[VISUAL] The process of dissolution of NaCl(s) in H2O(l) is represented in the diagram above. Which — Solutions Chemistry Question

STEPS:

1. Understand the three-step thermodynamic model for dissolution:
The dissolution of an ionic solute (like $\text{NaCl}$) in a polar solvent (like $\text{H}_2\text{O}$) can be broken down into three distinct steps:
* Step 1: Separating (breaking) solvent-solvent molecular interactions.
* Step 2: Separating (breaking) solute-solute ionic interactions.
* Step 3: Allowing the separated solute particles and solvent molecules to come together and form new solute-solvent interactions (hydration).

2. Analyze the thermodynamic signs for breaking solvent-solvent interactions (Step 1):
* Enthalpy ($\Delta H^\circ$): Hydrogen bonds between water molecules must be broken to create space for the incoming ions. Because breaking intermolecular forces always requires an input of energy (endothermic), $\Delta H^\circ$ is positive ($\mathbf{+}$).
* Entropy ($\Delta S^\circ$): Separating water molecules from their highly ordered, cohesive liquid network increases their spatial freedom and molecular disorder, so $\Delta S^\circ$ is positive ($\mathbf{+}$).

3. Analyze the thermodynamic signs for breaking solute-solute interactions (Step 2):
* Enthalpy ($\Delta H^\circ$): The strong electrostatic ionic bonds holding the $\text{Na}^+$ and $\text{Cl}^-$ ions together in the rigid crystal lattice must be broken (this is the lattice energy). Because disrupting these stable bonds requires an input of energy, $\Delta H^\circ$ is positive (\mathbf{+}\')).
* **Entropy (\(\Delta S^\circ): Dismantling a highly ordered, rigid three-dimensional crystal lattice into individual mobile ions significantly increases the freedom of motion and disorder of the particles, so $\Delta S^\circ$ is positive ($\mathbf{+}$).

4. Analyze the thermodynamic signs for forming solute-solvent interactions (Step 3):
* Enthalpy ($\Delta H^\circ$): Ion-dipole forces are established as water molecules surround the free $\text{Na}^+$ and $\text{Cl}^-$ ions. Because forming new chemical attractions is a stabilizing process that always releases energy (exothermic), $\Delta H^\circ$ is negative ($\mathbf{-}$).
* Entropy ($\Delta S^\circ$): As water molecules organize themselves into structured hydration shells around the charged cations and anions, they orient their molecular dipoles systematically. This restricts the rotational and translational freedom of these water molecules, which increases local order (decreases disorder). Thus, $\Delta S^\circ$ is negative ($\mathbf{-}$).

5. Combine the results to select the correct option:
* Breaking solvent-solvent: $\Delta H^\circ = +$, $\Delta S^\circ = +$
* Breaking solute-solute: $\Delta H^\circ = +$, $\Delta S^\circ = +$
* Forming solute-solvent: $\Delta H^\circ = -$, $\Delta S^\circ = -$
* This matching set corresponds perfectly to Option A.

*

WHY_OTHERS_WRONG:

• Option B is incorrect: This option claims that $\Delta S^\circ$ is positive ($+$) for forming solute-solvent interactions. However, the formation of highly structured hydration shells around $\text{Na}^+$ and $\text{Cl}^-$ restricts the motion of the water molecules, which decreases the entropy (making $\Delta S^\circ$ negative).
• Option C is incorrect: This option gets the signs of enthalpy reversed for all three steps. It claims that breaking interactions is exothermic ($\Delta H^\circ = -$) and forming interactions is endothermic ($\Delta H^\circ = +$). In chemistry, breaking bonds/IMFs always requires energy ($+$), and forming bonds/IMFs always releases energy ($-$).
• Option D is incorrect: Similar to Option C, this option incorrectly states that breaking solvent-solvent and solute-solute interactions is exothermic ($\Delta H^\circ = -$) and forming solute-solvent interactions is endothermic ($\Delta H^\circ = +$).