UPSC Combined Geo-Scientist, Mains · 2021 · Mains Paper I
Q1(j). Orbital contribution to magnetic moments
Question
Why does the μeff value for transition metal ions with ground state term S agree well with μs.o. value while it differs with metal ions having ground state term as D or F? Explain with reason.
Answer in brief
An S term has L = 0, so the leading orbital moment vanishes. D and F terms have L = 2 and 3 and can give residual orbital and spin-orbit contributions.
Key points
- For an S spectroscopic term, L = 0; the letter S here is not the numerical spin quantum number.
- The spin-only approximation is μs.o. = g√[S(S + 1)] μB ≈ √[n(n + 2)] μB, using g ≈ 2.
- D and F free-ion terms have nonzero orbital angular momentum. Spin-orbit coupling can alter the effective g value and hence μeff.
- A ligand field can partly or strongly quench orbital motion, so the deviation depends on the actual ligand-field ground state.
The full model answer
As written for the answer booklet: the diagram carries the working, the lines stay short. Read here only; it is not offered as a download.
Topic notes: crystal field theory, with every Civil Services, IFoS and Geo-Scientist PYQ →
Coversmagnetic momentspin-only formulaorbital contributionterm symbolquenching of orbital angular momentumspin-orbit coupling
All parts of Question 1
- 1(a) Quartz-type aluminophosphate5 marks
- 1(b) Fifth ammonia ligand on copper(II)5 marks
- 1(c) Why hydride and phosphine are strong-field ligands5 marks
- 1(d) Indicator choice for weak-acid titration5 marks
- 1(e) Postprecipitation versus coprecipitation5 marks
- 1(f) Role of mercuric chloride in iron titration5 marks
- 1(g) Chelators in complexometric titrations5 marks
- 1(h) Carbonyl C-O bond strengths5 marks
- 1(i) Balancing nuclear reactions5 marks
- 1(j) Orbital contribution to magnetic moments5 marks