So we're ready for our final learning objective of this unit in which we're going to learn how to do the electron configuration of any element in the periodic table. We're going to be doing it for ground state atoms, meaning all their electrons are in their lowest possible energy state. We'll be using a principle called the Aufbau principle. This is a building up principle where we'll start with putting the first electron in the lowest possible orbital and moving out, away, and higher in energy as we place more and more electrons into their appropriate orbitals. Now, when we looked at this diagram earlier, we noticed that all the, in the hydrogen atom, all of the sub levels in a level, have exactly the same energy. But this was not the case in the multi-electron system. In the multi-electron system we knew that they spread out, with s being lower energy than d, which is lower energy than, well, s, then p, then d, and then f. Because of this, it causes a complication of knowing how to place those electrons in from lowest to highest, because they end up up here as they get closer together. As the spreading occurs we end up with overlapping energies of sublevels. I want you to focus in on this region right here. If we were going through, we might think that we'd put electrons in the 2s, then the 3p and then the 3d, and then move to the 4s, 4p, 4d, 4f. But that's not the case because we see, after the 3p comes the 4s in energy. It has actually lower energy, it may be further away and then farther out, but it has lower energy than the 3d because of the splitting pattern that we see here. So it's important to know the order that they get filled in. And it'd be difficult to memorize just how that splitting pattern occurs. So there's a couple of ways that you can determine, and this is going to be one of them. So, what you can do is you can write the orbitals in their available, actually I'm going to write them as sublevels, okay? Write them down. So in the first, the first number I put down is the 1s, because in the first shell there's only the s subshell. And I go to the next level. It has s's and p's so I write 2s and 2p. As I move further away from the period, from the nucleus, we have s, p's, and d's. Then the fourth is s, p, d's, and f's. And the fifth would be s, p, d, f's, and g's, but I'll never need to write the g's down, because in a ground state atom will never have electrons in the g. Then we will go to 6s, 6p, 6d, 6f. And you've written them down, okay? The next thing you'll want to do is put diagonal arrows as I'm going to show you here through them. This is going to, the arrows are going to give you the order of filling. So we fill the 1s first, then we fill the 2s. And then we fill the 2p and the 3s. And then we fill the 3p and the 4s. And then, the next diagonal arrow, where would it go? 3d, 4p, 5s. The next diagonal err, err, shoot error. I can't say it! Arrow! Will be 4d, 5p, 6s, and so forth. So if you follow the green arrow that I've laid down here, you'll know what order they get filled in. In the Aufbau principle, we put electrons in filling up these sub-levels as we go. Knowing the order will be essential. [COUGH] Beginning is pretty easy. We don't have many electrons to work with, and what we're going to do here is we're going to do three things. We're going to write the orbital diagram. And then we're going to do what's called an electron configuration. I'll show you how that's different. And then I will talk about the quantum numbers associated with electrons that we're putting in. Let's begin with hydrogen. Hydrogen, the atom will have one electron. So we'll write the first available sub-level is the 1s sub-level. It has one orbital. I will represent my orbital with a line. I will place the electron in that orbital. And that arrow represents the spin of that electron. It could spin up or it could spin down. I always choose to spin it up first because I'm an up kind of gal and so I like, I always put an up arrow. But you could put a down arrow if you like. So that is the orbital diagram. The electron configuration. You write the name of the sublevel. And you put as a superscript the number of electons in it. So that would be 1s 1, written there. Now the quantum numbers, I could write the quantum numbers for that electron. The first three tell me the location. N comes from this number. It's in the first shell. L comes from the name of the subshell. It's an s, so that's a 0. There is only one choice for orbitals, and that is 0 for the s subshell. And since I chose to spin it up, I will put a plus one-half for the spin. Let's move to helium. Helium has two electrons. With those two electrons, I know that I can put the second one in that 1s subshell. So I can write electron configuration. It's a 1s and I would put a 2, two electrons in the 1s. So the orbital diagram always gives you a little bit more information, it shows you the electrons and their spin. Here I'm just showing that there is two electrons in that 1s, but I know they're spinning in opposite directions. So if I were to write the electrons of that second, of the quantum numbers of that second electron that I put in there, it would still be in the first shell. It would still be in the S sub shell. It's in the same orbital, but it's spinning in the opposite direction. So those two quantum numbers would represent both of those electrons in the helium ground state atom. Moving out to lithium, we have three electrons. So in the first shell in that orbital, it gets filled, and we have to move to the next. The next available is 2s. Okay? And we'll put that electron into the 2s. And the electron configuration would be 1s 2, 2s 1. I could give you the quantum numbers of that electron. Remember a set of numbers identifies the electron, where it's located, and what it's doing. So it is in the 2s subshell. It's in the orbital of the 2s subshell and it's spinning in a plus one-half. Moving to beryllium. We have four electrons. We put two electrons in the first shell. We go to the second shell with its 2s orbital, and we put both of those electrons there. That would be 1s 2, 2s 2 for the electron configuration. Now let's have you consider the quantum numbers of those two electrons. What would they be? Stop and choose an answer. If you said it was number 3, then you are correct. We have them both in the same orbital. If the first three numbers are exactly the same that represents they're in the same orbital but they are spinning in opposite directions. So that's the quantum numbers for that. We go to do Boron next. It has five electrons. We have filled up the 1s. We have filled up the 2s and we have to consider what is the next available sub level. Well it is the 2p. Now when we write 2p, whenever we write a p we know that there are three orbitals. So we're going to draw three lines for the three orbitals in the 2p sub shell. And we have that fifth electron to place in there. Where will we place it? Well all three of those orbitals are exactly the same energy, so it doesn't matter where I place them. Any one of those orbitals would be the same. We say that they're degenerate, they have the same energy. Because I read from left to right I always put them in from left to right. But there's nothing to say they can't go anywhere in there. And it can spin up or it can spin down. All would be correct. When we write the electron configuration, we do 1s 2, 2s 2, and now we're ready to write 2p. Now we don't separate out the orbitals. When you do an electron configuration, you're just saying the sub level is called 2p, and I have one electron in there. And if I did the quantum numbers and I want to do them just to this last one I placed in, what would they be? Well it's the second shell. So that's the 2. It's a p subshell, so we have a 1 here. It is in an orbital. Now what are my choices for numbers for those orbitals? We have a negative 1, a 0, or a 1. It could have gone in any one of them, so what number will I choose for my m sub l? Well I generally, just to stay consistent call, this first p negative one, then zero and one. Okay. There is nothing magical about that. I didn't choose any number. But since I put it in the first one I'll call it a negative one. And it has an up spin so I will give it a plus one-half. So there's the quantum numbers for that last electron I placed in to the 2p subshell. Okay, we're ready for our next rule to consider, and this is called Hund's Rule. Hund's Rule states the most stable arrangement of electrons in a subshell is the one with the greatest number of parallel spins. Let's consider what parallel spin means. Parallel spin means spinning in the same direction, 'kay? So if they're spinning in the same direction, that's a parallel spin. So the result of Hund's Rule is that you will always, in degenerate orbitals, half fill the sublevel, half fill the subshell orbitals, before you go back and fill them up. So let's go to the next element, which is carbon. Carbon has six electrons. So we did the 1s and put the two electrons in. We did the 2s and put the two electrons in. We moved to the 2p, in which there's three orbitals. We put one here, and when we're ready to put the next one in, Hund's rule drives for me that I want to put the next one here. We maximize parallel spins, so they're not going to share. The reason for this is because this is a lower energy state for the atom, than having the electrons in the same orbital spinning in opposite directions. When we write the electron configuration, it really doesn't, unless you know what's going on, tell you about them being at different orbitals because all you write is the sub level with how many electrons are in it. So the electron configuration would be 1s 2, 2s 2, 2p 2. You need to know that those two electrons are actually spread out as we see there. But you do not indicate that in electron configuration. We won't worry about those quantum numbers for now. Let's move on to nitrogen. Nitrogen has seven electrons, okay. So we have 1s with its two electrons, 2s with its two electrons. The 2p, we have the three orbitals, we spread them out, and there we go. So we have them maximizing the number of parallel spins according to Hund's rule. This gives me 1s 2, 2s 2, 2p 3 for the electron configuration. Now, I want us to write the three electrons, take those three electrons and do each of their quantum numbers. So we know, and I'll just spread this out in this space here, okay. We know it's in the second shell. We know it's a p subshell. We know that they are all in different orbitals. So I'm just going to worry about those first three numbers 2, 1, plus 1. So this tells me that those three electrons are all in different orbitals because this number is telling me the orbital location. But they're all spinning in the same direction. Since I chose to write them in the up spin, I will choose to write plus one-half for those. And those are the three electrons in the 2p sublevel. Let's move on out to oxygen. Now oxygen has eight electrons. If oxygen has eight electrons and we place them in, so far, writing my three orbitals for my 2p. I've done four. I go five, six, seven maximum parallel spins and now I have to start doubling them up. So I'll come back here and I'll put one electron there. That would be 1s 2, 2s 2, 2p 4. And again it's just how many electrons are in that 2p subshell. So let's have you answer a question for quantum numbers of oxygen. Which quantum numbers are different for the last two electrons placed into oxygen? So, which were the last two that I placed in? Okay? We placed them here and then we came and put it here. So I want you to consider those two electrons and consider which of these sets of quantum numbers would apply to those two electrons. Well if you said both the m sub l and m sub s would be different you would be correct. The m sub l tells me that they're in different orbitals. So those two numbers would have to be different. The m sub s, they're spinning in different directions. So they would have to be different as well. Notice as we go back to that, that this one is an up spin because I was maximizing parallel spins, and this one here was a down spin because I was coming back and I was doubling them up into that level. 'Kay, now let's go to the fluorine with it's nine electrons. And we have two here, two here and then we will go in my 2p. One, three, four. The remaining and that is fluorine. And then neon would be 1s 2, 2s 2, 2p, and they completely fill up. So its electron configuration would be 1s 2, 2s 2, 2p 6. Now, any time the s and p sublevel are completely filled like this, see how they're full up, right, there's a great stability to that element. So the noble gases will always have this situation. Their s and p sublevel will be completely filled. And because of that, the noble gasses are very stable elements. That's why they do not like to react with other elements, and that's why we call them noble. They stand apart from the other elements in that way. So we've got, the first indication of, what we see here as a filled up subshell. Now we're filling them up according to the diagonal arrows that we drew in that table that I set up for you. But I'm next going to show you how to use a periodic table to come up with the orbital and enable you to do the electron configuration of any element without having to use that diagonal. In order not to have to design and develop that chart every time we want to do an electron configuration, and by that chart, I mean writing 1s and then one, 2s, 2p, and then 3s, 3p, 3d. And doing the diagonal arrows through you know, that sort of thing. You don't want to have to do that every time. You can use a periodic table in order to come up with our electron configuration order. What order do we fill these things by? So let me erase this and get it out of the way. And let me show you how to use a Periodic Table to determine the order. You need to break it down according to these regions. This is the s-block. This is the p-block. This is the d-block. [SOUND] And this is the f-block. Okay, that's the first thing that you need to do. Now let's look at the size of these blocks and see how they make sense for us. In the s-block there's one orbital. Or I should say in the s sublevel there's one orbital. That orbital will hold two electrons and so this block is two-wide. In the p-block, there are three orbitals. In every p sublevel, there are three orbitals. Each orbital holds two electrons, so that would be a total of six electrons and this is six elements wide. In the d sublevel, there are five orbitals. Each orbital can hold two electrons. This is ten wide. In the f-block there are, or in the f sublevel, there are seven orbitals. Each orbital can hold two electrons. That would be 14 and this is 14 elements wide. So s in the green, p in the blue, purple, d in the orange and f in the blue. The next thing you need to do is number these. because we need to know the number of the sublevels. And to do this we think about the orbitals and where they exist. The first place your going to run across an s sublevel is in the first shell. So we start numbering this with number one. One, two, three, four, five, six, seven. Now there's no such thing as a 1p. The first place that you come across a p is in the second shell. So this is numbered with a two, three, four, five, six, seven. First place you come across a d is in the third shell. So you number these with three, four, five, six. And the first place that you come across an f is in the fourth shell. So we call this four and five. Okay, so that is the numbering system. Now let's do that all again but quickly. Okay so let's erase the whole screen, and we'll start all over. This is the s, s, p, d, f, number these with one, these with two, these with three, and these with four. Okay, so you can duplicate that, you can do the electron configuration of any element. This will give us the order. Now just follow along with my pen as we read through the order. What do we come across, we start at the very beginning. We come across a 1s, that gets us across the first one. Then we have the 2s, then the 2p, then the 3s, then the 3p, then the 4s, the 3d, the 4p, the 5s, the 4d, the 5p. I'm just rattling off the order that the sub levels are filled. We have the 6s, the 5d, the 6p. Oops, forgot something in there. This sneaks up inside of there. So we have thee hm, hm, hm, hm, we better erase an arrow there so let's get the eraser out. Let's erase that arrow, let's erase that arrow and let's remember to enter the f-block. So after the 6s then comes the, come on, then comes the 4f, then the 5d, and the 6p. And then the 7s and then the 5f and then the 6d and then the 7p. Okay. Now. As we get in, involve the f, and we get further and further away from the nucleus, these sublevels get closer and closer together. And there starts to be got a lot of exceptions, where an electron will want to go into the d before the f or do one in the f and then come back to the d. There's a lot of subtleties that we will not get into. I will teach you at least one exception to our regular order but there are, if you look at an electron configuration of the elements that have a lot more electrons in them. You'll see a lot of oddities that you won't know the answer to why they occur that way. For this class, I won't get into any exceptions except for one necessary one. And I will keep you away from the f-block because of all the exceptions to the rules that apply when you get into the f-block. So what we're going to do is what's called a, short hand electron configuration. Because if you have to start off, you're doing an element with a lot of electrons, and you have to draw the electron configuration of every element, I mean of every electron. It's gets kind of tedious, and the line gets kinds long. So there's a shorthand procedure that we can follow, and what it does is you locate the electron, or the noble gas that comes before the element that you're doing the electron configuration for, and you write that element in square brackets, okay? So for example, you, if you are doing an element that Helium comes before it, you would put Helium in square brackets. What you're saying with that little piece of information right there, is you're saying up to this point, the electron configuration is just like Helium, and then we'll add some more electrons beyond that. Okay? So then you use that breakdown of the Periodic Table that I show you. And it, until you get to the element of interest knowing which orbitals come next, so let's do some examples of this. Let's find Chlorine. Chlorine is located right here on our periodic table. What noble gas comes before Chlorine? Okay? Chlorine is element 17, argon is after it, but the one that comes just before is neon. So we write neon in square brackets. And what we're saying with that is, those ten electrons, the first ten, are just like neon's electron configuration. Which, as a side note, would have been 1s 2, 2s 2, 2p 6. because that's what we finished our electron configurations of when we were doing out those tables with all of their detail. But putting the square brackets around the neon says, up to this point, it's just like neon. And now we're ready to go from Neon. So we were sitting right here, and work our way til we get to chlorine. So we follow those arrows. Where are we here with the green elements? That was our s-block. And we count down one, two, three. This is a 3s. So we write 3s. And it takes two elements to get in and out of that block so it's 3s 2. We work our way until we get to the purple area here and we know that this is our p-block. And we number our p's two, three. So this is 3p. And now we have to count till we get to the chlorine. We count every block. One, two, three, four, five. So this is going to be 3p 5. Okay? So that is the electron configuration of chlorine. Let's do tin next, okay? So we find tin on the periodic table, Sn. Find the noble gas that comes before it. Krypton. So we write krypton. What we're saying with that square bracket, is that the electron configuration, 4-10, has everything in common with krypton, plus something. So now we have to go from krypton, which is 36, move over here to element 37 and our work our way over til we get to ten. And write down what we see. Well we know that this is our s-block. We count down one, two, three, four, five. So this is 5s and it takes two to get in and out of that area. Now we're entering the orange, which is the d-block. We count down. Well how do we count with d? Well s was with one, p's were with two, d's are with three, okay? So we are entering the 4d area. To get across that d area takes ten, ten to get in and out of there. And now we hit the purple zone, and that was our p-block. We've counted down, so that's 5p, and it takes one, two til we get to ten. We count the ten as well, so it's 5p 2. Okay, next is the vanadium. Let's find vanadium is right here and I want to do vanadium in a different color, just so we don't get it confused with the other. So vanadium is located right here. Find a noble gas that comes before Vanadium, that is Argon. [SOUND] Okay. Working away from Argon which is element 18, move over to 19. And work our way until we get to Vanadium. This right here, is the 4s. It takes two to get in and out. And then we enter the 3d and we count over one, two, three, so it's 3d 3. Okay, so I want you to go back and look at that periodic table in your notes. And figure out what is the electron configuration of antimony? Did you choose number 2? Then you are correct. Okay. So just to make sure if you want to move ahead that's fine, but for those of you who did not pick that, let's do it together. Antimony, antimony is Sb. 'Kay? So I'm going to change my pen color. Sb. Periodic table, all right. So the elements that come before, again is krypton, and then we move our way over until we get to antimony, so antimony is Krypton, 5s 2, 4d 10, 5p 3. The only wrong thing about number three is you have the wrong noble gas. Okay, here you didn't drop down from 5s to 4d, and that's it. Okay, transition metals are elements that have incompletely filled d subshells or readily give rise to cations that have incompletely filled d subshells. In the transition element region, there is an exception to learn. And that is, if you're one away from being half-full, what would that mean? To be half full you'd have five electrons, a d subshell has five orbitals. Okay? If you are half filled, that would be this. If you're one away from being half filled, then you would have to take one electron away, and that would be at d4, right? So, that would be this. So being one away from half full would be d with four electrons. Or, if you're one away from being full. Well what does full look like? Full looks this, d sublevel, any d having ten. So it you're one away from being full, you'd have a nine. Now, whenever that happens, it will be a lower energy state to take an electron out of an s, and promote it up to actually get it half-full, or get it filled. There is stability in having it completely half-filled, or completely full. Okay? So, it's an energy argument. It is more energetically favorable. It relaxes it down lower energy state. It can take a little bit of energy to promote up the electron, then everything relaxes down and you end up with a lower energy state in the end. Okay? So if we were to do chromium's electron configuration, this is what you would have come up with if you followed the rules. Chromium is argon 4s 2, 3d 4. Now I'm looking for that four to show up. When it shows up, I know I am not finished with it. I have to instead promote one electron up. And this would be the correct electron configuration of, of chromium. So if I droze, drew the orbital diagram of these last electrons that we put in, in the 4s there's one electron. In the 3d, 'kay, we have maximized the number of parallel spins, and this is a lower energy state than what you see before that. 'Kay? I would like for you to the electron configuration of silver and see what you have there. And if you do its electron configuration, I'll give you a minute here. Stop and if you need more time and resume. You would have come up with this originally. When you see that 9 right there, a flag has to go up and say, you can't. It's not finished yet. And so we end up with this as the electron configuration instead. This is actually the correct. So this is one place where we see exceptions to the octet rule. So if you were doing the orbital diagram of those electrons, 5s, this is what you would have come up with originally. There's your four d with it's one, two, three, four, five, six, seven, eight, nine, 'kay. Instead of having this as the electron configuration, what we will have instead is to take one electron out of here and then promote it up to here. Okay, so this is what it would look like instead, completely filling that d sub-shell because this is lower in energy. Now we shouldn't leave this lesson without defining what a Lanthanide is. A Lanthanide has an incompletely filled 4f subshells or readily give rise to cations that have incompletely filled 4f subshells. And your actinides well they would have incompletely filled 5fs. But they are not found in nature. You'll never have to do an electron configuration of those guys. Now before we finish that I want to move back to a Periodic Table for just a moment, and let us talk about those definitions of the transition elements. Transition elements, having completely filled d subshells, or readily give rise to cations with incompletely filled d subshells. We are only talking about these elements. It's not the entire d-block. These are not transition elements. And let's see why, zinc, okay. Zinc's electron configuration would be, noble gas that comes before it is argon. Move over here and we are at 4s 2 keep working till we get to zinc, we have 3d 10. It does not have an incompletely filled subshell, it's 3d 10. It doesn't readily give rise to ions that have an incompletely filled d subshell. We haven't talked about electron configurations of ions, but we will later. But this will not do that either. So we have these elements right here are not transition elements. The whole area is called the d-block, but only from here to here are your transition elements. So we finally got through this entire unit. We've started by understanding light, its connection with electrons. We saw the experiments that led to the understanding of those connections and how the electrons were laid out in an atom, how they are quantized and their energy. We saw the Schrodinger equations that defined our quantum numbers. And we finished by being able to identify the location of all the electrons in an atom. Now why do we spend so much time talking about the electrons' neuroelectronic structure? It's because it's the electrons that drive the chemistry. As we start bringing these elements together to form compounds, or maybe the elements turn into ions, whether they're cations or anions. The reason they do that is because of their electronic structure. So to move forward, and to start talking about these elements coming together to form these compounds, we're going to be examining how the electronic structure determines the compounds that can be formed, so be looking for those connections as you move forward in the lessons to come.