Crystal Field Stabilization Energy Calculator

Calculate crystal field stabilization energy (CFSE) for transition metal complexes.

Geometry

Number of d electrons

Fe2+=6, Co3+=6, Ni2+=8, Cu2+=9

Ligand field

Delta_o (crystal field splitting, kJ/mol) (kJ/mol)

Typical: 150-250 kJ/mol for 3d metals

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Why d Orbitals Split

In a free transition metal ion the five d orbitals have identical energy. Surround it with ligands and that degeneracy breaks, because d orbitals pointing directly at incoming ligands are destabilised more than those pointing between them.

Geometry	Higher set	Lower set	Splitting
Octahedral	e_g (d_z², d_x²−y²) at +0.6Δ_o	t_2g (three) at −0.4Δ_o	Δ_o
Tetrahedral	t₂ (three) at +0.4Δ_t	e (two) at −0.6Δ_t	Δ_t ≈ 4/9 Δ_o

The barycentre is preserved: the weighted average energy stays unchanged, which is why the two e_g orbitals rise by 0.6Δ while the three t_2g fall by 0.4Δ. Multiplying out, 2(+0.6) + 3(−0.4) = 0.

Tetrahedral splitting is smaller for two reasons: only four ligands rather than six, and none points directly at a d orbital. The resulting Δ_t of roughly four ninths of Δ_o is almost never large enough to force electron pairing, which is why tetrahedral complexes are essentially always high spin.

High Spin or Low Spin

For d⁴ through d⁷ octahedral complexes, electrons face a choice: occupy a higher orbital, costing Δ_o, or pair in a lower one, costing the pairing energy P.

Condition	Configuration	Ligand type	Example
Δ_o > P	Low spin — pair up first	Strong field: CN⁻, CO, NH₃	[Fe(CN)₆]⁴⁻
Δ_o < P	High spin — occupy singly first	Weak field: I⁻, Br⁻, F⁻, H₂O	[Fe(H₂O)₆]²⁺

The spectrochemical series ranks ligands by field strength: I⁻ < Br⁻ < Cl⁻ < F⁻ < OH⁻ < H₂O < NH₃ < en < CN⁻ < CO. This ordering also explains complex colour: Δ_o corresponds to a visible-light photon energy, so stronger field ligands shift absorption to shorter wavelengths and change the observed colour.

Worked Examples

Example 1: Fe2+ (d6) in strong field: Co(NH3)6

CFSE=-0.4x6=-2.4 delta_o

Result: CFSE=-2.4x200=-480 kJ/mol

Low spin: t2g^6 eg^0 fully filled

Example 2: Mn2+ (d5) high spin: octahedral

CFSE=0 (half-filled d5 high spin)

Result: CFSE=0 kJ/mol - no stabilization

Half-filled shell has no CFSE

Example 3: High spin d⁶

[Fe(H₂O)₆]²⁺: t_2g⁴e_g²

Result: CFSE = 4(−0.4) + 2(+0.6) = −0.4Δ_o

Water is a weak field ligand, so Δ_o < P and electrons spread out. Compare the low spin case at −2.4Δ_o — six times more stabilisation.

Example 4: Zero CFSE cases

d⁵ high spin and d¹⁰

Result: CFSE = 0 in both

Half-filled and fully filled shells distribute electrons symmetrically across both sets, so the stabilisation cancels exactly.

Example 5: Tetrahedral d³

e²t₂¹: CFSE = 2(−0.6) + 1(+0.4)

Result: −0.8Δ_t ≈ −0.36Δ_o

Because Δ_t is only 4/9 of Δ_o, tetrahedral CFSE values are much smaller — part of why octahedral geometry dominates.

Common Mistakes

⚠️

Applying the low spin option to tetrahedral complexes

Δ_t is only about 4/9 of Δ_o and essentially never exceeds the pairing energy. Tetrahedral complexes are high spin in practice.

⚠️

Forgetting pairing energy in CFSE calculations

Low spin configurations pair electrons that would otherwise be unpaired. A complete CFSE comparison must add the pairing energy cost, not just count orbital occupancy.

⚠️

Expecting d⁵ high spin to have CFSE

With one electron in each of the five orbitals, 3(−0.4) + 2(+0.6) = 0. High spin d⁵ and d¹⁰ both have zero CFSE.

⚠️

Assuming colour comes directly from the metal

Colour arises from the d–d transition energy, which depends on the ligands. The same metal ion gives different colours with different ligands — [Cu(H₂O)₆]²⁺ is pale blue while [Cu(NH₃)₄]²⁺ is deep blue.

Frequently Asked Questions

Why do crystal field effects matter?

CFSE explains: why certain metal-ligand combinations are preferred, colors (d-d transitions), magnetic properties (spin state), and thermodynamic stability. d5 high-spin and d0 and d10 have zero CFSE.

Strong vs weak field ligands?

Spectrochemical series: I- < Br- < Cl- < F- < OH- < H2O < NH3 < CO < CN-. Strong field ligands force pairing (low spin), give large delta_o, cause unusual magnetic properties.

Why do d orbitals split in a ligand field?

Orbitals pointing directly at ligands experience greater electrostatic repulsion and rise in energy, while those pointing between ligands are stabilised.

When is a complex low spin?

When Δ_o exceeds the pairing energy, which happens with strong field ligands such as CN⁻, CO and NH₃. Only d⁴ to d⁷ octahedral complexes have the choice.

Why are tetrahedral complexes always high spin?

Δ_t is roughly four ninths of Δ_o because there are fewer ligands and none points directly at a d orbital. It is rarely enough to overcome pairing energy.

Why do d⁵ high spin complexes have zero CFSE?

One electron occupies each of the five orbitals, so 3(−0.4) + 2(+0.6) = 0. The stabilisation of the lower set exactly cancels the destabilisation of the upper.

How does crystal field splitting explain colour?

Δ_o corresponds to visible photon energies. The complex absorbs light matching that gap and appears the complementary colour, which is why ligand changes alter colour.

Formula Explorer connections

Interpretation: This relationship uses electronic structure, bonding or molecular geometry to predict a chemical property or structural descriptor. Assumption: The model may be an approximation; resonance, solvent, coordination environment, conformation and experimental conditions can affect real molecules.

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