Chemistry - Coordination Chemistry Question with Solution | TestHub

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( ${{\mathrm{NH}}_3)}_6]{\mathrm{Cl}}_3$, 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 $t_{2g}$ level, diamagnetic. In $\left[\mathrm{Ni}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]{\mathrm{Cl}}_2,\mathrm{Ni}$ (II) has $3d^8$-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.