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UPSC Civil Services (Mains)

Chemistry optional

  • 2 papers
  • 250 marks each
  • 3 h per paper
  • 5 of 8 questions, Q1 and Q5 compulsory

Paper I covers physical and inorganic chemistry.

UPSC Civil Services (Mains), Chemistry optional · 2017 · Paper I

Q5(e). Why the magnetic moments of Ln³+ ions differ from spin-only values

  • 10 marks
  • Answered
Question

“Magnetic moment of Lanthanide (III) ions is generally different from the value predicted by spin state.” Explain.

Answer in brief

The 4f electrons are buried under 5s²5p⁶, so the orbital angular momentum is not quenched and strong spin–orbit coupling makes J the good quantum number: μ=g_J√(J(J+1)) μ_B, not √(n(n+2)) μ_B.

Key points

  • Spin-only formula μₛ=√(n(n+2)) μ_B works for 3d ions because the strong ligand field quenches the orbital contribution.
  • 4f electrons are shielded by the filled 5s²5p⁶ shells. The ligand field is very weak, so L is not quenched; the ions behave almost like isolated ions.
  • Spin–orbit coupling is strong (heavy atoms) and much larger than kT. L and S couple to J: J=L-S for less than half-filled (f¹–f⁶), J=L+S for more than half-filled (f⁸–f¹³). Only the ground J level is populated.
  • μ_(eff)=g_J√(J(J+1)) μ_B. Example Tb³+ (4f⁸, ⁷F₆: S=3, L=3, J=6, g_J=1.5): μ=1.5√(42)= 9.72 μ_B, against spin-only 6.93 μ_B; observed 9.69–9.81.

The full model answer

Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.

FORMULAμ_(eff)=g_J√(J(J+1)) μ_BLANDÉg_J=1+(J(J+1)+S(S+1)-L(L+1))/(2J(J+1))

  1. Spin-only formula μₛ=√(n(n+2)) μ_B works for 3d ions because the strong ligand field quenches the orbital contribution.
  2. 4f electrons are shielded by the filled 5s²5p⁶ shells. The ligand field is very weak, so L is not quenched; the ions behave almost like isolated ions.
  3. Spin–orbit coupling is strong (heavy atoms) and much larger than kT. L and S couple to J: J=L-S for less than half-filled (f¹–f⁶), J=L+S for more than half-filled (f⁸–f¹³). Only the ground J level is populated.
  4. Moment: μ_(eff)=g_J√(J(J+1)) μ_B. Example Tb³+ (4f⁸, ⁷F₆: S=3, L=3, J=6, g_J=1.5): μ=1.5√(42)= 9.72 μ_B, against spin-only 6.93 μ_B; observed 9.69–9.81.
  5. Agreement: the g_J values fit the observed moments for most ions (Ce³⁺ 2.54 vs 2.46; Dy³⁺ 10.65 vs 10.0–10.6). Only Gd³+ (4f⁷, L=0) and the f⁰/f¹⁴ ions match the spin-only value.
  6. Exceptions Sm³⁺ and Eu³⁺: the first excited J level lies close above the ground level (for Eu³+, ⁷F₁ only 300 cm⁻¹ above ⁷F₀) and is partly populated at room temperature. Eu³+ has J=0 (μ=0) but shows 3.3–3.5 μ_B; Sm³⁺ shows about 1.6 instead of 0.85.

Application. the large, unquenched moments make Dy³+ and Ho³+ compounds the strongest paramagnets and Nd–Fe–B and Sm–Co the strongest permanent magnets; Gd³+ (7 unpaired electrons, L=0) is used in MRI contrast agents.

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Model answer to Q5(e), 2017 Paper I: solution page 22
Solution page 22 of 38
Read the whole 2017 solution2017 question paper

All parts of Question 5