One mole of the octahedral complex compound  $$Co(NH_3)_5Cl_3$$  gives 3 moles of ions on dissolution in water. One mole of the same complex reacts with excess  $$AgNO_3$$ solution to yield two moles of  $$AgCl(s).$$  The structure of the complex is:

We need to determine the structure of the complex $$Co(NH_3)_5Cl_3$$ using experimental evidence.

The experimental observations are as follows. One mole of the complex gives 3 moles of ions on dissolution in water, and one mole of the complex reacts with excess $$AgNO_3$$ to give 2 moles of $$AgCl$$.

We begin by interpreting the $$AgCl$$ precipitation data. Silver nitrate precipitates only the free chloride ions (those outside the coordination sphere) as $$AgCl$$. Chloride ions directly coordinated to the metal ion inside the complex do not react with $$AgNO_3$$ under normal conditions. Since 2 moles of $$AgCl$$ are formed, there are 2 free $$Cl^-$$ ions outside the coordination sphere.

With a total of 3 chloride ions and 2 free, the number of coordinated chloride ions is $$3 - 2 = 1$$.

Assuming the complex has the formula $$[Co(NH_3)_5Cl]Cl_2$$, it dissociates as:
$$[Co(NH_3)_5Cl]Cl_2 \rightarrow [Co(NH_3)_5Cl]^{2+} + 2Cl^-$$. This gives 3 ions total (1 complex cation + 2 chloride anions), which matches the given data. ✓

In the cation $$[Co(NH_3)_5Cl]^{2+}$$, cobalt is coordinated to 5 $$NH_3$$ molecules and 1 $$Cl^-$$ ion, giving a coordination number of 6, which is consistent with the octahedral geometry stated in the question. ✓

The correct answer is Option 3: $$[Co(NH_3)_5Cl]Cl_2$$.