The superoxide ion, \( \mathrm{O}_{2}^{-} \), is a reactive species that may play a role in the chemistry of aging. Use a molecular orbital diagram to determine the bond order of the superoxide ion. Enter the number (no units).

To determine the bond order of the superoxide ion \((\mathrm{O}_{2}^{-})\), we can use the molecular orbital theory. The molecular orbital configuration for \(\mathrm{O}_{2}\) (neutral oxygen molecule) is \((\sigma_{1s})^2(\sigma_{1s}^*)^2(\sigma_{2s})^2(\sigma_{2s}^*)^2(\sigma_{2p_z})^2(\pi_{2p_x})^2(\pi_{2p_y})^2(\pi_{2p_x}^*)^1\), where \(\sigma\) and \(\pi\) refer to the types of bonds formed by the overlap of atomic orbitals, and the asterisk indicates an antibonding orbital. In \(\mathrm{O}_{2}^{-}\), an additional electron is added to one of the \(\pi^*\) antibonding orbitals, resulting in the configuration \((\pi_{2p_x}^*)^2\). The bond order can be calculated using the formula: \[ \text{Bond Order} = \frac{(\text{Number of bonding electrons}) - (\text{Number of antibonding electrons})}{2} \] For \(\mathrm{O}_{2}\), there are 10 bonding electrons and 2 antibonding electrons. In \(\mathrm{O}_{2}^{-}\), it becomes 10 bonding electrons and 3 antibonding electrons: \[ \text{Bond Order} = \frac{10 - 3}{2} = \frac{7}{2} = 3.5 \] Thus, the bond order of the superoxide ion \((\mathrm{O}_{2}^{-})\) is \(1.5\).