Valence bond theory (VBT)
In VBT, the metal ion mixes some empty orbitals to make equal hybrid orbitals. Each hybrid orbital takes one lone pair from a ligand.
| CN | Hybridisation | Shape | Example |
|---|---|---|---|
| 4 | sp³ | tetrahedral | [NiCl₄]²⁻ |
| 4 | dsp² | square planar | [Ni(CN)₄]²⁻ |
| 6 | d²sp³ (inner, 3d) | octahedral | [Co(NH₃)₆]³⁺ |
| 6 | sp³d² (outer, 4d) | octahedral | [CoF₆]³⁻ |
Inner orbital (low spin) complexes use (n−1)d orbitals; electrons pair up first. Outer orbital (high spin) complexes use nd orbitals; electrons stay unpaired.
Magnetic moment
Each unpaired electron is a tiny magnet. Spin-only moment μ = √(n(n+2)) BM, where n = number of unpaired electrons. From the measured μ we can find n and so the bonding.
Limits of VBT
- It does not explain colour.
- It cannot say which ligands are strong or weak.
- It does not give energy values, and it cannot separate weak and strong fields exactly.
Crystal field theory (CFT)
CFT treats the metal–ligand bond as purely electric: ligands are point negative charges (or dipoles). Free metal ion: all 5 d orbitals have equal energy (degenerate). When ligands approach, orbitals pointing at them are pushed up more.
Octahedral field
dx²−y² and dz² point straight at the ligands → go up by 0.6Δo (the e_g set). dxy, dyz, dxz point between them → go down by 0.4Δo (the t₂g set). Δo is the crystal field splitting energy.
Tetrahedral field
The order flips: e is lower, t₂ is higher, and the split is small: Δt ≈ (4/9)Δo. So tetrahedral complexes are almost always high spin.
Spectrochemical series
Ligands in order of increasing field strength (increasing Δ): I⁻ < Br⁻ < SCN⁻ < Cl⁻ < S²⁻ < F⁻ < OH⁻ < C₂O₄²⁻ < H₂O < NCS⁻ < EDTA⁴⁻ < NH₃ < en < CN⁻ < CO.
High spin and low spin
For d⁴ to d⁷ there is a choice. Let P = pairing energy (cost of putting two electrons in one orbital).
- Δo < P (weak ligand): electron goes up → high spin, e.g. d⁴ = t₂g³e_g¹.
- Δo > P (strong ligand): electron pairs in t₂g → low spin, e.g. d⁴ = t₂g⁴e_g⁰.
Crystal field stabilisation energy (CFSE)
CFSE (octahedral) = (−0.4 × n(t₂g) + 0.6 × n(e_g)) Δo (+ pairing energy terms if extra pairs form).
Colour and magnetism explained by CFT
When white light falls on a complex, an electron in t₂g can absorb light of energy exactly Δo and jump to e_g. This is a d-d transition. The absorbed colour is removed; we see its complementary colour.
- [Ti(H₂O)₆]³⁺ absorbs green-yellow (~500 nm) → looks violet.
- No d electrons (d⁰, like Sc³⁺, Ti⁴⁺) or full d (d¹⁰, like Zn²⁺, Cu⁺): no jump → colourless.
- No ligand → no splitting → no d-d colour. That is why anhydrous CuSO₄ is white but CuSO₄·5H₂O is blue.
- Changing the ligand changes Δo, so the colour changes: [Ni(H₂O)₆]²⁺ is green, [Ni(en)₃]²⁺ is violet.
Limits of CFT: it treats ligands as point charges, so it cannot explain why a neutral ligand like CO splits more than anionic ones like F⁻, and it ignores the covalent part of the bond.
Bonding in metal carbonyls
Metal carbonyls are complexes of a metal (usually in 0 oxidation state) with CO: [Ni(CO)₄] (tetrahedral), [Fe(CO)₅] (trigonal bipyramidal), [Cr(CO)₆] (octahedral), [Mn₂(CO)₁₀] and [Co₂(CO)₈] (with metal–metal bonds).
The M–C bond has two parts:
- σ bond: the carbon of CO gives its lone pair into an empty metal orbital.
- π back bond: a filled metal d orbital gives electrons back into the empty antibonding π* orbital of CO.
Each part helps the other (synergic bonding): σ donation makes the metal richer in electrons, so it can give back more; back donation takes charge off the metal, so it can accept more. Result: a strong M–C bond, and a slightly weaker, longer C≡O bond.
Importance and uses of coordination compounds
- Biology: chlorophyll (Mg), haemoglobin (Fe), vitamin B12 (Co) are complexes.
- Analysis: EDTA titration finds Ca²⁺ and Mg²⁺ — the hardness of water. Dimethylglyoxime gives a red precipitate with Ni²⁺. Colour tests for Cu²⁺ and Fe³⁺ use complex formation.
- Metallurgy: gold and silver are extracted as cyanide complexes [Au(CN)₂]⁻, [Ag(CN)₂]⁻. Nickel is purified via [Ni(CO)₄] (Mond process).
- Medicine: cisplatin treats cancer; EDTA removes lead poisoning; D-penicillamine removes excess copper.
- Industry: Wilkinson's catalyst [(Ph₃P)₃RhCl] for hydrogenation; silver-plating and gold-plating baths use [Ag(CN)₂]⁻ and [Au(CN)₂]⁻ for smooth coats; photography fixing uses [Ag(S₂O₃)₂]³⁻.
Try it: predict, then check
Before you touch the 3D, predict: for Fe³⁺ (d⁵), how many unpaired electrons with F⁻ (weak) and with CN⁻ (strong)? Write your guess. Now in the last step set dⁿ = 5, choose weak, then strong. You should see 5 and 1 unpaired, and μ = 5.92 BM and 1.73 BM. At home: add a few drops of ammonia solution to a light blue copper sulphate solution (with an adult). It turns deep blue: NH₃ replaced water around Cu²⁺, Δo changed, so the colour changed.
Key formulas and definitions
- μ (spin-only) = √(n(n + 2)) BM, n = unpaired electrons
- Octahedral: t₂g at −0.4Δo, e_g at +0.6Δo
- CFSE (oct) = (−0.4 n_t₂g + 0.6 n_eg) Δo [+ P for each extra pair]
- Δt ≈ (4/9) Δo
- Δo < P → high spin · Δo > P → low spin
- E = hc/λ: absorbed wavelength gives Δo
Worked examples
1. Find the spin-only magnetic moment of [Fe(H₂O)₆]²⁺ (high spin).
Step 1: Fe²⁺ = 3d⁶. H₂O is weak → high spin: t₂g⁴e_g². Step 2: Unpaired: t₂g has 1 pair + 2 singles, e_g has 2 singles → n = 4. Step 3: μ = √(4 × 6) = √24 = 4.90 BM. Answer: 4.90 BM.
2. Find the hybridisation and magnetic nature of [Ni(CN)₄]²⁻.
Step 1: Ni²⁺ = 3d⁸. Step 2: CN⁻ is strong: the 8 electrons pair in 4 of the 3d orbitals, one 3d empty. Step 3: 1 d + 1 s + 2 p → dsp², square planar. Step 4: n = 0. Answer: dsp², square planar, diamagnetic.
3. [NiCl₄]²⁻ is paramagnetic but [Ni(CO)₄] is diamagnetic, though both are tetrahedral. Why?
Step 1: [NiCl₄]²⁻: Ni²⁺ = 3d⁸, Cl⁻ weak, no pairing → 2 unpaired → sp³, paramagnetic. Step 2: [Ni(CO)₄]: Ni(0) = 3d⁸4s². CO is strong: the 4s electrons move into 3d → 3d¹⁰, all paired → sp³. Answer: 2 unpaired vs 0 unpaired.
4. Calculate the CFSE of a d⁶ ion in a strong field (low spin) octahedral complex.
Step 1: Low spin d⁶ = t₂g⁶e_g⁰. Step 2: CFSE = 6 × (−0.4Δo) + 0 = −2.4Δo. Step 3: Compared with the free ion, 2 extra pairs are formed (high spin d⁶ has 1 pair, low spin has 3), so add +2P. Answer: −2.4Δo + 2P (often just written −2.4Δo).
5. A complex has μ = 3.87 BM. How many unpaired electrons does it have?
Step 1: √(n(n+2)) = 3.87 → n(n+2) = 15. Step 2: Try n = 3: 3 × 5 = 15. ✓ Answer: 3 unpaired electrons (for example Cr³⁺, d³).
6. [Ti(H₂O)₆]³⁺ absorbs light of wavelength 498 nm. Find Δo in kJ mol⁻¹. (h = 6.626 × 10⁻³⁴ J s, c = 3 × 10⁸ m s⁻¹, N_A = 6.022 × 10²³)
Step 1: E per ion = hc/λ = (6.626 × 10⁻³⁴ × 3 × 10⁸) / (498 × 10⁻⁹) = 3.99 × 10⁻¹⁹ J. Step 2: Per mole: 3.99 × 10⁻¹⁹ × 6.022 × 10²³ = 2.40 × 10⁵ J mol⁻¹. Answer: Δo ≈ 240 kJ mol⁻¹.
7. Write the electronic configuration of d⁵ in (a) high spin and (b) low spin octahedral complexes. Give n and μ for both.
Step 1: High spin (weak, e.g. [Fe(H₂O)₆]³⁺ or [MnCl₆]⁴⁻): t₂g³e_g², n = 5, μ = √35 = 5.92 BM. Step 2: Low spin (strong, e.g. [Fe(CN)₆]³⁻): t₂g⁵e_g⁰, n = 1, μ = √3 = 1.73 BM. Answer: 5.92 BM vs 1.73 BM.
Common mistakes
- Putting e_g below t₂g in an octahedral field. In octahedral, e_g points at the ligands, so it is higher. Only in tetrahedral is the order flipped.
- Using n = number of d electrons in μ = √(n(n+2)). n is the number of unpaired electrons only.
- Thinking tetrahedral complexes are often low spin. Δt is small (≈ 4/9 Δo), so they are almost always high spin.
- Saying a complex shows the colour it absorbs. It shows the complementary colour of the absorbed light.