1. Determine oxidation state and electronic configuration of the metal in each complex

All given complexes are octahedral because each has coordination number $6$.

We now check magnetic nature and stability.

2. Option A: $\mathrm{Na}_3[\mathrm{CoCl}_6]$

The complex ion is $[\mathrm{CoCl}_6]^{3-}$.

Let oxidation state of Co be $x$: $x+6(-1)=-3$ $x=+3$

So, Co is $\mathrm{Co}^{3+}$.

Atomic number of Co = 27 $\mathrm{Co}: [Ar]3d^74s^2$ $\mathrm{Co}^{3+}: 3d^6$

Ligand $\mathrm{Cl}^-$ is a weak field ligand, so octahedral complex is high spin: $t_{2g}^4e_g^2$

This has 4 unpaired electrons, hence it is paramagnetic, not diamagnetic.

So A is not correct.

3. Option B: $[\mathrm{Co}(\mathrm{H}_2\mathrm{O})_6]\mathrm{Cl}_2$

The complex cation is $[\mathrm{Co}(\mathrm{H}_2\mathrm{O})_6]^{2+}$.

Since water is neutral, oxidation state of Co is: $x=+2$

So, Co is $\mathrm{Co}^{2+}$: $\mathrm{Co}^{2+}=3d^7$

$\mathrm{H_2O}$ is a weak field ligand, so the octahedral complex is high spin: $t_{2g}^5e_g^2$

This has 3 unpaired electrons, so it is paramagnetic.

So B is not correct.

4. Option C: $\mathrm{K}_3[\mathrm{Co}(\mathrm{CN})_6]$

The complex ion is $[\mathrm{Co}(\mathrm{CN})_6]^{3-}$.

Let oxidation state of Co be $x$: $x+6(-1)=-3$ $x=+3$

So, Co is $\mathrm{Co}^{3+}$: $\mathrm{Co}^{3+}=3d^6$

$\mathrm{CN^-}$ is a strong field ligand, so octahedral splitting is large and the complex becomes low spin: $t_{2g}^6e_g^0$

All electrons are paired, so the complex is diamagnetic.

Also, $\mathrm{CN^-}$ forms very stable complexes because of:

• strong field nature,
• large crystal field stabilization energy,
• possible $\pi$-back bonding,
• low-spin $d^6$ configuration, which is especially stable.

So C is octahedral, diamagnetic, and the most stable.

5. Option D: $[\mathrm{Ni}(\mathrm{NH}_3)_6]\mathrm{Cl}_2$

The complex cation is $[\mathrm{Ni}(\mathrm{NH}_3)_6]^{2+}$.

Since $\mathrm{NH_3}$ is neutral: $x=+2$

So, Ni is $\mathrm{Ni}^{2+}$: $\mathrm{Ni}^{2+}=3d^8$

In octahedral field: $t_{2g}^6e_g^2$

This has 2 unpaired electrons, so it is paramagnetic.

So D is not correct.

6. Final comparison

Only option C is:

• octahedral,
• diamagnetic,
• highly stable.

Therefore, the correct answer is: $\boxed{\mathrm{K}_3[\mathrm{Co}(\mathrm{CN})_6]}$ which corresponds to Option C.