For a diamagnetic complex, there should not be any unpaired electron in the valence shell of central metral. In $\mathrm{K}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]$, Fe (III) has $d^5$-configuration (odd electrons), hence it is paramagnetic. In [Co( $\left.\left.\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_3$, Co (III) has $d^6$ configuration in a strong ligand field, hence all the electrons are paired and the complex is diamagnetic. In $\mathrm{Na}_3\left[\mathrm{Co}(\mathrm{ox})_3\right]$, $\mathrm{Co}$ (III) has $d^6$-configuration and oxalate being a chelating ligand, very strong ligand and all the six electrons remains paired in lower $t_{2 g}$ level, diamagnetic. In $\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right] \mathrm{Cl}_2, \mathrm{Ni}$ (II) has $3 d^8$-configuration and $\mathrm{H}_2 \mathrm{O}$ is a weak ligand hence


In $\mathrm{K}_2\left[\mathrm{Pt}(\mathrm{CN})_4\right]$, Pt (II) has $d^8$-configuration and $\mathrm{CN}^{-}$is a strong ligand, hence all the eight electrons are spin-paired. Therefore, complex is diamagnetic.
In $\left[\mathrm{Zn}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]\left(\mathrm{NO}_3\right)_2, \mathrm{Zn}$ (II) has ${ }_3 \mathrm{~d}^{10}$ configuration spin paired, hence diamagnetic.