Coordination Compounds · Lesson 7 of 12
Valence Bond Theory and Magnetism
“Connect formal d-electron counts, donor-pair bonding, geometry and approximate magnetic moments.”
• Obtain metal-ion d counts by removing outer s electrons first. • Use the elementary VBT orbital descriptions for common geometries. • Count unpaired electrons and calculate spin-only magnetic moments. • Explain where the VBT model needs experimental input.
Start with the metal, not the ligand reputation
Two complexes containing nickel can have different oxidation states, d counts and magnetic behaviour even if their geometries match. Determine oxidation state before discussing pairing. For first-row metal ions, remove the 4s electrons before 3d electrons. Thus Fe(II) is d⁶, Fe(III) d⁵, Co(III) d⁶, Ni(II) d⁸, Cr(III) d³ and Mn(II) d⁵. A neutral carbonyl has a different formal metal state from an anionic halido complex.
Valence bond theory, VBT, represents bonding by ligand electron pairs occupying metal–ligand bonding combinations of appropriate direction. Its elementary hybridisation labels are a useful geometrical bookkeeping model. Hybrid orbitals are mathematical combinations of wavefunctions, not separately observable physical containers that must form before a ligand approaches. Modern molecular orbital descriptions give a more complete account of covalency and orbital energies.
| CN | Geometry | Elementary label | Orbitals used in the model |
|---|---|---|---|
| 4 | Tetrahedral | sp³ | one s and three p |
| 4 | Square planar | dsp² | one d, one s and two p |
| 5 | Trigonal bipyramidal | sp³d | one s, three p and one d |
| 6 | Octahedral, inner orbital | d²sp³ | two (n−1)d, ns and three np |
| 6 | Octahedral, outer orbital | sp³d² | ns, three np and two nd |
Cobalt(III): same d count, different pairing
Problem
Explain the VBT descriptions of [Co(NH₃)₆]³⁺ and [CoF₆]³⁻.
- 1.Both contain Co(III): ammonia is neutral, whereas six F⁻ give x − 6 = −3. Co³⁺ is 3d⁶.
- 2.In [Co(NH₃)₆]³⁺, the low-spin arrangement has all six d electrons paired within three d orbitals. The elementary VBT scheme then uses two vacant 3d orbitals with 4s and 4p for d²sp³, an inner-orbital octahedral description. It is diamagnetic.
- 3.For [CoF₆]³⁻, a high-spin d⁶ arrangement has four unpaired electrons. In the VBT scheme 4d rather than 3d orbitals are included in sp³d², an outer-orbital octahedral description. It is paramagnetic.
- 4.The pairing difference is experimental input in basic VBT. Crystal field theory will explain it through a competition between splitting and pairing energies.
Nickel(II): distinguish square plane from tetrahedron
Both [Ni(CN)₄]²⁻ and [NiCl₄]²⁻ contain Ni(II), d⁸, and have CN 4. The cyanido complex is square planar and diamagnetic. In the elementary VBT scheme d electrons pair to leave one inner d orbital available for dsp². The chlorido complex is tetrahedral and has two unpaired electrons; it receives the sp³ description. These differences cannot be deduced from CN or oxidation state alone.
Ni(CO)₄ is tetrahedral but diamagnetic. CO is neutral, so the metal has oxidation state zero and formal d¹⁰ count in ligand electron bookkeeping, rather than the d⁸ count of Ni(II). All ten d electrons are paired. It is therefore incorrect to claim that all tetrahedral nickel complexes must have the same magnetism. The free atom configuration and the formal d count in a bound complex are not interchangeable descriptions; bonding reorganises orbital occupations.
Magnetic moments from unpaired electrons
Unpaired electrons give a paramagnetic contribution and attraction to a magnetic field. A species with all electrons paired is diamagnetic in the elementary classification and is weakly repelled. Real magnetic data can include orbital contributions and interactions between centres.
Each unpaired electron has spin 1/2. For n parallel unpaired electrons the total spin is S = n/2. Using μ ≈ g√[S(S+1)]μB with g ≈ 2 gives μ ≈ √[n(n+2)]μB. This estimate does not predict n; first determine the electron arrangement. It is also not an exact equality for every metal, especially when orbital contributions, spin–orbit coupling or magnetic exchange matter.
| n | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| μso / BM | 0 | 1.73 | 2.83 | 3.87 | 4.90 | 5.92 |
Problem
[MnBr₄]²⁻ has a moment near 5.9 BM. Explain the likely geometry.
- 1.Mn +2 follows from x − 4 = −2. Mn²⁺ is d⁵ and CN is 4.
- 2.Five unpaired electrons give √(5×7) = √35 = 5.92 BM, matching the observation. Weak-field bromido and the high-spin d⁵ pattern support a tetrahedral entity.
- 3.The basic VBT label is sp³. This is a chemically supported inference using ligand identity and magnetism together; magnetic moment alone is not a universal geometry-measuring instrument.
A comparison set that tests the method
| Entity | Metal / d count | Unpaired electrons | Description |
|---|---|---|---|
| [Fe(CN)₆]⁴⁻ | Fe(II), d⁶ | 0 | Low spin, octahedral, d²sp³ |
| [FeF₆]³⁻ | Fe(III), d⁵ | 5 | High spin, octahedral, sp³d² |
| [Co(C₂O₄)₃]³⁻ | Co(III), d⁶ | 0 | Low spin, octahedral, d²sp³ |
| [CoF₆]³⁻ | Co(III), d⁶ | 4 | High spin, octahedral, sp³d² |
| [Mn(CN)₆]³⁻ | Mn(III), d⁴ | 2 | Low spin, octahedral |
| [MnCl₆]³⁻ | Mn(III), d⁴ | 4 | High spin, octahedral |
| [Cr(NH₃)₆]³⁺ | Cr(III), d³ | 3 | Octahedral, t₂g³ |
| [Ni(NH₃)₆]²⁺ | Ni(II), d⁸ | 2 | Octahedral, outer-orbital VBT label |
For d¹ to d³ octahedral ions there are already at least two unoccupied d orbitals in the elementary free-ion filling picture. For d⁴ to d⁶ the inner-orbital description may require pairing. An inner-orbital complex need not be diamagnetic: low-spin d⁵ [Fe(CN)₆]³⁻ still has one unpaired electron. Conversely square-planar [Pt(CN)₄]²⁻ has Pt(II), 5d⁸, with all electrons paired in the established square-planar field.
What VBT does not settle
VBT can organise known shapes and electron pairings, but its elementary form does not calculate absorption colours, quantitatively predict ligand field strengths, reliably choose every four-coordinate geometry, or explain thermodynamic and kinetic stability. Calling a ligand “strong” without specifying metal and geometry is insufficient. Use experimental observations and a more informative orbital-energy model when these questions matter.
Quick Quiz
Quiz
What is the formal d count of Fe(III)?
Neutral Fe loses its two 4s electrons and one 3d electron to form Fe³⁺.
What is μso for three unpaired electrons?
Insert n = 3 into the spin-only expression: √[3(3+2)] = √15 = 3.87 BM.
Which comparison correctly explains two tetrahedral nickel complexes?
Their different oxidation states give different d counts despite the shared tetrahedral geometry.
Which description can be paramagnetic?
Low-spin d⁵ leaves one unpaired electron.
What is an important limitation of elementary VBT?
An energy-level model is needed to discuss absorption and splitting quantitatively.
Practice Problems
- Compare [Fe(CN)₆]³⁻ and [Fe(H₂O)₆]³⁺ using oxidation state, d count and μso.
- Explain why [Co(NH₃)₆]³⁺ is described as inner orbital but [Ni(NH₃)₆]²⁺ as outer orbital.
- Find μso for [Cr(NH₃)₆]³⁺, [Ni(CN)₄]²⁻ and [Pt(CN)₄]²⁻.
- Discuss [Fe(CN)₆]⁴⁻, [FeF₆]³⁻, [Co(C₂O₄)₃]³⁻ and [CoF₆]³⁻ in the elementary VBT model.
- Among [Cr(H₂O)₆]³⁺, [Fe(H₂O)₆]²⁺ and [Zn(H₂O)₆]²⁺, which has the largest spin-only moment?
- What uncertainty remains when an observed moment is compared with a spin-only value?
1. Both have Fe +3 and d⁵. 2. The cyanido entity is low spin with one unpaired electron, giving √3 = 1.73 BM. 3. The aqua entity is high spin with five unpaired electrons, giving √35 = 5.92 BM. 4. Inner/outer VBT labels are d²sp³ and sp³d² respectively; ligand field splitting explains why their pairings differ.
Key Takeaways
• Obtain oxidation state before assigning a d count. • First-row ions lose 4s electrons before 3d electrons. • Hybridisation labels summarise a limited bonding model. • Inner-orbital does not always mean diamagnetic. • Geometry and ligand field influence electron pairing. • Spin-only moments follow μso = √[n(n+2)] BM. • Magnetism supports structural inference but does not determine every structure alone. • Elementary VBT does not explain absorption colours quantitatively.