Among the following complexes (K-P), K_3[Fe(CN)_6](K) , [. Co...

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.