In this video, we're going to focus on electron configuration and how to write the orbital diagrams for elements. So let's start with the element sulfur. If you look at the periodic table, sulfur has an atomic number of 16 and a mass number of 32. The atomic number tells you the number of protons. So sulfur has 16 protons. And as an atom, it has 16 electrons.
Atoms have equal number of protons and electrons. That's why they're neutral. Ions have unequal number of protons and electrons.
So because we're dealing with a sulfur atom, sulfur has 16 electrons. Now, how can we write the ground state electron configuration for the element sulfur? This is what you can do. In the first energy level, we only have one sublevel, S.
In the second energy level, you have two sublevels, 2S and 2P. In the third level, you have 3S, 3P, and 3D. In the fourth level, you have 4S, 4P, 4D, and 4F.
But we don't need to go this far for sulfur. Now the S sublevel can hold up to two electrons. P can hold up to six, and D can have a maximum of ten electrons. So we're going to start with 1S.
So it's going to be 1S2. And then after 1s, the next sublevel is 2s. So 2s2.
And then after 2s, it's 2p then 3s. Now, p can hold up to 6, so 2p6. Right now, we have a total of 10. 2 plus 2 plus 6 is 10, and we need to get up to 16. So let's continue. After 2P, we have 3S.
So that's going to be 3S2. So we have a total of 12, so we need 4 more. Now 3P can hold up to 6, but we don't need 3P6.
Otherwise the total will be 18. We want to stop at 3P4. So that the total is equal to the number of electrons in this sulfur atom. So this is the ground state electron configuration for sulfur. Now how can we represent it using the orbital diagrams? So if you want to write the orbital diagram for sulfur, here's what you can do.
So it's going to be... We're going to draw boxes. This is for the 1s orbital, and then 2s. The s sublevel only has one orbital.
p has three orbitals, so this is 2p, and then 3s, and then 3p. So 1s is completely filled. There's two arrows.
Each of these arrows will have opposite spins. So in every orbital, you can have a maximum of two electrons. The 3s is filled.
Now the 3p is not filled, and according to Hund's rule, you need to add one arrow at a time with parallel spins. So 1, 2, 3. After they're completely unpaired, then the fourth one has to be paired, but with the opposite spin. So you have to add it one at a time according to Hund's rule.
So that's how you can write the orbital diagram for the element sulfur. Now how can you represent the ground state electron configuration using noble gas notation? So if you go to the periodic table, look for sulfur, which has the atomic number 16, and pick a noble gas that has an atomic number 16. less than 16 but close to 16 as possible. There's only two noble gases that have an atomic number less than 16. That's helium which has an atomic number of 2 and neon which has an atomic number of 10. But neon is closer to sulfur. So we're going to write neon.
Now neon has an atomic number of 10 which which is equivalent to 1s2 2s2 2p6. That's the configuration for neon and those numbers add up to 10. 2 plus 2 plus 6 is 10. So to write the configuration using normal gas notation, replace 1s2 2s2 2p6 with neon. and then write everything after that.
So it's 3s2 3p4. So you can also write the electron configuration for sulfur in this way. Now, how would you write the electron configuration for the sulfide ion?
Feel free to pause the video and try that example. So we have the electrical configuration for sulfur. We know it's 1s2, 2s2, 2p6, 3s2, and 3p4. And the reason why it's like that is because sulfur has a total of 16 electrons. Now, how many electrons does the sulfide ion have?
To find the number of electrons, it's equal to the atomic number minus the charge. The atomic number of sulfur is 16, and as an atom, it has a charge of 0. So, a sulfur atom has 16 electrons, but for an ion, it's different. As we said before, ions have an unequal number of protons and electrons.
In this case, the charge is negative 2, which is 16 plus 2. So, the sulfide ion has 18 electrons. So to write the electron configuration for sulfide, we need to add 2 to the 3P4. So the configuration is going to be 1s2, 2s2, 2P6, 3s2, and 3P6.
As you can see, if you add up all the exponents, you should get 18. 2, 2, and 6 is 10. 2 and 6 is 8. 10 and 8 gives you 18. So that's how you can write the electron configuration of an anion that's a negatively charged ion. Whatever the charge is, simply add that number to the electron configuration. So if you have a minus 2 charge, add 2 electrons. If it's a minus 3, add 3, and so forth. For the sake of practice, go ahead and write the ground state electron configuration for nitrogen and the nitride ion.
So feel free to pause the video as you work through these examples. So we have 1s, 2s, 3s, 2p, 3p, 3d. For this example, I don't think we need more than this.
Now remember, s can hold up to 2 electrons, p can have up to 6. And D can have a maximum of 10. So let's start with nitrogen. Nitrogen has an atomic number of 7. So we need the exponents to go up to 7. So it's going to be 1s2 and then 2s2. So we have a total of 4. 7 minus 4 is 3, so we need 3 more.
Instead of going to 2p6, we're going to stop at 2p3. Because 2 plus 2 plus 3 is 7. And that's the electron configuration for nitrogen. Now if you want to write it using noble gas notation, we can't use neon because that's an atomic number of 10. We have to use helium because it's less than 7. Now helium, which has an atomic number of 2, will replace the 1s2.
So it's going to be helium 2s2 2p3. So that's the electron configuration for nitrogen using noble gas notation. Now what about for the nitride ion? What is the electron configuration? So let's calculate the number of electrons in the nitride ion.
It's going to be the atomic number minus the charge. So we've got to add 3 to the number of electrons. So the nitride ion has 10 electrons. So we need to write the configuration until we get up to 10. So it's going to be 1s2, 2s2.
Now instead of writing 2p3, let's add 3 to it. So we're going to stop at 2p6. Because 2 plus 2 plus 6...
equals 10. So for anions, you're going to have to extend the electron configuration. So this is what you need to do to write it for nitrogen and the nitride ion. Now, how can we write the orbital diagram for the nitrogen atom?
How would you do it? So this is for the 1s orbital, 2s orbital, and p has three orbitals. So 1s and 2s is filled, but the 2p orbital is not filled. It's half filled. And according to Hund's Rule, we need to add it one at a time with parallel spins.
So, according to Hund's Rule, you don't want to do this. That's not the proper way to fill an empty or... What's the word I'm looking for?
If you don't have, if it's not 2B6, you don't want to fill this orbital with two electrons and leave this one empty. According to Hund's rule, you want to add it one at a time. Electrons prefer to have their own orbitals.
If there's an empty orbital, they prefer not to share. They rather... they prefer to have their own orbitals because whenever you have two light charges next to each other, light charges, they repel each other. And so when two electrons share an orbital, it's... it's a less stable situation, so to speak.
Opposite charges attract. So, this configuration is less stable than this configuration because electron repulsion is minimized. So, according to Hund's rule, make sure you add it one at a time before you pair up electrons in a single orbital.
Now here's another question for you relating to nitrogen. Elemental nitrogen, is it paramagnetic or is it diamagnetic? Sometimes you may see a question like that on the test. And what you need to know is that a paramagnetic substance is one that contains unpaired electrons.
And therefore they're attracted to an external magnetic field. So nitrogen is paramagnetic. It contains three unpaired electrons. A substance that is diamagnetic is weakly repelled by a magnetic field and it contains no unpaired electrons. So the nitride ion has no unpaired electrons.
It's 2p6 at the end, so it's completely paired. So nitride will be considered diamagnetic, whereas the nitrogen atom is paramagnetic. Now sometimes they may ask you, how many S electrons are in this element, or how many P electrons?
If you get a question like that, simply write the configuration. In the case of nitrogen, nitrogen has three P electrons. In terms of S electrons, nitrogen has four S electrons. Now, how many valence electrons does nitrogen have?
Let's say if you get a question like that. The valence electrons are simply the electrons in the last energy level. The last energy level for nitrogen is the second energy level, where you see this number of two in front of S and P. So, 2S2 and 2P3. Those electrons represent valence electrons.
That's electrons in the outermost energy level, or in this case, the second energy level. So nitrogen has five valence electrons. Seven minus five is two.
So this is the core electrons. It has two core electrons. Now what about sulfur? We already wrote the configuration for sulfur.
How many p electrons and s electrons does sulfur have? And how many valence electrons and core electrons does it contain? So let's write the configuration.
So we know it's 1s2, 2s2, 2p6, 3s2, 3p6. I mean, not 3p6. I went too far. 3p4. So the number of p electrons is based on 2p6 and 3p4.
So you can add 6 and 4. That means that sulfur has 10 p electrons. Now, how many s electrons do you see? 1s2 2s2 3s2.
2 plus 2 plus 2 three times is the same as 2 times 3 which is 6. So sulfur has 6s electrons. Now how many valence electrons does it have? Now remember the valence electrons is defined as the electrons in the outermost energy level. Those are the electrons that participate in chemical reactions. So the highest energy level is not the first energy level, it's not the second energy level, but in the case of sulfur, it's the third energy level.
So only these numbers have a 3 in front of it, so it's 3s2 3p4. sulfur has six valence electrons. And also you could find this based on the group number.
Sulfur is in group 6A of the periodic table, and group 6A elements have the configuration NS2 and P4. Notice that these numbers add up to 6. So all of the calcogens like sulfur, selenium, oxygen, their configuration always end in NS2NP4. They have 6 valence electrons.
To find the core electrons, you can count everything else. 2 plus 2 plus 6 is 10. Or if you have a very large atom where the configuration is very long, simply take 16, subtract it. by the valence electrons and you'll get the core electron 16 minus 6 is 10 giving you this number now sulfur is it paramagnetic or diamagnetic to answer it realize that everything before 3p4 is going to be completely filled so to answer the question you just need to look at the last part the 3p4 so this is going to be 3p1234 notice that we have two unpaired electrons so since we have unpaired electrons, it's paramagnetic. Now here's a question for you, if we have two unpaired electrons, how many are paired?
Well we know the total is 16, so if two are paired, I mean if two are unpaired, then 16 minus 2, the other 14 must be paired. So sulfur has two unpaired electrons, 14 paired electrons. Now let's move on to another element.
Let's try aluminum. What is the electron configuration for aluminum and the aluminum plus 3 cation? Feel free to pause the video and work out this example. So aluminum has an atomic number of 13. So following the same trend, we know we have the 1s orbital, then 2s. 2s, 2p, 3s, and 3p.
So it's going to be 1s2, 2s2, 2p6. So we have a total of 10 at this point. 3s2, which is 12. We need one more to get to 13, so we're going to stop at 3P1.
Now what about the aluminum plus 3 ion? How many electrons does it have? To count the electrons, it's going to be the atomic number minus the charge.
So the atomic number of aluminum is 13 and the charge is positive 3. So it's 13 minus 3, therefore the aluminum plus 3 ion has 10 electrons. So we need to write the configuration until it gets to 10. So we got to stop at 2p6. So for the most part, just take off these three electrons and that will give you the electron configuration for the aluminum plus 3 cation. So it's 1s2 2s2 2p6.
That is the ground state electron configuration. If you ever wonder what is the difference between ground state and the excited state, The ground state is simply the state where the electrons have their lowest possible energy levels. If an electron absorbs energy, it can jump to a higher energy level, and it's in the excited state.
So for example, aluminum has this configuration. But let's say if you see the configuration 1s2, 2s2, 2p6, 3s2, and instead of 3p1, let's say if you see 4s1. What that means is that the 3P1 electron, they jump to the force level. So this could still represent aluminum, but it's no longer the ground state of electron configuration.
This would be an excited state. It's still aluminum because the total still adds to 13. 2, 2, and 6 is 10, plus 2 and 1 is still 13. It simply represents an aluminum atom in the excited state. So if you ever see an electron configuration where... There's a jump. If you're missing 3p, if it goes from 3p to 4s, or if it jumps to a high level, you know it's in the excited state.
Now, how would you write the ground state electron configuration for aluminum used in normal gas notation? So, this time, we're going to have to use neon, because neon has an atomic number of 10, which is less than 13, but still close to it. So it's going to be neon, and neon is going to replace everything up to 2p6, and then we're going to write what's after that.
So neon 3s2 3p1. So that's the configuration using noble gas notation for aluminum. Now how can you write the orbital energy level diagram for aluminum?
Not the orbital diagram that we did before, but the orbital energy level for aluminum. So sometimes, you may see a chart that looks like this. This is 1s, 2s, 3s, and above 2s, we have 2p and also 3p.
The 1s orbital is closest to the nucleus, so it's going to have the lowest energy. The 3p sublevel is further away from the nucleus, so it has the highest energy compared to all the sublevels that are listed here. Now, according to the off-ball principle, we need to fill the lower energy levels first with electrons before we fill the higher energy levels.
So, we need to fill the 1s level first, and then the 2s level. Now, once we get to 2p, we have three orbitals of equal energy. So, in this case, Hund's rule applies, which states that whenever you're adding electrons to degenerate orbitals, that is, orbitals of equal energy, you need to add each electron one at a time to an orbital with parallel spins. So, once you add the first electron, you don't want to add the second one in the same orbital.
the empty orbital but with a parallel spin so if the first arrow is going up the second arrow must also be going up and this is going to be 3 4 5 6 now that we're finished with 2p orbital we need to go up to the next orbital 3s so that's filled and then 3p1 now this substance aluminum is it paramagnetic or diamagnetic Since it has an unpaired electron, it's paramagnetic. Now how many S electrons does aluminum have, and how many P electrons does it contain? So aluminum has 6 S electrons, 1s2, 2s2, and 3s2.
In terms of P electrons, it has 7, 2p6, and 3p1. Now how many valence electrons does aluminum have, and how many core electrons does it contain? So the valence electrons are the electrons in the highest energy level, in this case the third energy level.
So we have 3s2, 3p1. 2 plus 1 is 3, so aluminum contains three valence electrons. If you look at the periodic table, aluminum is in group 3a, and elements in group 3a contain three valence electrons. Now the core electrons is going to be 13 minus the three valence electrons, which is 10, which correlates to this part, 1s2, 2s2, 2p6.
Now how many paired and unpaired electrons does aluminum contain? So we can clearly see that it has one unpaired electron, and 13 minus 1 is 12, so it has 12 paired electrons. 2, 4, 6, 8, 10, and 12. Now let's try an example using a transition metal. We're going to use cobalt as our example.
Cobalt has an atomic number of 27. So go ahead and write the ground state electron configuration for cobalt and also the noble gas notation. So we have 1s, 2s, 3s, 4s, and then 2p, 3p, 4p, 3d, and 4d. So as you mentioned before, S can have a maximum of two electrons. It has one orbital. P has three orbitals, which can contain up to six electrons.
D has five orbitals, which can hold a maximum of ten electrons. So let's start with 1S. After 1S, it's going to be 2S, 2P, and then 3S.
After the 3s orbital, it's 3p, then 4s. And then after 4s, we have 3d, then 4p. So we said s can hold up to 2. So it's 1s2, 2s2. p can hold up to 6, so 2p6.
Right now, we have a total of 10 electrons. Our goal is to get up to 27. So then it's 3s2, that's 12. 3p6, 12 plus 6 is 18, 4s2, that's 20. So we need 7 more to get to 27, so we're going to stop at 3d7. So that is the ground state electron configuration for cobalt. Now what is it, what is the ground state configuration using noble gas notation? Which noble gas will you use?
Now we don't want to use neon, which has an atomic number of 10. We want to use argon, which is 18. Krypton is too high. Krypton has an atomic number of 36, which exceeds 27. So we're going to use argon. So argon has an atomic number of 18, which equates to everything up to 3P6.
For Neon, it's 2p6. Argon, 3p6. Krypton, 4p6.
So everything up to 3P6 will add up to 18. 2, 2, and 6 is 10. 2 and 6 is 8, so that's 18. And then we're going to write everything that remains. So it's argon, 4s2, 3d7. So that is the electron configuration for cobalt using noble gas notation.
so in cobalt how many p electrons are there and how many s and d electrons do we have so let's start with s so we have two four six eight s electrons in the cobalt atom now how many p electrons are there so it's 2p6 and 3p6. 6 and 6 is 12 so there are 12 p electrons in cobalt. Now how many d electrons? We only have 3d and it goes up to 7 so it has 7 3d electrons.
Now, cobalt, is it paramagnetic or diamagnetic? And how many paired and unpaired electrons does it have? So to answer this, let's write...
the orbital energy level so I'm going to rewrite the electric configuration it was 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 7 so we're going to have the 1s level 2s 3s and 4s and then after 2s we have 2p 3p and after 4s we have 3d the d sub level has five orbitals So 1s2 is completely filled, and then 2s, so we've got to fill the lower energy levels first according to the off-ball principle, and then 2p6, and then 3s2, 3p6. 4s2 and then 3d7 so according to Hunts rule we need to add one electron at a time with parallel spins so 3d3 4 5 6 7 So cobalt has a total of 27 electrons as an atom. And as you can see, it has 3 unpaired electrons. So if 3 are unpaired, then 27 minus 3, the other 24... are paired electrons and you can count it this is 2 4 6 8 10 12 14 16 18 20 22 24 now is it paramagnetic or diamagnetic since it contains unpaired electrons overall cobalt is a paramagnetic substance Now that you know the electron configuration for cobalt, write the electron configuration for the cobalt plus 2 ion and the cobalt plus 3 ion.
Now be careful when you're dealing with transition metals, because the way you remove electrons from transition metals might differ from that of representative elements. So first let's calculate the number of electrons. The cobalt plus 2 is the number of electrons.
ion has 25 electrons. To acquire a charge of plus 2, it must lose 2 electrons. For cobalt plus 3, it lost 3 electrons, so it has 24 electrons. Now we already know the configuration for the cobalt atom. It's 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d7.
So using that, what is the configuration for the cobalt 2 plus ion? Now we know that we need to remove two electrons to get 25 electrons. Should we remove two electrons from the last 3D sublevel? Or should we remove it from the 4S sublevel? rule is that you want to remove the electrons from the highest energy level first.
Now, you don't want to remove it from the 3D, you want to remove it to the 4S, because 4 has a higher number than 3. So generally speaking, remove the electrons from the fourth energy level before you remove it from the third energy level. So we're going to take off two electrons from the 4s2 sublevel. So it's going to be 1s2, 2s2, 2p6, 3s2, 3p6, 4s0.
Sometimes they won't even write 4s0, but I'm going to write it. 4s0 and then 3d7. Now since 4s is empty you can ignore 4s. You don't have to write it if you don't want to. But just so you can see what happened I'm going to leave it there.
Now for cobalt plus 3 we need to take one more electron from the cobalt plus 2 ion. So there's no more electrons to remove from the 4s level. So the next best thing is to remove it from the 3d sublevel. So this is going to be 1s2. 2s2, 2p6, 3s2, 3p6, and then 3d6.
So that's how you can write the configuration for a transition metal cation. There's one more topic that we need to cover for electron configurations, and that is the exceptions. The exceptions that you need to be aware of is chromium, and below chromium, molybdenum.
And then to the right, copper, silver, and gold. We're going to cover these two exceptions, chromium and copper. They're very common on a typical chemistry test, so if you see them, know that they're exceptions. So let's start with chromium. Chromium has an atomic number of 24. So go ahead and write the ground state configuration for chromium.
So starting with 1s, it's going to be 1s2, 2s2, 2p6. So right now we have a total of 10. And then after 2p6... comes 3s. So that's going to be 3s2.
And then after the 3s level, it's 3p then 4s. So 3p6, 4s2. So at this point, we have a total of 20. So we need 4 more.
After 4S, we have the 3D sublevel. So then it's 3D4. So this is the ground state configuration.
Now what is the configuration used in noble gas notation? So we're going to have to use argon. Argon ends at 3P6, so argon covers everything up to 3P6. So it's argon, 3P, I mean argon, 4S2, and then 3D4.
So right now, it appears as if that's the configuration for chromium. But we have to adjust it, because that's not the final answer. So, if we write the orbital energy level diagram for just the force 2 3d4 level, notice what happens. So, if we write the orbital energy level diagram for just the force 2 3d4 level, notice what Now, initially, it appears as if the force level is completely filled, and the 3D level is not. So notice that we have this empty orbital.
This electron, it doesn't want to share the force orbital with the other electron. Because of electron... repulsion there's a lack of stability so this electron it sees the empty orbital and it wants to jump to this orbital now it can jump because the forest level and the 3d look sub level they're not too far apart in terms of energy.
They're very close to each other. So the amount of energy that's needed to jump from the 4s to the 3d level is compensated by the stability gain when it reduces electron repulsion by having its own orbital. So it's more stable for this electron to jump to a higher energy level and be unpaired rather than remain paired in the force level.
So it does it. And so this is going to be the orbital energy level diagram. It's going to look like this. So what we're going to have is 4s1, 3d5.
So this is not the correct configuration. It's going to be argon, 4s1, 3d5. So all you've got to do is take an S-electron and add it to the 3d sublevel. And the same is true for copper as well.
You can do the same thing. Now. Now how many unpaired electrons do you see that chromium has?
So be careful, if you're not aware of the exception, you'll pick 4. But the answer is not 4. Chromium has 6 unpaired electrons, so it's very paramagnetic, so to speak. So if you get a question that asks you which element has the most number of unpaired electrons, chances are it's going to be chromium if it's not an element. element if it's not an inner transition element, because that can have up to seven unpaired electrons, perhaps even more. But if you don't have any elements such as the lanthanides and antonides, for the most part, chromium or...
molybdenum, they're going to have a very high number of unpaired electrons. Now, if chromium has 6 unpaired electrons, how many paired electrons does it have? So then all you have to do is 24 minus 6, so it has 18 paired electrons. Now let's write the configuration for copper, but using noble gas notation. So copper has an atomic number of 29, and we're going to start with argon.
We know argon. contains 18 electrons and it's everything up to 3p6 so after 3p6 it's 4s then 3d so 4s2 would give us a total of 20 we need nine more so 4s2 3d9 now we know this is an exception so the real answer is going to be argon 4s1 3d10 So, looking at the 4S and the 3D sublevel for the first one, initially the 4S is completely filled, and the 3D sublevel is not. But for some reason, this electron wants to jump to the 3d sub-level and it just works out that way so that's something you'll have to know for the test so for copper it's not going to be forced to 3d9 it's going to be 4s1 3d10.
For some reason, the 3d sub-level prefers to be paired rather than the 4s sub-level. So make sure you're aware of that for copper. And the same is true for the elements below copper, silver and gold. The s-electron is going to jump to the 3d sub-level.
So that is it for this video. Thanks for watching and have a great day.