![\[{{[\mathbf{Cr}{{(\mathbf{N}{{\mathbf{H}}_{\mathbf{3}}})}_{\mathbf{6}}}]}^{\mathbf{3}+}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-39b74154417ca7572b9b77e89bc26feb_l3.png "Rendered by QuickLaTeX.com")
In [ Ni ( CN)4 ] 2−, Ni has an oxidation state of +2. Thus, it has d 8configuration.
Ni 2+ :
Because CN– is a strong field ligand, electrons in 3d orbitals couple. Ni 2+ undergoes dsp2 hybridization as a result of dsp2 hybridisation.
Since all the electrons are paired, it is diamagnetic in nature.
The oxidation state of Cr is +3. It has a d3 arrangement as a result. The electrons in the 3d orbital do not pair because NH3 is not a strong field ligand.
Cr3+:
The 3d orbital electrons remain unpaired after d2sp3 hybridization. As a result, [ Ni ( CN)4 ]2 is paramagnetic.