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JEE Advanced 2023 Paper 1 · Chemistry · Coordination Compounds · medium

Match the complex ions in List-I with their properties in List-II. Fe=26, Co=27, Ni=28.) List-I: (P) [Fe(CN)6]^3-; (Q) [CoF6]^3-; (R) [Ni(CN)4]^2-…

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Match the complex ions in List-I with their properties in List-II. (Atomic numbers: Fe=26\mathrm{Fe}=26, Co=27\mathrm{Co}=27, Ni=28\mathrm{Ni}=28.)

List-I: (P) [Fe(CN)6]3−\mathrm{[Fe(CN)_6]^{3-}}; (Q) [CoF6]3−\mathrm{[CoF_6]^{3-}}; (R) [Ni(CN)4]2−\mathrm{[Ni(CN)_4]^{2-}}; (S) [NiCl4]2−\mathrm{[NiCl_4]^{2-}}

List-II: (1) spin-only magnetic moment ≈1.73 BM\approx1.73\ \mathrm{BM}; (2) diamagnetic and square planar; (3) four unpaired electrons; (4) tetrahedral with spin-only magnetic moment ≈2.83 BM\approx2.83\ \mathrm{BM}; (5) t2g3eg0t_{2g}^{3}e_g^{0} configuration
  1. A. P→1, Q→3, R→2, S→4
  2. B. P→3, Q→1, R→2, S→4
  3. C. P→1, Q→3, R→4, S→2
  4. D. P→5, Q→3, R→2, S→4
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Answer: (A) P→1, Q→3, R→2, S→4

(P) Fe3+\mathrm{Fe^{3+}} is d5d^5; CN−\mathrm{CN^-} is a strong-field ligand, so the configuration is low spin t2g5t_{2g}^5 with one unpaired electron: μ=3=1.73 BM\mu=\sqrt{3}=1.73\ \mathrm{BM} ⇒\Rightarrow 1.
(Q) Co3+\mathrm{Co^{3+}} is d6d^6; F−\mathrm{F^-} is a weak-field ligand, so the complex is high spin t2g4eg2t_{2g}^4e_g^2 with four unpaired electrons ⇒\Rightarrow 3.
(R) Ni2+\mathrm{Ni^{2+}} is d8d^8; with the strong-field CN−\mathrm{CN^-} it forms a square planar dsp2dsp^2 complex with all electrons paired, hence diamagnetic ⇒\Rightarrow 2.
(S) With the weak-field Cl−\mathrm{Cl^-}, Ni2+\mathrm{Ni^{2+}} (d8d^8) forms a tetrahedral sp3sp^3 complex with two unpaired electrons: μ=8=2.83 BM\mu=\sqrt{8}=2.83\ \mathrm{BM} ⇒\Rightarrow 4.
No complex in List-I has t2g3eg0t_{2g}^3e_g^0, so (5) is a distractor. Correct code: P→1, Q→3, R→2, S→4.
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