# Why is Δo the double of Δt ?

The crystal field splitting in octahedral complexes (Δ_{o}) is double than that in tetrahedral complexes (Δ_{t}) (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 d_{xy}, d_{yz} and d_{zx} orbitals are at 45^{o}, 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 d_{x2 - y2} and d_{z2} 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, Δ_{t} = 2/3 x 2/3 x Δ_{o} or Δ_{t} = 4/9 Δ_{o}

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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