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"?
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