Crystal Field Theory Explained: d-Orbital Splitting, CFSE and High vs Low Spin
Short answer: In an octahedral complex the five d orbitals split into t2g (−0.4Δo) and eg (+0.6Δo). If Δo is larger than the pairing energy P, electrons pair in t2g (low spin); if smaller, they spread out (high spin). CFSE = (−0.4 n(t2g) + 0.6 n(eg)) Δo + mP, and μ = √(n(n + 2)) BM. Example: [Fe(CN)₆]⁴⁻ is low-spin d⁶ with CFSE −2.4Δo + 2P and no unpaired electrons. The free crystal field splitting diagram does this for any common metal, geometry and ligand.
Key terms in one place
| Crystal field splitting (Δ) | The energy gap between d-orbital sets caused by the ligands. Δo for octahedral, Δt for tetrahedral. 1 Δo = 10 Dq. |
|---|---|
| t2g and eg | Octahedral sets: t2g (dxy, dxz, dyz) at −0.4Δo and eg (dz², dx²−y²) at +0.6Δo. |
| Pairing energy (P) | The energy cost of putting two electrons in the same orbital. |
| High spin and low spin | High spin: Δ < P, maximum unpaired electrons. Low spin: Δ > P, electrons pair in the lower set. |
| CFSE | Crystal field stabilisation energy: the energy gained by the d electrons compared with a spherical field. |
| Spin-only moment | μ = √(n(n + 2)) BM, where n is the number of unpaired electrons. |
The rules step by step
- Find dⁿ. dⁿ = group number − oxidation state. Fe²⁺: 8 − 2 = d⁶. Cu²⁺: 11 − 2 = d⁹. Mn in MnO₄⁻ (+7): d⁰.
- Pick the splitting pattern. Octahedral: t2g below eg. Tetrahedral: e below t2, with Δt ≈ 4/9 Δo. Square planar: dx²−y² far above the rest.
- Decide the spin state. Only octahedral d⁴, d⁵, d⁶ and d⁷ have a choice. Strong-field ligands (CN⁻, CO, NO₂⁻, en, bipy) and 3+ ions favour low spin. Weak-field ligands (I⁻, Br⁻, Cl⁻, F⁻, H₂O) favour high spin. Tetrahedral complexes are always high spin. 4d and 5d metals are low spin.
- Fill the orbitals. High spin: one electron in each of the five orbitals first. Low spin: fill the lower set completely first.
- Calculate. CFSE = (−0.4 n(t2g) + 0.6 n(eg)) Δo + mP, where m is the number of extra pairs compared with the free ion. Then count unpaired electrons and use μ = √(n(n + 2)).
Spectrochemical series (weak to strong): I⁻ < Br⁻ < SCN⁻ < Cl⁻ < F⁻ < OH⁻ < C₂O₄²⁻ < H₂O < NCS⁻ < py < NH₃ < en < bipy < phen < NO₂⁻ < PPh₃ < CN⁻ < CO.
How to use the crystal field splitting diagram tool
- Open the crystal field splitting diagram.
- Choose the Metal and the Oxidation state. The menu shows the dⁿ count next to each state.
- Choose the Geometry: octahedral, tetrahedral, square planar, trigonal bipyramidal or square pyramidal.
- Choose the Ligand. For a mixed complex, set a Second ligand (optional) and How many of it.
- Leave Spin state on Automatic (Δ vs pairing energy), or choose High spin or Low spin to compare.
- Or press an example such as [Fe(CN)₆]⁴⁻, [CoF₆]³⁻ or [Ni(CN)₄]²⁻.
- Read the configuration, CFSE, unpaired electrons, μ, Jahn–Teller note and colour note. Save with Download SVG or Download PNG.
Worked examples
1. [Fe(CN)₆]⁴⁻ versus [Fe(H₂O)₆]²⁺
Both contain Fe²⁺ (d⁶). With water, Δo ≈ 10 400 cm⁻¹ is below P ≈ 17 600 cm⁻¹, so the complex is high spin: t2g⁴ eg², CFSE = −0.4Δo, 4 unpaired electrons, μ = 4.90 BM. With cyanide, Δo ≈ 32 800 cm⁻¹ is above P, so it is low spin: t2g⁶, CFSE = −2.4Δo + 2P, 0 unpaired, diamagnetic.
2. [CoF₆]³⁻ versus [Co(NH₃)₆]³⁺
Co³⁺ is d⁶. Fluoride is weak-field: high spin, 4 unpaired electrons, μ = √24 = 4.90 BM. Ammonia gives Δo ≈ 22 900 cm⁻¹: low spin, 0 unpaired. Co³⁺ is low spin with almost every ligand except F⁻.
3. [Ni(CN)₄]²⁻ versus [NiCl₄]²⁻
Ni²⁺ is d⁸. Cyanide gives a square planar complex: the eight electrons fill dxz, dyz, dz² and dxy, and dx²−y² stays empty. It is diamagnetic. Chloride gives a tetrahedral complex: e⁴ t2⁴, 2 unpaired electrons, μ = 2.83 BM, CFSE = −0.8Δt.
4. [Cu(H₂O)₆]²⁺ and the Jahn–Teller effect
Cu²⁺ is d⁹: t2g⁶ eg³. The eg set is unevenly filled, so the octahedron stretches along one axis (Jahn–Teller elongation). CFSE = 6(−0.4) + 3(0.6) = −0.6Δo, 1 unpaired electron, μ = 1.73 BM.
5. Colour of [Ti(H₂O)₆]³⁺
Ti³⁺ is d¹ with one d–d band at Δo = 20 300 cm⁻¹. λ = 10⁷ / 20 300 ≈ 493 nm. Green-yellow light is absorbed, so the solution looks purple. CFSE = −0.4Δo ≈ −97 kJ/mol.
6. [MnO₄]⁻ has no d electrons
Mn is +7, so the ion is d⁰ and CFSE = 0. Its deep purple colour comes from ligand-to-metal charge transfer, not from d–d transitions.
Summary table: octahedral complexes
| dⁿ | High spin (unpaired, CFSE) | Low spin (unpaired, CFSE) | Example |
|---|---|---|---|
| d¹ | 1, −0.4Δo | same | [Ti(H₂O)₆]³⁺ |
| d³ | 3, −1.2Δo | same | [Cr(H₂O)₆]³⁺, 3.87 BM |
| d⁴ | 4, −0.6Δo | 2, −1.6Δo + P | [Cr(H₂O)₆]²⁺ HS, [Mn(CN)₆]³⁻ LS |
| d⁵ | 5, 0 | 1, −2.0Δo + 2P | [Fe(H₂O)₆]³⁺ HS, [Fe(CN)₆]³⁻ LS |
| d⁶ | 4, −0.4Δo | 0, −2.4Δo + 2P | [CoF₆]³⁻ HS, [Co(NH₃)₆]³⁺ LS |
| d⁷ | 3, −0.8Δo | 1, −1.8Δo + P | [Co(H₂O)₆]²⁺ HS |
| d⁸ | 2, −1.2Δo | same | [Ni(H₂O)₆]²⁺, 2.83 BM |
| d⁹ | 1, −0.6Δo | same | [Cu(H₂O)₆]²⁺, Jahn–Teller |
| d¹⁰ | 0, 0 | same | [Zn(H₂O)₆]²⁺, colourless |
Common mistakes
- Taking electrons from 3d before 4s. Transition metal ions lose 4s electrons first: Fe²⁺ is 3d⁶, not 3d⁴4s².
- Treating tetrahedral complexes as low spin.
- Leaving out the pairing energy term, or counting all pairs instead of the extra pairs.
- Writing tetrahedral CFSE in Δo without converting: −0.8Δt is about −0.36Δo.
- Using the octahedral labels for tetrahedral complexes. In Td the lower set is e and the upper set is t2 (no g, because there is no centre of inversion).
- Saying d⁰ and d¹⁰ ions are coloured because of d–d transitions.
Exam tips
- JEE and NEET: magnetic moment questions reduce to counting unpaired electrons. Memorise 1.73, 2.83, 3.87, 4.90 and 5.92 BM for n = 1 to 5.
- JEE and NEET: Ni(II) with CN⁻ is square planar and diamagnetic (dsp² in valence bond terms). Ni(II) with Cl⁻ is tetrahedral and paramagnetic (sp³).
- A-level: colour comes from d–d transitions. The colour you see is the complementary colour of the light absorbed. Changing the ligand changes Δ and so the colour.
- AP Chemistry: you may meet crystal field ideas in colour and magnetism questions; link a larger Δ to absorption at shorter wavelength.
- University: include the pairing energy in CFSE, explain the Jahn–Teller effect for d⁹ and high-spin d⁴, and use ligand field theory (π donors lower Δ, π acceptors raise it).
Test any complex with the crystal field splitting diagram and download the diagram for your notes.
Questions and answers
What is crystal field splitting?
Crystal field splitting is the energy gap between groups of d orbitals caused by the ligands around a metal ion. In an octahedral complex, t2g (dxy, dxz, dyz) drop by 0.4Δo and eg (dz², dx²−y²) rise by 0.6Δo.
How do I calculate CFSE?
For an octahedral complex, CFSE = (−0.4 × electrons in t2g + 0.6 × electrons in eg) Δo + mP, where m is the number of extra electron pairs compared with the free ion. Low-spin d⁶ gives −2.4Δo + 2P.
What decides high spin or low spin?
The size of Δo compared with the pairing energy P. If Δo < P the complex is high spin; if Δo > P it is low spin. Strong-field ligands such as CN⁻ and CO, 3+ charges and 4d or 5d metals favour low spin.
Why are tetrahedral complexes always high spin?
Δt is only about 4/9 of Δo for the same metal and ligand, because there are fewer ligands and none point straight at the d orbitals. Δt is almost always smaller than the pairing energy.
How many unpaired electrons does [CoF₆]³⁻ have?
Four. Co³⁺ is d⁶ and F⁻ is weak-field, so the configuration is t2g⁴ eg². The spin-only magnetic moment is √(4 × 6) = 4.90 BM.
Why does [Cu(H₂O)₆]²⁺ show a Jahn–Teller distortion?
Cu²⁺ is d⁹, t2g⁶ eg³. The three eg electrons cannot fill dz² and dx²−y² evenly, so the complex lowers its energy by stretching along one axis.
Why is [Ti(H₂O)₆]³⁺ purple?
It is d¹ with Δo = 20 300 cm⁻¹, so it absorbs light near 493 nm (green-yellow). The transmitted light looks purple.