cases, n≠0n \neq 0. Hence, it is paramagnetic.

6. [Co(en)3]3+[\text{Co}(\text{en})_3]^{3+}

  • Central metal ion: Co3+\text{Co}^{3+} (3d63d^6)
  • Ligand: en\text{en} (ethylenediamine) is a strong field chelating ligand (low-spin complex).
  • Configuration: t2g6eg0t_{2g}^6 e_g^0
  • Unpaired electrons: n=0n = 0
  • Hence, it is diamagnetic.

Conclusion

Only 11 complex, [Co(en)3]3+[\text{Co}(\text{en})_3]^{3+}, is diamagnetic.

The total number of diamagnetic species is 1.