Draw an MO energy diagram and predict the bond order of Be2+ and Be2- . Do you expect these molecules to exist in the gas phase?

Video transcript

Welcome back everyone to another video, draw an mo energy diagram and predict the bond order of brilliant two with a positive charge and brilliant two with a negative charge. Do you expect these molecules to exist in the gas phase? Our first step in this problem is to calculate the number of a selections for each species. And we're going to start with beryllium with a positive charge, right. So we're going to state that we have diatomic beryllium with a positive charge. We know that bellum is in group two A. So we have two V electrons multiplied by two atoms and we're also subtracting one because we have a positive charge and we end up with three electrons. And similarly the number of neon select transfer brilliant to with a negative charge which corresponds to the same calculation. But we would add one instead because we have a negative charge, this gives us five von electrons. Our next step is to draw a vertical axis which represents an increase in energy. And we have to recall that whenever we draw a molecular orbital diagram, we specifically want to draw the same number of molecular orbitals as the number of atomic orbitals. So now what we're going to do here is just draw the overall shape of our diagram and indicate that we are taking our two S orbitals for each beryllium atom. And we are forming sigma two s molecular orbital and sigma two S star, which is the antibody orbital. Now, we have a total of 3 billion electrons. So we can draw two electrons for one bellum and one for the other according to the B principle. And the hunts rule, we are first of all going to pair two electrons at the sigma two S level, right, there is only one orbital available. So we immediately pair the two electrons and now the remaining electron goes towards the sigma star two S level. And this is how we get our first molecular orbital diagram. Now for Berli two with a negative charge, the diagram might seem a bit more complex. So let's go ahead and draw it out. Now, first of all, we're going to draw a very similar diagram to the one that we have drawn previously. So let's go ahead and do that. So now once again, what did we have? Well, essentially we had two s energy level right now in total, we have five von electrons. That means we are going to take two electrons from each bellum first and we are going to have an axis of one electron. So now we have two S for each atomic orbital And as we said before, we are forming sigma to s and sigma star to s which is higher in energy. And in this case, because we have two available molecular orbitals, we're going to completely fill them. Now, we still have to understand that there is one more electron available. And for that purpose, the only way to incorporate it is to add E orbitals, we know that we have three P orbitals available. We will add our two P level. OK. So we will have our two P atomic orbitals. And now we're essentially forming our molecular orbital tribe. Now let's draw two P for the atomic orbitals on the right as well. And we know that we have one extra electron. So let's just add that one extra electron in the atomic orbital. Now, we also want to understand that the diagram is a bit more complex. So let's add our remaining levels. We know that first of all, we start with two molecular orbitals at the very bottom, we will have P two P. So let's go ahead and fix our diagram slightly. So we will have two orbitals, let's separate them. As we said, that would be by two P. After that, we have sigma two P. So let's go ahead and draw one orbital which will be sigma two P. Let's recall that afterwards we once again have two orbitals and they will correspond to P star two P, which is the P anti bonding orbital. And finally, we have one orbital left which is sigma star two P. OK. And if we only have one electron, that means we want to add it to the pi two P level. And that's because this is the energy level that is the lowest energy level available. So we're going to draw one electron which will be our unpaired electron well done. So we have our mo diagrams and we are ready to calculate the bond order for each. So now for the first species be to a positive charge, the bond order is calculated as one half the difference between the bonding and anti bonding electrons. Now we have two bonding electrons, right? And we're going to subtract one anti bonding electron which gives us one half and out the bond order or brilliant. See what a negative charge would be one half. We can neglect our first molecular orbital diagram and just move on to the second one. Now, once again, we have a one bonding electron and zero anti bonding electrons which gives us one half again, meaning they both can exist in a gaseous phase because the bond order is not zero, it's actually positive, right? So whenever we have a positive bond order, those structures can exist in a gas phase, that would be it for today. And thank you for watching.