[cchave3]

How do I know if an organometallic Ni complex will be paramagnetic or diamagnetic?  It seems as though in the literature, some Ni(II) complexes are diamagnetic and some are paramagnetic.

[dipesh747]

With regard to Ni (II) it is a d8 atom. The crystal field will split into a square planar complex so you will get the following orbitals

Dx2-y2
Dxy
Dz2
Dxz,Dyz (e)

So if you have 8 electrons and fill up from the bottom, you will not have any unpaired electrons so it must be diamagnetic in its ground state. In an excited state it can be paramagnetic.

The only other way a Ni complex will be paramagnetic at ground state is if it isn't Ni(II). Ni(I) will contain unpaired electrons (paramagnetic) as it will be d9.

[darko]

With regard to Ni (II) it is a d8 atom. The crystal field will split into a square planar complex so you will get the following orbitals

Dx2-y2
Dxy
Dz2
Dxz,Dyz (e)

So if you have 8 electrons and fill up from the bottom, you will not have any unpaired electrons so it must be diamagnetic in its ground state. In an excited state it can be paramagnetic.

The only other way a Ni complex will be paramagnetic at ground state is if it isn't Ni(II). Ni(I) will contain unpaired electrons (paramagnetic) as it will be d9.

sorry for my english, but there is some things to be explained about Ni²⁺ complexes.

d8 for Ni(II) is ion, not atom, it is correct for Ni(II) complexes for week cristal field, but in strong field dz² is below dxz,dyz.

for Ni(II) complexes is tipical that in solutions (alcohol, aqueos,amines..) they are becoming in structure of tetragonal bipyramid so two aditional ligands from solvent are acepted as trns-ligands, as a result is week splitting of e_g and dx2-y2 and dz2 are close so each will acept one electron and complexes will be paramagnetic. It will maintan so also if complexes are cristalized from solutions.

The sam efect will become with polimerisation of squeare complexes, with octahedral coordination