Why is Δo the double of Δt ?


Why is Δo the double of Δt

The crystal field splitting in octahedral complexes (Δo) is double than that in tetrahedral complexes (Δt) (i.e., Δo is double of Δt) because:

(i) The number of ligands is 2/3 of the octahedral complex, therefore, the splitting of the d-orbitals is also two-thirds.

(ii) None of the ligands point directly towards any of the five d-orbitals. Therefore, the splitting is reduced by roughly two-third. In tetrahedral and octahedral complexes, the dxy, dyz and dzx orbitals are at 45o, i.e., at equal distance from the ligands. Therefore, in both the complexes, each ligand repels these orbitals in equal amounts. In octahedral complexes, the ligand approach directly to dx2 - y2 and dz2 orbitals and repel them strongly. But in tetrahedral complexes, these orbitals are at (√2/2) l = 0.71 l (it approximately equal to the 2/3) as compared to that of octahedral complexes. Therefore, in tetrahedral complexes, the ligands repel these orbitals 2/3 times than that of octahedral complexes. Thus, due to this fact, splitting decreases by 2/3rd in tetrahedral complexes as compared to octahedral complexes.

Therefore, Δt = 2/3 x 2/3 x Δo or Δt = 4/9 Δo

Thus, Δo is double of Δt or we can say that the crystal field splitting in tetrahedral complexes is smaller than that in octahedral complexes.

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