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Step-by-Step Solution
Step 1: Understand the Complexes and Their Formulas
We have three complexes:
1. $[Co(NH_3)_6]Cl_3$
2. $[Co(NH_3)_5Cl]Cl_2$
3. $[Co(NH_3)_4Cl_2]Cl$
Each complex has a different arrangement of chloride ions ($Cl^-$) inside and outside the coordination sphere. Only the chloride ions present outside the coordination sphere will be free to react with $AgNO_3$ to give $AgCl$ precipitate.
Step 2: Identify Which Chloride Ions are Outside the Coordination Sphere
• In $[Co(NH_3)_6]Cl_3$, all three chlorides are outside the coordination sphere (since none of the ligands inside the bracket is $Cl$). Hence, all 3 chlorides are free to react with $AgNO_3$.
• In $[Co(NH_3)_5Cl]Cl_2$, one chloride ligand is inside the coordination sphere, and two chlorides are outside. Hence, only 2 chlorides are free to react with $AgNO_3$.
• In $[Co(NH_3)_4Cl_2]Cl$, two chlorides are inside the coordination sphere, and one chloride is outside. Hence, only 1 chloride is free to react with $AgNO_3$.
Step 3: Determine the Stoichiometry of $AgCl$ Formed
When $AgNO_3$ is in excess, the number of moles of $AgCl$ formed directly corresponds to the number of chloride ions outside the coordination sphere:
1. $[Co(NH_3)_6]Cl_3$ + 3 $AgNO_3$ → 3 $AgCl$↓ + $[Co(NH_3)_6](NO_3)_3$
2. $[Co(NH_3)_5Cl]Cl_2$ + 2 $AgNO_3$ → 2 $AgCl$↓ + $[Co(NH_3)_5Cl](NO_3)_2$
3. $[Co(NH_3)_4Cl_2]Cl$ + $AgNO_3$ → $AgCl$↓ + $[Co(NH_3)_4Cl_2](NO_3)$
Step 4: Conclude the Order of AgCl Precipitate
From the above reactions, the correct order of the stoichiometries of $AgCl$ formed (for CoCl3.6NH3, CoCl3.5NH3, and CoCl3.4NH3 respectively) is
3AgCl, 2AgCl, 1AgCl.
Step 5: Final Answer
Hence, the correct option that represents the stoichiometries of $AgCl$ formed is
3AgCl, 2AgCl, 1AgCl.
