Step 1: Freezing Point Depression FormulaThe freezing point depression is directly proportional to the number of dissociated particles in the solution. The formula is:
$$ \Delta T_f = K_f \cdot m \cdot i $$
where:
- $ \Delta T_f $ represents the freezing point depression,
- $ K_f $ is the cryoscopic constant (specific to the solvent),
- $ m $ is the molality of the solution,
- $ i $ is the van 't Hoff factor, indicating the number of particles formed upon dissociation.
As freezing point depression is inversely related to the number of dissociated particles, the solution with the smallest $ i $ (fewer particles) will exhibit the highest freezing point.
Step 2: Determine the $ i $ for Each CompoundWe will now determine the van 't Hoff factor $ i $ for each compound:
- $ [\text{Co}(\text{H}_2\text{O})_6]\text{Cl}_3 $ dissociates into 4 particles: $ [\text{Co}(\text{H}_2\text{O})_6]^{3+} $ and 3 $ \text{Cl}^- $. Thus, $ i = 4 $.
- $ [\text{Co}(\text{H}_2\text{O})_5\text{Cl}]\text{Cl}_2 \cdot \text{H}_2\text{O} $ dissociates into 3 particles: $ [\text{Co}(\text{H}_2\text{O})_5\text{Cl}]^{2+} $ and 2 $ \text{Cl}^- $. Thus, $ i = 3 $.
- $ [\text{Co}(\text{H}_2\text{O})_4\text{Cl}_2]\text{Cl} \cdot 2\text{H}_2\text{O} $ dissociates into 2 particles: $ [\text{Co}(\text{H}_2\text{O})_4\text{Cl}_2]^{+} $ and $ \text{Cl}^- $. Thus, $ i = 2 $.
- $ [\text{Co}(\text{H}_2\text{O})_3\text{Cl}_3] \cdot 3\text{H}_2\text{O} $ does not dissociate, remaining as one particle: $ i = 1 $.
Step 3: ConclusionSince freezing point depression is inversely proportional to $ i $, the compound with the lowest $ i $ will have the highest freezing point. In this scenario, $ [\text{Co}(\text{H}_2\text{O})_3\text{Cl}_3] \cdot 3\text{H}_2\text{O} $, possessing $ i = 1 $, will exhibit the highest freezing point.
Final Answer:
$$ \boxed{[\text{Co}(\text{H}_2\text{O})_3\text{Cl}_3]\cdot3\text{H}_2\text{O}} $$