Why complex [Cu(NH3)4] 2+ is square planar and not tetrahedral?? need the answer very urgentlyy!!
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The shapes of compounds with metals containing d electrons is not explained by VSEPR theory and lone pair repulsions. They are better explained by Crystal field theory, Ligand field theory and Jahn-Teller effect.
Cu^2+ has : 3p6 3d9 4s0 4p0
The hybridization that happens is dsp2.
NH3 is a strong ligand (compared to Cl- for example). So when they come near the d orbitals with electrons, due to repulsion the d orbitals split into higher and lower energy levels than before. Their shapes remain as before.
In Cu with d9 config, there is degeneration of eg set orbitals due to lone electron. So eg set : dx2-y2 and dz2 orbitals point towards the ligands. So distortion occurs in order that lower energy situation results. That results in more stability. The t2g orbitals do not point towards the ligands and they do not affect the shape.
The shape is not tetrahedral as there is very little influence of the dxy,dyz,dzx orbitals. The dx2-y2 is at very high energy level (more than dxy) and dz2 is at a high level (more than dx,dyz).
Cu^2+ has : 3p6 3d9 4s0 4p0
The hybridization that happens is dsp2.
NH3 is a strong ligand (compared to Cl- for example). So when they come near the d orbitals with electrons, due to repulsion the d orbitals split into higher and lower energy levels than before. Their shapes remain as before.
In Cu with d9 config, there is degeneration of eg set orbitals due to lone electron. So eg set : dx2-y2 and dz2 orbitals point towards the ligands. So distortion occurs in order that lower energy situation results. That results in more stability. The t2g orbitals do not point towards the ligands and they do not affect the shape.
The shape is not tetrahedral as there is very little influence of the dxy,dyz,dzx orbitals. The dx2-y2 is at very high energy level (more than dxy) and dz2 is at a high level (more than dx,dyz).
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