Answer:
Iodine (I) has a lower Ionization Energy.
Step-by-step explanation:
Our Periodic Trend for Ionization Energy is up and to the right.
We know that Fluorine and Iodine are both in Group 17. However, Fluorine is in Period 2 and Iodine is in Period 5.
According the Coulomb's Law, the greater the distance between the electrons and the nucleus, the less Forces of Attraction (FOA) there are. Since Iodine has a lot more electrons than Fluorine, and those electrons are spread out farther from the nucleus, they will have less FOA acting on them. Therefore, it will require less energy to remove those electrons from Iodine.
Since Iodine has so many electrons, it will also have more core electrons than Fluorine. This will create the Shielding Effect, where the core electrons "shield" the nucleus' FOA. With the reduced FOA, the outer electrons aren't held as closely to the nucleus as the core electrons. Since higher sublevels also means more energy, those outer electrons contain a lot of energy. All electrons want to have the lowest energy level possible, and since the FOA is reduced, less energy is needed to remove the electrons of Iodine rather than Fluorine.