give reason for the following

1. tetrahedral Ni(2) complexes are paramagnetic but square planar Ni(2) complex are diamagnetic

2. only transition metals are know to form pie complexes ..

Ni in 2+ oxidation state has following electronic configuration.


To form square planar complex it need dsp² hybridization for which it require one vacant d orbital. The vacant d orbital would be available only in case of pairing of all the electrons. By acquiring dsp² hybridization​ , all the electrons get pairied hence complexes are diamagnetic in nature as shown below:


To form tetrahedral complex it use its outer 4s and 4p orbital to form sp³ hybridization. So the electrons in d-orbitals remains unpaired , hence tetrahedral complexes show paramagnetic behaviour.

 

Pie complexes are formed by ligands such as CO, CN^-, NO⁺ etc. These ligands have vacant Π orbitals. The transition metals form pie complexes because they have t_2g orbitals of correct symmetry which allows them to overlap with the vacant Π orbitals of these ligands, forming Π bonds. This bonding is known as back bonding. 

Due to the presence of d-orbitals of proper symmetry in transition metals, they are favourable to form pie complexes.