# Is [PdCl2(PMe3)2] diamagnetic or paramagnetic?

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"?

• The splitting pattern for a square planar complex is simply different from that of a tetrahedral or octahedral complex. This is not about Jahn–Teller distortions. Jan 21 '18 at 11:48
• Possible duplicate of Why is [PdCl4]2- square planar whereas [NiCl4]2- is tetrahedral? Jan 21 '18 at 12:03

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.

Image source: Wikimedia Commons

Image source:Wikimedia Commons

• Would there be any coordination complex of Pd that is tetrahedral for CN=4?