1. Use the ionization data

The compound gives 3 moles of ions on dissolution in water.

So, in water it must dissociate as: $\text{complex compound} \rightarrow 1\text{ complex ion} + 2\text{ counter ions}$

Hence the compound should have the form: $[\text{coordination entity}]X_2$ which gives total ions $= 3$.

So among the options, we check which one produces 3 ions.

• A: $[\mathrm{Co}(\mathrm{NH}_3)_5\mathrm{Cl}]\mathrm{Cl}_2$ $[\mathrm{Co}(\mathrm{NH}_3)_5\mathrm{Cl}]\mathrm{Cl}_2 \rightarrow [\mathrm{Co}(\mathrm{NH}_3)_5\mathrm{Cl}]^{2+} + 2\mathrm{Cl}^-$ Total ions $=3$ ✅

• B: $[\mathrm{Co}(\mathrm{NH}_3)_4\mathrm{Cl}]\cdot \mathrm{Cl}_2 \cdot \mathrm{NH}_3$ This is not a proper octahedral coordination formulation as written, and effectively does not match the required 6 coordinated ligands around Co in the coordination sphere.

• C: $[\mathrm{Co}(\mathrm{NH}_3)_3\mathrm{Cl}_3]\cdot 2\mathrm{NH}_3$ Here all 3 chlorides are inside the coordination sphere, so no free $\mathrm{Cl}^-$ ions are present. It would not give 3 ions. ❌

• D: $[\mathrm{Co}(\mathrm{NH}_3)_4\mathrm{Cl}_2]\cdot \mathrm{Cl} \cdot \mathrm{NH}_3$ This would have only one chloride outside the coordination sphere, so it would give only 2 ions in solution: $[\mathrm{Co}(\mathrm{NH}_3)_4\mathrm{Cl}_2]^+ + \mathrm{Cl}^-$ Total ions $=2$ ❌

So from ion count alone, A fits.

2. Use the $\mathrm{AgNO_3}$ test

Excess $\mathrm{AgNO_3}$ gives 2 moles of $\mathrm{AgCl}$.

This means there are 2 free chloride ions outside the coordination sphere, because only ionizable chloride ions precipitate immediately with $\mathrm{Ag}^+$: $\mathrm{Ag}^+ + \mathrm{Cl}^- \rightarrow \mathrm{AgCl}(s)$

Therefore the complex must contain 2 counter chloride ions outside the bracket, and 1 chloride ligand inside the coordination sphere.

That structure is: $[\mathrm{Co}(\mathrm{NH}_3)_5\mathrm{Cl}]\mathrm{Cl}_2$

3. Check octahedral requirement

For an octahedral complex, coordination number must be 6.

In option A:

• $5$ ligands are $\mathrm{NH_3}$
• $1$ ligand is $\mathrm{Cl^-}$

So total coordinated ligands $= 6$, hence octahedral. ✅

4. Final conclusion

The correct structure is: $\boxed{[\mathrm{Co}(\mathrm{NH}_3)_5\mathrm{Cl}]\mathrm{Cl}_2}$

So, Option A is correct.