Animated Solution for Chemistry - Solutions: For 1 molal aqueous solution of the following compounds, which one will show the highest freezing point?

This is a brilliant problem that tests your understanding of two distinct chapters: Coordination Compounds and Solutions (Colligative Properties). Let's break down the logic step-by-step to see how the structure of a complex directly dictates its physical properties.

When a non-volatile solute is added to a pure solvent, it disrupts the solvent's ability to form a solid lattice, causing the freezing point to drop. This phenomenon is known as Freezing Point Depression, and it is mathematically expressed as:

Here, $i$ is the van't Hoff factor (the number of particles the solute breaks into), $K_f$ is the molal depression constant of the solvent, and $m$ is the molality of the solution.

The actual freezing point of the solution ($T_f$) is calculated by subtracting this depression from the freezing point of the pure solvent ($T_f^{\circ }$):

From this equation, a crucial insight emerges: To get the highest freezing point ($T_f$), we must have the smallest possible depression ($\Delta T_f$). Since the molality ($m=1$) and the solvent (water, hence constant $K_f$) are identical for all options, the depression depends entirely on the van't Hoff factor $i$. Therefore, we are looking for the compound with the minimum $i$.

According to Werner's Coordination Theory, only the ions present outside the square brackets (the ionization sphere) will dissociate in an aqueous solution. The species inside the square brackets (the coordination sphere) are tightly bound to the central metal atom and do not ionize.

Let's analyze the dissociation of each option:

Option (a): $[Co(H_{2}O{)}_6]C{l}_3$ This complex has three chloride ions in the ionization sphere. It dissociates as: Total particles = 1 complex ion + 3 chloride ions = 4. So, $i=4$.

Option (b): $[Co(H_{2}O{)}_{5}Cl]C{l}_2\cdot H_{2}O$ Here, two chloride ions are outside the bracket. The water of hydration simply mixes with the solvent and does not act as an ion. Total particles = 1 complex ion + 2 chloride ions = 3. So, $i=3$.

Option (c): $[Co(H_{2}O{)}_{4}C{l}_2]Cl\cdot 2H_{2}O$ Only one chloride ion is outside the bracket. Total particles = 1 complex ion + 1 chloride ion = 2. So, $i=2$.

Option (d): $[Co(H_{2}O{)}_{3}C{l}_3]\cdot 3H_{2}O$ All three chloride ligands are trapped inside the coordination sphere. This means the complex is entirely neutral and will not dissociate into ions at all. Total particles = 1 intact neutral molecule. So, $i=1$.

Comparing the van't Hoff factors, Option (d) has the lowest value ($i=1$). Consequently, it will cause the least depression in the freezing point, resulting in the highest freezing point among the four solutions.