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JEE Advanced 2024 Paper 1 · Chemistry · Coordination Compounds · medium

Match the complex ions in List-I with their geometry and magnetic character in List-II (use atomic numbers: Fe 26, Co 27, Ni 28; spin-only formula…

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Match the complex ions in List-I with their geometry and magnetic character in List-II (use atomic numbers: Fe 26, Co 27, Ni 28; spin-only formula, μ=n(n+2)\mu=\sqrt{n(n+2)} BM).

| List-I | List-II |
|---|---|
| P. [Fe(H2O)6]2+\mathrm{[Fe(H_2O)_6]^{2+}} | 1. Square planar, diamagnetic |
| Q. [Co(NH3)6]3+\mathrm{[Co(NH_3)_6]^{3+}} | 2. Tetrahedral, μ≈2.83\mu\approx2.83 BM |
| R. [NiCl4]2−\mathrm{[NiCl_4]^{2-}} | 3. Octahedral, μ≈4.90\mu\approx4.90 BM |
| S. [Ni(CN)4]2−\mathrm{[Ni(CN)_4]^{2-}} | 4. Octahedral, diamagnetic |
| | 5. Octahedral, μ≈1.73\mu\approx1.73 BM |
  1. A. P→3, Q→4, R→2, S→1
  2. B. P→3, Q→5, R→2, S→1
  3. C. P→4, Q→3, R→1, S→2
  4. D. P→3, Q→4, R→1, S→2
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Answer: (A) P→3, Q→4, R→2, S→1

P: Fe2+\mathrm{Fe^{2+}} is d6d^6; H2O\mathrm{H_2O} is a weak-field ligand, so high spin, t2g4eg2t_{2g}^4e_g^2, 4 unpaired electrons, μ=24=4.90\mu=\sqrt{24}=4.90 BM, octahedral: 3. Q: Co3+\mathrm{Co^{3+}} is d6d^6; NH3\mathrm{NH_3} is strong enough for low spin t2g6t_{2g}^6, 0 unpaired, diamagnetic octahedral: 4. R: Ni2+\mathrm{Ni^{2+}} is d8d^8; chloride is a weak-field ligand, giving tetrahedral geometry with 2 unpaired electrons, μ=8=2.83\mu=\sqrt8=2.83 BM: 2. S: CN−\mathrm{CN^-} is a strong-field ligand which pairs the d8d^8 electrons, giving dsp2dsp^2 square planar, diamagnetic: 1. Thus P→3, Q→4, R→2, S→1. Option 5 (1.731.73 BM, one unpaired electron) matches none of the four.
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