Question

Give reason :
$\left[Cr\right(N{H}_3{)}_6{]}^{3+}$ is paramagnetic while $\left[Ni\right(CN{)}_4{]}^{2-}$ is diamagnetic

Answer 1

Cr is in the +3 oxidation state i.e.,$d^3$ configuration. Also, $N{H}_3$ is a weak field ligand that does not cause the pairing of the electrons in the $3d$ orbital. and the electrons in the $3d$ orbitals remain unpaired. Hence, it is paramagnetic in nature.

In $\left[Ni\right(CN{)}_4{]}^{2-}$, Ni exists in the $+2$ oxidation state i.e., $d^8$ configuration. $C{N}^{-}$ is a strong field ligand. It causes the pairing of the 3d orbital electrons. Then, $N{i}^{2+}$ undergoes dsp2 hybridization. As there are no unpaired electrons, it is diamagnetic.