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JEE Main 2019 · Chemistry · Coordination Compounds · medium

The ratio of spin-only magnetic moments of [Mn(CN)6]^3- (strong-field ligand) to [MnCl6]^3- (weak-field ligand) is

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The ratio of spin-only magnetic moments of [Mn(CN)6]3−[\mathrm{Mn(CN)_6}]^{3-} (strong-field ligand) to [MnCl6]3−[\mathrm{MnCl_6}]^{3-} (weak-field ligand) is
  1. A. 32\dfrac{\sqrt3}{2}
  2. B. 23\dfrac{2}{3}
  3. C. 12\dfrac{1}{2}
  4. D. 13\dfrac{1}{\sqrt3}
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Answer: (D) 13\dfrac{1}{\sqrt3}

Mn(III) is d4d^4. With strong-field CN−\mathrm{CN^-}, the electrons pair as t2g4eg0t_{2g}^4e_g^0, giving 2 unpaired electrons: μ=2⋅4=8 BM\mu=\sqrt{2\cdot4}=\sqrt8\ \mathrm{BM}. With weak-field Cl−\mathrm{Cl^-}, high spin t2g3eg1t_{2g}^3e_g^1 has 4 unpaired electrons: μ=4⋅6=24 BM\mu=\sqrt{4\cdot6}=\sqrt{24}\ \mathrm{BM}. Ratio =8/24=1/3=0.577=\sqrt{8/24}=1/\sqrt3 = 0.577.
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