9.19 [Cr(NH3)6]3+ is paramagnetic while [Ni(CN)4]2– is diamagnetic. Explain why?

Step 1:

In [Cr(NH3)6]3+Cr is in the +3 oxidation state i.e., 3${d}^{3}$ configuration. Also, $N{H}_{3}$ is a weak field ligand that does not cause the pairing of the electrons in the 3d orbital. $C{r}^{3+}$

Therefore, it undergoes ${d}^{2}s{p}^{3}$ hybridization and the electrons in the 3d orbitals remain unpaired. Hence, it is paramagnetic in nature.

Step 2:

In [Ni(CN)4]2-, Ni oxidation is +2 and its electronic configuration is 3d8 configuration.

$C{N}^{-}$ is a strong field ligand. Hence, it paired the 3d electrons.  Then, $N{i}^{2+}$ undergoes $ds{p}^{2}$ hybridization.

As there are no unpaired electrons, it is diamagnetic in nature.