In this lecture, we're going to dig more deeply into our understanding of atomic structure by looking more closely at this electron shell model that we have developed for describing the arrangement of the electrons about the nucleus. Let's reiterate a bit of what we've learned about this electron shell model. We know that what we mean by an electron shell is a group of electrons which are a certain distance away from the nucleus about equal to each other, that there are multiple sub shells, that each shell can only accommodate a certain number of electrons. We can't put all the electrons into the lowest energy shell. Once a shell is filled we have to add subsequent electrons to shells which are larger. And by larger, we mean at a radius farther away from the nucleus, meaning they are less strongly attracted to the nucleus. They're at a higher energy level. At some point, we reach the highest energy shell for each atom, and we call that shell the valence shell, what we mean by that is it's the most important shell. The electrons in that shell are most important to the chemistry of the each individual atom. And we call those electrons the valence electrons. One of the things we'll noticed that the atoms in the same group of the periodic table has the same number of a valence electrons, for example, say, fluorine and chlorine, which both have seven valence electrons. And we imagine that that's the reason why they have very similar kinds of chemical properties. For example, somewhat similar ionization energies so that's for ionization. All the remaining electrons that are not in the valence shell we're going to refer to those as core electrons. Now this model you recall, actually it grows out of examination of this data having to do with the ionization energies of the elements. And there were really two aspects to this data that we were focused on. Aspect one, was this rather dramatic rise in the ionization energies as we move across the periodic table. We accounted for this by saying the electrons are about the same distance away from the nucleus. In other words, in the same shell. But that there's an increased nuclear charge as we move from lithium across the periodic table to oxygen, fluorine, and finally, neon. And therefore, those electrons about the same distance from the nucleus are more strongly attracted. And therefore, are harder to ionize. We also noted, this rather precipitous drop in the ionization energies as we went from the farthest right element on the periodic table, neon, to the next element, sodium, all the way on the left side of the periodic table. And we accounted for this by saying that the next electron added when we increased the nuclear charge on sodium, we increased the number of electrons, that next electron must be in a new shell farther from the nucleus, and therefore less strongly attracted. But we should think a little bit more carefully about this. Because, let's examine this particular data here, this drawing that I've made to compare fluorine and neon. If I take a little bit of a close examination here, I noticed that neon has a higher ionization energy than fluorine does, and our model accounts for that as follows. Here is a drawing of the valence electrons for fluorine. Here's the +9 nuclear charge on fluorine, the two electrons filling the first shell, and then seven subsequent electrons in the valence shell of fluorine. Here's neon with its +10 nuclear charge. There are ten electrons all together, two in the inner core, and eight in the valence shell. And what we've been imagining here is that, for example, this electrons attraction to the +9 nucleus is not as great as this electrons, attraction to the +10 nucleus because of the greater nuclear charge. But when we stare at these diagrams, there should be something else that jumps out at us. Electrons are attracted to positive charges, but they're repelled by negative charges. Let's count the numbers of negative charges. For this neon electron here, it's attracted to a +10 nucleus. But it gets repelled by, one, two, three, four, five, six, seven, eight, nine, other electrons. And, if I look at the fluorine electron, this one here, it's attracted to a +9 nuclear charge that is weaker. But how many electrons is it being repelled by? One, two, three, four, five, six, seven, eight, not nine. In other words, adding the extra nuclear charge, of course, also means that we have added an extra electron for repulsion amongst the electrons in the valence shell. So the question is, then, as the atomic number increases, why isn't it the case that the increase in the attraction to the nucleus is all set by an increase in electron-electron repulsion. We need an answer to these question that we're going to be able to understand our ionization energy data. And thereby, understand our shell model. To do that, we need some more data. Here's the data we're going to look at. We're going to call these successive ionization energies. In our previous data, we have examined only the first ionization. We're moving a single electron, and we'll call that energy the first ionization energy, IE1. The energy required to remove the first electron. But of course, we could remove a second electron. That's this process here, taking electron out of the ion which is already formed. The energy required to remove the electron from the ion already formed will call the second ionization energy. Core sector remove the third electron, fourth electron, fifth electron etc., and measure all of those energies. Let's analyze that data by taking a look at sodium, and thinking about what the ionization would just going to look like for sodium. And to do this, now, let's take our data, and here are our successive ionization energies. That's a lot of data in this table. So we need to analyze it slowly. First, let's make sure we know what we're looking at here. These are the elements sodium through argon. Those are all elements in the third row of the periodic table. Let's pop up the periodic table here. You can see them all in the same row, sodium going across to argon there. And then in this column, to the left, what we see is in fact the successive ionization energies starting with IE1, and working our way down all the way we've removed maybe as many as seven electrons. What do we see in this data? Well, I mentioned we're going to focus on sodium first, so let's analyze this data here. What do we see? Two things, one is that each successive ionization energy is larger than the one before it. But secondly, that one particular comparison here is particularly important. We see as we go from ionizing the first electron to ionizing the second electron, a gigantic increase a factor of almost ten, actually about nine. Subsequently, if we ionize a third electron, the comparison of the third ionization energy to the second ionization energy, yes, it goes up, but it only goes up by about 50%. It doesn't go up by a factor of nine. So there's something unique in sodium about ionizing the first electron. But our diagram for sodium, back over here, explains this. We have sodium, which has a plus 11 nuclear charge, and has 11 electrons as a neutral atom. Two in the center core here, eight in the next shell in the core, and then one valence electron out here. If I remove that one valence electron, its ionization energy is going to be determined by its attraction to the +11 nucleus. And by its repulsion from the ten other electrons which are here. If I remove a second electron, that electron is now closer to the nucleus and we expect it to have a higher ionization energy, and that's exactly what we see. So the fact that there's a significant jump as we go from the first ionization to the second ionization tells me that there's only one valence electron in the sodium atom. We could actually use the same kind of analysis to find the number of valence electrons in each of these elements, look for the large increases, there is a large increase, there is a large increase, here is the large increase, here, and here. And in each circumstance for example, what I can see is that the large increase for magnesium occurs here between the second ionization energy and the third ionization energy. That means after I've removed two electrons from magnesium, it's much harder to remove the third. That tells me there are two valence electrons in magnesium. Working over here to sulfur, the large increase occurs between the sixth ionization energy, and the seventh ionization energy. That means that sulfur has six valence electrons. So that all looks pretty darn good, except if we analyze this data again, and notice the changes here now. And notice the changes, that change is quite a bit larger than we might have expected. Go back and look at our diagram here again. As they go from this electron and removing it, to removing this electron. Yes, it is closer to the nucleus, but we wouldn't expect an increase of a factor of nine as we move one step closer into the nucleus. That's two grade of change. What's accounting for this has to be the repulsion of this election by all of the other electrons as well. That is, if we consider what the attractions are, the attraction is the attraction of the electron 2 a +11 nucleus. But if we consider what the repulsions are, then the repulsions actually repulsion from the electron 2. Let's see, one, two, three, four, five, six, seven, eight, nine, ten core electrons. Now what apparently must be happening is that the attraction of the electron to the nucleus is being at least partially offset by repulsion from the core electrons. Consequently, we're going to describe something now that we're going to call the core charge. The core charge is a number which attempts to take into account simultaneously, the attraction of the electron to the nucleus, and the repulsion of the electrons from the inner shell electrons. In the next slide here, we describe the core charge as being the difference between the nuclear charge +11, and the number of core electrons, 10 in the case of sodium. So we'll describe the core charge now as being +1. That means this electron is effectively feeling an attraction to the nucleus of only net +1 representing its attraction to the plus 11, and it's repulsion from the -10. Well, now let's go back and look at our data again. Our data tell us, that we get a dramatic increase, here, when we try to remove the second electron. Let's think about what might be causing that, by looking at a model for what the sodium ion looks like, instead of the sodium atom. Here's a model of the sodium ion. Notice, I've removed the electron out here which was a valence electron, leading behind the ten core electrons. But now we recognize that the sodium ion, the outer most shell is a shell closed rim. Now let's imagine what we have here. We have the attraction of the electron, let's say that one, to the nucleus is due to the attraction of the electron to the +11 nuclear charge. The repulsion, if we follow the same model that we used a little while ago, is the electron to now only two core electrons. Notice for this electron here, as it looks in towards the nuclear area, there are only two core electrons. Like go back and I looked at my definition of the core charge, then the core charge for the electrons in the valence shell of the sodium ion, are +11- 2+9. An enormous increase in the attraction of the electrons in this new shell of the sodium ion to the nucleus. This accounts for a gigantic increase in the ionization energy. Let's go back again and look at the data. And in this data, what do we see? We see this factor of nine here, as we go from the first ionization to the second ionization. Why then don't we see such a precipitous increase as we go to the third ionization? Well, the answer is, that this electron, let's say I've already pulled this one out, this electron is still in the same shell, it still feels the same +11 charge, it still feels the same repulsion from the two core electrons. The core charge for pulling that second electron out is a +9, the same as it was before. As a consequence, we don't expect to see a dramatic increase in the ionization energy, because the core charge has not gone up. So, this idea of a core charge here actually can account for an enormous amount of what we see in the variations. There is still one more bit of data to analyze, though. Which is if we look at this data little bit more closely as well, we'll notice that even once we have gone past ionizing the valence electrons, in each case, there continued to be significant increases in the ionization energy. It's just not as great as the big jump we get when we removed the last valence electron. What's causing that? Well again, it makes sense to us that if this electron were missing from the sodium ion, the number of total electrons-electron repulsions felt by this electron is now fewer. Let's think about how many electrons this one is repelled by. One, two, three, four, five, six, seven, eight, nine. How many repulsions are there from this one, once this electron has been removed. The answer is eight, one, two, three, four, five, six, seven, eight. Consequently, electron-electron repulsion within the valence shell also matters and accounts for the increases that we see, for example, once we remove all of the valence electrons in sodium, or for example, in silicon here. Those increases are arising from valence electron-valence electron repulsion. That also make sense. Once we've remove an electron from an atom what remains behind is a positively charged ion. It should be harder to remove a negatively charged electron from a positively charged ion than from a neutral atom. And the greater the positive charge, in other words the more electrons we've already removed, the harder it will be to remove any further ones because of greater attraction of the electron to the remaining positive charge. So the effective electron-electron repulsion in our shell model here is two-fold. One, is repulsion of the valence electrons by the core electrons which we take into account with this core charge model such as we've applied to sodium. But, there's also, we should not forget, electron-electron repulsion amongst the valence electrons, and that has a lot to do with the properties of the atoms as well. In the end, then, we've got a model, now, which really well accounts for the variation in the ionization energies of the elements, by understanding this idea of the valence shell, and the core charge felt by that electrons when they are in the valence shell. The ionization energies could be understood in terms of this. And it turns out as we will see over the course of the rest of this semester, much of chemistry can be understood by understanding the ionization interests of the atoms because they're related to how strongly attracted the electrons are to each atom. And we'll pick that up a bit further in the next lecture.
