[Ni(Cl)4]2- is paramagnetic while [Ni(Co)4]is diamagnetic though both are tetrahedral. Why?


In [Ni(Cl)4]2- ion, Cl- is a weak field ligand so it will not pair the unpaired electrons of Ni+2 ion.

Electronic configuration of Ni is: [Ar]3d84s2 where [Ar] = 1s22s22p63s23p6


Electronic configuration of Ni+2 = [Ar]3d8


Outer electronic configuration of Ni+2 = 3d8


Overall charge balance:


X + 4(-1) = -2


X = + 2.


Therefore it undergoes sp3 hybridization. So it will have tetrahedral geometry.



Since there are 2 unpaired electrons in the d orbital so it is a paramagnetic compound.


In [Ni(Co)4]:


Overall charge is neutral and oxidation state of Ni can be calculated as:


X + 4(0) = 0


X = 0


Ni is in zero oxidation state.



Co is a strong field ligand it causes pairing of the 4 unpaired electrons in d orbital. Also, it causes the 4s electrons to shift to the 3d orbital, thereby undergoes sp3 hybridisation forming tetrahedral geometry. Since it has no unpaired electron, therefore it is diamagnetic compound.


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