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JEE Advanced 2025 Paper 2 · Chemistry · Coordination Compounds · medium

The complexes [Ni(CN)4]^2-, [Fe(CN)6]^3-, [NiCl4]^2-, [Cr(H2O)6]^3+ and [Fe(H2O)6]^2+ are arranged in order of increasing spin-only magnetic moment.…

Single correct option

The complexes [Ni(CN)4]2−[\mathrm{Ni(CN)_4}]^{2-}, [Fe(CN)6]3−[\mathrm{Fe(CN)_6}]^{3-}, [NiCl4]2−[\mathrm{NiCl_4}]^{2-}, [Cr(H2O)6]3+[\mathrm{Cr(H_2O)_6}]^{3+} and [Fe(H2O)6]2+[\mathrm{Fe(H_2O)_6}]^{2+} are arranged in order of increasing spin-only magnetic moment. The correct order is
  1. A. [Ni(CN)4]2−<[Fe(CN)6]3−<[NiCl4]2−<[Cr(H2O)6]3+<[Fe(H2O)6]2+[\mathrm{Ni(CN)_4}]^{2-}<[\mathrm{Fe(CN)_6}]^{3-}<[\mathrm{NiCl_4}]^{2-}<[\mathrm{Cr(H_2O)_6}]^{3+}<[\mathrm{Fe(H_2O)_6}]^{2+}
  2. B. [Ni(CN)4]2−<[NiCl4]2−<[Fe(CN)6]3−<[Cr(H2O)6]3+<[Fe(H2O)6]2+[\mathrm{Ni(CN)_4}]^{2-}<[\mathrm{NiCl_4}]^{2-}<[\mathrm{Fe(CN)_6}]^{3-}<[\mathrm{Cr(H_2O)_6}]^{3+}<[\mathrm{Fe(H_2O)_6}]^{2+}
  3. C. [Ni(CN)4]2−<[Fe(CN)6]3−<[Cr(H2O)6]3+<[NiCl4]2−<[Fe(H2O)6]2+[\mathrm{Ni(CN)_4}]^{2-}<[\mathrm{Fe(CN)_6}]^{3-}<[\mathrm{Cr(H_2O)_6}]^{3+}<[\mathrm{NiCl_4}]^{2-}<[\mathrm{Fe(H_2O)_6}]^{2+}
  4. D. [Ni(CN)4]2−<[NiCl4]2−<[Cr(H2O)6]3+<[Fe(H2O)6]2+<[Fe(CN)6]3−[\mathrm{Ni(CN)_4}]^{2-}<[\mathrm{NiCl_4}]^{2-}<[\mathrm{Cr(H_2O)_6}]^{3+}<[\mathrm{Fe(H_2O)_6}]^{2+}<[\mathrm{Fe(CN)_6}]^{3-}
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Answer: (A) [Ni(CN)4]2−<[Fe(CN)6]3−<[NiCl4]2−<[Cr(H2O)6]3+<[Fe(H2O)6]2+[\mathrm{Ni(CN)_4}]^{2-}<[\mathrm{Fe(CN)_6}]^{3-}<[\mathrm{NiCl_4}]^{2-}<[\mathrm{Cr(H_2O)_6}]^{3+}<[\mathrm{Fe(H_2O)_6}]^{2+}

Use μ=n(n+2)\mu=\sqrt{n(n+2)} B.M. [Ni(CN)4]2−[\mathrm{Ni(CN)_4}]^{2-}: Ni2+\mathrm{Ni^{2+}} is d8d^8; strong-field CN−\mathrm{CN^-} gives square planar, n=0n=0, μ=0\mu=0. [Fe(CN)6]3−[\mathrm{Fe(CN)_6}]^{3-}: d5d^5 low spin t2g5t_{2g}^5, n=1n=1, μ=1.73\mu=1.73. [NiCl4]2−[\mathrm{NiCl_4}]^{2-}: weak-field Cl−\mathrm{Cl^-} gives tetrahedral d8d^8 (e4t24e^4t_2^4), n=2n=2, μ=2.83\mu=2.83. [Cr(H2O)6]3+[\mathrm{Cr(H_2O)_6}]^{3+}: d3d^3, n=3n=3, μ=3.87\mu=3.87. [Fe(H2O)6]2+[\mathrm{Fe(H_2O)_6}]^{2+}: d6d^6 high spin (water is weak field for Fe2+\mathrm{Fe^{2+}}), n=4n=4, μ=4.90\mu=4.90. The increasing order is 0<1.73<2.83<3.87<4.900<1.73<2.83<3.87<4.90, option 1. Option 2 treats NiCl42−\mathrm{NiCl_4^{2-}} as diamagnetic like the cyanide; option 4 treats the cyanoferrate as high spin (5.925.92).
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