[ NiCl4] 2– is paramagnetic while [ Ni(CN) 4 ] 2– is diamagnetic

Why is [ NiCl4 ] ^2– paramagnetic while [ Ni(CN) 4 ] ^2– is diamagnetic ?

The Ni(II) ion is 3d⁸ system. According to Hund’s rule the outer electronic configuration of Ni(II) ion is [ Ar ] 3d⁸.  From electronic configuration of Ni(II) ion , it is shown  that it has two unpaired electrons.

Now, depending upon the hybridization, there are two types of possible structure of Ni(II) complex are formed with co-ordination number 4.

If the complex involves ‘sp3’ hybridization, it would have tetrahedral structure . Again, If the complex involves ‘dsp2’ hybridization, it would have square planar  structure .

Consequently, for the formation of tetrahedral structure through the ‘sp3’ hybridization, the 3d-orbital of nickel atom remain unaffected.

Therefore, 3d-orbital of nickel(II) ion possessed two unpaired electrons and hence the concern complex would be paramagnetic.

On the other hand, for the formation of square planar structure through the dsp2 hybridization, one of the 3d-orbital of nickel atom should be empty and available for hybridization.

But, it is possible if the two unpaired 3d-electrons are paired up due to the energy made available by the approach of ligands.

Under this condition, all the 3d-electrons of Ni(II) ion are paired and hence the complex would be diamagnetic.

Now, in case of [ NiCl4 ] ^2–complex ion, Ni(II) ion with co-ordination 4 involves ‘sp3’ hybridization. Hence the geometry of, [ NiCl4 ] ^2–complex ion would be tetrahedral.

Under this condition, the electronic arrangement of Ni(II) ion is evidently shown that the 3d-orbitals of Ni(II) ion have '2' unpaired electrons and hence[ NiCl4 ]^2–complex ion is paramagnetic.

On the other hand, in case of [ Ni ( CN ) ₄ ] ^2– complex ion, Ni(II) ion with co-ordination number 4 involves ‘dsp2’ hybridization. 

Hence the geometry of, [ Ni ( CN ) ₄ ] ^2– complex ion would be square planar.

The empty hybrid orbitals of metal overlap with fully filled orbitals of cyanide ion to form metal-ligand co-ordinate bond.

Under this condition, the electronic arrangement of Ni(II) ion is evidently shown that there is no unpaired electrons present in 3d-orbitals of Ni(II) ion and hence [Ni ( CN ) ₄ ] ^2– complex ion is diamagnetic.

• Why is [ NiCl4 ] ^2– ion paramagnetic while [ Ni(CN) 4 ] ^2– ion is diamagnetic ?
• Why is [ NiCl4 ] ^2– ion paramagnetic in nature?
• Why is [ Ni(CN) 4 ] ^2–ion paramagnetic in nature?
• Why is [Fe(CN)₆]^3–a low spin complex ? 
• Why is [Fe(H₂O)₆]³⁺a high spin complex ?

Why is [Ni(NH₃)₆]Cl₂  paramagnetic but [Co(NH₃)₆]Cl₃ is a diamagnetic?