Among the following complexes (K to P),
$\begin{array}{l}{\mathrm{K}}_3\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]\left(\mathrm{K}\right),\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_6\right]{\mathrm{Cl}}_3\left(\mathrm{L}\right),\\ {\mathrm{Na}}_3\left[\mathrm{Co}{\left(\text{oxalate}\right)}_3\right]\left(\mathrm{M}\right),\left[\mathrm{Ni}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]{\mathrm{Cl}}_2\left(\mathrm{N}\right),\\ {\mathrm{K}}_2\left[\mathrm{Pt}{\left(\mathrm{CN}\right)}_4\right]\left(\mathrm{O}\right),\text{ and }\left[\mathrm{Zn}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]{\left({\mathrm{NO}}_3\right)}_2\left(\mathrm{P}\right)\end{array}$
The diamagnetic complexes are:

detailed solution

Correct option is B

[K] : K3FeCN6: Fe3+ has 3d5 configuration; CN- is strong field ligand and thus four electrons are paired leaving one unpaired electron with d2sp3 hybridization and thus paramagnetic.[L]: CoNH36Cl3:Co3+ has configuration; NH3 is strong field ligand thus all six electrons are paired with d2sp3 hybridization and therefore diamagnetic.[M]: Na3Coox3: Co3+ has 3d6 configuration: C2O42− is strong field ligand and thus all the six electrons are paired with d2sp3 hybridisation and therefore diamagnetic.[N]: NiH2O6Cl2:Ni2+ has 3d8 configuration with two unpaired electrons. H2O is weak field ligand and thus sp3d2 hybridisation and paramagnetic.[O]: K2PtCN4: Pt2+ has 3d8 configuration; CN- is strong field ligand and thus all the eight electrons are paired with dsp2 hybridisation and therefore diamagnetic. [P]: ZnH2O6NO32:Zn2+ has 3d10 configuration and thus all paired showing sp3d2 hybridization and so diamagnetic.