Which of the following best helps to explain why the value of ΔH° for the dissolving of CaF2 in wate — Thermodynamics Chemistry Question

STEPS:

1. Understand the thermodynamics of the dissolution process: Dissolving an ionic solid in water can be modeled as three thermodynamic steps:
* Solute separation: Breaking the electrostatic attractions holding the ions in the crystal lattice. This step is always endothermic ($\Delta H > 0$) because it requires energy to overcome the ionic bonds.
* Solvent separation: Breaking the hydrogen bonds between water molecules to make room for the solute ions. This is also endothermic ($\Delta H > 0$).
* Solute-solvent attraction (hydration): Forming new ion-dipole interactions between the separated ions and the polar water molecules. This step is always exothermic ($\Delta H < 0$) because forming attractions stabilizes the system and releases energy.
2. Analyze the overall sign of $\Delta H^\circ_{\text{solution}}$: The net enthalpy change ($\Delta H^\circ_{\text{solution}}$) is the sum of these steps:
$$\Delta H^\circ_{\text{solution}} = \Delta H_{\text{solute separation}} + \Delta H_{\text{solvent separation}} + \Delta H_{\text{hydration}}$$
If $\Delta H^\circ_{\text{solution}}$ is positive (endothermic), the energy required to break the existing attractions (solute-solute and solvent-solvent) must be greater than the energy released when forming the new ion-dipole attractions.
3. Relate the positive $\Delta H^\circ$ to calcium fluoride's structure: Calcium fluoride ($\text{CaF}_2$) contains $\text{Ca}^{2+}$ cations with a $+2$ charge and $\text{F}^-$ anions with a $-1$ charge. According to Coulomb's Law, the high charge density of $\text{Ca}^{2+}$ and the small ionic radius of $\text{F}^-$ produce exceptionally strong ion-ion electrostatic attractions within the crystal lattice.
4. Select the correct explanation: Because these ion-ion interactions in the crystal lattice are so strong, they require a very large input of energy to break ($\Delta H_{\text{solute separation}} \gg 0$). This massive endothermic step dominates the hydration energy, resulting in a positive net enthalpy change. This makes Option C the correct answer.

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WHY_OTHERS_WRONG:

• A is incorrect: While $\text{CaF}_2(s)$ is indeed relatively insoluble in water, "insolubility" is a macroscopic equilibrium observation (governed by $\Delta G^\circ > 0$), not a molecular-level explanation for why the specific enthalpy term ($\Delta H^\circ$) is positive. Furthermore, some highly soluble salts also dissolve endothermically.
• B is incorrect: As an ionic compound, $\text{CaF}_2$ dissociates into individual aqueous ions ($\text{Ca}^{2+}(aq)$ and $\text{F}^-(aq)$) rather than dissolving as intact, neutral molecular "$\text{CaF}_2(aq)$" particles.
• D is incorrect: Strong ion-dipole interactions between the $\text{Ca}^{2+}$ ions and water are exothermic. If the ion-dipole interactions were the dominant factor in determining the overall enthalpy sign, they would release a large amount of energy and make $\Delta H^\circ$ *negative* (exothermic), rather than positive.