Valence bond theory and magnetic properties
In valence bond theory the metal hybridises empty orbitals to accept ligand electron pairs. Count the unpaired electrons to predict the shape and the magnetic moment.
Coordination number, hybridisation and shape
| Coordination number | Hybridisation | Shape | Examples |
|---|---|---|---|
| 4 | sp³ | Tetrahedral | [Ni(CO)₄], [NiCl₄]²⁻ |
| 4 | dsp² | Square planar | [Ni(CN)₄]²⁻ |
| 5 | sp³d | Trigonal bipyramidal | [Fe(CO)₅] (uses an inner 3d orbital, so often written dsp³) |
| 6 | d²sp³ (inner orbital) | Octahedral | [Co(NH₃)₆]³⁺, [Fe(CN)₆]³⁻ |
| 6 | sp³d² (outer orbital) | Octahedral | [CoF₆]³⁻, [FeF₆]³⁻ |
Worked example 1: [Co(NH₃)₆]³⁺, an inner orbital complex
Co: Z = 27, [Ar] 3d⁷ 4s². In the complex, cobalt is +3, so Co³⁺ is [Ar] 3d⁶. Arrows are the metal's electrons; ×× is a pair given by a ligand.
- 1Free Co³⁺ ion: 3d ↑↓ ↑ ↑ ↑ ↑ | 4s _ | 4p _ _ _4 unpaired electrons.
- 2In this complex the 3d electrons pair up: 3d ↑↓ ↑↓ ↑↓ _ _Six electrons now fill three 3d orbitals, leaving two 3d orbitals empty.
- 3Two 3d + one 4s + three 4p orbitals mix: d²sp³ hybridisationSix equal hybrid orbitals point to the corners of an octahedron.
- 4Six NH₃ donate six pairs: 3d ↑↓ ↑↓ ↑↓ ×× ×× | 4s ×× | 4p ×× ×× ××
- 5★Octahedral, inner orbital (low spin) complex with 0 unpaired electrons: diamagnetic.
★ key step
Worked example 2: [CoF₆]³⁻, an outer orbital complex
Co³⁺ again: [Ar] 3d⁶.
- 1F⁻ is a weak ligand, so there is no pairing: 3d ↑↓ ↑ ↑ ↑ ↑4 unpaired electrons remain.
- 2No 3d orbital is free, so the outer 4s, 4p and 4d orbitals mix: sp³d²
- 3Six F⁻ donate six pairs: 4s ×× | 4p ×× ×× ×× | 4d ×× ××
- 4★Octahedral, outer orbital (high spin) complex, 4 unpaired electrons, paramagnetic: μ = √[4(4 + 2)] = √24 ≈ 4.90 BM.
★ key step
More worked electron counts
| Complex | Metal ion, d electrons | 3d after ligands bond | Hybridisation, shape | Unpaired; μ (BM) |
|---|---|---|---|---|
| [Ni(CN)₄]²⁻ | Ni²⁺, 3d⁸ | ↑↓ ↑↓ ↑↓ ↑↓ _ (CN⁻ forces pairing) | dsp², square planar | 0; diamagnetic |
| [NiCl₄]²⁻ | Ni²⁺, 3d⁸ | ↑↓ ↑↓ ↑↓ ↑ ↑ (no pairing) | sp³, tetrahedral | 2; 2.83 |
| [Ni(CO)₄] | Ni(0), 3d⁸ 4s² | ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ (4s electrons move into 3d) | sp³, tetrahedral | 0; diamagnetic |
| [Cr(NH₃)₆]³⁺ | Cr³⁺, 3d³ | ↑ ↑ ↑ _ _ (two 3d already empty) | d²sp³, octahedral | 3; 3.87 |
| [Mn(CN)₆]³⁻ | Mn³⁺, 3d⁴ | ↑↓ ↑ ↑ _ _ | d²sp³, octahedral | 2; 2.83 |
| [Fe(CN)₆]³⁻ | Fe³⁺, 3d⁵ | ↑↓ ↑↓ ↑ _ _ | d²sp³, octahedral | 1; 1.73 |
| [Fe(CN)₆]⁴⁻ | Fe²⁺, 3d⁶ | ↑↓ ↑↓ ↑↓ _ _ | d²sp³, octahedral | 0; diamagnetic |
| [FeF₆]³⁻ | Fe³⁺, 3d⁵ | ↑ ↑ ↑ ↑ ↑ | sp³d², octahedral | 5; 5.92 |
From a magnetic moment to the type of complex
The spin-only magnetic moment of [Fe(H₂O)₆]²⁺ is found to be 4.90 BM. How many unpaired electrons does it have? Is it an inner or an outer orbital complex?
- Formula: μ = √[n(n + 2)] BM.
- Try n = 4: √[4 × 6] = √24 = 4.90 BM. This matches, so n = 4.
- Fe: Z = 26, [Ar] 3d⁶ 4s². Fe²⁺: 3d⁶.
- Four unpaired electrons means no pairing: 3d ↑↓ ↑ ↑ ↑ ↑. No 3d orbital is free.
- So the outer 4s, 4p and 4d orbitals are used: sp³d².
4 unpaired electrons; an octahedral outer orbital (high spin) complex, sp³d².
Limitations of valence bond theory
- It involves a number of assumptions.
- It gives no quantitative explanation of magnetic data.
- It does not explain the colour of coordination compounds.
- It gives no quantitative idea of the thermodynamic or kinetic stability of complexes.
- It cannot predict exactly whether a 4-coordinate complex will be tetrahedral or square planar.
- It does not distinguish between weak and strong ligands.
Hospital MRI scans often use contrast agents that are chelate complexes of gadolinium(III), which has seven unpaired electrons. Their strong paramagnetism makes soft tissues show up more clearly.
3 marks: “Using valence bond theory, explain the hybridisation, shape and magnetic behaviour of [Ni(CN)₄]²⁻.” Ni²⁺ is 3d⁸. Strong CN⁻ pairs the electrons, freeing one 3d orbital. One 3d + 4s + two 4p give dsp²: square planar, no unpaired electrons, diamagnetic.
3 marks: “[Fe(CN)₆]³⁻ is weakly paramagnetic but [FeF₆]³⁻ is strongly paramagnetic. Explain.” Both have Fe³⁺, 3d⁵. CN⁻ forces pairing: 1 unpaired electron (1.73 BM), d²sp³, inner orbital. F⁻ does not: 5 unpaired electrons (5.92 BM), sp³d², outer orbital.
2 marks: “Write two limitations of valence bond theory.” It does not explain the colour of complexes, and it does not distinguish weak from strong ligands (also: no quantitative magnetic data).
A strong ligand squeezes electrons into pairs: inner, d²sp³, low spin. A weak ligand leaves them alone: outer, sp³d², high spin. μ for 1 to 5 unpaired: 1.73, 2.83, 3.87, 4.90, 5.92.