Electron configurations with the periodic table | Chemistry | Khan Academy

Let's explore electronic configurations. It's basically arranging electrons of different elements in various shells and subshells. Let me quickly show you some examples. Yes, this will look overwhelming, but for now, focus on these numbers: 1, 2, 3, 4, 5, 6. These numbers represent the shells, the energy shells. If you look at the letters like s, p, d, and there can be f as well, that represents the subshells.

So look, we've arranged the electrons of various elements in their shells and subshells. That's what electronic configuration is all about. But the big question is, how do we write it? I mean, I used to find it super hard. Well, by the end of this video, we would have learned a cool way to actually write the electronic configuration of any element just by looking at the periodic table. So if you're excited, let's begin.

This is the periodic table where we arrange the elements according to their atomic numbers. So hydrogen is 1, then helium is 2, 3, 4, 5, 6, 7, 8, 9, 10, and then 11, 12, just like reading a book. All right, now we need to familiarize ourselves with two important terms. The first one is periods. Period represents the horizontal rows, so we call this the first period, second period, and so on and so forth until the seventh period.

I'm sure you must be wondering, what about these elements? We'll get to them in a second. But the second one are the vertical groups. So the vertical columns are called groups. So this is group one element, group two element, and so on and so forth up till group 18 elements. Okay, what about these elements over here? Why are they separated out? Well, if you look at the elements over here, if we zoom in, you can see this one is 57, and the next one is 72, which means there's a big gap. That gap is filled by these elements.

So these elements on the top over here belong to period six, and the elements over here belong to period seven in between these two. Now, if we were to write them out all over here, our periodic table would become too wide, and it'll be hard to fit in a page. And that's why we are writing it out separately over here. Don't worry too much about them because for now, what's important for us is to think about how do the periods and the groups help us figure out the electronic configuration.

So here's the thing: the period represents the highest energy shell or the valence shell of the elements. For example, hydrogen and helium are in period one, which means the first shell is their valence shell. That's the highest energy shell. If you look at sodium, for example, it's in period three, which means the highest energy shell is the third shell, and so on and so forth.

So let's quickly look at some examples. If you look at the electronic configuration of barium, don't focus on anything else; just look at the highest shell. It's two because it's in the second period. Look at nitrogen: highest shell two, it's in the second period. Look at sodium, as we said, three, third period. Look at sulfur: don't look at anything else, but the highest shell is three; it's in the third period.

Okay, what about the groups? What do they tell us? Well, they tell us the subshell in which the outermost electrons are. For example, if you look at group one and group two elements, the outermost electrons of those elements will be in the s subshell, so we call them the s block elements. If you look at groups 13 to 18, the outermost electrons of these elements will be in the p subshell, so we call them the p block. Similarly, over here, group 3 to group 12, their outermost electrons will be in the d subshell, so we call them d block elements. We also call them transition elements because their properties are in between that of s block and p block, so sort of like transitioning from here to there.

But anyways, finally, if you look at the elements over here, their outermost electrons are in the f subshell, and therefore we call them f block elements. And again, these are also called inner transition elements. We'll not get into the technical reason behind why it's called so, but for our purposes, we can think that they are on, at least on the periodic table, literally inner transition elements.

Anyways, again going back to the same example that we saw earlier, this time focus on the electrons in the outermost subshell. Which subshell is that for these two? Look, it's the s subshell because they belong to the s block. And for these two, look at the p subshell because they belong to the p block. Why is it 2s and 2p? Oh, that's because they belong to the second period, so the valence shell is two. And why is it 3s and 3p? Oh, that's because they belong to the third period, so their valence shell is three.

It all makes sense, right? Of course, the final question for us would be, how do we figure out the number of electrons in that outermost subshell? Like, how do we know it's 2, 1, 3, 4? Well, for that, you can just count how many squares to the right that particular square sits in that block. Okay, here's what I mean: if you look at beryllium, it sits one, two—it's the second square—and therefore there will be two electrons. If you look at sodium, it's the first square in this block, and therefore there's only one electron.

Look here, nitrogen sits on 1, 2, 3 squares to the right, and therefore three electrons. And sulfur, 1, 2, 3, 4 squares to the right, and therefore there are four electrons. So look, if I know the location of the elements in the periodic table, I know exactly what the last term is. The period tells me the valence shell, the blocks tell me which subshell the outermost electrons belong to, and by counting the number of squares in that block to the right, we get the number of electrons in that subshell.

Now, of course, the same thing applies to d block and f block as well. But now the final question is, how do we write the entire electronic configuration? For that, we'll just write down all the subshells on the periodic table itself. So this makes sense, right? This is the 2s, 2p, 3s, 3p, and so on and so forth. I'm not writing down the d block and the f block right now because there's a small caveat over there, so we'll tackle them a little later.

You can see helium is an exception. It actually belongs to the s block, but it's written over here because it's a noble gas, so it has similar properties to all the other noble gases which come over here. That's why it's written over here, so it's an exception. It's not a p block, and that's why it's 1s, but from here, you have 2p, 3p, and all of that.

Okay, so first, let's get some practice for light elements. So let's start with hydrogen, the simplest element, which has only one electron. Okay, the way to do it is you just locate it on the periodic table. Hydrogen is over here, and therefore I know it is 1s, and it's the first square in my s block, so 1s¹. And that's it, we're done. And you can see hydrogen has one electron, and there is one electron over here. Perfect.

Let's do a few more; it'll make sense, you'll get the hang of it. Okay, helium. Helium is over here, so helium would be 1s, but look, it is the second square to the right in the s block. This is an exception, right? It's the second square: 1, 2—second square to the right in the s block, so it will be 1s². And that's it, and you can verify that there are two electrons in helium.

Okay, let's try lithium. Lithium is over here. Now, things will get interesting, so how do we write it? Well, we always start from here. You start from here, and you start reading like a book. So we start with 1s². So we know 1s² gets completely filled, and then we come here. So 1s²—let's do that—1s². And then we come here, 2s. But lithium is the first square in my s block, so 2s¹. 2s¹, and we are done. And again, we can verify: 2 + 1 = 3; lithium has three electrons in them.

Okay, let's try carbon. Carbon is over here. Again, all we have to do is locate it on the periodic table. Okay, so where do we start? We always start from here, from the top. So 1s², 1s is completely done, 1s². So we are here, then 2s²; 2s also is completely done, so 2s². And then we get to 2p, and notice in the p block, carbon is one, two squares to the right in the p block, so it'll be 2p². That's the last one, and we are done. And again, we can check: 2 + 2 = 6; carbon has six electrons in them.

Okay, why don't you try chlorine? Chlorine is over here. So let me erase all of these. All right, great! Are you to pause the video and see if you can try it yourself first? All right, let's do it. We always start from the top, so we have 1s²; 1s² is done. Then we come to 2s; that's also done, 2s². 2s², then we go to 2p; that also gets done, so 2p: 1, 2, 3, 4, 5, 6. So 2p⁶. Then we come down, 3s; it also gets done, 3s². And then finally, we come to 3p. 3p: 1, 2, 3, 4, 5. It's five squares to the right in the p block, so it's 3p⁵. And now we can check: chlorine has a total of 17 electrons.

So let's check that. So 5 + 2 = 7 + 6 = 13 + 4 = 17, and there you have it. All right, now let's look at what happens for d block and f block. All right, you ready? What do you notice? You see a 3d over here, not 4d. This is a 4d, not 5d; it's one less. Why is that? That's because the electrons of the d orbital are never in the valence shell; they go into the core.

So the important thing is that the electrons in the d block are always one level below the valence shell, and that's why this number is one less than the period. So that's the major difference, but the procedure and everything else stays the same. So let's take an example; let's consider vanadium. Vanadium lies over here. Okay, so let me erase this. And now that I know it, that's over here, everything else, the procedure stays the same.

So we start from here. We have 1s², 1s². Then we go here, 2s², 2s². Then 2p⁶: 1, 2, 3, 4, 5, 6, so 2p⁶. Then 3s², 3s². Then 3p⁶, 3p⁶. This is getting exciting, right? Then 4s², 4s². And then finally, we'll be in 3d, and it is 1, 2, 3 squares to the right in the d block, and therefore it is 3d³. And if we add them all up, vanadium has 23 electrons, so we'll just check it: 3 + 2 = 5 + 6 = 11 + 2 = 13 + 6 = 19 + 4 = 23. There you have it.

And again, look, electrons in the d subshell are one level below the valency, so it's not 4d³, even though it's in the fourth period; it's not 4d³; it is 3d³. Okay, you try considering tin. Tin lies over here, and it has 50 electrons. So why don't you pause and give it a shot? All right, let's do this from the top: 1s². Let's do it a little quickly now. Then 2s², 2p⁶, 2s², 2p⁶, 3s², 3p⁶. 3s², 3p⁶. Then 4s², 4s². Then we have 3d: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, so 3d¹⁰. Then we have 4p⁶, 4p⁶. Then we go to 5s², 5s². Then we have 4d¹⁰, 4d¹⁰, and finally 5p², 5p². If we add these all up, we should get 50, and if you check that, we'll get 50.

And just like with the d block, if you go into the f block elements, then you will see the electrons are not in the valence shell; in fact, they are valence shell minus two. That's why if you look at the sixth period, you're going into 4f, and for the seventh period, you get 5f. All right, we're pretty much done because we can now write the electronic configuration of any element that you can locate on the periodic table. Right? But when you get to heavier elements, look, the electronic configuration becomes so huge.

So we invent a shortcut, and the shortcut is try to locate a noble gas, the nearest noble gas that comes before the element. So for example, when it comes to tin, the nearest noble gas that comes before the element is krypton. And so now what we do is we say, "Look, the electronic configuration would be everything that krypton has." So we write krypton in the bracket, which basically means everything that krypton has, and then some more. And then you start from krypton, and then you go here. So we say 5s², 4d¹⁰, and then 5p².

So 5s², 4d¹⁰, and 5p². See, this is the part that we have written over here, and all the rest over here, we say that's basically krypton's electronic configuration. So we write that in the bracket. This is a shorthand notation of what we wrote over here. So let's consider one last example, which is scandium, which comes over here. So can you try writing with the shortcut? Okay, so we locate the nearest noble gas that comes before it; that is argon in our case. So we write argon, and then after argon, we have 4s², and then we have 3d¹. And there you have it.