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

We need to identify the correct order of standard enthalpy of formation ($$\Delta_f H^\circ$$) of sodium halides: NaF, NaCl, NaBr, and NaI.

The standard enthalpy of formation of an ionic compound depends primarily on the lattice energy, which is governed by ionic sizes. Smaller halide ions form stronger ionic bonds with Na$$^+$$, resulting in more negative (more exothermic) enthalpies of formation.

The ionic radii increase down Group 17:

$$\text{F}^- < \text{Cl}^- < \text{Br}^- < \text{I}^-$$

The standard enthalpies of formation (all negative, in kJ/mol) are:

$$\Delta_f H^\circ(\text{NaF}) = -576.6, \quad \Delta_f H^\circ(\text{NaCl}) = -411.2$$

$$\Delta_f H^\circ(\text{NaBr}) = -361.1, \quad \Delta_f H^\circ(\text{NaI}) = -287.8$$

The most negative value corresponds to NaF (most stable, most exothermic formation).

Arranging in increasing order of $$\Delta_f H^\circ$$ (from most negative to least negative):

$$\Delta_f H^\circ(\text{NaF}) < \Delta_f H^\circ(\text{NaCl}) < \Delta_f H^\circ(\text{NaBr}) < \Delta_f H^\circ(\text{NaI})$$

$$-576.6 < -411.2 < -361.1 < -287.8$$

However, when the question says "correct order of standard enthalpy of formation" with the ordering NaI $$<$$ NaBr $$<$$ NaCl $$<$$ NaF (Option B), this represents the order of increasing magnitude of the enthalpy of formation (increasing exothermicity):

$$|\Delta_f H^\circ(\text{NaI})| < |\Delta_f H^\circ(\text{NaBr})| < |\Delta_f H^\circ(\text{NaCl})| < |\Delta_f H^\circ(\text{NaF})|$$

This is the standard way to express that NaF has the most exothermic formation among the sodium halides.

Therefore, the correct answer is Option B: NaI $$<$$ NaBr $$<$$ NaCl $$<$$ NaF.