Example 2: Applying Molecular Orbital Theory

In this example problem, we apply molecular orbital theory to various diatomic species that contain elements from 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. F2 1- has a shorter bond than F2. B. N2 1+ has a bond order of 2.5. C. B2 2- is paramagnetic. D. C2 2+ has a weaker bond than C2. E. B2 2+ should be stable enough to exist. Let's discuss each of these using molecular orbital theory. To compare bond strength between F2 1- and F2, we'll need to compare bond order. As the bond order increases, bonds get shorter and stronger. Think about comparing a single bond, which would correspond to a bond order of 1, to a double bond, a bond order of 2. A double bond is shorter and stronger. To calculate bond order, we need to fill in the molecular orbital diagram. We'll start with F2 and make adjustments for F2 1-. There are 2 fluorine atoms and each has 7 valence electrons in the second shell. F2 has no charge, so there are a total of 14 valence electrons. We fill in the molecular orbital diagram starting with the lowest energy orbitals and working our way up, filling in 14 electrons. 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 anti-bonding molecular orbitals, and divide by 2. The bonding molecular orbitals do not have an asterisk and are circled in blue; we have 8 electrons in bonding molecular orbitals. The anti-bonding molecular orbitals do have an asterisk and are circled in red; we have 6 electrons in anti-bonding molecular orbitals. So our bond order is 1. To adjust this molecular orbital diagram for F2 1-, we need to add 1 more electron for the 1charge. This electron goes into the sigma 2P anti-bonding molecular orbital. To calculate bond order, we still have 8 electrons in bonding molecular orbitals, but now we have 7 electrons in anti-bonding molecular orbitals. So our bond order is 1/2. Adding an extra electron to an anti-bonding molecular orbital decreases the bond order and makes the bond weaker and longer. So the first statement is false. F2 1- should have a longer and weaker bond than F2. To calculate the bond order of N2 1+, we need to fill in the molecular orbital diagram. Let's start by calculating valence electrons in the second shell. Each nitrogen has five valence electrons, and there are two nitrogen atoms. We remove one electron because of the 1+ charge, so we have nine valence electrons. We will fill in the molecular orbital diagram from the lowest energy up, so we start by adding 2 electrons to the sigma 2S bonding molecular orbital. Then 2 electrons to the sigma 2S anti-bonding molecular orbital. And we can fully fill the two pi 2P bonding molecular orbitals with 4 electrons. We'll put the last 1 electron in the sigma 2P bonding molecular orbital. To calculate bond order, we take the number of electrons in bonding molecular orbitals, subtract the number of electrons in anti-bonding molecular orbitals, and divide by 2. The bonding molecular orbitals do not have an asterisk and are circled in blue. We have 7 electrons in bonding molecular orbitals. The anti-bonding molecular orbitals do have an asterisk and are circled in red; we have 2 electrons in anti-bonding molecular orbitals. So our bond order is 2.5. The second statement is true. To determine whether B2 2- is paramagnetic or diamagnetic, we need to fill in the molecular orbital diagram and determine if there are any unpaired electrons. If there are unpaired electrons, then it should be paramagnetic; if all electrons are paired, it should be diamagnetic. We'll start by finding the number of valence electrons. There are two boron atoms with three valence electrons each; the charge adds two more electrons for a total of eight valence electrons. Filling in the molecular orbital diagram from the bottom up, we will fully fill the sigma 2S bonding, sigma 2S anti-bonding, and the pi 2P bonding molecular orbitals completely. This means all of our electrons are paired. So the third statement is false. B2 2- is diamagnetic because it does not have unpaired electrons. To determine if C2 2+ has a weaker bond than C2, we need to compare the bond order. We'll start by finding the bond order of C2 and make adjustments for C2 2+. We have two carbon atoms with four valence electrons each. C2 is not charged, so there are a total of eight valence electrons. We fill the lowest energy molecular orbitals first, starting with the sigma 2S bonding molecular orbital. Then the sigma 2s anti-bonding molecular orbital. Then we completely fill both pi 2P bonding molecular orbitals, and we've used all 8 electrons. To calculate the bond order for diatomic molecules, we take the number of electrons in bonding molecular orbitals, subtract electrons in anti-bonding molecular orbitals, and divide by 2. The bonding molecular orbitals do not have an asterisk and are circled in blue; there are 6 electrons in bonding molecular orbitals. The anti-bonding molecular orbitals do have an asterisk and are circled in red. There are 2 electrons in anti-bonding molecular orbitals. So our bond order for C2 is 2. To make adjustments for C2 2+, we need to remove 2 electrons. Since the two pi 2P bonding molecular orbitals are the same energy, we need to remove one electron from each, leaving two unpaired electrons. To calculate bond order, we now have four electrons in bonding molecular orbitals and we still have two electrons in anti-bonding molecular orbitals. So our bond order for C2 2+ is 1. Our fourth statement is true. C2 2+ does have a weaker bond than C2 because C2 2+ has a lower bond order than C2. To determine if B2 2+ is stable enough to exist, we need to calculate the bond order. If the bond order is greater than zero, it should be stable enough to exist under certain conditions. If the bond order is zero, it should not be stable enough to exist. To fill in the molecular orbital diagram, we'll need the number of valence electrons. We have 2 boron with 3 valence electrons each. We need to remove 2 electrons because of the 2+ charge, so we have 4 valence electrons. We will fill in the sigma 2S bonding and anti-bonding molecular orbitals with these 4 electrons. To calculate the bond order for diatomic molecules, we take the number of electrons in bonding molecular orbitals, subtract the number of electrons in anti-bonding molecular orbitals, and divide by 2. We have two electrons in the bonding molecular orbitals, which are circled in blue. We also have two electrons in the anti-bonding molecular orbitals, which are circled in red. Since the bond order is zero, B2 2+ should not be stable enough to exist. The fifth statement is false.