so what can we do with these molecular orbital diagrams in terms of predicting our properties so once we have this middle part of the mo diagram we can actually start to fill them out so if we have something like beryllium two we know that beryllium each one of these beryllium um has two electrons so two electrons and then there's two beryllium so we should end up getting four electrons so we can go ahead and take our mo diagrams and start filling up from the bottom up and we follow all the same rules that we filled up with our atomic orbitals we follow the same rules that we fill up with the molecular orbitals so we go ahead and fill this up we end up getting one the next orbital has to go in the same molecular orbital but it has to have opposite spin and then we end up getting uh those are two electrons here's number three and here is number four and this is now the mo diagram and the electron configuration for beryllium two so we can do this mo diagram for the entire row two periodic of the periodic table so it will look something like this so the lithium beryllium um uh orbitals all will all look the same and then what you have to watch out for is right at oxygen we have a switching of the two orbitals so we end up getting a swap of these two uh this uh sigma 2px actually goes down in energy and your book describes this as there's a there's a some orbital mixing between the s um the the sigma and the the 2p of the pi orbital so there's a little bit of orbital mixing in there that swaps those two but for our class what we have to do is know that there's a transition in terms of the order and you should have this molecular orbital diagram available to you for anything that you should need to do with this so we can start to fill up any of these these orbitals and look at this so we can go ahead and look at for example nitrogen we know that nitrogen has five electrons in each one so we should end up getting 10 electrons uh on there so we end up getting 10 electrons so we're gonna go ahead and fill up at the bottom first one two three four then we're gonna follow our rules for filling up when they're degenerate they're gonna go in parallel spins first so this is five six 7 8 9 10. so we can fill up the orbitals just like we do with atomic orbitals to get the molecular orbital configuration now what we can also do is calculate what we call the bond order so your bond order is going to be the number of bonding electrons subtracted by the number of antibonding electrons and we're going to divide that value by two so if we go back and look at the nitrogen we know that this is a sigma over here this is a sigma star and then these are our pi and this is also over here this is our sigma on here so if we look at this we look at the number of bonding uh electrons we have one two three four five six seven eight eight minus two in the antebonding and then we're gonna go ahead and divide this by two equals three so we get a bond order of three and guess what that also matches that matches what we draw in our lewis structure it also matches what we think of in our valence bond theory kind of picture so that means all of our models are matching up with each other so it's a great place to kind of think about our bonding and the higher the bond order the stronger the interactions are between the two atoms and then if we end up going back to our beryllium example remember we had four electrons uh in here we end up getting one two three four our bond order equals zero so that means there is no net bonding in there so we would expect that something like beryllium is going to be not likely to form as a compound so let's go ahead and think about this in terms of uh stability in terms of molecular hydrogen and why is hydrogen in in h2 and helium which is also a first row is also only just in helium so if we end up going so remember our only bonding orbitals that are available are our valence electrons uh and then those are the 1s so that is why we are going to work with the s orbitals right now we end up getting a bonding and an antibonding sigma sigma star h2 we should anticipate two electrons our bond order is going to be equal to one on here and if we end up looking at um helium so we end up getting a hydrogen uh bond order one our helium has four electrons for helium two so we end up getting two more electrons and our bond order equals zero so that explains why helium 2 doesn't actually exist however we can go ahead and look at something like helium-2 positive so an ion of helium-2 that's going to have three electrons because we start off with four electrons to make it a positive charge it's a minus one and then what we can do when we fill this out we end up getting one two three our bond order equals one half that means there is a slight bonding interaction and so in reality there is helium plus two helium two plus that exists as a stable species and that happens quite a bit in our sun so that is one of the interactions that is quite um prominent in the sun itself