Crystal Field Splitting in Tetrahedral and Square Planar Complexes
The octahedral case is standard. The other two geometries are where questions test whether the reasoning is understood or the diagram memorised.
The tetrahedral field
Four ligands approach along alternate corners of a cube, so none points directly at any d orbital. The d orbitals that point toward the cube edges — the t2 set — come closer to the ligands than those pointing at the cube faces, the e set.
Why tetrahedral complexes are always high spin
Two factors reduce the splitting. There are only four ligands instead of six, and none points directly at a d orbital, so the interaction is weaker in both respects.
The resulting splitting is almost never large enough to exceed the pairing energy, so electrons occupy the upper set rather than pairing. That is why tetrahedral complexes are essentially always high spin, and it is one of the most reliable generalisations in coordination chemistry.
The square planar case
Start from an octahedron and remove the two ligands along the z axis. The orbitals with a z component are no longer repelled by those ligands and fall in energy; the orbital in the xy plane pointing directly at the remaining ligands rises sharply.
The result is a large gap between the highest orbital and the rest. Deriving square planar splitting as an extreme axial elongation of the octahedron is both easier and more instructive than memorising the diagram.
Why d8 is the characteristic configuration
With eight d electrons, the four lower orbitals are filled and the highest one is empty. All electrons are paired, so the complex is diamagnetic, and it gains substantial stabilisation because the destabilised orbital is unoccupied.
That is why square planar geometry is so strongly associated with d8 metals, and why such complexes are diamagnetic while their tetrahedral d8 counterparts are paramagnetic. Distinguishing the two geometries by a magnetic measurement is a standard question.
Which geometry forms
| Favours tetrahedral | Favours square planar |
|---|---|
| Weak field ligands | Strong field ligands |
| Bulky ligands | Small ligands |
| Metals with little CFSE preference | d8 metals, especially heavier ones |
| Higher coordination entropy | Large crystal field stabilisation |
Heavier d8 metals show a stronger square planar preference than lighter ones, because their larger d orbitals interact more strongly with ligands and give a bigger splitting. A first-row d8 metal may form either geometry depending on the ligand, which is why the same metal can give a paramagnetic tetrahedral complex with one ligand and a diamagnetic square planar one with another.
Consequences for colour
Tetrahedral complexes are usually more intensely coloured than octahedral ones, because the absence of a centre of symmetry removes the Laporte restriction. Their smaller splitting also shifts absorption to longer wavelength, so the colours differ noticeably from octahedral complexes of the same metal.
Frequently asked questions
Why is tetrahedral splitting smaller?
Because there are fewer ligands and none points directly at a d orbital, so the electrostatic interaction is weaker on both counts.
Why are tetrahedral complexes always high spin?
Because the splitting almost never exceeds the pairing energy, so occupying the upper set costs less than pairing in the lower one.
How can square planar and tetrahedral d8 be distinguished?
By magnetism. Square planar is diamagnetic with all electrons paired; tetrahedral is paramagnetic with two unpaired electrons.
Why are tetrahedral complexes more intensely coloured?
Because they lack a centre of symmetry, so the Laporte selection rule does not forbid d–d transitions and they gain much greater intensity.
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