UPSC Combined Geo-Scientist, Mains · 2021 · Mains Paper I
Q1(c). Why hydride and phosphine are strong-field ligands
Question
Both H⁻ and P(C₆H₅)₃ are ligands of similar field strength, high in spectrochemical series. Discuss the orbital factors that account for the field strength of each ligand.
Answer in brief
H⁻ gives strong σ bonding; PPh₃ combines σ donation with π acceptance. Different orbital interactions can therefore produce large ligand-field splitting.
Key points
- Hydride has a filled 1s orbital that overlaps strongly with metal orbitals of σ symmetry.
- Strong σ donation raises the metal-dominated σ-antibonding eg level; hydride has no suitable occupied pπ set to raise t₂g by π donation.
- PPh₃ donates its phosphorus lone pair through a σ bond; metal dπ back-donation into P-C σ* acceptor combinations can also stabilise t₂g.
- Raising eg and/or lowering t₂g increases Δ; a large field splitting need not imply the same bonding mechanism for both ligands.
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 →
Coversspectrochemical seriesstrong-field ligandhydride ligandtriphenylphosphineπ back-bondingsigma donor
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