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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"?

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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

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  • $\begingroup$ Would there be any coordination complex of Pd that is tetrahedral for CN=4? $\endgroup$
    – ADP
    Commented Jan 21, 2018 at 13:48
  • $\begingroup$ 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: pubs.rsc.org/en/content/articlelanding/1999/cc/… $\endgroup$
    – Tan Yong Boon
    Commented Jan 21, 2018 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

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  • $\begingroup$ 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? $\endgroup$
    – ADP
    Commented Jan 21, 2018 at 10:39
  • $\begingroup$ 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 $\endgroup$
    – Subir Podder
    Commented Jan 21, 2018 at 10:42
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    $\begingroup$ How is this a Jahn–Teller distortion from octahedral geometry when there are only four ligands to begin with? $\endgroup$
    – orthocresol
    Commented Jan 21, 2018 at 11:47
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    $\begingroup$ 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. $\endgroup$
    – orthocresol
    Commented Jan 21, 2018 at 11:52
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    $\begingroup$ No, the Jahn-Teller theorem does not cover changes in molecular composition, it only covers changes in the arrangement of the atoms. $\endgroup$
    – orthocresol
    Commented Jan 21, 2018 at 11:57

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