For 1 molal aqueous solution of the following compounds, which one will show the highest freezing point?

First, recall the colligative‐property relation for depression of the freezing point

$$\Delta T_f = i\,K_f\,m$$

where $$\Delta T_f$$ is the lowering of the freezing point, $$K_f$$ is the cryoscopic constant of water, $$m$$ is the molality, and $$i$$ is the van’t Hoff factor (the total number of solute particles produced per formula unit on complete dissociation).

In this problem every solution is 1 molal, and the solvent (water) is the same. Hence $$K_f$$ and $$m$$ are identical for all the options. The only variable is $$i$$. A larger $$i$$ gives a larger $$\Delta T_f$$, which means a bigger lowering of the freezing point and therefore a smaller final freezing temperature. Conversely, the compound with the smallest $$i$$ will have the least depression and thus the highest freezing point.

We must therefore calculate $$i$$ for each compound by writing its dissociation in water. Inside the square brackets of a coordination complex the ions are bound covalently and stay together; only species outside the brackets (counter-ions and waters of crystallisation) can separate in solution.

Option A   [Co(H$$_2$$O)$$_3$$Cl$$_3$$].3H$$_2$$O

The entire unit inside brackets is electrically neutral, and there are no counter-ions. Waters of crystallisation merely mix with the solvent and do not create additional solute particles. Therefore

$$[{\rm Co(H_2O)_3Cl_3}] \; \longrightarrow \; [{\rm Co(H_2O)_3Cl_3}]$$

Number of particles = 1, so $$i_A = 1$$.

Option B   [Co(H$$_2$$O)$$_6$$]Cl$$_3$$

The complex has charge +3; three Cl$$^-$$ ions are outside the bracket. Dissociation:

$$[{\rm Co(H_2O)_6}]^{3+} + 3\,{\rm Cl}^-$$

Particles = 1 + 3 = 4, so $$i_B = 4$$.

Option C   [Co(H$$_2$$O)$$_5$$Cl]Cl$$_2$$.H$$_2$$O

The complex has charge +2; two Cl$$^-$$ ions are outside. Dissociation:

$$[{\rm Co(H_2O)_5Cl}]^{2+} + 2\,{\rm Cl}^-$$

Particles = 1 + 2 = 3, so $$i_C = 3$$.

Option D   [Co(H$$_2$$O)$$_4$$Cl$$_2$$]Cl.2H$$_2$$O

The complex has charge +1; one Cl$$^-$$ ion is outside. Dissociation:

$$[{\rm Co(H_2O)_4Cl_2}]^{+} + {\rm Cl}^-$$

Particles = 1 + 1 = 2, so $$i_D = 2$$.

Now we compare:

$$i_A = 1 \lt i_D = 2 \lt i_C = 3 \lt i_B = 4$$

Since $$i_A$$ is the smallest, the freezing-point depression $$\Delta T_f$$ is the least for Option A, giving it the highest actual freezing point among the four solutions.

Hence, the correct answer is Option A.