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Crystal Field Splitting Diagram (CFSE and Spin State)
In short: Crystal field splitting diagram tool for octahedral, tetrahedral, square planar, trigonal bipyramidal and square pyramidal complexes. It finds the d-electron count, fills t2g and eg (or e and t2) for high or low spin, and gives CFSE with pairing energy, unpaired electrons, spin-only magnetic moment, a Jahn–Teller hint and a colour note. It is free, works in your web browser and needs no download.
Choose the metal ion, geometry and ligands. The diagram shows the free-ion d level, the split d orbitals with their labels and the electrons as arrows. The result explains the spin state with the Δ versus pairing energy rule.
Examples
Lower levels are teal and higher levels red. Energies are drawn roughly to scale around the barycentre (the average energy of the five d orbitals); very close levels are spaced apart so they stay readable. Δ values are measured where known, otherwise rough estimates.
How to use the crystal field diagram
- Choose the Metal and its Oxidation state. The tool works out the d-electron count: dⁿ = group number − oxidation state (Fe²⁺: 8 − 2 = d⁶).
- Choose the Geometry and the Ligand. For mixed complexes such as [Co(NH₃)₅Cl]²⁺, add a Second ligand and set How many of it.
- Leave Spin state on Automatic to apply the Δ versus pairing energy rule, or set high or low spin yourself.
- Read the configuration, CFSE, unpaired electrons, magnetic moment, Jahn–Teller hint and colour note. Save the diagram with Download SVG or Download PNG.
Crystal field theory in brief
In a free metal ion the five d orbitals have the same energy. Ligands approaching the metal repel the d electrons. Orbitals that point at the ligands rise in energy and the others fall, while the average energy (the barycentre) stays the same. The energy gap is the crystal field splitting, Δ.
Splitting patterns
| Geometry | Lower set | Upper set | Energies |
|---|---|---|---|
| Octahedral | t2g (dxy, dxz, dyz) | eg (dz², dx²−y²) | t2g −0.4Δo, eg +0.6Δo (Δo = 10 Dq) |
| Tetrahedral | e (dz², dx²−y²) | t2 (dxy, dxz, dyz) | e −0.6Δt, t2 +0.4Δt, Δt ≈ 4/9 Δo |
| Square planar | dxz, dyz < dz² < dxy | dx²−y² (highest) | −5.14, −4.28, +2.28, +12.28 Dq |
| Trigonal bipyramidal | e″ (dxz, dyz) < e′ (dxy, dx²−y²) | a1′ (dz²) | −2.72, −0.82, +7.07 Dq |
| Square pyramidal | e (dxz, dyz) < b2 (dxy) < a1 (dz²) | b1 (dx²−y²) | −4.57, −0.86, +0.86, +9.14 Dq |
Spectrochemical series
I⁻ < Br⁻ < SCN⁻ < Cl⁻ < F⁻ < OH⁻ < C₂O₄²⁻ < H₂O < NCS⁻ < py < NH₃ < en < bipy < phen < NO₂⁻ < PPh₃ < CN⁻ < CO. Ligands on the left are weak-field (small Δ); ligands on the right are strong-field (large Δ). Δ also grows with metal charge (M³⁺ > M²⁺) and down a group (5d > 4d > 3d).
High spin or low spin
For octahedral d⁴ to d⁷, compare Δo with the pairing energy P. If Δo < P, electrons spread out first: high spin. If Δo > P, they pair in t2g: low spin. Tetrahedral complexes are always high spin because Δt is small. 4d and 5d metals and square planar d⁸ complexes are low spin.
CFSE (octahedral) = (−0.4 × n(t2g) + 0.6 × n(eg)) Δo + m P μ(spin-only) = √(n(n + 2)) BM
Here m is the number of electron pairs formed in addition to those already present in the free ion, and n is the number of unpaired electrons. For tetrahedral complexes, CFSE = (−0.6 × n(e) + 0.4 × n(t2)) Δt.
Octahedral CFSE for d¹ to d¹⁰
| dⁿ | High spin | CFSE (high spin) | Unpaired (HS) | Low spin and CFSE | Unpaired (LS) |
|---|---|---|---|---|---|
| d¹ | t2g¹ | −0.4Δo | 1 | same as high spin | 1 |
| d² | t2g² | −0.8Δo | 2 | same as high spin | 2 |
| d³ | t2g³ | −1.2Δo | 3 | same as high spin | 3 |
| d⁴ | t2g³ eg¹ | −0.6Δo | 4 | t2g⁴ eg⁰: −1.6Δo + P | 2 |
| d⁵ | t2g³ eg² | 0 | 5 | t2g⁵ eg⁰: −2.0Δo + 2P | 1 |
| d⁶ | t2g⁴ eg² | −0.4Δo | 4 | t2g⁶ eg⁰: −2.4Δo + 2P | 0 |
| d⁷ | t2g⁵ eg² | −0.8Δo | 3 | t2g⁶ eg¹: −1.8Δo + P | 1 |
| d⁸ | t2g⁶ eg² | −1.2Δo | 2 | same as high spin | 2 |
| d⁹ | t2g⁶ eg³ | −0.6Δo | 1 | same as high spin | 1 |
| d¹⁰ | t2g⁶ eg⁴ | 0 | 0 | same as high spin | 0 |
Jahn–Teller distortion
A non-linear complex with an unevenly filled degenerate set distorts to lower its energy. The effect is strong when eg is unevenly filled: d⁹ (Cu²⁺), high-spin d⁴ (Cr²⁺, Mn³⁺) and low-spin d⁷. Six-coordinate Cu²⁺ complexes are usually elongated along one axis.
Worked examples
[Fe(H₂O)₆]²⁺
Fe²⁺ is d⁶. Water is weak-field: Δo ≈ 10 400 cm⁻¹ < P ≈ 17 600 cm⁻¹, so high spin t2g⁴ eg².
CFSE = 4(−0.4) + 2(0.6) = −0.4Δo.
4 unpaired, μ = √24 = 4.90 BM
[Fe(CN)₆]⁴⁻
d⁶ with strong-field CN⁻: low spin t2g⁶ eg⁰. Three pairs versus one in the free ion, so 2 extra pairs.
CFSE = 6(−0.4)Δo + 2P.
−2.4Δo + 2P, 0 unpaired, diamagnetic
[CoF₆]³⁻ versus [Co(NH₃)₆]³⁺
Both are Co³⁺ d⁶. F⁻ gives Δo ≈ 13 000 cm⁻¹ (below P): high spin. NH₃ gives Δo ≈ 22 900 cm⁻¹: low spin.
[CoF₆]³⁻: 4 unpaired, 4.90 BM. [Co(NH₃)₆]³⁺: 0 unpaired
[Ni(CN)₄]²⁻ versus [NiCl₄]²⁻
Both are Ni²⁺ d⁸. CN⁻ gives square planar, dx²−y² empty. Cl⁻ gives tetrahedral e⁴ t2⁴.
[Ni(CN)₄]²⁻ diamagnetic; [NiCl₄]²⁻ 2 unpaired, 2.83 BM, CFSE −0.8Δt
[Cu(H₂O)₆]²⁺
d⁹: t2g⁶ eg³. The eg set is unevenly filled.
CFSE = 6(−0.4) + 3(0.6) = −0.6Δo.
1 unpaired, 1.73 BM, strong Jahn–Teller elongation
[Ti(H₂O)₆]³⁺ colour
d¹ with Δo = 20 300 cm⁻¹. λ = 10⁷ / 20 300 = 493 nm, so green-yellow light is absorbed.
CFSE = −0.4Δo = −0.4 × 243 kJ/mol = −97 kJ/mol.
The solution looks purple
Common mistakes
- Wrong d count. Remove the 4s electrons first: Fe is [Ar]3d⁶4s², Fe²⁺ is 3d⁶ and Fe³⁺ is 3d⁵.
- Making tetrahedral complexes low spin. Δt ≈ 4/9 Δo is almost always below the pairing energy, so tetrahedral complexes are high spin.
- Forgetting the pairing energy term. Low-spin d⁶ CFSE is −2.4Δo + 2P, not just −2.4Δo. Count only pairs that are extra compared with the free ion.
- Mixing Δt and Δo. −0.8Δt for [NiCl₄]²⁻ equals about −0.36Δo, not −0.8Δo.
- Swapping the tetrahedral labels. In a tetrahedral field the e set (dz², dx²−y²) is lower and t2 is higher, the reverse of octahedral.
- Expecting colour from d⁰ or d¹⁰. Zn²⁺ is colourless. The purple of MnO₄⁻ (d⁰) comes from charge transfer, not d–d transitions.
Drawing coordination complexes? Draw ligands, metal complexes and reaction schemes in MolDraw, a free online chemical structure editor.
Open MolDraw editorFrequently asked questions
What is the CFSE of [Fe(CN)₆]⁴⁻?
Fe²⁺ is d⁶ and CN⁻ is strong-field, so the complex is low spin t2g⁶ eg⁰. CFSE = 6 × (−0.4Δo) + 2P = −2.4Δo + 2P, with 0 unpaired electrons.
How do I know if a complex is high spin or low spin?
Compare Δo with the pairing energy P. Weak-field ligands such as H₂O, F⁻ and Cl⁻ usually give high spin; strong-field ligands such as CN⁻ and CO give low spin. Tetrahedral complexes are always high spin, and 4d and 5d metals are low spin.
Why is [CoF₆]³⁻ high spin but [Co(NH₃)₆]³⁺ low spin?
Both contain Co³⁺ (d⁶). F⁻ is a weak-field ligand, so Δo (about 13 000 cm⁻¹) is below the pairing energy: 4 unpaired electrons and μ = 4.90 BM. NH₃ gives Δo about 22 900 cm⁻¹, so all six electrons pair in t2g.
How do I calculate the spin-only magnetic moment?
Use μ = √(n(n + 2)) BM, where n is the number of unpaired electrons. n = 1 gives 1.73 BM, 2 gives 2.83, 3 gives 3.87, 4 gives 4.90 and 5 gives 5.92 BM.
Why is [Ni(CN)₄]²⁻ square planar and diamagnetic?
Ni²⁺ is d⁸. With strong-field CN⁻ the square planar arrangement leaves the very high dx²−y² orbital empty, so all eight electrons pair in the lower four orbitals. [NiCl₄]²⁻ is tetrahedral with 2 unpaired electrons.
Which complexes show a Jahn–Teller distortion?
A strong Jahn–Teller distortion occurs when the eg set is unevenly filled in an octahedral complex: d⁹ (Cu²⁺), high-spin d⁴ (Cr²⁺, Mn³⁺) and low-spin d⁷ (Co²⁺ with strong ligands).
How is the colour of a complex related to Δo?
The complex absorbs light of energy Δo, so λ (nm) ≈ 10⁷ / Δo (cm⁻¹). [Ti(H₂O)₆]³⁺ has Δo = 20 300 cm⁻¹, absorbs near 493 nm and looks purple, the complementary colour.
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