Final answer:
The main group atom with the lowest second ionization energy would typically be a Group 13 atom. This is because after the first electron is removed from a p-orbital in Group 13 elements, it is generally easier to remove a second electron compared to Group 2 elements where electrons are removed from lower energy s-orbitals. Therefore, correct option is c.
Step-by-step explanation:
The main group atom expected to have the lowest second ionization energy is Group 1. Ionization energy generally increases across a period and decreases down a group due to electron shielding and atomic size.
Group 1 atoms, or alkali metals, have a single electron in their outermost shell. The second ionization energy refers to the energy needed to remove a second electron.
For Group 1 elements, after the first electron is removed, they attain a noble gas configuration, which is very stable. Therefore, removing a second electron requires significantly more energy, making their second ionization energy much higher than that of other groups where electrons are still being removed from a valence shell.
For the comparison among main group atoms, while the second ionization energy for Group 1 is high, it is the first ionization energy that is the lowest across all groups because of the ease of removing the single outermost electron.
Group 13 (boron group) would generally be expected to have a lower second ionization energy than Group 2 (alkaline earth metals) because the first electron from Group 13 is removed from a p-orbital, which is at a higher energy level compared to the s-orbital electrons of Group 2.