Identification of Diamagnetic Transition Metal Complexes

To determine which of the given complexes are diamagnetic, we find the oxidation state, the $d$-electron configuration of the central metal ion, and the crystal field splitting (pairing of electrons):

1. Complex $\mathbf{K}$: $\mathrm{K}_3[\mathrm{Fe}(\mathrm{CN})_6]$

   • The central metal is iron, with oxidation state:
     $x + 6(-1) = -3 \implies x = +3\ (\mathrm{Fe}^{3+})$
   • The electronic configuration of $\mathrm{Fe}^{3+}$ is $[\mathrm{Ar}]\,3d^5$.
   • Since cyanide ($\mathrm{CN}^-$) is a strong field ligand, it causes pairing of electrons resulting in a low-spin octahedral complex:
     $t_{2g}^5\,e_g^0$
   • It possesses $1$ unpaired electron ($n = 1$).
   • Therefore, complex $\mathbf{K}$ is paramagnetic.

2. Complex $\mathbf{L}$: $[\mathrm{Co}(\mathrm{NH}_3)_6]\mathrm{Cl}_3$

   • The central metal is cobalt, with oxidation state:
     $x + 6(0) = +3 \implies x = +3\ (\mathrm{Co}^{3+})$
   • The electronic configuration of $\mathrm{Co}^{3+}$ is $[\mathrm{Ar}]\,3d^6$.
   • Ammonia ($\mathrm{NH}_3$) acts as a strong field ligand with $\mathrm{Co}^{3+}$, causing pairing ($\Delta_o > P$):
     $t_{2g}^6\,e_g^0$
   • There are no unpaired electrons ($n = 0$).
   • Therefore, complex $\mathbf{L}$ is diamagnetic.

3. Complex $\mathbf{M}$: $\mathrm{Na}_3[\mathrm{Co}(\text{oxalate})_3]$

   • The central metal is cobalt, with oxidation state:
     $x + 3(-2) = -3 \implies x = +3\ (\mathrm{Co}^{3+})$
   • The electronic configuration of $\mathrm{Co}^{3+}$ is $[\mathrm{Ar}]\,3d^6$.
   • For $\mathrm{Co}^{3+}$, even oxygen-donor chelating ligands like oxalate ($\text{ox}^{2-}$) provide a crystal field splitting large enough to overcome the pairing energy ($\Delta_o > P$), yielding a low-spin state:
     $t_{2g}^6\,e_g^0$
   • There are no unpaired electrons ($n = 0$).
   • Therefore, complex $\mathbf{M}$ is diamagnetic.

4. Complex $\mathbf{N}$: $[\mathrm{Ni}(\mathrm{H}_2\mathrm{O})_6]\mathrm{Cl}_2$

   • The central metal is nickel, with oxidation state:
     $x + 6(0) = +2 \implies x = +2\ (\mathrm{Ni}^{2+})$
   • The electronic configuration of $\mathrm{Ni}^{2+}$ is $[\mathrm{Ar}]\,3d^8$.
   • In an octahedral field, the $d^8$ configuration is:
     $t_{2g}^6\,e_g^2$
   • It has $2$ unpaired electrons ($n = 2$).
   • Therefore, complex $\mathbf{N}$ is paramagnetic.

5. Complex $\mathbf{O}$: $\mathrm{K}_2[\mathrm{Pt}(\mathrm{CN})_4]$

   • The central metal is platinum, with oxidation state:
     $x + 4(-1) = -2 \implies x = +2\ (\mathrm{Pt}^{2+})$
   • Platinum is a $5d$ transition metal with a $5d^8$ configuration. Due to the very large crystal field splitting for $5d$ metals, 4-coordinate $\mathrm{Pt}^{2+}$ complexes are square planar ($dsp^2$ hybridization) and low-spin.
   • All 8 electrons are paired ($n = 0$).
   • Therefore, complex $\mathbf{O}$ is diamagnetic.

6. Complex $\mathbf{P}$: $[\mathrm{Zn}(\mathrm{H}_2\mathrm{O})_6](\mathrm{NO}_3)_2$

   • The central metal is zinc, with oxidation state:
     $x + 6(0) = +2 \implies x = +2\ (\mathrm{Zn}^{2+})$
   • The electronic configuration of $\mathrm{Zn}^{2+}$ is $[\mathrm{Ar}]\,3d^{10}$.
   • All $d$ orbitals are completely filled ($t_{2g}^6\,e_g^4$), leaving no unpaired electrons ($n = 0$).
   • Therefore, complex $\mathbf{P}$ is diamagnetic.

Thus, the diamagnetic complexes are $\mathbf{L}$, $\mathbf{M}$, $\mathbf{O}$, and $\mathbf{P}$.

Correct Answer: C (or $\mathbf{L}, \mathbf{M}, \mathbf{O}, \mathbf{P}$)