Q. Crystal Field Splitting in Tetrahedral Field?

A tetrahedron can be viewed as a part of cube where the metal ion occupies the centre of the cube and the four ligands are placed at four alternate comers of the cube. 

It is obvious from the figure that none of the orbitals d_xy d_yz:, d_x or d_x2  y2 and d,_z2 point  directly at the ligands but the first set of three orbitals are nearer to the ligands as compared  to the last two orbitals. Hence, in this case the d orbital splitting will be as shown in Fig.

Note that the lower set of orbitals is denoted by a symbol 'e' while the upper is designated as 't' which are different from a those used in case of octahedral complexes. This is so because a tetrahedral molecule does not possess a centre of symmetry. The difference in energy between the two sets of orbitals is denoted by' At. It is found that if the metal ion and the ligands remain same and the distances between metal and ligands are also same in octahedral and tetrahedral compounds then.

Δ_t= 4/9 Δ₀

Following the arguments parallel to those used in case of octahedral complexes we find the  each orbital of e set is lowered by -3/5Δ_t and each orbital belonging to the 't' set is raised in  energy by +2/54Δ_t.