Question

Why complex Cu(NH3)4]2+ is square planar and not tetrahedral
need the answer veryy urgentlyy!!

Answer 1

It is very convinient to use crystall field theory to discuss that matter.It is usually assumed, that in octahedral coordination energy level of d-orbitals is split, with two orbitals (dz2 and dx2−y2) well above three others. Let's assume we have the split large enough to overcome electron pairing energy.Then, first six electrons will populate three lower orbitals (dxy, dxz, dyz) The next two electrons should occupy dz2 and dx2−y2. However, if this two electrons occupy dz2, two corresponding opposite ligands become well shielded and leave. This is the common case of d8 complexes, such asNi+2, Pd+2, Pt+2, Rh+1, Cu+3 and some more.When we add one more electron, it merely makes remaining ligands to bound 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 ad easily dissociate.However, Cu+2 usually adopt distorted octahedral neighborhood, with two ligands farther than four other. These two are very weakly bound and exchange quickly. For this reason NH3 complex is written with only four molecules, the two other are so weakly bound it isn't even funny.This type of distortion is a case of Jahn–Teller distortion, very common in chemistry.