In this example problem, we apply molecular orbital theory to various diatomic species which have elements in the second row of the periodic table. The problem reads, "Which of the following is or are true? Select all that apply. Use the molecular orbital diagrams given." A. C2 2+ is diamagnetic. B. N2 1- has a bond order of 2.5. C. F2 2- should be stable enough to exist. D. B2 1- should have a stronger and shorter bond than B2. Let's discuss each of these using molecular orbital theory. To determine whether C2 2+ is diamagnetic or paramagnetic, we need to fill in the molecular orbital diagram for C2 2+. The diagram on the left shows the order of filling for B2, C2, and N2 and their charged species. The diagram on the right shows the order of filling for O2, F2, and NE2 and their charged species. So we'll fill in the molecular orbital diagram on the left. After we fill in the diagram, if it shows all electrons are paired, then C2 2+ should be diamagnetic. If there are any unpaired electrons, then C2 2+ should be paramagnetic. To fill in the molecular orbital diagram, we'll first need to determine how many electrons to add. The molecular orbital diagram shown only includes the second shell, so we're focusing on the valence electrons in the second shell. Each carbon has four valence electrons in the second shell based on its position in the periodic table, and we have two carbons. There is a two plus charge, so we need to remove two of these valence electrons for a total of six valence electrons. We add these six valence electrons to the molecular orbital diagram starting with the lowest energy orbital, which is the sigma 2S bonding molecular orbital. Then we add two electrons to the sigma 2s antibonding molecular orbital. There are two pi 2p bonding molecular orbitals with the same energy, so we fill them singly first with the same spin, similar to filling degenerate orbitals in electron configurations using Hund's rule. Since we have unpaired electrons in the molecular orbital diagram, C2 2+ should be paramagnetic: our first statement is false. To calculate bond order, we first need to fill in the molecular orbital diagram. Let's calculate the number of valence electrons. Each nitrogen has five electrons in the second shell according to its position in the periodic table. The one minus charge adds one electron, so we have a total of 11 valence electrons. We fill in the molecular orbital diagram starting with the lowest energy molecular orbital and working our way up. Two electrons should first be put in the sigma 2S bonding molecular orbital. Then two electrons in the sigma 2S antibonding molecular orbital. Then four electrons to fill up the two pi 2P bonding molecular orbitals. Then two electrons to fill in the sigma 2P bonding molecular orbital. And finally, one electron in the pi 2P antibonding molecular orbitals. To calculate bond order for a diatomic molecule or ion, we take the number of electrons in bonding molecular orbitals, subtract the number of electrons in antibonding molecular orbitals, and divide by 2. Our bonding molecular orbitals do not have an asterisk. They are the sigma 2S, pi 2P, and sigma 2P bonding molecular orbitals circled in blue. We have a total of 8 electrons in these orbitals. Our antibonding molecular orbitals do have an asterisk; they are circled in red. We have a total of three electrons in these orbitals. So 8 minus 3 is 5, divided by 2 is 2.5. The second statement is true: the bond order of N2 1- is 2.5. To determine whether a species is stable enough to exist, it should have a bond order greater than 0. If it has a bond order of 0, then it is not stable enough to exist. To determine bond order, we need the total number of valence electrons first. Each fluorine has 7 electrons in the second shell. We need to add 2 more electrons given the two minus charge, for a total of 16 valence electrons. We'll fill in the molecular orbital diagram on the right since we have a charged species of F2. When we fill in the orbital diagram, we start at the bottom and work our way up. 16 electrons fill all the molecular orbitals in the diagram. To calculate bond order for a diatomic molecule or ion, we take the number of electrons in bonding molecular orbitals, subtract the number of electrons in antibonding molecular orbitals, and divide by 2. The bonding molecular orbitals do not have an asterisk; they're circled in blue. We have 8 electrons in these bonding molecular orbitals. The antibonding molecular orbitals do have an asterisk; they're circled in red. We also have 8 electrons in the antibonding molecular orbitals. Since our bond order is 0, F2 2- should not be stable enough to exist. We have the same number of electrons in bonding molecular orbitals as antibonding molecular orbitals, so creating F2 2- would not lower the overall energy of our system. To compare bond strength or length between B2 1- and B2, we can compare their bond orders. Let's start by creating the molecular orbital diagram for B2 and then adjust it for B2 1-. For total valence electrons, each boron has 3 electrons in the second shell. B2 isn't charged, so we don't need to make an adjustment. That means we have 6 valence electrons. We fill in the sigma 2S bonding molecular orbital first. Then the sigma 2S antibonding molecular orbital. Since there are two pi 2P bonding molecular orbitals, we fill them singly first with the same spin before doubling up. To calculate bond order for a diatomic molecule or ion, we take the number of electrons in bonding molecular orbitals, subtract the number of electrons in antibonding molecular orbitals, and divide by 2. The bonding molecular orbitals do not have an asterisk and are circled in blue; we have 4 electrons in the bonding molecular orbitals. The antibonding molecular orbitals do have an asterisk and are circled in red. We have 2 electrons in antibonding molecular orbitals, so the bond order of B2 is 1. If we want to make adjustments to this molecular orbital diagram to make B2 1-, we need to add one more electron to take into account the 1- charge. This electron goes into the pi 2P bonding molecular orbital, so we should expect that the bond order will increase and the bond will get stronger and shorter. To calculate bond order, we now have 5 electrons in bonding molecular orbitals and we still have 2 electrons in antibonding molecular orbitals. So the bond order is now 1.5. Since B2 1- has a larger bond order than B2, it should have a stronger and shorter bond. Our fourth statement is true.