Principle involved

For dilute solutions, depression in freezing point is given by: $\Delta T_f = iK_f m$ where:

• $i$ = van't Hoff factor
• $K_f$ = cryoscopic constant
• $m$ = molality

Since all solutions are 1 molal aqueous solutions, $K_f$ and $m$ are same for all.

Therefore, the solution with smallest $i$ will have the smallest depression in freezing point, hence the highest freezing point.

Find the number of ions produced by each compound

Water of crystallization does not affect dissociation in solution as free ions; only the coordination sphere and counter ions matter.

Option A: $[Co(H_2O)_3Cl_3]\cdot 3H_2O$

Here, all 3 chloride ions are inside the coordination sphere: $[Co(H_2O)_3Cl_3]$ This is a neutral complex.

So in water, it behaves essentially as: $[Co(H_2O)_3Cl_3] \rightarrow \text{no ions}$

Number of particles: $i = 1$

Option B: $[Co(H_2O)_6]Cl_3$

Here, 3 chloride ions are outside the coordination sphere: $[Co(H_2O)_6]Cl_3 \rightarrow [Co(H_2O)_6]^{3+} + 3Cl^-$

Total ions produced: $1 + 3 = 4$ So, $i = 4$

Option C: $[Co(H_2O)_5Cl]Cl_2\cdot H_2O$

Here, 1 chloride is inside the coordination sphere, 2 are outside: $[Co(H_2O)_5Cl]Cl_2 \rightarrow [Co(H_2O)_5Cl]^{2+} + 2Cl^-$

Total ions produced: $1 + 2 = 3$ So, $i = 3$

Option D: $[Co(H_2O)_4Cl_2]Cl\cdot 2H_2O$

Here, 2 chlorides are inside the coordination sphere, 1 is outside: $[Co(H_2O)_4Cl_2]Cl \rightarrow [Co(H_2O)_4Cl_2]^+ + Cl^-$

Total ions produced: $1 + 1 = 2$ So, $i = 2$

Compare freezing points

Since: $\Delta T_f \propto i$ smaller $i$ means smaller freezing point depression and hence higher freezing point.

The values are:

• A: $i=1$
• B: $i=4$
• C: $i=3$
• D: $i=2$

Therefore, Option A will show the highest freezing point.

Final answer

$\boxed{A}$

Comparison with stored correct answer

Stored correct answer: A

My derived answer: A

Hence, they agree.