Problem 19 - A Complete Resource Book in Chemistry for JEE Main

Question

Maximum freezing point will be for 1 molal solution of, assuming equal ionization in each case: (a) \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{3}\) (b) \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2} \cdot \mathrm{H}_{2} \mathrm{O}\) (c) \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{Cl} .2 \mathrm{H}_{2} \mathrm{O}\) (d) \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{3} \mathrm{Cl}_{3}\right] .3 \mathrm{H}_{2} \mathrm{O}\)

Step-by-Step Solution

Answer

The maximum freezing point will be for option (d): \([\mathrm{Fe(H_2O)_3\,Cl_3}]\cdot 3\mathrm{H}_2\mathrm{O}\).

Step 1: Understand Freezing Point Depression

The freezing point of a solution is decreased by the presence of solute particles. The relationship is given by the formula: \( \Delta T_f = i \cdot K_f \cdot m \), where \( \Delta T_f \) is the freezing point depression, \( i \) is the van't Hoff factor (number of particles the solute dissociates into), \( K_f \) is the cryoscopic constant, and \( m \) is the molality.

Step 2: Evaluate Ionization of Each Compound

For each complex provided, determine its dissociation in water:- (a) \([\mathrm{Fe(H_2O)_6}]\mathrm{Cl}_3\): This compound dissociates into \([\mathrm{Fe(H_2O)_6}]^{3+}\) and 3 \(\mathrm{Cl}^-\) ions, \(i = 4\).- (b) \([\mathrm{Fe(H_2O)_5Cl}]\mathrm{Cl}_2\cdot \mathrm{H}_2\mathrm{O}\): This molecule breaks into \([\mathrm{Fe(H_2O)_5Cl}]^{2+}\) and 2 \(\mathrm{Cl}^-\) ions, \(i = 3\).- (c) \([\mathrm{Fe(H_2O)_4Cl_2}]\mathrm{Cl} \cdot 2 \mathrm{H}_2\mathrm{O}\): The dissociation is into \([\mathrm{Fe(H_2O)_4Cl_2}]^{+}\) and 1 \(\mathrm{Cl}^-\) ion, \(i = 2\).- (d) \([\mathrm{Fe(H_2O)_3Cl_3}]\cdot 3\mathrm{H}_2\mathrm{O}\): This does not dissociate further in water, \(i = 1\).

Step 3: Determine Maximum Freezing Point

The solution with the smallest depression in freezing point will have the smallest value of \( \Delta T_f \), meaning it has the smallest \(i\)-value. Therefore, the solution remains closest to the original freezing point of the solvent if it dissociates the least.

Step 4: Compare Ionization (Van't Hoff Factor)

Among the compounds, option (d) \([\mathrm{Fe(H_2O)_3\,Cl_3}]\cdot 3\mathrm{H}_2\mathrm{O}\) has the smallest \(i\) value of 1. This compound ionizes the least, leading to less depression of the freezing point compared to the other options.

Step 5: Conclusion

Therefore, the compound that results in the maximum freezing point is option (d) because it shows the least ionization in solution.

Key Concepts

Van't Hoff Factor

The Van't Hoff factor, represented as \( i \), is essentially the measure of the number of particles a compound dissociates into when dissolved in a solution. It plays a crucial role in calculating the freezing point depression, among other colligative properties. The more a compound dissociates, the larger its Van't Hoff factor, which directly impacts the freezing point of the solution.
For example, in the case of

• \([\mathrm{Fe(H_2O)_6}]\mathrm{Cl}_3\): this dissociates into one \([\mathrm{Fe(H_2O)_6}]^{3+}\) ion and three \(\mathrm{Cl}^-\) ions, resulting in \(i = 4\).
• \([\mathrm{Fe(H_2O)_5Cl}]\mathrm{Cl}_2\cdot \mathrm{H}_2\mathrm{O}\): this breaks down into one \([\mathrm{Fe(H_2O)_5Cl}]^{2+}\) ion and two \(\mathrm{Cl}^-\) ions, giving \(i = 3\).
• \([\mathrm{Fe(H_2O)_4Cl_2}]\mathrm{Cl} \cdot 2 \mathrm{H}_2\mathrm{O}\): dissociates into one \([\mathrm{Fe(H_2O)_4Cl_2}]^{+}\) ion and one \(\mathrm{Cl}^-\) ion, thus, \(i = 2\).
• \([\mathrm{Fe(H_2O)_3Cl_3}]\cdot 3\mathrm{H}_2\mathrm{O}\): does not dissociate further in water, so \(i = 1\).

The larger the Van't Hoff factor, the greater the freezing point depression.

Cryoscopic Constant

The cryoscopic constant, \( K_f \), is a proportionality factor that relates the freezing point depression of a solvent to the molality of the solution and the Van't Hoff factor. It is unique for each solvent and provides insight into how sensitive a solvent is to the solute it dissolves.
For instance, water has a cryoscopic constant of approximately 1.86 °C kg/mol, affecting how compounds in aqueous solutions depress the freezing point. The equation to determine freezing point depression \( \Delta T_f \) is given by:\[\Delta T_f = i \cdot K_f \cdot m\]
Here, \( m \) represents the molality of the solution. The product of \( i \) and \( K_f \) offers the potential amount by which the freezing point of the solvent is lowered per molal of the solution.
Understanding \( K_f \) is vital for predicting how much a solution will freeze lower than its pure solvent, directly related to the Van't Hoff factor and molality.

Chemical Ionization

Chemical ionization involves the process of a compound splitting into ions when dissolved in a solvent. This is fundamental to understanding freezing point depression, because the dissociation affects the Van't Hoff factor directly. Compounds like coordination complexes often exhibit different levels of ionization depending on their structural properties.
In the exercise:

• \([\mathrm{Fe(H_2O)_6}]\mathrm{Cl}_3\) ionizes the most, forming four ions, including three \(\mathrm{Cl}^-\) ions.
• \([\mathrm{Fe(H_2O)_3Cl_3}]\) barely ionizes, thus maintaining the maximum freezing point close to that of pure solvent.

The less a compound ionizes, the less impact it has on the freezing point depression, which is directly linked to the Van't Hoff factor. Monitoring ionization is crucial in predicting the behavior of solutions and the compounds' effects on physical properties like freezing point.

Coordination Compounds

Coordination compounds are structures formed by central metal atoms or ions bonded to surrounding molecules or ions called ligands. These have unique properties due to the metal-ligand interactions, impacting their chemical behavior, including ionization and solubility.
Such compounds often hold ions in place, reducing the number of free ions when dissolved.

• In \([\mathrm{Fe(H_2O)_3Cl_3}]\cdot 3\mathrm{H}_2\mathrm{O}\), the chlorine atoms are more tightly held, reducing its dissociation in water.
• In contrast, \([\mathrm{Fe(H_2O)_6}]\mathrm{Cl}_3\) creates multiple free ions, leading to a greater Van't Hoff factor.

Thus, understanding coordination compounds is key in determining how they might affect physical properties like freezing point depression. These compounds illustrate the balance between complex formation, ionization degree, and solution behavior, highlighting their importance in physical chemistry.