Crystal Field Stabilization Energy Calculator – d-Orbital Splitting, Spin States and CFSE
Crystal field theory treats the ligands around a transition-metal ion as point charges. In a free ion the five d orbitals have the same energy. Once ligands surround it, the orbitals that point at the ligands are pushed up and the ones that point between them drop. This crystal field stabilization energy calculator works out the d-electron count, draws the d-orbital splitting diagram for the geometry, places the electrons high spin or low spin, and reports the CFSE in terms of Δ and the pairing energy P. When Δ is known it also gives cm⁻¹ and kJ/mol.
The barycentre rule and the octahedral splitting
In an octahedron the ligands sit on the x, y and z axes. The dz² and dx²−y² orbitals (eg) point straight at them, and dxy, dxz and dyz (t₂g) point between them. The total energy of the five orbitals stays at the spherical-field average, the barycentre. So the three t₂g orbitals drop by 0.4 Δo and the two eg orbitals rise by 0.6 Δo, since 3 × (−0.4) + 2 × 0.6 = 0. The CFSE is simply electrons times orbital energy:
CFSE = (−0.4 × n(t2g) + 0.6 × n(eg)) Δo + (pairing term) P
d electrons = group number − oxidation stateFor Fe(II) in group 8, 8 − 2 = d⁶. In low-spin [Fe(CN)₆]⁴⁻ all six sit in t₂g, so the Δ term is 6 × (−0.4 Δo) = −2.4 Δo. Note that complexes use this ionic count, not the free-atom configuration: Cr(0) in Cr(CO)₆ is d⁶, even though the chromium atom is 3d⁵ 4s¹.
High spin, low spin and pairing energy
From d⁴ to d⁷ an octahedral ion has a choice. Putting the next electron in eg costs Δo, while pairing it in t₂g costs the pairing energy P. When Δ < P the complex is high spin; when Δ > P it is low spin. For octahedral and tetrahedral complexes the crossover is exactly at Δ = P, and the "Which spin state wins?" chart shows it as two lines crossing.
The spectrochemical series: what makes Δ large
Ligands rank by field strength: I⁻ < Br⁻ < Cl⁻ < F⁻ < H₂O < NH₃ < en < bipy < NO₂⁻ < CN⁻ < CO. Δ also grows with the metal's oxidation state and down a group (3d < 4d < 5d), which is why nearly every 4d and 5d complex is low spin. When no measured value is available, the calculator estimates Δo ≈ f × g using Jørgensen's ligand and metal factors. It can also take Δ from an absorption maximum, using Δ (cm⁻¹) = 10⁷ / λ (nm). [Ti(H₂O)₆]³⁺ absorbing at 493 nm gives Δo ≈ 20,300 cm⁻¹. The Photon Energy Calculator handles other wavelength-to-energy conversions.
Tetrahedral and square-planar complexes
A tetrahedron inverts the pattern: e lies at −0.6 Δt and t₂ at +0.4 Δt. With only four ligands, none pointing at an orbital, Δt ≈ 4/9 Δo, so tetrahedral complexes are almost always high spin. Square-planar fields push dx²−y² very high. A strong-field d⁸ ion such as Ni(II) with cyanide, or Pd(II) and Pt(II) with almost anything, gains about −2.457 Δo by leaving that orbital empty. That is why [Ni(CN)₄]²⁻ and [PtCl₄]²⁻ are square planar and diamagnetic. The calculator also covers square-pyramidal, trigonal-bipyramidal, linear, trigonal-planar and pentagonal-bipyramidal geometries, using point-charge model energies.
Jahn–Teller distortion
An unevenly filled eg level (d⁹, high-spin d⁴, low-spin d⁷) makes a regular octahedron unstable. The complex distorts, usually by stretching along z, which splits eg and t₂g further. [Cu(H₂O)₆]²⁺ is the classic tetragonally elongated example. Uneven t₂g filling gives only a weak effect.
Where CFSE shows up
High-spin octahedral CFSE traces a double hump across d⁰–d¹⁰, with peaks at d³ and d⁸. The hydration enthalpies of the M²⁺ ions from Ca²⁺ to Zn²⁺ follow the same pattern once the smooth trend is removed. The octahedral site preference energy, CFSE(oct) − CFSE(tet), also predicts whether a spinel AB₂O₄ is normal or inverse. Ni²⁺ (d⁸) strongly prefers octahedral holes, so NiFe₂O₄ is inverse. Cr³⁺ (d³) holds its octahedral sites, so FeCr₂O₄ is normal. Ion size and lattice energy decide the cases where CFSE barely differs.