The depression in freezing point $\Delta T_f = i \cdot K_f \cdot m$, where $i$ = van't Hoff factor (number of particles from ionization).

The compound with the fewest ions in solution gives the smallest $\Delta T_f$ and therefore the highest freezing point.

• $[\text{Co(H}_2\text{O)}_4\text{Cl}_2]\text{Cl}$: ionizes to give 2 particles → $i = 2$
• $[\text{Co(H}_2\text{O)}_3\text{Cl}_3]$: non-electrolyte → $i = 1$ (no ions) — smallest depression
• $[\text{Co(H}_2\text{O)}_6]\text{Cl}_3$: gives 4 particles → $i = 4$
• $[\text{Co(H}_2\text{O)}_5\text{Cl}]\text{Cl}_2$: gives 3 particles → $i = 3$

Highest freezing point → smallest $i$ → Option B ($i=1$).