Identify the set of coordination complexes in which all the species...

In $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$, $\mathrm{Fe}$ is in $+2$ oxidation state with a $3d^6$ configuration. Since $\mathrm{CN}^-$ is a strong field ligand, electrons pair up to give $t_{2g}^6 e_g^0$. It has zero unpaired electrons and is diamagnetic.

In \left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$,$\mathrm{Co}$is in$+3$oxidation state with a$3d^6$configuration.$\mathrm{NH}_3$acts as a strong field ligand for$\mathrm{Co}^{3+}$, leading to$t_{2g}^6 e_g^0$. It has zero unpaired electrons and is diamagnetic.<br><br>In \left[\mathrm{PtCl}_4\right]^{2-}$, $\mathrm{Pt}$ is in $+2$ oxidation state with a $5d^8$ configuration. For $4d$ and $5d$ series metals, the crystal field splitting is large, resulting in a square planar geometry ($dsp^2$ hybridization) where all electrons are paired. It is diamagnetic.

In \left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$,$\mathrm{Fe}$is in$+3$oxidation state ($3d^5$). With the strong field ligand$\mathrm{CN}^-$, the configuration is$t_{2g}^5 e_g^0$, having one unpaired electron (paramagnetic).<br><br>In \left[\mathrm{CoF}_6\right]^{3-}$, $\mathrm{Co}$ is in $+3$ oxidation state ($3d^6$). With the weak field ligand $\mathrm{F}^-$, the configuration is $t_{2g}^4 e_g^2$, having four unpaired electrons (paramagnetic).

In \left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$,$\mathrm{Fe}$is in$+2$oxidation state ($3d^6$). With the weak field ligand$\mathrm{H}_2\mathrm{O}$, the configuration is$t_{2g}^4 e_g^2$, having four unpaired electrons (paramagnetic).<br><br>Thus, only the first set contains all diamagnetic species.<br><br>Answer:$\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}, \left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}, \left[\mathrm{PtCl}_4\right]^{2-}