To solve this problem, we need to determine the number of chloride ions available for precipitation as AgCl when each complex is treated with excess $AgNO_3. \newline$The complexes given are:$\newline$1. $$CoCl$_3 \cdot 6$NH$_3 \newline$2. $$CoCl$_3 \cdot 5$NH$_3 \newline$3. $$CoCl$_3 \cdot 4$NH$_3 \newline$The formation of AgCl depends on the number of free chloride ions in the solution.$\newline$Let's analyze each complex:$\newline$1. $$CoCl$_3 \cdot 6$NH$_3: \newline$• This complex is known as hexaamminecobalt(III) chloride.$\newline$• All 6 NH$_3$ ligands are coordinated to the cobalt ion, leaving all 3 chloride ions free.$\newline$• Therefore, 3 moles of AgCl will be formed.$\newline$2. $$CoCl$_3 \cdot 5$NH$_3: \newline$• This complex is known as pentaamminechlorocobalt(III) chloride.$\newline$• 5 NH$_3$ ligands are coordinated, and 1 chloride ion is also coordinated, leaving 2 chloride ions free.$\newline$• Therefore, 2 moles of AgCl will be formed.$\newline$3. $$CoCl$_3 \cdot 4$NH$_3: \newline$• This complex is known as tetraamminechlorocobalt(III) chloride.$\newline$• 4 NH$_3$ ligands are coordinated, and 2 chloride ions are also coordinated, leaving 1 chloride ion free.$\newline$• Therefore, 1 mole of AgCl will be formed.$\newline$Thus, the correct order of stoichiometries of AgCl formed is:$\newline 3$ AgCl, $2$ AgCl, $1$ AgCl.$\newline$This corresponds to Option 2.