Problem 6 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

For 1 molal aqueous solution of the following compounds, which one will show the highest freezing point? (a) \(\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{3}\) (b) \(\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2} \cdot \mathrm{H}_{2} \mathrm{O}\) (c) \(\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{Cl} .2 \mathrm{H}_{2} \mathrm{O}\) (d) \(\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{3} \mathrm{Cl}_{3}\right] .3 \mathrm{H}_{2} \mathrm{O}\)

Step-by-Step Solution

Answer

(d) \( \left[\text{Co}\left(\text{H}_{2} \text{O}\right)_{3} \text{Cl}_{3} \right] .3 \text{H}_{2} \text{O}\) has the highest freezing point.

Step 1: Understanding Freezing Point Depression

The freezing point depression of a solution depends on the number of solute particles in the solution, which is determined by the van 't Hoff factor \(i\). The greater the number of particles, the greater the freezing point depression and thus the lower the freezing point.

Step 2: Identify the van 't Hoff Factor for each Compound

To determine \(i\), identify the dissociation of each compound in water:- (a) \(\left[\text{Co}\left(\text{H}_{2} \text{O}\right)_{6}\right] \text{Cl}_{3} \) dissociates into 4 ions: 1 complex ion \(\left[\text{Co}\left(\text{H}_{2} \text{O}\right)_{6}\right]^{3+}\) and 3 \(\text{Cl}^{-}\).- (b) \(\left[\text{Co}\left(\text{H}_{2} \text{O}\right)_{5} \text{Cl}\right] \text{Cl}_{2} \cdot \text{H}_{2} \text{O}\) dissociates into 3 ions: 1 complex ion \(\left[\text{Co}\left(\text{H}_{2} \text{O}\right)_{5} \text{Cl}\right]^{2+}\) and 2 \(\text{Cl}^{-}\).- (c) \(\left[\text{Co}\left(\text{H}_{2} \text{O}\right)_{4} \text{Cl}_{2}\right] \text{Cl} .2 \text{H}_{2} \text{O}\) dissociates into 2 ions: 1 complex ion \(\left[\text{Co}\left(\text{H}_{2} \text{O}\right)_{4} \text{Cl}_{2}\right]^{+}\) and 1 \(\text{Cl}^{-}\).- (d) \(\left[\text{Co}\left(\text{H}_{2} \text{O}\right)_{3} \text{Cl}_{3}\right] .3 \text{H}_{2} \text{O}\) does not dissociate in water (\(i=1\)).

Step 3: Compare the Dissociation Outcomes

To find which solution has the highest freezing point, we look for the smallest \(i\) value, as this will have the minimal reduction in freezing point. Hence, the compound that dissociates into the fewest particles should have the highest freezing point.

Step 4: Conclusion

Among all the options, compound (d) \(\left[\text{Co}\left(\text{H}_{2} \text{O}\right)_{3} \text{Cl}_{3} \right] .3 \text{H}_{2} \text{O}\) does not dissociate, resulting in \(i=1\), and hence it has the highest freezing point.

Key Concepts

van 't Hoff factor

The van 't Hoff factor, abbreviated as \( i \), is crucial in understanding how solutes affect the freezing point of a solvent. It reflects the number of particles a compound dissociates into when it dissolves in water. This concept is incredibly important in the context of colligative properties, such as freezing point depression.

When a compound dissolves and splits into multiple ions, the van 't Hoff factor increases. The more ions produced, the larger the value of \( i \). This larger number means a greater effect on the freezing point of the solution. For example:

• A compound breaking down into 2 ions has \( i = 2 \).
• A compound breaking down into 4 ions has \( i = 4 \).

The connection is straightforward: a higher van 't Hoff factor indicates a more substantial freezing point depression. This happens because more particles in the solution disrupt the formation of the solid lattice structure required for freezing, thereby lowering the freezing point.

solution chemistry

Solution chemistry involves understanding how substances dissolve and interact in a liquid medium. When discussing freezing point depression, we view how solutes impact the physical properties of solvents.

The freezing point of a pure solvent decreases when a non-volatile solute is added. This is because the dissolved particles interfere with the solvent molecules’ ability to form an organized lattice needed in the solid phase. As a result, a lower temperature is required to freeze the solution, exhibiting freezing point depression.

This phenomenon is a part of the solutions' colligative properties. These properties depend on the presence of solute particles, not their identity. Factors like the type of solvent and the amount of solute impact the solution's behavior, but it's the number of dissolved particles that are paramount in determining effects like freezing point depression and boiling point elevation.

ionic dissociation

Ionic dissociation refers to the process where a compound breaks down into ions when dissolved in a solvent, often water. This is fundamental in predicting how salts and complex compounds will behave in solution chemistry.

Upon dissolution, ionic compounds separate into their constituent cations and anions. For instance, when table salt (NaCl) dissolves, it splits into sodium ions \((Na^+)\) and chloride ions \((Cl^-)\).

The extent of dissociation affects the van 't Hoff factor \( i \), directly impacting the solution’s colligative properties. Fully dissociated compounds yield an \( i \) corresponding to the total number of ions produced, enhancing colligative effects like the freezing point depression. Conversely, compounds that do not dissociate (or dissociate less completely) will have a smaller \( i \), leading to lesser impacts on these properties.

The process of dissociation is essential for understanding many concepts in solution chemistry, including conductivity and the qualitative analysis of ions in a solution.