## Chemistry and Chemical Reactivity (9th Edition)

The electrostactic attraction force, by Coulomb's Law is $F=-k\frac{q_1q_2}{d^2}$, The distance, in this case, is the sum of the ionic radii: $NaF$: $116\ pm + 119\ pm = 235\ pm$ $NaI$: $116\ pm + 206\ pm = 322\ pm$ Since in both cases one electron is exchanged, in NaF the force of attraction is bigger because the distance between the charges is smaller.