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. 


[![enter image description here](https://i.stack.imgur.com/IJ3Is.png)](https://i.stack.imgur.com/IJ3Is.png)


Image source: [Wikimedia Commons](https://commons.wikimedia.org/wiki/File:Electron-config-d8-s2.png)


[![enter image description here](https://i.stack.imgur.com/YjJzG.png)](https://i.stack.imgur.com/YjJzG.png)


Image source:[Wikimedia Commons](https://upload.wikimedia.org/wikipedia/commons/a/a6/Chem507f09sqvstet2.png)


 4

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](https://i.stack.imgur.com/6NqTh.jpg)](https://i.stack.imgur.com/6NqTh.jpg)


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