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Period trend for ionization energy

Explaining trends in ionization energy across a period using nuclear charge, shielding, and distance of electrons from nucleus. Created by Jay.
Video transcript
In the last video, we looked at the trend for ionization energy going down a group on the periodic table. In this video, we're going to look at the general trend for going across a period. And so here we are in period two. And the general trend as you move to the right across the period is for there to be an increase in the ionization energy. And once again, in this video, we're going to talk about the first ionization energy. So if we look at lithium, lithium has an ionization energy of positive 520 kilojoules per mole. And then we go to beryllium and that number increases. And you can see, in general, the trend is to increase. So carbons increases from that. Nitrogens increases from that. And so you can see these numbers are getting bigger and bigger. So there are a couple exceptions to this general trend that we can see. We can see that from beryllium to boron, there's actually a drop in the ionization energy. And also from nitrogen to oxygen there's a drop in the ionization energy. And so we will talk about those exceptions to the general trend. But first, let's analyze lithium and beryllium and see if we can figure out why, in general, as you move across a period there's an increase in the ionization energy. And so let's look at these diagrams down here. And let's go ahead and draw in lithium and beryllium. So first, lithium with an atomic number of three. So three protons in the nucleus. And the electron configuration for lithium is 1s2, 2s1. So two electrons in the first energy level and an s orbital. So those are the two electrons in the first energy level occupying an s orbital. And in the second energy level, one electron. So let's go ahead and put in that one electron there. For beryllium, atomic number of four. So we have four protons in the nucleus. And so for beryllium, the electron configuration is 1s2, 2s2. And so in the first energy level, we have our two electrons. And in the second energy level, we also have two electrons. So a neutral atom of beryllium. So let's go ahead and put in those two electrons in the second energy level. And again, these diagrams are not perfect, but they're just going to help us try to understand the simple concepts for figuring out this trend here. And so let's analyze the ionization energy for these two atoms. And so let's use the same factors that we talked about in the last video. So the first factor was nuclear charge. So let's go ahead and write nuclear charge here. So the nuclear charge, of course, is referring to how many protons you have in your nucleus. And the more protons you have, the more an attractive force there will be for that outer electron. So if we look at lithium, with a charge of plus three in the nucleus, that's going to attract this outer electron. So there's a plus three charge. For beryllium, with a plus four charge, it's going to attract this electron right here. And since it's plus four, you would think that it would have more of an attractive force for one of these outer electrons. So just thinking about nuclear charge, we would expect there to be a greater attractive force for one of the outer electrons in beryllium than lithium. So therefore, you would think that it would take more energy to pull that outer electron away. And so therefore, an increase in the ionization energy. So just thinking about nuclear charge, we would say they would need to be an increase in the ionization energy. Next, let's think about electron shielding or electron screening. So electron shielding or electron screening. So the idea is that inner shell electrons will screen the outer electron from the effect of the nucleus. So if you look at lithium, this electron right here is going to repel this outer electron. And this electron over here on the right is also going to repel this electron. So they shield that outer electron from the pole of the nucleus. They screen it from the positive charge. And if we look at beryllium, so once again, these inner shell electrons are going to do the exact same thing. They're going to shield this outer electron from the effects of the nucleus. And so we're going to have approximately the same amount of electron shielding here. Now for beryllium, beryllium does have an electron right here which could repel this electron. And that probably does contribute some to it, but in general, the inner shell electrons do more of the shielding or the screening than this electron because this electron is also in the 2s orbital. So these inner shell electrons have more of an effect on screening. So we're going to say that the screening or the shielding is approximately the same for these two atoms. And so, therefore, there's really no effect on the ionization energy. And then, finally, distance, that was the third factor that we talked about. So distance, that outer electron is for the nucleus. Well this electron and lithium is in the 2s orbital. So on average, it's a certain distance from the nucleus. And this electron that we're talking about in beryllium is also in a 2s orbital. So it's on average about the same distance from the nucleus. And so the distance has really not much of an effect at all or no effect on the ionization energy. And so distance has no effect, and electron screening or electron shielding has no effect, really all we have to think about is the nuclear charge here. And let's go ahead and calculate the effective nuclear charge for both of these atoms really quickly. So for lithium, the effective nuclear charge would be the number of protons. So that's plus 3. So we're calculating effective nuclear charge for lithium. So it would be plus 3, minus number of shielding electrons. So that would be this one and this one. So plus 3 minus 2. So we get an effective nuclear charge of close to plus 1. In reality, it's about 1.3 when you do the more complicated calculation. For beryllium, the effective nuclear charge would be the number of protons which would increase to plus 4. And the shielding electrons, once again, we're only going to consider these inner shell electrons for our very simplified calculation of effective nuclear charge. And so we would get an effective nuclear charge of close to positive 2. In reality, it's approximately 1.9 when you do the more complicated calculation. And so we can see that we have an increased effective nuclear charge from plus 1 to plus 2. So as you move across a period, you're going to have an increase in the effective nuclear charge. And that's the reason that's given for the trend of an increase in ionization energy as you move across a period on the periodic table. So let's next talk about exceptions to that. And so the next one we see is we went from 520 for lithium to an increase for beryllium because of the increase nuclear charge. And then we go from beryllium to boron we see this drop. Right? We go from 900 to 800. So that's not following our general trend. Let's see if we can explain it though. So let's draw a picture for boron over here. So boron has one more proton in the nucleus. So there's five protons in the nucleus, so plus 5. In a neutral atom of boron there'd be five electrons. So let's go ahead and write the electron configuration for boron. It would be 1s2, 2s2. And then we would go to p1. So for boron, we'd go ahead and show the two electrons in the 1s orbital, the two electrons in the 2s orbital. And then this one electron in the p orbital. On average, an electron in a p orbital is a little bit further away from the nucleus. So I'm going to represent this electron right here. That would be the outermost electron. So this p electron right here. So that's the one that's going to be taken away in terms of ionization. And so let's, once again, think about our three factors, right? So first, nuclear charge. Well compared to beryllium, that had a plus 4 charge nucleus. Boron has a plus 5 charge nucleus. So we have an increased nuclear charge so we could think about this electron as experiencing a greater attraction for the nucleus. And so just thinking about nuclear charge, we might think, oh there's an increase in the ionization energy to follow our general trend. But we have these other two factors to think about too. So if we think about electron shielding, we know this outer electron this electron in the 2p orbital is shielded by these inner electrons in the 1s. So there's definitely some shielding due to those electrons in the 1s orbital like that. But since, on average, this electron in the p orbital is a little bit further away than these electrons here-- these electrons here are in the 2s orbital-- so they're all in the second energy level. But on average, the electrons in the s orbital are a little bit closer to the nucleus. And so there's a little bit of shielding that's going to go on here. And so there's a little bit of extra shielding for that electron in the p orbital. And so because of that extra shielding or screening, it's not going to feel as much of a pull from the nucleus from that increased nuclear charge. And so just thinking about electron shielding, that's screening that outer electron from the attraction of the nucleus and, therefore, should be easier to take that electron away. So shielding says a decrease in ionization energy. Finally, distance. So for thinking about distance, this, again on average, this outer electron here is a little bit further away from the nucleus than this electron over here in beryllium. And so since it's further away, there's less of an attractive force for the nucleus. And so therefore, it takes a little bit less energy to pull that electron away. And so just thinking about distance, we could say there's a decrease in the ionization energy. And so we add up all three of these factors. So we think about a distance as decrease in ionization energy. Electron shielding says decrease in ionization energy. Those two turned out to be a little bit more important than this increase to ionization energy due to the increased nuclear charge. And so that's the reason for this drop in ionization energy that you see from beryllium to boron. So that explains this first exception to the general trend. Next, let's talk about this other exception here. So this drop in ionization energy, that would be from nitrogen to oxygen. And so let's go ahead and write the electron configurations for nitrogen and oxygen. So first, let's write the electron configuration for nitrogen. Well that would be 1s2, 2s2, and then 2p3. So let's go ahead and do that in orbital notation. So we have the 1s orbital, the 2s orbital, and then our p orbitals like that. And so let's go ahead and put in these electrons in blue I guess. So 1s2, that would take care of the first energy level. And then second energy level, 2 electrons. So that takes care of those electrons right here. And then the p orbital, we have 1, 2, and 3. Next, let's go ahead and write the electron configuration for oxygen. One more electron to think about. So 1s2, 2s2, and then 2p4. So when we write our orbital notation, we'll go ahead and fill in those electrons. I'll use blue again here. So if we think about those electrons, so 1s2, that takes care of these. And then 2s2. And then we have 2p4. So we go 1, 2, 3, and then that fourth electron, that's going to just that's going to pair up spins. That's going to go into this orbital right here. And because that electron in magenta is repelled by that electron in blue, that means that it's a little bit easier to remove that electron. So a little bit of electron electron repulsion there means it's easier to remove the electron, therefore, it takes less energy to remove that last electron than it would to remove an electron from over here for nitrogen. And so that's the reason that you usually see given for this decrease in the ionization energy from nitrogen to oxygen. So we've talked about the two exceptions, but once again, in general, as move across a period in the periodic table, you're going to get an increase in the ionization energy. And that's due to the increased number of protons in the nucleus pulling the outer electron, having a greater attractive force for that outer electron. And therefore, making it harder to pull that electron away, resulting in an increase in ionization energy.