Which of the following complex is octahedral, diamagnetic and the most stable?

We need to identify the complex that is octahedral, diamagnetic, and most stable.

First, we analyze $$\mathrm{Na}_3[CoCl_6]$$, where cobalt is in the $$3+$$ oxidation state with a $$d^6$$ configuration and chloride is a weak field ligand. In an octahedral weak field, this gives a high spin arrangement of $$t_{2g}^4e_g^2$$ with four unpaired electrons, making the complex paramagnetic and therefore not the answer.

Next, we consider $$[Ni(NH_3)_6]Cl_2$$, in which nickel is $$2+$$ with a $$d^8$$ configuration and ammonia as a ligand. In an octahedral geometry, $$d^8$$ invariably leads to $$t_{2g}^6e_g^2$$ with two unpaired electrons, resulting in paramagnetism and ruling it out.

Then, examining $$K_3[Co(CN)_6]$$, cobalt is again $$3+$$ with $$d^6$$, but cyanide is a strong field ligand that induces a large crystal field splitting. This leads to a low spin configuration of $$t_{2g}^6e_g^0$$ with no unpaired electrons, rendering the complex diamagnetic. Its octahedral geometry combined with high crystal field stabilization energy makes this the most stable candidate that meets all criteria.

Meanwhile, $$[Co(H_2O)_6]Cl_2$$ features cobalt in the $$2+$$ oxidation state with $$d^7$$ and water as a weak field ligand, which yields high spin $$t_{2g}^5e_g^2$$ with three unpaired electrons, again paramagnetic and thus unsuitable.

Therefore, the correct choice is Option 3: $$K_3[Co(CN)_6]$$.