9.6 [NiCl4 ] 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.

9.6 [NiCl₄ ]^2– is paramagnetic while [Ni(CO)₄ ] is diamagnetic though both are tetrahedral. Why?

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Vishal Baghel | Contributor-Level 10

10 months ago

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)