Use the MO model to determine which of the following has the smallest ionization energy: $\mathrm{N}_{2}, \mathrm{O}_{2}, \mathrm{N}_{2}^{2-}, \mathrm{N}_{2}^{-}, \mathrm{O}_{2}^{+} .$ Explain your answer.

Using MO theory, we first determine the bond order and electron configuration of each molecule/ion to understand its electronic structure. Bond order can be determined with this formula: Bond Order = \(\frac{1}{2} \times (\mathrm{Number\,of\,electrons\,in\,bonding\,orbitals} - \mathrm{Number\,of\,electrons\,in\,antibonding\,orbitals})\) We will list the occupancy of bonding orbitals (BO) and antibonding orbitals (ABO) for \(\sigma\), \(\sigma^{*}\), \(\pi\) and \(\pi^{*}\). For \(\mathrm{N}_2\) (14 electrons): BO: \(\sigma_1\) (2), \(\sigma_2\) (2), \(\pi_3\) (4) ABO: \(\sigma^{*}_1\) (0), \(\sigma^{*}_2\) (0), \(\pi^{*}_3\) (0) For \(\mathrm{O}_2\) (16 electrons): BO: \(\sigma_1\) (2), \(\sigma_2\) (2), \(\pi_3\) (4), \(\sigma_4\) (2) ABO: \(\sigma^{*}_1\) (0), \(\sigma^{*}_2\) (0), \(\pi^{*}_3\) (2), \(\sigma^{*}_4\) (0) For \(\mathrm{N}_2^{2-}\) (14 + 2 = 16 electrons): BO: \(\sigma_1\) (2), \(\sigma_2\) (2), \(\pi_3\) (4), \(\sigma_4\) (2) ABO: \(\sigma^{*}_1\) (0), \(\sigma^{*}_2\) (0), \(\pi^{*}_3\) (2), \(\sigma^{*}_4\) (0) For \(\mathrm{N}_2^{-}\) (14 + 1 = 15 electrons): BO: \(\sigma_1\) (2), \(\sigma_2\) (2), \(\pi_3\) (4), \(\sigma_4\) (1) ABO: \(\sigma^{*}_1\) (0), \(\sigma^{*}_2\) (0), \(\pi^{*}_3\) (2), \(\sigma^{*}_4\) (0) For \(\mathrm{O}_2^{+}\) (16 - 1 = 15 electrons): BO: \(\sigma_1\) (2), \(\sigma_2\) (2), \(\pi_3\) (4), \(\sigma_4\) (1) ABO: \(\sigma^{*}_1\) (0), \(\sigma^{*}_2\) (0), \(\pi^{*}_3\) (2), \(\sigma^{*}_4\) (0)

Now, we will identify the highest occupied molecular orbitals (HOMO) for each molecule/ion. HOMO for \(\mathrm{N}_2\): \(\pi_3\) HOMO for \(\mathrm{O}_2\): \(\sigma_4\) HOMO for \(\mathrm{N}_2^{2-}\): \(\sigma_4\) HOMO for \(\mathrm{N}_2^{-}\): \(\sigma_4\) HOMO for \(\mathrm{O}_2^{+}\): \(\sigma_4\)

Now that we have identified the HOMO for each molecule/ion, we need to compare their energy levels. A molecule/ion with the highest HOMO energy level will have the smallest ionization energy. Based on the HOMO energy levels mentioned above, we can deduce the following ordering of HOMO energy levels: \(\pi_3 < \sigma_4\) We can now conclude that \(\mathrm{N}_2\) has the highest HOMO energy level. As a result, \(\mathrm{N}_2\) has the smallest ionization energy among the given molecules and ions.