Explain briefly why [Ni(CO)4] is tetrahedral but [Ni(CN)4]2- is square planaralthough both of them are diama- gnetic.?

[Ni(CO)4] is tetrahedral because the ligands (CO) are strong-field ligands that cause the d orbitals to split in such a way that the electrons fill the lower energy orbitals first, resulting in a tetrahedral geometry.

[Ni(CN)4]2- is square planar because the ligands (CN) are weak-field ligands that cause the d orbitals to split in such a way that the electrons fill the higher energy orbitals first, resulting in a square planar geometry. Despite both complexes being diamagnetic, the different ligands and resulting orbital splitting lead to different geometries.

[Ni(CO)4] is tetrahedral due to its sp3 hybridisation of Ni. The four CO ligands occupy the four corners of a tetrahedron, resulting in a tetrahedral geometry. The absence of unpaired electrons (diamagnetic) indicates that all the electrons are paired in the d-orbitals, supporting the sp3 hybridisation.

[Ni(CN)4]2- is square planar due to its dsp2 hybridisation of Ni. The four CN ligands occupy the four corners of a square, resulting in a square planar geometry. The absence of unpaired electrons (diamagnetic) indicates that all the electrons are paired in the d-orbitals, supporting the dsp2 hybridisation.

The difference in geometry despite both species being diamagnetic arises from the different sizes and strengths of the ligands. CO is a stronger ligand than CN-, which results in a stronger metal-ligand bond in [Ni(CO)4] and a shorter Ni-C bond length. This stronger bond favors the formation of a tetrahedral geometry to maximize the distance between the ligands.

On the other hand, CN- is a smaller and weaker ligand, leading to a longer Ni-C bond length and a less constrained geometry. This weaker bond allows the ligands to adopt a square planar arrangement to minimize steric hindrance.