[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.