Cu1 is tetrahedral while cu2 is in square planar why
Answers
It is very convenient to use crystal field theory to discuss this.
It is usually assumed that in octahedral coordination the energy levels of the five d-orbitals are split, with two orbitals (dz2 and dx2−y2) well above the other three. The splitting is assumed to be large enough to overcome electron pairing energy.
The first six electrons will populate three lower orbitals (dxy, dxz, dyz). The next two electrons should occupy dz2 and dx2−y2. However, if these two electrons occupy dz2, two corresponding opposite ligands become well shielded and leave. This is the common case of d8 complexes, such as Ni+2, Pd+2, Pt+2, Rh+1, Cu+3 etc
Adding one more electron makes the remaining ligand's bonding weaker. The charge of bivalent cation Cu2+ is large enough to bond four charged ligands if available, such as in [CuCl4]2−, but the ligands are bound weakly and easily dissociate.
However, Cu+2 ions usually adopt a distorted octahedral geometry, with two ligands having a longer bond length than the four others. These two are very weakly bound and exchange quickly. For this reason the NH3 complex is written with only four molecules; the two other are so weakly bound.
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