Identify the correct order of standard enthalpy of formation of sodium halides.

🧠 Smaller anion → stronger lattice energy → more negative formation enthalpy For sodium halides, lattice energy is the main driver of $\Delta H_f$. The cation is fixed ($\ce{Na+}$), so the anion size decides everything. Smaller anion sits closer to $\ce{Na+}$, stronger ionic attraction, more energy released on formation.

🗺️ Order by anion size Anion size: $\ce{F^-} < \ce{Cl^-} < \ce{Br^-} < \ce{I^-}$. Lattice energy: $\ce{NaF} > \ce{NaCl} > \ce{NaBr} > \ce{NaI}$. $\Delta H_f$ (in kJ/mol): $\ce{NaF}$: $-576$ (most negative). $\ce{NaCl}$: $-411$. $\ce{NaBr}$: $-361$. $\ce{NaI}$: $-288$ (least negative).

Increasing order of $\Delta H_f$ (least negative to most negative): $\ce{NaI} < \ce{NaBr} < \ce{NaCl} < \ce{NaF}$.

⚠️ The trap Option (a) flips $\ce{NaF}$ and $\ce{NaCl}$. Students sometimes think "Cl is the standard halide for salts (like table salt)" and put $\ce{NaCl}$ at the top. But $\ce{F^-}$ is smaller than $\ce{Cl^-}$, so $\ce{NaF}$ has the highest lattice energy and the most negative $\Delta H_f$. Smallest anion always wins for the alkali halide series.

$\boxed{\text{Answer: (b)}}$