Final answer:
Among Fe+, Fe2+, Fe3+, and Fe4+, the atom with the smallest ionization energy is Fe+, since the ionization energy increases as the ion's positive charge increases, with the remaining electrons held more strongly due to an increased effective nuclear charge.
Step-by-step explanation:
In each of the following sets, which atom of iron has the smallest ionization energy? Among Fe+, Fe2+, Fe3+, and Fe4+, the atom with the smallest ionization energy is Fe+. Ionization energy increases as the positive charge on the ion increases, because the remaining electrons are held more strongly by the fewer number of protons. Therefore, Fe+, which has lost one electron, has the lowest ionization energy compared to Fe2+, which has lost two electrons, Fe3+, which has lost three electrons, and Fe4+, which has lost four electrons.
The electron configuration of iron (Fe) is [Ar] 4s² 3d⁶. When iron loses an electron to become Fe+, it loses it from the 4s orbital, making the electron configuration [Ar] 3d⁶. As more electrons are removed (leading to Fe2+, Fe3+, and Fe4+), the ionization energy increases because the electrons are increasingly more tightly bound due to less electron-electron repulsion and increased effective nuclear charge. Therefore, it requires more energy to remove an electron from Fe2+, even more from Fe3+, and the highest energy to remove an electron from Fe4+.