Concept:Lattice energy is inversely proportional to the ionic radius (size) of the ions involved, assuming the charges are identical.
Formula:$$ \text{Lattice Energy} \propto \frac{1}{r_{\text{cation}} + r_{\text{anion}}} $$
Solution:- In all options, the cation is kept constant as Sodium (\( \text{Na}^+ \)) with a +1 charge.
- The anions belong to the halogen group, all with a -1 charge. Therefore, we must compare the sizes of the halide anions: \( \text{F}^- \), \( \text{Cl}^- \), \( \text{Br}^- \), \( \text{I}^- \).
- Down the group, ionic size increases: \( \text{F}^- < \text{Cl}^- < \text{Br}^- < \text{I}^- \).
- Because the Fluoride ion (\( \text{F}^- \)) has the smallest ionic radius, it allows the oppositely charged ions to pack the closest together.
- This shorter internuclear distance maximizes electrostatic attraction, giving \( \text{NaF} \) the greatest lattice energy.
Why other options are incorrect:- A, B, C: As the size of the anion increases (Cl, Br, I), the internuclear distance increases, weakening the electrostatic bond and lowering the lattice energy.
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