- 2069 Bhadra · 2+3 marks
Low vs high spin; VBT of [Ni(CO)₄]
Short answer
Low spin: strong-field ligands, Δo > pairing energy, few unpaired electrons; high spin: weak-field ligands, Δo < pairing energy, maximum unpaired electrons. [Ni(CO)₄]⁰: CO forces pairing to 3d¹⁰, Ni uses sp³ hybridization, tetrahedral, diamagnetic.
Differentiate between low spin and high spin complexes. Explain the geometry and magnetic behaviour of [Ni(CO)₄]⁰ on the basis of valence bond theory.
Low spin vs high spin complexes
| Basis | Low spin | High spin |
|---|---|---|
| Ligand field | Strong-field ligands (CN⁻, CO, NH₃) | Weak-field ligands (F⁻, Cl⁻, H₂O) |
| Δo vs pairing energy | Δo > P | Δo < P |
| Unpaired electrons | Few or none (pairing occurs) | Maximum number |
| Orbitals used | Inner (d²sp³, dsp²) | Outer (sp³d², sp³) |
| Magnetism | Weakly paramagnetic or diamagnetic | Strongly paramagnetic |
| Example | [Fe(CN)₆]³⁻ | [FeF₆]³⁻ |
[Ni(CO)₄]⁰
Ni is in the 0 state: 3d⁸ 4s². CO is a strong-field ligand, so the two 4s electrons shift into the 3d orbitals, giving 3d¹⁰ (all paired). The empty 4s and three 4p orbitals form four sp³ hybrid orbitals, which accept the lone pairs from four CO.
3d: [↑↓][↑↓][↑↓][↑↓][↑↓] 4s [ ] 4p [ ][ ][ ] → sp³
- Geometry: tetrahedral.
- Magnetism: no unpaired electron, so diamagnetic.
✓ Answer checked by Claude Opus 5.5 (high effort)
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