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