Relation between ∆(tetrahedral) and ∆(octahedral), ∆t = 4/9 ∆o [duplicate]








**This question already has answers here**:
 
 




[Why do tetrahedral complexes have approximately 4/9 the field split of octahedral complexes?](/questions/60262/why-do-tetrahedral-complexes-have-approximately-4-9-the-field-split-of-octahedra)

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[Why do octahedral metal ligand complexes have greater splitting than tetrahedral complexes?](/questions/42257/why-do-octahedral-metal-ligand-complexes-have-greater-splitting-than-tetrahedral)

 (2 answers)
 

Closed 6 years ago.



I remember my teacher saying that ∆(tetrahedral) being 4/9 times the ∆(octahedral) for the same ligands and the same central metal ion due to some analogy by number of ligands, but I do not seem to remember it.


