On treatment of 100 mL of 0.1 M solution of CoCl$$_{3}$$.6H$$_{2}$$O with excess AgNO$$_{3}$$; $$1.2 \times 10^{22}$$ ions are precipitated. The complex is:

We have been given a 100 mL sample of a $$0.1\ \text{M}$$ aqueous solution of $$\text{CoCl}_3\cdot6\text{H}_2\text{O}$$. First, let us calculate how many moles of the cobalt compound are present.

The definition of molarity is

$$\text{Molarity (M)}=\dfrac{\text{moles of solute}}{\text{volume of solution in litres}}.$$

So, rearranging and substituting the given numbers,

$$\text{moles of solute}=0.1\ \text{M}\times0.100\ \text{L}=0.010\ \text{mol}.$$

Next, the solution is treated with a large excess of $$\text{AgNO}_3$$. Any chloride ion that is outside the coordination sphere will be free in solution and will therefore precipitate as $$\text{AgCl}$$. We are told that the number of chloride ions that actually precipitate equals $$1.2\times10^{22}$$.

Avogadro’s number is

$$N_A=6.022\times10^{23}\ \text{ions mol}^{-1}.$$

The moles of chloride that precipitate are therefore

$$n_{\text{Cl}^-}=\dfrac{1.2\times10^{22}\ \text{ions}}{6.022\times10^{23}\ \text{ions mol}^{-1}} \approx1.99\times10^{-2}\ \text{mol}\approx0.020\ \text{mol}.$$

Let $$y$$ be the number of chloride ions present outside the coordination sphere for each mole of the complex. Because every mole of complex produces $$y$$ moles of free chloride, the relation is

$$y\times(\text{moles of complex})=\text{moles of Cl}^- \text{ precipitated}.$$

Substituting the known values,

$$y\times0.010\ \text{mol}=0.020\ \text{mol}.$$

Dividing both sides by $$0.010\ \text{mol}$$, we obtain

$$y=2.$$

Hence, two chloride ions per formula unit are outside the coordination sphere (counter-ion chlorides).

Originally every formula unit contains three chloride ions. If two are outside, then

$$3-2=1$$

chloride ion must be inside the coordination sphere.

Now, cobalt(III) commonly has coordination number 6, so the total number of ligands in the inner sphere must be six. We already have one $$\text{Cl}^-$$ inside; therefore the remaining

$$6-1=5$$

positions are occupied by water molecules, which are neutral ligands.

Thus the coordination entity is

$$\left[\text{Co}(\text{H}_2\text{O})_5\text{Cl}\right]^{2+}.$$

The charge check is straightforward:

Co(III) contributes $$+3$$, one coordinated chloride contributes $$-1$$, giving $$+2$$ overall for the complex cation. Exactly two external chloride ions $$\text{Cl}^-$$ are required to balance this $$+2$$ charge, producing a neutral compound.

Finally, the original formula has six water molecules in all. We have already placed five of them in the inner sphere, so one water molecule remains as water of crystallisation outside the bracket.

Combining everything, the complete formula is

$$\left[\text{Co}(\text{H}_2\text{O})_5\text{Cl}\right]\text{Cl}_2\cdot\text{H}_2\text{O}.$$

This corresponds exactly to Option C.

Hence, the correct answer is Option C.