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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 dxy dyz:, dx or dx2 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.
Δt= 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 -3/5Δt and each orbital belonging to the 't' set is raised in energy by +2/54Δt.
Glucose and fructose are epimers so they are differ in their second carbon position and remaining is same so when they react with phenylhydrazeen they give same osazone
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