Concept:Lattice energy is governed by Coulomb's law. It increases as the charge on the ions increases and as the size of the ions decreases.
Formula:$$ E \propto \frac{q^+ q^-}{r^+ + r^-} $$
Solution:- All the options listed (\( \text{NaF} \), \( \text{LiCl} \), \( \text{NaI} \), \( \text{KI} \)) consist of singly charged ions (+1 and -1). Thus, charges are identical.
- We must evaluate the ionic radii. The smallest cation and smallest anion will yield the shortest internuclear distance and the highest lattice energy.
- Comparing \( \text{NaF} \) and \( \text{LiCl} \): Fluoride (\( \text{F}^- \)) is significantly smaller than Chloride (\( \text{Cl}^- \)). While Lithium (\( \text{Li}^+ \)) is slightly smaller than Sodium (\( \text{Na}^+ \)), the extremely small size of the fluoride ion dominates, giving \( \text{NaF} \) an exceptionally high charge density.
- Therefore, \( \text{NaF} \) has the highest lattice energy among the given choices.
Why other options are incorrect:- B, C, D: These compounds contain larger anions (\( \text{Cl}^- \), \( \text{I}^- \)) or larger cations (\( \text{K}^+ \)), which increases the distance between the ions and lowers the lattice energy.
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