I am not entirely sure if this question has been answered before, but I was wondering why when copper forms a complex ion with chlorine there are only 4 chlorine ligands. I asked my teacher and he said that it was because chlorine is a weak field ligand. I am not particularly satisfied with this answer as tetrahedral, octahedral and other geometries can be formed by weak field ligands.

So my question is essentialy why is it $\ce{[CuCl4]^{2-}}$ instead of something else like $\ce{[CuCl6]^{4-}}$

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    $\begingroup$ 4- is a lot of charge to pack into a small volume $\endgroup$ – Andrew Jun 28 '19 at 15:40
  • $\begingroup$ Is it possible to explain it using LFT or CFT at all? I would like a constructive way to determine whether a certain metal-ligand complex is feasible or not. $\endgroup$ – sab hoque Jun 29 '19 at 8:18
  • $\begingroup$ In general, CFT and LFT lead to "Structure preference energy" (SPE) estimations, which predict the likely preferred coordination number and structure based just on the number of d electrons and the strength of the ligand (ie whether complex is high or low spin). However, in the case of $d^9$ (like the Cu(II) you have here), the difference in energy between octahedral and square planar by SPE is 0, so it doesn't tell us which is more likely. SPE calculations also do not account for sterics or crystal packing considerations. Instead, we have to rely on observed trends or more complex math. $\endgroup$ – Andrew Jun 29 '19 at 11:18
  • $\begingroup$ @Andrew correct me if I'm mistaken but $\ce{[CuCl4]^{2-}}$ is tetrahedral not square planar? Also could you point me in the direction of some websites or books I could read in order to learn more about calculating SPE estimations? $\endgroup$ – sab hoque Jul 1 '19 at 14:17
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    $\begingroup$ Yes and no. In solution it's a slightly distorted tetrahedron, but in crystal structures it is more flattened and sometimes completely square planar depending on the counterion. $\endgroup$ – Andrew Jul 1 '19 at 17:28

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