Question

\([Cr(NH_3)_6]^{3+}\) is paramagnetic while \([Ni(CN)_4]^{2-}\) is diamagnetic. Explain why?

Answer 1

\(Cr \) is in the \(+3\) oxidation state i.e., \(d^{3}\) configuration. Also, \(NH_3\) is a weak field ligand that does not cause the pairing of the electrons in the \(3d\) orbital.
\(Cr^{3+}\)
 
Therefore, it undergoes \(d^{2}\) \(sp^{3}\) hybridization and the electrons in the \(3d\) orbitals remain unpaired. 
Hence, it is paramagnetic in nature. 
In \([Ni(CN)_4] ^{2-}\) , Ni exists in the \(+2\) oxidation state i.e., \(d^{8}\) configuration. 
\(Ni^{2+}\): 

\(CN^{-}\) is a strong field ligand. It causes the pairing of the \(3d\) orbital electrons. 
Then, \(Ni^{2+}\) undergoes \(dsp^{2} \)hybridization.
 
As there are no unpaired electrons, it is diamagnetic