Class 11 Chemistry (India)
- Introduction to electron configurations
- Electron configurations article
- Noble gas configuration
- Electron configurations for the first period
- Electron configurations for the second period
- Electron configurations for the third and fourth periods
- Electron configurations of the 3d transition metals
- Electron configurations
- Electron configurations of ions
- Atomic structure and electron configuration
How to write the electron configurations for elements in the third and fourth periods of the periodic table. includes using noble gas notation. Created by Jay.
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- I tried finding the electron configuration of copper using this method, but it does not match the one shown in Wikipedia. The link to the Wikipedia article on copper is: http://en.m.wikipedia.org/wiki/Copper(18 votes)
- Copper is one of about 19 elements that does not follow the usual pattern. Here is a list of the exceptions:
According to the usual pattern, the following elements should have two electrons in their outermost s, but one of those electrons has been "stolen" by the d subshell they are working on filling:
Cu, Cr, Nb, Mo, Ru, Rh, Ag, Pt, Au
The same thing happens with Pd, only it is both of the outermost s electrons that have been "stolen" by the d subshell, leaving what should have been its outermost shell completely empty. This is the only element where this phenomenon occurs.
Thus, Pd's configuration is: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ (the 5s is empty)
The following elements of the f-block fill in irregular ways, but you probably won't be asked to know these: La, Ce, Gd, Ac, Th, Pa, U, Np, Cm, Lr
There may be additional exceptions for elements after Lr, but these are not currently studied well enough to know for sure.
Note: in one of his videos, Sal does the configuration for La, but he gets it wrong because he didn't know it does not follow the usual pattern.
I can explain why these exceptions occur, but it is a rather long and complicated. If you are interested, let me know and I post the details.(52 votes)
- My instructor is telling me that the electron config for Mg2+ is [He], or Na+ is [He]...why is that, why do you not use Ne...I am confused!(2 votes)
- I hate to tell you this, but the information you are getting is incorrect.
The electron configuration for Mg is 1s² 2s²2p⁶ 3s² = [Ne] 3s².
The electron configuration for Mg²⁺ is 1s² 2s²2p⁶ = [Ne].
The electron configuration for Na is 1s² 2s²2p⁶ 3s = [Ne] 3s.
The electron configuration for Na⁺ is 1s² 2s²2p⁶ = [Ne].(29 votes)
- Why is it that the 4s orbital has a lower energy than the 3d orbitals?(9 votes)
- It's complicated, much more than is often explained. For K and Ca the 4s orbital is lower in energy than the 3d orbitals, but the energy of orbitals isn't fixed. When protons are added to the nucleus the energy of each orbital decreases but not at the same rate. As you move to Sc and across the transition metals the 3d orbitals are actually lower in energy than the 4s. But there are significant repulsions between the electrons in 3d orbitals which can be reduced promoting electrons into the higher energy 4s orbitals.(12 votes)
- Why do we fill the 3d orbitals when we are in the 4th line?(7 votes)
- The orbitals fill in relation to their energy levels. We fill the 4s orbital before the 3d because the electrons are at a lower energy, even though they are in a higher shell. A great way to remember which orbitals fill when is by looking at the periodic table. you fill them in the order they appear in the table.(7 votes)
- Hi,I have a couple of questions
(1)Why isn't the orbitals consistent. For an example, the s orbital comes again in the 4th shell when it was already in the first and second shell. This is confusing for me.
(2)Why does orbitals like s,p,d,f must be differ from each other and how can it contain different electron numbers in the one orbitals itself ? For an example, only two electrons can consist of s orbitals and p orbitals can hold up to 6 electrons. Why is there a difference in this ? Thanks in advance !(2 votes)
- (1) There is an s orbital in every shell. There are 1s, 2s, 3s, 4s, 5s, etc. orbitals.
(2) The s, p, d, and f orbitals are different from each other because they correspond to different types of vibration of the electron wave.
(3) Each type of orbital belongs in its own energy subshell. An s subshell contains one s orbital, a p subshell contains three p orbitals, a d subshell 5 d orbitals, and an f subshell 7 f orbitals.
(4) No orbital can contain more than two electrons. But a p subshell contains three p orbitals, so a p subshell can hold six electrons.(9 votes)
- So does this mean that electrons don't "revolve" around the atom like planets do? Do they just vibrate in place (that is, are they more or less stationary?)(1 vote)
- You're right that electrons don't revolve like planets around the sun, but they also don't vibrate in place. Rather than thinking of an electron as a particle zooming around, think of there being an area around the atom (an orbital) where the electron is likely to be found. The orbitals you usually see drawn enclose the area where there is a 95% chance of finding the electron at any given time. The electron itself can "jump" from place to place within this orbital, and doesn't need to travel continuously like we expect in the macroscopic world. This is because an electron has wave-like properties (just like light can be thought of as both a particle and a wave).(6 votes)
- when you do the process for Neon; n=1: 2, n=2: 8 and add these up, it is equal to 10.
when you do the same process for Argon; n=1: 2, n=2: 8, n=3: 18, add these up and you got 28 which isn't equal to Argon's atomic number and 10 more than it.
So this 10 more is for elements from Scandium to Zinc. I hope I am right otherwise I may not sleep tonight.(2 votes)
- n=3 is also 8 and not 18. Scandium to Zinc may have 3d configuration of highest energy level, but actually are in 4th period as energy level of 4s is lesser than the 3d orbital. So Ar=2+8+8=18, and Kr=Ar+18=2+8+8+18=36(3 votes)
- Can I represent Argon in noble gas notation like this?: [Ar] or do I have to write the whole configuration out like this: [Ne]3s23p6(2 votes)
- You have to write the whole thing out because the [Ne]3s23p6 is more descriptive then just [Ar]. I think what the point is, is basically figuring out where a certain element is on the periodic table. And if you don't know the configuration for the previous noble gas, then keep working backwards until you get to one you know, then go back down the table. I hope this was informative enough.(1 vote)
- Why would Krypton be a noble gas if the 4d shell isn't filled?(1 vote)
- When you go to the 4th quantum row of elements the d shell for that area goes to 3d this is because of the schrodinger equation, because quantum number 3 is the first instance of the d orbital the 4th row on the periodic tables transition metals actually occupy a 3d orbital and not a 4d. Since the 3rd quantum level had not been filled these metals fill the 3d orbital and then resume with the 4p orbital as we get to Kr.(4 votes)
- The configuration for phosphorus is [Ne]3s2 3p3. So its valency should be 3 (coz it needs 3 more to fill the 3 p orbitals). But I got to know that its valency is 3 and 5. Why is it so? Why also 5?(2 votes)
- [Voiceover] We ended the previous video with the electron configuration for neon. So, 1s2 2s2 2p6. I showed you how you could look at the periodic table and kind of run through these electron configurations. For example, this would be 1s1, 1s2 and then 2s2 would be here and then we had six. So 2p6 brings you all the way over to neon. And so for an electron configuration for the elements in the third period, so this would be the first period, second period, the third period. So let's do sodium. Sodium has 11 electrons so one more than neon but the second shell is full. The second shell is completely full. So for sodium's 11th electron we need to go into the third shell, into the third energy level. And so, n is equal to three for the third shell. Possible values of l include zero, one and two. In the third shell we're talking about an s orbital, right, one of them. P orbitals, right, 3p orbitals and when l is equal to two we're talking about a d orbital. If you do your magnetic quantum number you get five values for that. So we're talking about five d orbitals in the third shell, in the third energy level. And so if we plot those orbitals, all right, so let's go and plot those in terms of increasing energy. This side we'll put increasing energy going this way and the 3s orbital is here. So that's what we're talking about. We're talking about the third shell. We're talking about the s orbital in the third shell and there's one of them. All right next, we have 3p orbitals, in the third shell. Let's go ahead and draw in those p orbitals, so there's three of them. So one, two and three. Those are our 3p orbitals. And then finally we have some d orbitals, we have five of them in the third shell. So five d orbitals. Let me go ahead and draw those in. Those are our higher energy. So one, two, three, four and five. Here's the 3d orbitals like that. All right, let's do sodiums. Let's go back to sodium down here. 11 electrons. All right, so we need... Sodium has 11 electrons and so the first 10, we could put the first 10 electrons in just like neon. 1s2 2s2 2p6. And remember your superscripts tell you the total number of electrons, that's two, four and then 10. So that takes care of 10 electrons. We have one more, one more to account for and so, the 11th electron for sodium is going to go into the third shell and the lowest energy here will be the 3s orbital. So we go ahead and put the 11th electron for sodium into the 3s orbital so we can complete the electron configuration for sodium and we have to add on 3s1. Because we have one electron and an s orbital in the third shell. The complete electron configuration for sodium becomes 1s2 2s2 2p6 and 3s1. Notice that all of these, 1s2 2s2 2p6, this is the same electron configuration as neon and so we could represent all of that, we could represent all of these right here. We put neon in brackets and then we could write 3s1. And this is another way to write an electron configuration for sodium. So we call this noble gas notation because we're using the noble gas that precedes sodium. So if you just work backwards in the periodic table, you could go backwards from sodium, the first noble gas that you hit here is neon. And so we do neon's electron configuration is the same as neon's and then we have to add on 3s1. Neon 3s1 is our noble gas notation for sodium. Let's just do another element in the third period. Let's just go all the way over here to aluminum. So 13 electrons for aluminum. We can use noble gas notation to save us some time so we're saying the electron configuration is the same as neon. And that puts us right here. Let me use a different color. All right, so that puts us right here. And then we have sodium, sodium would be 3s1, magnesium would be 3s2 and then let's go ahead and put those in. Magnesium would be 3s2, so we fill that in here and then we need one more electron for aluminum. We need one more electron and of course that electron goes into one of these p orbitals here. And so let's go ahead and write the electron configuration. Just look at what we have on our orbital notation here. So we have two electrons in the 3s orbital so 3s2. And one electron in one of the 3p orbitals so we write 3p1. So, brackets neon 3s2 3p1 is the electron configuration for aluminum. Let's go ahead and do... Let's jump all the way over to argon here. So let's go all the way over to argon. Let's write the electron configuration for argon using noble gas notation. The noble gas before argon is neon. So we put neon in brackets and then once again we think, we could think this is 3s1, we could think this is 3s2. So we have 3s2. We go over here this is 3p1, 3p2, 3p3, 3p4, 3p5 and 3p6. We can go ahead and write 3s2 3p6 here for argon. And if we want to put in those electrons and how they fill, we could put in those electrons following Hund's Rule and we talked about in the last video. And then we put in all of those electrons so we can see that we've now filled the 3p orbitals like that. So that's argon. Next up is potassium. So now we're at the fourth period on the periodic table. So if we go one more element here we hit the fourth period of the periodic table and we get to potassium. You might think, all right, so one more electron than argon, so you might think potassium's electron would go into into a d orbital, because that makes sense. We have this d orbital here but that's not what happens. So let's go ahead and take that electron out of that d orbital. We are now, we're in the fourth period and we can actually open up a new shell here. We can go to n is equal to four. We can go to n is equal to four, let me go ahead and write this down. And in the fourth shell, I won't draw, I won't write in all the values for l but we definitely have an s orbital. So there's an s orbital in the fourth energy level, one of them and it turns out the energy for that 4s orbital is a tiny bit lower than the 3d orbitals. Let me draw that in with a different color here. The 4s orbital turns out to be a little bit lower in energy. Let me use a different blue so we don't confuse it here. This is representing our 4s orbital. For potassium, potassium this actually... The 4s orbital is a little bit lower in energy so the next electron that we add for potassium is going to go into this 4s orbital. We can go ahead and write the electron configuration for potassium. If we're using noble gas notation, we go backwards. So what's the noble gas that precedes potassium? Just go backwards in the periodic table and that is of course argon, that's argon right here. We could say that the electron configuration for potassium is the same as argon's and then we have one more electron to worry about here. That extra electron, that 19th electron for potassium is gonna go into this 4s orbital here. We write the same electron configuration as argon and we write 4s and one electron in that 4s orbital so we write 4s1. For potassium, one more electron. We can go ahead and write... Sorry, for calcium I should say. So for calcium, one more electron to worry about so the noble gas that precedes calcium is once again argon. So we say it's the same electron configuration as argon, one more electron and we know that there's space in our 4s orbital here. We can add another electrons pair spins up and so we can say 4s2 here for the electron configuration for calcium. Once again, we notice this pattern on the periodic table. This right here, let me use a blue here. We could say this is 4s1, we could say this is 4s2. All of these over here on the left we have called... Thought about them as being s orbitals anyway. So this is the s block on the periodic table. All right, that takes care of calcium and the next ones that you're going to hit are all of these elements in here. So all of these elements. We know that in the third energy level there are d orbitals. So let's go back up to here. Let me ho ahead and mark this so we can see. We know there are five d orbitals in the third energy level. Each d orbital can hold a maximum of two electrons so five times two is 10. How many spots do we have here? One, two, three, four, five, six, seven, eight, nine, 10. We have these 10 spots here. I'm actually gonna do a whole separate video on the d orbitals so I'm not gonna worry about them in great detail right now but we have 10 spots. This is the third shell and we have 10 spots for electrons. So 3d10 would fill these d orbitals. 3d10 fills these d orbitals here. And this is why the periodic table is shown the way it is here. It just helps you to think about writing your electron configurations. For example, let's just go ahead to gallium here. Let's write electron configuration for gallium. So if we're using noble gas notation, the noble gas that precedes gallium if we work backwards would be argon. So we can go ahead and write that in here. That takes us to this area. That would be 4s1 4s2. So we write 4s2 right here and then we have all of our d's. We can just go ahead and put the d's in there, 3d10. So we put in 3d10 here and this takes us to... We're in the fourth shell here and it's equal to four so l is equal to one is another possibility. And so those are the p orbitals. Once again, we have 3p orbitals in the fourth shell. Of course right here, this is where we hit the p orbitals. We start to fill the p orbitals. We can go ahead and say this is a p orbital in the fourth energy level. A p orbital in the fourth shell and we have one electron so we can write 4p1. And so this is one way to represent the electron configuration for gallium using noble gas notation. So argon 4s2 3d10 4p1. Sometimes you might see the 3d10 and the 4s switched in terms of their orders. You might see 3d10, 4s2, 4p1. It doesn't really matter how you do it. You'll see both ways done. All right, finally let's do krypton here. Let's do krypton. Our last electron configuration here. The preceding noble gas would be argon and let me use a color that we can see here. Argon will take us to here and then we have 4s2. So we fill 4s2 and then we have all of our d orbitals here. So 3d10. So 3d10 takes us to here. Now we're in the fourth shell in the p orbitals so one, two, three, four, five and then six. So 4s2 3d10 4p6 would be one way to represent the electron configuration for krypton. And once again, you could switch the 3d10 and the 4s2 if you wanted to. You could say for krypton the electron configuration, the noble gas notation would be argon 3d10 4s2 4p6. Just do whichever way your professor wants you to do it. That's covering the third and the fourth period but we've ignored the d orbitals. Again, that will be a whole separate video because it gets a little bit complicated when you get into d orbitals.