Question

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

Short answer

The correct order of ionization energies for the given species, arranged from lowest to highest, is: \(\mathrm{O}_{2}<\mathrm{O}_{2}^{-}<\mathrm{O}<\mathrm{O}_{2}^{+}\). This order is determined by analyzing the molecular orbital diagrams and considering the stability and electron configurations for each species. The ionization energy is highest for O2+ because removing an electron from it would decrease the bond order, making the process less energetically favorable.

Step-by-step solution

Molecular Orbital Diagrams for O, O2, O2-, O2+

The molecular orbital diagrams can be created by following the aufbau principle, which states that electrons fill molecular orbitals in order of increasing energy. Electron configurations for species: 1. O: \(1s^2 2s^2 2p^4\) 2. O2: \(1s^2 2s^2 2p^4 1s^2 2s^2 2p^4\) => \(\sigma_{1s}^2 \, \sigma_{1s*}^2 \, \sigma_{2s}^2 \, \sigma_{2s*}^2 \, \sigma_{2p}^2 \, \pi_{2p}^4 \, \pi_{2p*}^2\) 3. O2- (gained an extra electron): \(\sigma_{1s}^2 \, \sigma_{1s*}^2 \, \sigma_{2s}^2 \, \sigma_{2s*}^2 \, \sigma_{2p}^2 \, \pi_{2p}^4 \, \pi_{2p*}^3\) 4. O2+ (lost an electron): \(\sigma_{1s}^2 \, \sigma_{1s*}^2 \, \sigma_{2s}^2 \, \sigma_{2s*}^2 \, \sigma_{2p}^2 \, \pi_{2p}^4\, \pi_{2p*}^1\)

Arrange in Order of Ionization Energy

Now, let's use these molecular orbital diagrams to rank the species from lower to higher ionization energy: 1. O2: Removing an electron from the antibonding orbital \(\pi_{2p*}\) would require less energy since it isn't strongly bound, and removing an electron would increase the bond order as well (from 2 to 2.5). 2. O2-: The next species is O2- because as it gained an electron, its last electron goes into the antibonding orbital, reducing the bond order compared to O2 but still requiring more energy to remove as compared to O2. 3. O: Ionization energy of the elemental oxygen is higher than O2 and O2-, as oxygen has half-filled \(p\) orbitals, which makes it more stable and difficult to remove electrons. 4. O2+: The ionization energy of O2+ is the highest, as removing an electron from its antibonding orbital would decrease the bond order (from 1.5 to 1), making this process less favorable energetically. So the correct order of ionization energies is: \[\mathrm{O}_{2}<\mathrm{O}_{2}^{-}<\mathrm{O}<\mathrm{O}_{2}^{+}\]