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Please tell me where I am wrong. The complex will have $\ce{Pd^{2+}}$ ion, which has a $\mathrm{d^8}$ configuration. So, it will have $2$ unpaired electrons. Unpaired electrons will mean that it is paramagnetic.

My reference book has this line:

$\ce{[PdCl2(PMe3)2]}$ is a diamagnetic complex of Pd(II)

Shouldn't it say "paramagnetic"?

andselisk
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ADP
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2 Answers2

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You have not considered the d orbital splitting which has occurred due to the presence of the bonded ligands. Yes, the $\ce{Pd^2+}$ ion adopts the $\mathrm{4d^8}$ electronic configuration with two unpaired electrons, as shown below. However, when it forms the square planar complex, the d orbitals split in energy levels and the electrons now occupy the new energy levels differently, still abiding by Hund's rule and the Aufbau principle. These d orbitals no longer possess any unpaired electrons and thus, the complex is not paramagnetic, but diamagnetic.

enter image description here

Image source: Wikimedia Commons

enter image description here

Image source:Wikimedia Commons

Tan Yong Boon
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  • Would there be any coordination complex of Pd that is tetrahedral for CN=4? – ADP Jan 21 '18 at 13:48
  • Tetrahedral complexes for palladium are rather rare (As for the explanation as to why they are less preferred for Pd, the post Orthocresol linked above more offer some insight). Upon searching, I did manage to find one: http://pubs.rsc.org/en/content/articlelanding/1999/cc/a904016a#!divAbstract – Tan Yong Boon Jan 21 '18 at 15:24
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this is a special case of jahn teller distortion....in pd(II), weak ligands can remove the degeneracy of eg orbitals & as the energy difference between two eg orbitals are high, the ekectrons pair up in the dz2 orbitals. Thus form a diamagnetic complex JT distortion in d8 configuration

Subir Podder
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  • Okay.Is this correct:- So this complex is dsp2, hence square planar . For square planar there's that weird splitting. So configuration would be dzx(2) dyz(2) , dz2(2) , dxy(2), dx2-y2 (0) . No unpaired electrons! And dx2-y2 is the d in dsp2. Am i correct? – ADP Jan 21 '18 at 10:39
  • I think it will be dz2. See the pic I post above or the diagram for square planar complex. dz2 has low energy than dx2-y2 – Subir Podder Jan 21 '18 at 10:42
  • Would [NiCl2(PMe3)2] be paramagnetic? With two unpaired electrons (as shown in the figure on left side in the image you sent as answer)? – ADP Jan 21 '18 at 10:45
  • For nickel(II), JT distortion is only obtained by using strong field ligand. so the corresponding Ni complex will be paramagnetic(left diagram). But if any stong field ligand is present in complex, then it will be diamagnetic(right diagram) – Subir Podder Jan 21 '18 at 10:48
  • But dz2 would be filled as it has lower energy. So dx2-y2 would be used for coordinate bonding. – ADP Jan 21 '18 at 10:48
  • Also there would not be any Pd coordinate complex with tetrahedral structure for CN=4 , would there be? – ADP Jan 21 '18 at 11:35
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    How is this a Jahn–Teller distortion from octahedral geometry when there are only four ligands to begin with? – orthocresol Jan 21 '18 at 11:47
  • In the pic that I post, see how a octaderal complex has been transformed into square plannar in the Ni(CN)6 case.....An square plannar complex is formed from octahedral complex due to extreme jt distortion – Subir Podder Jan 21 '18 at 11:50
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    No this is not correct use of the term. A JT distortion leads to a change in molecular geometry, it does not lead to a complete loss of two ligands. – orthocresol Jan 21 '18 at 11:52
  • I agree that the orbital diagram is an extreme limiting situation of the JT distortion observed in d9 complexes, but using the term to refer to d8 square planar complexes is not something I’ve ever seen before and I don’t think it’s applicable. – orthocresol Jan 21 '18 at 11:54
  • It can leads to loss of one or two ligands. If it is not so, then how will you explain that [Ni(CN)4]2- is square planar not tetrahedral on the basis of CFT? – Subir Podder Jan 21 '18 at 11:56
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    No, the Jahn-Teller theorem does not cover changes in molecular composition, it only covers changes in the arrangement of the atoms. – orthocresol Jan 21 '18 at 11:57
  • CFT is a theory with limitations. You have to argue that in CFT the lower four orbitals are sufficiently stabilised (in turn due to the strong field ligand - note that CFT cannot explain why cyanide behaves strong field) such that populating then is more favourable than the tetrahedral case. See eg https://chemistry.stackexchange.com/a/40881/16683 – orthocresol Jan 21 '18 at 12:00
  • Now ans this, how to draw the splitting pattern of d-orbitals in square planar environment – Subir Podder Jan 21 '18 at 12:03
  • splitting pattern of square planar environment is drawn by considering extreme jt distortion & removing two axial ligands from oct. complex. This is similar to that. – Subir Podder Jan 21 '18 at 12:09
  • If you’re talking about CFT, you derive it based on qualitative electron repulsion arguments. And your picture is perfectly alright. There is some debate about the exact energy ordering of the lowest three orbitals, but it doesn’t affect the conclusion. Yes it is similar to the JT distortion, and as we agreed the orbital diagram can be derived logically from the JT distortion, but the state of only having four ligands is not a JT distortion. – orthocresol Jan 21 '18 at 12:09