Place the following molecules and ions in order from smallest to largest bond order: $\mathrm{N}_{2}{ }_{2}^{2+}, \mathrm{He}_{2}{ }^{+}, \mathrm{Cl}_{2} \mathrm{H}_{2}^{-}, \mathrm{O}_{2}{ }^{2-}$.

Molecular orbital (MO) theory is a suitable method for determining the bond order. We will use the molecular orbital diagrams for \(1\sigma, 1\sigma^{*}, 2\sigma, 2\sigma^{*}, 1\pi, 1\pi^{*}, 2\pi, \text{ and } 2\pi^{*}\) orbitals.

The bond order equation is given by:\(BO = \frac{(number\ of\ bonding\ electrons - number\ of\ antibonding\ electrons)}{2}\). We should determine the bond order for each of the given species.

To determine the bond order for each species, we have to calculate the electron count according to their electronic configurations. 1. \(\mathrm{N}_{2}{ }_{2}^{2+}\): \(N_2\) has a total of \(10\) electrons, and the \(2+\) means that it loses \(2\) electrons; thus, there are \(8\) electrons. Bond order for this is: \(BO_{N_{2}^{2+}}=\frac{(6-2)}{2}=2\) 2. \(\mathrm{He}_{2}{ }^{+}\): He has \(2\) electrons, and \(He_2\) will have \(4\) electrons. However, the \(+\) means that one electron is lost, leaving \(3\) electrons. So, the bond order for this is: \(BO_{He_{2}^{+}}=\frac{(2-1)}{2}=0.5\) 3. \(\mathrm{Cl}_{2}\): Cl has \(17\) electrons, and \(Cl_2\) will have \(34\) electrons. Bond order for this is: \(BO_{Cl_{2}}=\frac{(20-14)}{2}=3\) 4. \(\mathrm{H}_{2}^{-}\): H has \(1\) electron, and \(H_2\) would have \(2\) electrons. The negative charge implies an extra electron, making it \(3\) electrons. Bond order for this is: \(BO_{H_{2}^{-}}=\frac{(2-1)}{2}=0.5\) 5. \(\mathrm{O}_{2}{ }^{2-}\): \(O\) has \(8\) electrons, and \(O_2\) will have \(16\) electrons. The \(2-\) implies addition of \(2\) electrons, leaving \(18\) electrons. So, the bond order for this is: \(BO_{O_{2}^{2-}}=\frac{(12-6)}{2}=3\)

Now that we have the bond orders, we can arrange the species from lowest to the highest bond order: \[ \mathrm{He}_{2}{ }^{+}, \mathrm{H}_{2}^{-} < \mathrm{N}_{2}{ }_{2}^{2+} < \mathrm{O}_{2}{ }^{2-}, \mathrm{Cl}_{2} \]