In this example problem, we’ll use molecular orbital theory to examine the bond order, bond strength, and magnetic properties of different molecules. The problem reads, “Use molecular orbital theory to determine (a.) the bond order of N2, (b.) whether N2 is paramagnetic or diamagnetic, and (c.) order of increasing bond strength for N2, N2 2-, and N2+.” First we’ll calculate the bond order of N2 by creating a molecular orbital diagram. To create the molecular orbital diagram, we’ll start with the orbital diagram for both nitrogen atoms, focusing on the valence electrons. The 2s atomic orbitals of both of our nitrogen atoms will combine to create two molecular orbitals. These molecular orbitals are the sigma 2s orbitals, bonding and antibonding. The antibonding orbital is higher in energy and the bonding orbital is lower in energy than the 2s atomic orbitals of nitrogen. The antibonding orbital is indicated with the asterisk. Next we’ll combine the three 2p orbitals of both of our nitrogen atoms to create six new molecular orbitals. When the p orbitals combine, they create both sigma and pi molecular orbitals. The correct order of the lowest energy to highest energy orbitals are shown here and are determined computationally. Now we’ll fill our molecular orbitals. We’ll start by using the four electrons in the 2s orbitals of both nitrogen atoms. These electrons will fill the bonding and antibonding sigma 2s molecular orbitals. Next we use the six electrons from the 2p orbitals of both nitrogen atoms. Those six electrons will occupy the lowest available molecular orbitals, which are the pi 2p and sigma 2p molecular orbitals. To calculate bond order, we’ll take the number of electrons in bonding orbitals subtract the number of electrons in antibonding orbitals and divide that quantity by 2. In our N2 molecule we have 8 electrons in bonding orbitals and we have 2 electrons in antibonding orbitals. So 8 – 2 is 6, divided by 2 is equal to 3. For part b, we want to know whether N2 is paramagnetic or diamagnetic. In a diamagnetic molecule, all electrons are paired in orbitals; in a paramagnetic molecule there’s at least one electron that’s not paired. Although each nitrogen atom individually has 3 unpaired electrons in its 2p orbital, in the N2 molecule all of our electrons are paired. So our N2 molecule is diamagnetic. In part c, we want to find the order of increasing bonds strength for N2, N2 2-, and N2+. Bond strength increases with increasing bond order, so we’ll want to determine the bond order of all three. The charge of N2 2- means we have 2 additional electrons to place in our molecular orbitals. This gives us a total of 12 electrons. So we can fill our molecular orbitals with those 12 electrons. Then to calculate the bond order, we’ll take the number of electrons in bonding orbitals subtract the number of electrons in antibonding orbitals divided by 2. We still have 8 electrons in our bonding orbitals, but now we have 4 electrons in our antibonding orbitals. So we have 8 subtract 4 divided by 2, which will equal 2. So our bond order of N2 2- is 2. Now we can calculate the bond order of N2+. The charge of N2+ means that we have one less electron, which means we should have a total of 9 electrons. We can fill our molecular orbitals using these 9 electrons starting with the lowest energy and working our way up. To calculate the bond order of N2+, we have 7 electrons in bonding orbitals subtract our 2 electrons in antibonding orbitals, which is 5, divided by 2 which is 2.5. So the bond order of N2+ is 2.5. Our bond strength increases with increasing bond order, so the order increasing bond strength is N2 2-, then N2+, and then N2.