Using the molecular orbital model, write electron configurations for the following diatomic species and calculate the bond orders. Which ones are paramagnetic? a. \(\mathrm{Li}_{2}\) b. \(\mathrm{C}_{2}\) c. \(\mathrm{S}_{2}\)

Based on the periodic table, we have the atomic number for Li, C, and S as 3, 6, and 16 respectively. Hence, the number of electrons in their diatomic species will be 6, 12, and 32 respectively. a. \(\mathrm{Li}_{2}\) has 6 electrons: Electron configuration: \(\mathrm{1}\sigma_{g}^{2}\) \(\mathrm{1}\sigma_{u}^{2}\) \(\mathrm{2}\sigma_{g}^{2}\) b. \(\mathrm{C}_{2}\) has 12 electrons: Electron configuration: \(\mathrm{1}\sigma_{g}^{2}\) \(\mathrm{1}\sigma_{u}^{2}\) \(\mathrm{2}\sigma_{g}^{2}\) \(\mathrm{2}\sigma_{u}^{2}\) \(\mathrm{1}\pi_{u}^{4}\) \(\mathrm{1}\pi_{g}^{2}\) c. \(\mathrm{S}_{2}\) has 32 electrons: Electron configuration: \(\mathrm{1}\sigma_{g}^{2}\) \(\mathrm{1}\sigma_{u}^{2}\) \(\mathrm{2}\sigma_{g}^{2}\) \(\mathrm{2}\sigma_{u}^{2}\) \(\mathrm{1}\pi_{u}^{4}\) \(\mathrm{1}\pi_{g}^{4}\) \(\mathrm{3}\sigma_{g}^{2}\) \(\mathrm{1}\pi_{u}^{4}\) \(\mathrm{3}\sigma_{u}^{2}\) \(\mathrm{1}\pi_{g}^{4}\) \(\mathrm{2}\pi_{u}^{2}\)

The bond order is given by the following formula: Bond order = (number of electrons in bonding MOs - number of electrons in antibonding MOs)/2 a. Bond order of \(\mathrm{Li}_{2}\): Bond order = (4 - 2) / 2 = 1 b. Bond order of \(\mathrm{C}_{2}\): Bond order = (8 - 4) / 2 = 2 c. Bond order of \(\mathrm{S}_{2}\): Bond order = (16 - 16) / 2 = 0

A species is paramagnetic if it has unpaired electrons in its molecular orbitals. a. \(\mathrm{Li}_{2}\): No unpaired electrons in its molecular orbitals (all paired). Therefore, it is diamagnetic. b. \(\mathrm{C}_{2}\): Two unpaired electrons in the \(\pi_{*}\) molecular orbitals. Therefore, it is paramagnetic. c. \(\mathrm{S}_{2}\): No unpaired electrons in its molecular orbitals (all paired). Therefore, it is diamagnetic. In conclusion, the bond orders for \(\mathrm{Li}_{2}\), \(\mathrm{C}_{2}\), and \(\mathrm{S}_{2}\) are 1, 2, and 0, respectively. Out of the given diatomic species, only \(\mathrm{C}_{2}\) is paramagnetic.