Question

Why d orbital splitting is inverted in tetrahedral field splitting as in octahedral one?

Answer 1

rbitals are nearer to the direction of approach of the ligands than the e orbitals. There are only 4 ligands in the tetrahedral complex, and hence the ligand field is roughly 2/3 of the octahedral field. As a result, all tetrahedral complexes are high-spin since the CFSE is normally smaller than the paring energy.

Answer 2

Splitting of d-orbitals octahedral complexes

In octahedral complexes the metal ion is at the centre of the octahedron and the six ligands represented by point charges lie at the six corners along the three axis x, y and z. Let suppose that the metal ion, $M^{n+}$ has a single d-electron. In the free ion when there are no ligands, the electron can occupy any of the d-orbitals because all are of same energy and are called degenerate orbitals. However, in the octahedral complex, $MX_6$, all the five d-orbitals will not remain of same energy. This can be easily understood by considering the shape of these orbitals. The shapes of these five d-orbitals in octahedral field of ligands is shown in the attachment 1. From the attachment 1, it is clear that the lobes of ${d_x}^2$ and ${d_x}^2{_{-y}}^2$ orbitals point directly towards the corners of the octahedron where the negative charges of the ligands are present. The remaining three orbitals ($d_{xy}, d_{yz}$ and $d_{xz}$) have their lobes not towards ligands but in between the point charges. This means that the electrons in $d_{xy}, d_{yz}$ and $d_{xz}$ orbitals will be repelled less by the point charges of the ligands than those in d-orbitals (${d_x}^2$ and ${d_x}^2{_{-y}}^2$). As a result the three d-orbitals ($d_{xy}, d_{yz}$ and $d_{xz}$) experience less repulsions from the ligands and are stable and of lower energy then the two d-orbitals (${d_x}^2$ and ${d_x}^2{_{-y}}^2$). Thus, the five d-orbitals split up into two sets : ($d_{xy}, d_{yz}$ and $d_{xz}$) and (${d_x}^2$ and ${d_x}^2{_{-y}}^2$) in octahedral field.

Extra

Similarly, it can be shown from the geometry, that in tetrahedral complexes none of the d-orbitals point exactly towards the ligands and therefore, the splitting of energy will be less than that in octahedral field. The three d-orbitals ($d_{xy}, d_{yz}$ and $d_{xz}$) pointing close to the direction in which the ligands are approaching while the two d-orbitals (${d_x}^2$ and ${d_x}^2{_{-y}}^2$) are lying in between ligands. Therefore, the energies of the three orbitals will be raised by the energy of two orbitals will be lowered. Thus, in the presence of tetrahedral field the degeneracy of d-orbitals split up as :

a. The two d-orbitals (${d_x}^2$ and ${d_x}^2{_{-y}}^2$) become more stable and their energies are lowered. These are designated as ‘e’ orbitals.

b. The three orbitals ($d_{xy}, d_{yz}$ and $d_{xz}$) become less stable and their energies are raised. These are designated as ‘$t_2$’ orbitals. This splitting of d-orbital is shown in the attachment 2.

The energy difference between these two sets is called crystal field splitting in tetrahedral field and is abbreviated as $∆\sf _t$. It has been observed that the splitting of tetrahedral complexes is considerably in less than in octahedral complexes. It has been found that : $∆\sf _t ≈ \dfrac{4}{9} ∆\sf _o$

It may be noted that the splitting of d-orbitals in square planar complexes is complicated.