Arrange the following from lowest to highest ionization energy: \(\mathrm{O}, \mathrm{O}_{2}, \mathrm{O}_{2}^{-}, \mathrm{O}_{2}^{+} .\) Explain your answer.

We will start by examining the stability of each species, which depends on factors such as bond strength and molecular orbital configurations. - O: For a single oxygen atom, there are 6 electrons in the valence shell and we know that half-filled or completely filled shells tend to be stable. In this case, O is not at its most stable state. - O2: Diatomic oxygen forms a double bond between the two O atoms, leading to a relatively high bond strength and a stable molecular configuration. - O2- (Oxygen anion): Adding an electron to O2 results in the anion O2-. The additional electron occupies an antibonding molecular orbital, reducing the bond strength between the two O atoms; though, O2- is still more stable than a single O atom. - O2+ (Oxygen cation): Removing an electron from O2 results in the cation O2+. This removal of one electron increases the bond strength between the O atoms, making it even more stable compared to O2 and O2-.

Based on the stability analysis, we can arrange the given species in ascending order of ionization energy as follows: - O (least stable) ⟶ O2- ⟶ O2 ⟶ O2+ (most stable) Since species with higher stability have higher ionization energies, the order from lowest to highest ionization energy is: \(\mathrm{O} < \mathrm{O}_{2}^{-} < \mathrm{O}_{2} < \mathrm{O}_{2}^{+}\)