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Step-by-Step Solution
Step 1: Analyze the ionization behavior
When 1 mole of the complex Co(NH3)5Cl3 dissolves in water, it produces 3 moles of ions in total. This observation implies that the complex dissociates into one complex cation and two chloride ions. Symbolically, we can write:
$ \text{Co(NH}_3)_5 \text{Cl}_3 \;\longrightarrow\; \bigl[\text{Co(NH}_3)_5\text{Cl}\bigr]^{2+} \;+\; 2 \text{Cl}^-$
This shows that two chlorides must be present outside the coordination sphere (responsible for giving two Cl- ions in solution), while one chloride is within the coordination sphere, bound to cobalt.
Step 2: Reaction with AgNO3
The complex, upon reaction with AgNO3, gives two moles of AgCl per mole of complex. This indicates that only two chloride ions are free or “ionizable” in solution, as these two chloride ions can be precipitated out as AgCl. The chloride bound inside the coordination sphere does not react with AgNO3 and remains part of the complex cation.
$ \bigl[\text{Co(NH}_3)_5\text{Cl}\bigr] \text{Cl}_2 \;+\; 2\,\text{AgNO}_3
\;\longrightarrow\; \bigl[\text{Co(NH}_3)_5\text{Cl}\bigr](\text{NO}_3)_2 \;+\; 2\,\text{AgCl}\,(\text{s}) $
Step 3: Deduce the structure
From the ionization behavior and the silver nitrate test, it is clear that two chlorides are outside the coordination sphere (where they can ionize and form AgCl), and one chloride is directly bonded to the cobalt center inside the coordination sphere. Hence, the formula must be:
$ \bigl[\text{Co(NH}_3)_5\text{Cl}\bigr]\text{Cl}_2 $
Therefore, the correct structure of the complex is [Co(NH3)5Cl]Cl2.
