In this video, we're going to cover topics such as quantum numbers, NL, ML, MS. Also, the sub levels SPDF, things like that. We're going to go over electron configuration, how to determine how many paired or unpaired electrons are in an element, if it's paramagnetic, diiamagnetic, and a lot of other questions, too. So, you'll just have to watch and see how it's going to be. But let's start with electron configuration. So let's say if you want to write the electron configuration for florine, how do you do that? Now if you look at the periodic table, florine has an atomic number of 9 and the mass number is 19. A neutral atom of florine has an equal number of protons and electrons. The atomic number always represents the number of protons. But as a florine atom, it's going to have nine electrons. Atoms have equal number of protons and electrons. But for ions they differ in protons and electrons. Now in electron configuration you have things like 1 s, 2 s, 3s, 4s. In the first level one is the energy level. You only have one orbital s. In the second level two, there's two sub levels, 2 S and 2 P. In the third energy level, there's three sub levels, 3 S, 3 P, 3D. And in the fourth energy level, you have four sub levels, 4 S, 4 P, 4 D, and 4F. The S suble can hold up to two electrons. P can hold up to six. D can hold up to 10 and F can hold up to 14 electrons. So how can we use this to write the electron configuration? So the first level is going to be 1 S. Now S can hold two. So it's going to be 1 S2. After 1 S we have the 2S suble. So it's going to be 2S2. After 2S, we're going to move on to the P suble. So, it's 2 P, then it's 3S, but we're going to stop at 2 P5. Notice that if you add the exponents 2 + 2 + 5, that is equal to the atomic number 9. And that's where you want to stop. This is the electron configuration for an atom of florine. So, that's how you can find it for an element. So using that electron configuration, how can you represent it with an orbital diagram? In chemistry, you'll see boxes that looks like this. Each box represents an orbital. An orbital contains at most two electrons. S only has one orbital. So this is 1 S. This is 2 S. But P has three orbitals. Now remember we said that S can hold up to two electrons. Because it has one orbital, it can hold up to two electrons. You can only put two electrons in a single orbital. P can hold up to six electrons because it has three orbitals. Now let's fill up the orbital diagram with electrons. So 1 S has two electrons. One of them is going to be represented as an up arrow, the other one as a down arrow. And then 2s also has two electrons. And for 2p, we need to stop at five. But you should add it with parallel spins one at a time. So 2 p1, 2p 2, 2p3. After you add the up arrows, you can add the down arrows. 2p4 and 2p5. So that's how you can represent the electron configuration using an orbital diagram for the element florine. Now here's a question for you. Florine, would you consider it to be a paramagnetic substance or a diamagnetic substance? Elements that have unpaired electrons are paramagnetic. They are attracted to an external magnetic field. Elements that do not have unpaired electrons, if it's completely paired, then it's set to be diamagnetic, they are weakly repelled by a magnetic field. So let's try another example. Consider the element phosphorus. Phosphorus has an atomic number of 15 and a mass number of 31. Write the electron configuration for phosphorus. That is the ground state electron configuration and also determine how many unpaired electrons it has and how many paired electrons it has. And then predict if the substance is going to be paramagnetic or diamagnetic. It helps if you write this first. So we need to go up to 15. So let's start with 1s. It's going to be 1 s2 and then after 1 s we have the 2s orbital. So 2 S2 and then it's going to be 2 P then 3 S. So 2 P6 3 S2. Right? Now if you add the exponents 2 + 2 + 6 is 10 + 2 is 12. So to get to 15 we need three more. After 3S we have the 3 P suble. 3p can go up to six, but we're going to stop at three. So, we can get a total of 15. So, this is the electron configuration for phosphorus. That's how you can write it. But now, how would you write the orbital diagram for phosphorus? So, I'm just going to rewrite the ground state electron configuration. Actually before you do that, before we write the orbital diagram, how can you represent this ground state electron configuration using noble gas notation? How can you do that? So if you look at the periodic table, you want to look at the column that contains the noble gases such as helium, neon, argon, krypton, and you want to pick a noble gas that has an atomic number that's less to 15, but that's very close to 15. In this case, the one that you want to pick is neon. Neon has an atomic number of 10. So you can replace neon with the electron configuration the portion of it that adds up to 10 which is 1 s2 2 s2 2p6 and then after that you can write what's left over that is the 3s2 and 3p3. So this is the electron configuration for phosphorus using noble gas notation. Now how can we write the orbital diagram for phosphorus? So let's start with 1 S. So we have the 1 S box and 2 S. So remember S has only one orbital. Next we have 2 P. P has three orbitals and then it's 3 S and then 3 P. So we have 1 S2 which is going to be completely filled. Then 2 S2 2 P6 2 P1 2 3 4 5 6. Three acids filled and 3 P. This is going to be 3 P3. 1 2 3. Whatever you do, don't add it like this. You don't want to say one, two, three. You should add it one at a time with parallel spins. If the first one is up, the second is going to be up and the third is going to be up as well. So now, how many unpaired electrons does phosphorus have? As you can see, phosphorus has three unpaired electrons. Now, how many of its electrons are paired? So you have 2 4 10 12. Another way to calculate the number of paired electrons is to take the atomic number minus the unpaired electrons. The atomic number is 15. The number of unpaired electrons is 3. So 15 minus 3 will give you 12, which is the number of paired electrons. Finding the answer that way can be very helpful especially if you have an element with a lot of electrons. So what we have here is the orbital diagram. Now there's another diagram you can draw. It's the orbital energy levels and it looks like this. 1 s has the lowest energy. So it's going to be at the bottom since it's closest to the nucleus. And then after that you have the 2s suble. And then slightly higher than 2s you have 2p which contain three orbitals. Then after 2p it's 3s and then after that you have 3 p. Now according to the alphab principle you need to add electrons to these orbitals in order of increase in energy. So you have to start at the bottom and work your way to the top based on the alpha principle. So once we add the first electron, we can't add the second electron here unless it's in its excited state or something, but you don't want to do that. So after you add the first electron, add the second electron to the 1s uh level. Next, move up to the 2s and add the two electrons. So that's the off principle. You have to add electrons in order of increasing energy levels. So here's a pop quiz for you. According to the off principle, should we add the next electron to the 2p suble or the 3s suble? So based on the off principle, you should add it to the 2p suble because 2p is lower than 3s. So you got to add it to the 2p suble first. Now notice that we have three orbitals in a 2p suble that have the same energy. Orbitals that have the same energy are known as degenerate orbitals. So how should we add these electrons? I mean should we add it here first? Should we add it there? Should we add it like this? Because they have the same energy level. So the offbar principle doesn't apply here. A different principle applies. What do you think that is? when adding electrons to the same energy level. There's something called Huns rule. And Huns rule states that whenever you're adding electrons to degenerate orbitals or orbitals of the same energy like these, you need to add them one at a time with parallel spins before adding electrons with opposite spins. So after adding the first one, you need to add the second one with a parallel spin. So, if the first one's going up, the second one must be going up, too. And you have to add it to a different um orbital, not the same orbital. You don't want to add it like this. You want to add electrons to an orbital one at a time according to Hunter rule. So after adding the first one, you want to add the second one to a different orbital with a parallel spin and then add the third one with the same parallel spin. Now why do we have to do it like that? Why do we have to add electrons one at a time instead of adding the second electron in the same orbital? Why can't we just do it like this? Well, one reason is that electrons, they have negative charges. And whenever you have two particles with the same charge, they will repel. So electrons, they don't like to share the same room unless they have to. So if there's another room that's unoccupied, the electron prefers to occupy that room rather than share this room with another electron. Imagine if you have a brother or sister. Would you prefer to share rooms or would you prefer to have your own room? So, electrons, they like to have their own space. So, if it's up to them, they prefer to have their own orbitals because it minimizes electron repulsion. Now, once you add the fourth electron, it has a choice. Do you think it wants to go to the 3s level or do you think it wants to share a room with another electron? What would you say? So when dealing with different orbitals with different energy levels, in this case 3s and 2p, the off principle takes into effect. And remember, we need to add electrons to orbitals that have lower energy levels. So we can't add it to the 3s yet because 3s is significantly higher than 2p. So the fourth electron will occupy the lower energy level. And this time it has to share a room with another electron. It has no choice. But it's going to have the opposite spin. It's not going to have the same spin. Two electrons within an orbital will always have opposite spins. So now we can add the other two electrons to complete the 2p suble. So you can see why it says 2p6. There's a total of six electrons in the 2p suble. And as you can see each orbital can hold up to two electrons. So now we need to move on to 3s. According to the off principle we need to add electrons to 3s before we add it to 3p since 3s is lower in energy than 3p. Now once we move on to 3p since these orbitals have the same energy level they're horizontal to each other. Now Hun's rule takes into effect. So we have to add electrons to orbitals one at a time with parallel spins. So one 2 3. Now that we reach 3p3, we're going to stop there. So we can see that this element has three unpaired electrons and 12 paired electrons. So is it paramagnetic or is the substance diamagnetic? If you have at least one unpaired electron, it's going to be paramagnetic. The more unpaired electrons an element has, the stronger the interaction between that element and a magnetic field. Now, why is it that elements that contain unpaired electrons are said to be paramagnetic, whereas those that contain only paired electrons are said to be diamagnetic? Well, physics 101, a moving charge will create its own magnetic field. Within an atom, you have electrons that are in constant motion. And these electrons, they revolve around the nucleus and they can also spin about their own axis. And as they're moving, they create their own magnetic field. So they act as tiny barb magnets with a north and south pole. So here's a typical barb magnet. The magnetic field leaves the north pole and it enters the south pole. And it creates a pattern that looks like this. Well, the electron is the same way for the electron that we have in this particular example. This is going to be the south pole and this is going to be the north pole. So the magnetic field is facing the upward direction. So it looks like that. So this particular electron has an electron spin of positive one2. Notice that we have an up arrow. Now for an electron that has the opposite rotation, it's going to have the opposite spin. So this is now going to be the south pole and this is the north pole. And so the spin will be negative a half. So these spins represent the direction of the magnetic field created by the moving electron. Now what happens if you take two magnets and you put the north poles facing each other? Based on what you know, will these two magnets will they attract each other and connect or will they repel? We know that the magnetic field leaves the north pole of the magnet. Notice that these magnetic fields, they're opposite to each other. So, they repel. They go against each other. And so, what happens is these two magnets will move apart. They will feel a force that's going to push them away. Now what happens if you face the north pole of one magnetic field with the south pole of the other. The magnetic field from the first magnet will leave the north pole and from the second magnet will enter uh the south pole. So notice that these two magnetic fields are aligned. they're in the same direction and so they strengthen each other. What happens is you have a force of attraction. So in this case these two they attract to each other and they're going to connect. Now if you go back to physics whenever you have two vectors that are parallel to each other let's say if you have the first vector which is 100 newtons and another one which is 150. These two vectors are additive. they will create a larger force vector. If you have two vectors that are opposite to each other, they sort of cancel out so to speak. And in this case, the net force is the difference between 150 and 80. It's 70. So when the magnetic fields are aligned, it creates an additive effect. It creates a stronger magnetic field. when they oppose each other, when the vectors are opposite to each other, they sort of cancel out. Now, let's go back to our example with phosphorus where we drew the orbital diagram and it ended in 3p3. So, notice that the electron spins are parallel. And whenever you have this, the element is set to be paramagnetic. You could think of the electron spin as the direction of the magnetic field. And because they're going in the same direction, these three arrows add up to form a bigger arrow. And so the more unpaired electrons that an element has, the stronger its interaction will be with a magnetic field. Now in contrast, let's say if we have the element helium which has only two electrons, it has the electron configuration 1 s2. So notice that it doesn't contain any unpaired electrons. And notice that the spins are opposite. So for one electron, the magnetic field is in the upward direction and for the other one is in the downward direction. So these two they cancel and so this substance will not be attracted to a magnetic field. So therefore it's diiamagnetic. But this one is paramagnetic. The spins of those three electrons they can align itself with a magnetic field. And so paramagnetic substances are attracted to a magnetic field whereas diaminetic substances they're weakly repelled by it. Which atom or ion contains the greatest number of unpaired electrons? Is it a b c d or e? Which one would you pick? So let's write the configuration for each. So first we have the phosphide ion. Now we know the electron configuration for phosphorus using noble gas notation. It's a neon which equates 1 s2 2 s2 2p6 and then it's 3 s2 3 p3. Now for the phosphide ion it has ag3 charge which means that it has three more electrons. Phosphorus has a total of 15 electrons. So you have 10 for neon plus 2 + 3 that's 15. The phosphide ion which has a -3 charge that means that there's three more electrons than protons. So it doesn't have 15 electrons it has 18 electrons. Remember ions have unequal number of protons and electrons. The atomic number for phosphorus is 15. So phosphorus whether as an atom as P or as an ion as P minus 3 it has 15 protons but P minus3 has 18 electrons. It gained three more electrons. So we need to add three to the electron configuration. So it's not going to be 3 P3. It's going to be 3 P6 which adds up to 18. Now if you draw the orbital diagram for the 3p orbital the very last one everything before it's going to be filled. So we don't have to look at that 3p6 is completely filled. So therefore the phosphide ion has no unpaired electrons. So let's write a zero for it. Now what about Fe iron metal? How would you write the ground state electron configuration and also the configuration using noble gas notation in order to determine how many unpaired electrons it has. So Fe has an atomic number of 26. We don't have to worry about the mass number. And if you're ever unsure about how to write it, use this technique. Make a list of all the orbitals in this order. So in the first energy level, you only have the 1 s orbital. In the second energy level, you have 2 p. In the third, you have 3 p and 3d. And in the fourth, you have 4 p, 4d, 4f. So remember, s can only hold two electrons. P can have up to six. D can have up to 10. And f can hold 14. So starting with 1 S it's going to be 1 S2 and then after that 2 S2 and then 2 P6. So right now we have a total of 10. 2 + 2 + 6 is 10. After 2 P6 is 3 S2. And then after that we have 3 P and 4S. So 3 P6 and then 4 S2. After that we have the 3d suble. Right now if you add the electrons this is 10 plus another 10 we have a total of 20 but we need to stop at 26. So we need six more. So we're at the 3d level. D can hold up to 10 but we just need six. So this is the ground state electron configuration for an atom of fee. Now, how can we write the electron configuration using noble gas notation? So, once you have the ground state configuration, here's what you could do. So, look at the periodic table. Pick a noble gas that has an atomic number that's less than 26, but very close to 26. Helium has an atomic number of two. That's too far away. Then we have neon, which is 10. Argon is 18. Krypton is 36. Krypton is too much because that exceeds 26. So the closest one that's just under 26 is argon. So we're going to have argon. Now the noble gases usually end in P6 with the exception of helium which ends at 1 s2. Neon ends at 2p6 because neon has an atomic number of 10. Argon ends at 3 p6 because its atomic number is 18. 2 + 2 + 6 + 2 + 6 is 18. So argon is going to replace everything up to 3p6. So then we need to write the stuff that comes after 3p6 which is 4s2 3d6. So that is the electron configuration for fee using noble gas notation. Now once we have this we can write the orbital diagram or you can represent it using the orbital energy level. Let's draw the orbital energy level for Fe. So we have the 4s level and we have the 3d level. 3d has five orbitals. So you can draw boxes if you want. Now according to the off principle we need to fill up the lower energy levels with electrons first. So once we get to 3D where we have five orbitals of the same energy puns rule apply. So we'll need to add electrons one at a time with parallel spins. So that's 3d1 2 3 4 5 6. Now we don't have to worry about any orbitals that is less than 3d. So like 4s, 2p, 3s, all of the sub levels below 3d, it's going to be completely filled with electrons. So it's going to be completely paired. So if you're looking for the unpaired electrons, simply look at the last suble. So notice that we have four unpaired electrons. So therefore, Fe is a paramagnetic substance. Phosphide as an ion is diamagnetic since it has no unpaired electrons. So now let's move on to Fe plus2. Now that we have the electron configuration for Fe, how can we write it for Fe plus2? Now if you remember for the phosphide ion we had to add three to the number of electrons since it has a negative charge. So we increase the electron configuration by three. Now for the Fe plus2 ion because it has a plus2 charge that tells us that there's two more protons than electrons. Now Fe and Fe plus2 they both contain 26 protons since the atomic number is always 26 for Fe. However, when you see a plus2 charge, what that means is that the ion, in this case, the cation lost two electrons. Cations are ions that contain positive charges. Annions are ions that contain negative charges. So, P minus3 is an annion. Fe plus2 is a cation. So Fe plus2 contains 24 electrons whereas Fe contains 26 electrons. So the configuration is going to be argon and then something after that. So we need to remove two electrons from one of these subs. Should we remove it from the 4s suble or the 3d suble? Typically, you want to remove the electrons first from the higher energy level. In this case, the 4s suble. So, it's going to be 4s0 3d6. Which means that you don't have to write the 4 zero. So, you could simply say 3d6. So, that's how you can write the electron configuration of a positively charged ion. simply remove electrons from the electron configuration of the parent atom based on the charge that you see. So if you see a plus2 charge, take away two electrons starting from the higher energy level. So now that we have the configuration for the Fe plus2 cation, let's draw the orbital diagram for the 3D suble. So in this case, we're going to have five boxes and it's just going to be like Fe. Once we add the first five electrons, the next one is going to be unpaired. I mean paired. So as we can see, this one still has four unpaired electrons. So the Fe plus2 cation is paramagnetic. Now what about the Fe plus3 cation? it lost a total of three electrons. So 26 - 3 is 23. So it contains a total of 23 electrons. Now the first two electrons that we're going to remove is going to come from the 4s. Now we need to take another electron. So we're going to take the third electron from 3d. So it's going to be 3d5. Argon is 18 plus the five you get a total of 23. So if we draw the orbital diagram, we're going to have a total of five unpaired electrons. So this substance is also paramagnetic, but right now it's in the lead. So now the last thing we need to look at is chromium. How many unpaired electrons does chromium have? For the sake of practice, go ahead and write the configuration for the chromium atom. Chromium has an atomic number of 24 and it has a mass number 52. The atomic number tells you the number of protons. So it has 24 protons. And the mass number is the sum of the number of protons and neutrons. So if you do 52 minus 24, that will give you 28. So, chromium has 28 protons and it also has 24 electrons. An atom has an equal number of protons and electrons. So, let's use the same technique to find the ground state configuration for chromium. I don't think we need 4F. So, it's going to be 1 S2 and then after 1 S we have 2 S2 and then it's 2 P then 3S. So, 2 P6. So we have a total of 10 so far and then 3 S2 and then after that 3 P then 4S 3 P6 4 S2 so at this point we have a total of 20 so we need four more and then 3d4 now if you want to write the noble gas notation we're going to use argon again which is going to cover everything up to 3p6. Now you need to be careful because chromium is one of the exceptions. So it appears as if the electron configuration is argon 4s2 3d4. But it turns out that the actual configuration is 4s1 3d5. The question is why? Let's explain it using orbital energy levels as opposed to orbital diagrams. Now, what you need to know is that the 4S suble is very close in energy to the 3D suble. The 3D suble is slightly higher, but not by that much. So, according to the off bar principle, we want to fill the lowest energy level first. So we want to fill the 4s suble before we fill the 3d suble. And it makes sense because atoms they want to arrange themselves in such a way to have the lowest energy configuration possible. And so that's why they do what they do. Now this electron is repelled by the other electron in the 4s orbital. And so there's some electron repulsion going on there. And that creates instability. So this electron wants to have its own room. So it sees this vacant orbital and it wants to go there. And so it does. It jumps to this empty 3d orbital. And for the electron to jump to a higher energy level, it requires energy. And so it's a less stable configuration. The energy loss to jump to that higher energy level is compensated by the stability of being unpaired. By jumping to the 3D suble, it becomes more stable because it minimizes electron repulsion and that stability gained from minimizing electron repulsion compensates for jumping to the higher energy level. Now granted, the only reason why this is possible is because the 4S suble is very close in energy to the 3D suble. If the 3D suble was significantly higher in energy, the electron would have no desire to jump to this higher energy level. It would rather be paired and be in a lower energy level than jump to a very high energy level. So only because the 3D suble and the four suble are very close to energy and that's why it jumps higher to become more stable and to be unpaired. Now, let's see if we can come up with an illustration to explain this concept. Imagine if you and a group of friends are staying at a hotel on the first floor. And on this day, the hotel is packed and there's not many vacant rooms. So, you're packed in this hotel room and there's not enough space and it's just uncomfortable and you want your own space. So, you wish to rent another room and there's only one more room left over and the elevator in the hotel is not working and it's a very very big hotel with many floors. Now, if the vacant room, the last vacant room is in the second floor, would you mind climbing up one flight of stairs to go to the second floor to get that room to have your own space and be comfortable? Or let's say if the vacant room is in the 50th floor, would you prefer to climb 50 floors every day that you're staying at that hotel to be comfortable or would you prefer to stay in the first floor with your friends where it's packed and be uncomfortable? What would you do? Now, most people, they wouldn't mind going up one flight of stairs to the second level to have their own comfortable space, but to climb 50 stairs, it's not worth the effort. You might as well stay on the first floor. And the same is true with electrons. To go from the 4s to 3D, it doesn't require much work. So, it's going to do it to be unpaired. But if the 3D level was very, very high, it's not going to jump. it's going to stay at the force level because it requires too much energy or too much effort to get up to that high level. So in this case since the 3D level is very close to the force level it doesn't take much effort for the electron to get there. So that's why we have this exception. So the orbital energy level for chromium is going to look like this. 4s1 3d5. Notice that we have six unpaired electrons. So therefore the answer is E. Chromium has the greatest number of unpaired electrons because it's an exception. Next question. Sometimes on a test you might be given the electron configuration and you may need to calculate or identify the element that corresponds to it. So which element has the following electron configurations? Let's say 1 s2 2 s2 and 2p3. Now while you work on that one I'm going to write a few more examples. So in the meantime, see if you can figure out which element corresponds to the following ground state electron configurations. So let's start with the first one. If you wish to identify the element, all you need to do is add the exponents. 2 + 2 + 3 is 7. And that's going to be the atomic number of the element. So if you go to the periodic table, look for the element with an atomic number of seven. The atomic number is always the smaller of the two numbers. So this is nitrogen. Nitrogen has an atomic number of seven. Now what about for the next one? If you add up the exponents 2 + 2 + 6 is 10 + 2 and 5, that's 17. So chlorine has an atomic number of 17. Now, how would you identify it if it's given to you in noble gas notation? So, for argon, look at the atomic number of argon. It's 18. So, if you add 18 to that's 20 + 8, you get 28. And using the periodic table, nickel has an atomic number of 28. Now for the last one we have is uh krypton which has an atomic number of 36 plus 2 that's 48 I mean that's 38 + 10 gives us a total of 48 and it turns out that cadmium has an atomic number of 48. So that's how you can identify the element if you're given the electron configuration. Now, sometimes you might receive a multiple choice problem where you need to identify which configuration is a ground state configuration and which one is an excited state. So, I'm going to give you a list. Determine if it's excited or if it's in a ground state. So looking at the first one, after 1 S comes 2S and after 2S comes 2 P. So there's no jump. So this is a ground state electro configuration. All of the electrons are in their lowest energy levels. Now looking at the next one, we have 1 S2, 2 S2. Before we get to 3S, the 2P suble should be filled. It should be 2p6 before we get to 3S. So we have a jump. Instead of having 2 P4, it jumped to 3S1. So this represents an excited state. Looking at the next one, 1 S2, 2 S2, 2 P6, 3 S2, 3 P3. It follows the appropriate order. So it's a ground state configuration. And for the last one after 2 P6 you should have 3 S2 or 3 S1 but instead we have 4S1. So there's a jump and it's in the excited state. So let's say if an atom has three electrons one, two and three. If we fill it like this, this is the ground state configuration. However, if we put the third electron in a higher level in a 2p suble as opposed to the 2s, it's in the excited state and that electron has a lot of extra energy if it can jump to a higher energy level. And so this is not a ground state configuration. For it to be in a ground state configuration, all the electrons have to fill the lower energy levels as much as they can. If they jump to a higher level, then they're in the excited state. Now let's move on into quantum numbers. There are four quantum numbers that you need to know. The first one is n, which is called the principal quantum number. N is associated with the energy level within an atom. And then there's L, which is the aumathol or the angular momentum quantum number and it's associated with the suble. And then you have ML, the magnetic quantum number, and that is associated with the specific orbital within the suble. And then there's MS, the electron spin, which can be up or down. So let's talk about N first. So if you think of the bore atom, let's say hydrogen, the first shell is the first energy level. This is the second and then this is the third. So this is n= 1, n= 2, n= 3 and so forth. So n represents the main energy level. L represents the suble. When L is zero, you have the S suble. When L is one, you have the P suble. And when L is 2, it's associated with the D suble. And when L is three, you have the F suble. S forms the shape of a sphere. P is a dumbbell. D usually is a clover leaf. And F is some complicated shape that is not always easy to draw. So what is the relationship between N and L? Now this is where you want to understand the electron configuration pretty well. So when N is one, L can only be zero. So you only have the 1s suble. There's no one p. When n is two, you have the 2s suble and the 2p suble. When n is two, there are two sub levels. That means when n is two, l can be zero or it can be one. Anytime you have the s suble, l is always zero. For p, l is one. Now when n is three, how many sub levels do we have? Now if you know the electron configuration pretty well, you know that we have a 3s, 3p and a 3d suble. So the number of sub levels is equal to the energy level. So if n is three, there's three sub levels. If if n is seven, there's seven suble. If n is 20, there's 20 sub levels. So when n is three, l can be three numbers. It can be zero for s, one for p and two for d. And finally when n is four, let's see if we can make some space. When n is four, there's four sub levels. So we have the 4S suble, 4 P, 4 D and 4 F. So for 4S, L is zero. For P L is 1. For D, L is 2. For F, L is three. Notice that L can at most be one less than N. So if N is four, L could be anywhere from 0 to 3. If N is 10, L could be anywhere from 0 to 9. If n is 100, L could be anywhere from 0 to 99. So therefore, we could say that L is less than or equal to N minus one. So if N is 500, L is less than or equal to 499. That means it could be zero up to 499. But L can't be 500. Now what about ML? When L is zero, ML can only be zero. When L is 1, ML can be 1, 0, or 1. When L is 2, ML could be -2, 1, 0, 1, or 2. And finally, when L is 3, ML could be -3, -2, 1, 0, 1, 2, and 3. Now we said ML is associated with the specific orbital within a suble. So when L is zero, we have the S suble and S only contains one orbital. ML is the number associated with that orbital which is zero. Now when L is one we have the P suble and for P there's three boxes or three orbitals and notice that when L is one ML could be -1 0 or one. So ML describes which orbital the electron is in. So it describes the orbital within the suble. Now what about when L is 2? When L is 2, we have the D suble and we know that for D, there's five orbitals. So if L is 2, it's going to be from -2 all the way to two. Finally, when L is three, we have the F suble and F contains seven orbitals. 1 2 3 4 5 6 7. So in LS3, ML could be anywhere from -3 to three. So every value for ml corresponds to a particular orbital within a suble. Now in each orbital the spin could be up or it could be down. MS is the electron spin. As we mentioned before, it's plus a half if you have an up arrow and it's minus a half if you have a down arrow. So let's say if you want to find the four quantum numbers of an electron, let's say if we're dealing with the 3p2 electron, what are the four quantum numbers for that electron? Every set of four quantum numbers specifies a specific electron within an atom. It's basically the address of an electron within an atom. According to Paulie's exclusion principle, no two electrons can have the same force set of quantum numbers. Every unique set of quantum numbers, every unique four sets or four quantum numbers specifies only one electron within an atom. So what is the value of n for this 3 p2 electron? N is the number that you see outside. That's three. Now what about l? Whenever you have the p suble l is one. For s l is zero. For d l is two. for f L is three. Now what is the value of ML and MS? So for the 3P suble there are three boxes and since L is one ML could be anywhere from -1 to 1. So if we are concerned with a 3p2 electron this is 3p1 this is 3p2. Notice that 3p2 is in the second orbital. So ml is zero. And because we have an up arrow, it's positive a half. If the arrow was down, it would be negative a half. Let's try some more examples. So let's say if we have the 4d9 electron and go ahead and write the four quantum numbers that corresponds to this electron. So let's start with n the principal quantum number which is associated with the energy level. This electron is in the fourth energy level. So therefore n is four. So now let's move on to the suble or the angular momentum quantum number. What is L for the D suble? So for S, L is zero. For P L is 1. For D, L is 2. So now let's find the values of ML and MS. So at this point, you want to draw the orbital diagram for the 4D suble. D has five orbitals. 1 2 3 and four five. So we need to get to 49. So this is going to be 1 2 3 4 5. We got to add it one at a time according to Hun's rule. And then 6 7 8 9. The ninth electron lands in the in this orbital where ML is one. So that's how you can find the value of ML. And because it's a down arrow, MS is negative a half. So that's how you can identify the four quantum numbers given the electron configuration or the configuration for that specific electron. Try this one. 5F7. go ahead and find the five I mean the four quantum numbers. So we can see that n is equal to five and for the f suble l has to be three. It's zero for s one for p two for d three for f. Now the f suble has seven orbitals. 1 2 3 4 5 6 and 7. And since L is three, ML could be anywhere from -3 to 3. So we're looking for the 5 F7 electrons. This is 1 2 3 4 5 6 7. So the seventh electron lands in the orbital where ML has a value of positive 3. And since the arrow is in the upward direction, the spin is positive a half. Now let's try this one. What about the 3s1 electron? So we can see that n is three for s. L is zero and s only has one orbital. So therefore ml is zero. Now there's only one orbital and that one electron it can be an up arrow or it can be a down arrow if you want to be technical. It can be either. So therefore the electron spin could be plus or minus a half. It could be positive a half or negative half. So the answer could be 30 0 1/2 or 30 0 negative half. Now let's talk about the relationship between L and ML. We said that if L is 2 for example, ML could be anywhere from -2 to 2. So you want to have or be familiar with this equation. ML can be anywhere between L and L. So if L is 2, we can therefore say ML is greater than or equal to -2 and it's less than or equal to positive2. Now ML has to be an integer. It won't be a decimal value like 1.3 or something like that. It has to be an integer. Now let's get back to the electron configuration. Consider the following electron configuration. Let's say if it's 1 s2 2 s2 2 p6 3 s2 3 p4. Given this ground state electron configuration, how can you determine the number of veence electrons and core electrons? What do you think the answer is to that question? The veence electrons are simply the number of electrons in the highest energy level. The highest energy level is the third energy level. So we have 2 + 4. So there's a total of six veence electrons. The core electrons are the electrons found in the lower energy levels. In this case, the 1 s2, the 2s2, and the 2p6. So this particular element has 10 core electrons. The total number is 16. This represents sulfur which has an atomic number of 16 and an atomic mass of 32. Elements like sulfur, oxygen, selenium, these are known as calcens in the periodic table. These elements tend to have six aence electrons. They're found in group 6A or group 16 of the periodic table. So let's work on another example. So let's say if you have the configuration neon 3s2 3p2 how many veence electrons do you see here? So first let's identify the element. Neon has an atomic number of 10 and if we add two and two it's 14. So we know the element is silicon and we could see that there's four veence electrons and the highest energy level and 14 - 4 is going to give us the number of core electrons. So silicon has four veence electrons and it has 10 core electrons and it's found in group 4 A of the periodic table. Now what if you get a question that asks you how many p electrons are in the element florine? What is your answer? So how many p electrons does florine contain? The best way to answer these types of problems is to write the ground state electron configuration for florine. It's 1 s2 2 s2 2p5. Since 2 + 2 + 5 adds up to 9. So as you can see florine has five p electrons and notice that it has four s electrons. It has no d electrons. So let's work on another example. How many p electrons are in the element bromine? Bromine has an atomic number of 35. and a mass number of 80. So how many p electrons does broine contain? Now there's an easy way to get the answer. I mean you can always write the configuration and see it that way but if you have the periodic table in front of you um you can use it as well. So locate bromine in a p block. The P block is the upper right section of the periodic table. And a P block. Maybe I need to make this bigger. In the P block, you'll see elements like boron, carbon, nitrogen, oxygen, florine, neon. That's the uh 2p suble. This is 3 p. This is 4 p. Bromine is somewhere over here. Bromine ends in 4 p5. This is 1 2 3 4. So 4 p5. So if you want to count the number of p electrons that bromine has, here's what you can do. Count the number of elements in a p block up to bromine. This is 1 2 3 4 5 6 7 8 9 10 11 12 all the way up to 17. So bromine has 17 p electrons. That's how you can get the answer the fast way. Now granted, you can always do it the oldfashioned way. So if we write the configuration all the way to 35. Once we get to 2p6 that's 10 electrons. Then 3 s2 3 p6 4s2 that's 20. After 4s2 is 3d10. So we have a total of 30. And then comes 4 p5 which adds up to 35. So we have 6 + 6 + 5. So that tells us that we have 17 p electrons. Now how many s electrons does bromine have? You can see it's 2 4 6 8. So bromine has eight s electrons and it also has 10 d electrons. So that's how you can count the different types of electrons in a particular element. So let's go over one more example. How many electrons how many s electrons does strontium have? Strontium is atomic number 38. So how can we figure this one out? So using the periodic table, here's what you can do. So this is the s block and helium is also an s electron. So this is helium. Here we have hydrogen. Below hydrogen you have lithium, sodium, potassium, ribidium. And to the right you have burillium, magnesium, calcium and strontium. Hydrogen represents 1 s1, helium is 1 s2. Lithium is 2 S1 2 S2. This is 3 S1 3 S2 4 S1 4 S2 5 S1 5 S2. So the quick way to find the number of s electrons is to count the elements in the s block. So helium is part of the s block. So that's 2 3 4 5 6 7 8 9 10. So strontium has 10 ss electrons. Let's say if you get a test question that asks you what is the maximum number of electrons that can occupy the n equ= 10 energy level. How would you answer that? If you're not sure, let's talk about how it works. So first, how many electrons are in the first energy level? Now remember in the first energy level you only have the one s suble and s can only have two electrons. So there's a maximum of two electrons that can occupy the first energy level. Now what about in the second energy level? How many electrons can be in the second level? In the second level, you only have 2 s and 2 p. We know S can hold two, P can hold six. 2 + 6 is 8. So there's eight electrons that can occupy the second energy level. Now, what about the third energy level? So we have 3 S, 3 P, and 3d. So it's going to be 2, six, and D can hold up to 10. So there's up to 18 electrons in the third energy level. In the fourth, you have 4 S, 4 P, 4 D, and 4 F. So that's going to be 2 + 6 + 10, which is 18 + 14. F can hold up to 14 electrons. It has seven orbitals. So what we now have is 32. Now what about the fifth energy level? There's 5 S, 5 P, 5 D, 5 F. After 5 F comes 5G. And if N is six, after 6G comes 6H. So it's 2 6 10 14 and after 14 is 18. As you can see, for every new suble that you add, you always add four electrons or two orbitals to it. So 2 6 10 14 18, you keep adding four. So 32 + 18 is 50. So is there an equation that relates the energy level to the maximum number of electrons that can occupy the energy level? It turns out that there is. The equation is 2 and2. 2 and2 represents the maximum number of electrons in a given energy level. So when n is 1, it's 2 * 1^ 2. 1^ 2 is 1 * 2 gives you 2. When n is 2, 2 ^ 2 is 4 * 2. That gives you 8. When n is 3, 3 2 is 9 * 2 is 18. When n is 4, 4 2 is 16 * 2 is 32 and when is 5, 5 * 5 is 25 * 2 gives you 50. So therefore, how many electrons or what is the maximum number of electrons that can occupy the 10th energy level? So using equation 2 n^ 2 it's going to be 2 * 10^ 2 which is 10^ 2 that's 10 * 10 that's 100 * 2 that corresponds to 200 electrons. Now how many orbitals are in the 10th energy level? The number of orbitals is equal to n^2. Remember every orbital can hold two electrons. So if there's 200 electrons, you have a maximum of 100 orbitals. So it's simply 10^ squ, which will give you 100 orbitals. So the max number of electrons is equal to 2 n^2 and the max number of orbitals in an energy level is simply equal to n^2. So make sure you add these equations to your list in addition to the other equations already mentioned. Now let's talk more about orbitals and energy levels. So let's say if n is three, we know that the number of orbitals is going to be n^2, which is equal to 9. So we have the equation, but sometimes on a test, you might forget the equation, but if you understand it, you'll know what steps to take to find the right answer. So let's understand why the maximum number of orbitals is n^2. In the third energy level, we know that there's three sub levels. 3s, 3 p, and 3d. Now, we know the s suble only has one orbital. It has one box. P has three orbitals, three boxes. And d has five orbitals or five boxes. So if you add the total number of boxes, it's 1 + 3 + 5, which is equal to 9. So in the third level, there's a total of 9 orbitals. Let's try one more example. In the fourth level, we know that the number of orbitals is n^2 or 4^2, which will give us 16. Now in a fourth level, there's four sub levels. 4 S, 4 P, 4 D, and 4 F. So S has one orbital, P has three, D has five and F has seven. So if you add the numbers, this is 1 3 5 and 7. 1 + 3 is 4. 4 + 5 is 9. 9 + 7 is 16. So you can draw it out and you can literally count how many orbitals are in a given energy level. Now let's say if you're given two quantum numbers. Let's say if you're given a value of n and l. So let's say if n is three and l is zero and let's say n is four and l is 1. n is three, l is 2 and n is 5 when l is four or three. So for these different cases, identify the suble calculate the maximum number of electrons that can have these two quantum numbers and also determine the number of orbitals that can have those two numbers. So let's start with the suble. When n is three and when l is zero, what suble does it correspond to? When L is zero, it corresponds to S. So this is the 3s suble. S only has one orbital. So that's it for the number of orbitals. And for the 3s suble, since there's only one orbital, the most number of electrons you can have is two electrons. So the answer is 3s. That's the type of suble. We have one orbital and a max of two electrons. Now, what about the next one? What suble corresponds to a quantum number of n= 4 and an L value of one. So, L is one for the P suble. So, this is the 4 P suble. and P has three orbitals and there's a max of six electrons that can occupy those three orbitals. So how many electrons can occupy or can have the quantum numbers three and two for N and L? So when L is two, we have the D suble. So this is going to be 3d. And for D there's five orbitals, two electrons per orbital. So the most number of electrons that can have these two quantum numbers is 10. And what about the last one? When N is 5 and L is equal to three. L is three for the F suble. So this is 5 F. And we know that F can hold up to 14 electrons and there's seven orbitals. So that's it for that one. Now how many electrons can have the following three quantum numbers? Let's say when n is three, l is 2 and when ml is one. Now, since we have a specific type of orbital within the suble, the answer has to be two because this identifies only one box out of the five boxes that are in the d suble and you can only have two electrons in a given orbital. So, if you have n, l and ml, the answer is always three. I mean, it's always two electrons. So for the two when L is two we have the D suble. So this is 3D and there's five boxes 1 2 3 4 5. ML can vary between -2 and two. And so there's two electrons that have an ML value of one. So the answer is two. Now, how many electrons can have the following four quantum numbers? According to Paulie's exclusion principle, there's only one electron that can have four quantum numbers. So for four and three we have the 4F suble and we know F has seven orbitals which will vary between -3 and positive3. So the electron of interest is in this orbital and the spin tells us that it's the one that has the down arrow and there's only one electron that fits this criteria. Now let's say if you have the value of n and the spin, what is the maximum number of electrons that can have these two quantum numbers? What is your answer? Now remember in every orbital the spin could be either plus a half or minus a half. So that means that half of the electrons will have a down arrow and the other half can have an up arrow. Now we know the maximum number of electrons in the third edge level. It's equal to 2 n ^ 2 which is 2 * 3^ 2 and that's 18. So since there's a total of or a maximum of 18 electrons in a third energy level, nine of those electrons will have a spin of negative a half. So if you have 3s, 3 p and 3d, here's the 3s block. This is 3 p and this is 3d. There's no 3f for the third level. So each orbital can have a down arrow. Each orbital can only have one down arrow, not two, because the other one has to be up. So, as you can see, there's nine down arrows. So, it's going to be half of this number. So, there's nine electrons that can have these two quantum numbers. Now how many electrons can have the following quantum numbers? So if we focus on N and L, L is two for the D suble. So we have the 4D suble and we know that D can hold up to 10 electrons because D has five orbitals. So if D can hold up to 10 electrons, half of those 10 electrons is going to have the up arrow of positive a half. So the answer must be five. So it's going to be 1 2 3 4 5. So once you add MS to the equation, if you're looking for the maximum number of electrons with those quantum numbers, it's always going to be half of whatever the previous uh two numbers will be. So for example, let's say if we have the quantum numbers n and ml. Let's say n is three and ml is one. How many electrons will have these two quantum numbers? So nistry for well when nistry you can have the 3s suble 3p and 3d for 3s there's one orbital for 3p there's three and for 3d there's five for s l is zero so ml has no choice but to be zero for p l is 1 so ml could be1 to one and for D L is 2. So ML varies between -2 and 2. Now notice that there's only two boxes with an ML value of one. So there's only four electrons that can be in these two boxes. So the answer is four. There's four electrons that have the quantum numbers three and one where n is three and ml is one. Now let's say if we want to find the number of electrons that has these two quantum numbers in addition to a spin of negative a half then the answer is going to be half of four. It's going to be two. Which are these two? It has to be a down arrow. Let's try another problem like that. So let's say n is five and ml is equal to zero. How many electrons will have these two quantum numbers? So in the fifth level we have 5s, 5 p, 5d, 5 and also 5g. So for the s suble l is zero which means ml0. For the p suble l is 1 which means that ml will vary between -1 and 1. And for the D suble L is 2. So ML will vary from -2 to 2. Now for the F suble L is three and F has seven orbitals. So ML will vary from -3 to 3. And for the G suble L is four. So ML can vary from -4 to 4. Now we need to identify every orbital that has a zero value for ML. So this is one, two, three, four, five. So there's five orbitals with these two quantum numbers. And each of those orbitals can hold up to two electrons. So there's a total of 10 electrons with the quantum numbers n= 5 and ml= z. Now let's say if we add the spin to it. Let's say if the spin is negative a half then the answer is going to be half of what we have here. So half of the 10 electrons will have a down arrow. So the final answer is five that has these three quantum numbers but 10 electrons have the first two quantum numbers. Now let's say if you have a multiple choice question and it asks you which of the following configurations corresponds to a halogen would you say it's NS1 NS2 NS2 NP4 NS2 NP5 which one is it the hilogens correspond to NS2 SP5 5. The calcent correspond to NS-2 and P4. The alkaline metals correspond to NS1 and the alkaline earth metals NS-2. So let's talk about the alkaline metals. So you have lithium, sodium, potassium, rubidium, cesium, and so forth. All of these elements have the configuration ns1. For lithium, it's 1 s2 2 s1. For sodium, it ends in 3s1. Potassium 4s1. So all of the alkaline metals, they will end in ns1. And then you have the next group, the alkaline earth metals. So that includes burillium, magnesium, calcium, strontium and berium. So these elements they end in ns2, burillium is 2s2, calcium 4s2, strontium 5s2. And then you have the calcens like oxygen, sulfur, selenium. These they have the configuration ns2 and p4. If you write the electron configuration for oxygen it's 1 s2 2 s2 2 p4. Sulfur ends in 3 p4. For sulfur will be 3s2 3p4. So all the calcens have the configuration ns2p4. The hilogens like florine, chlorine, bromine, iodine it's ns2p5. Now for the noble gases like neon, argon, krypton, it's ns2 np6. The exception is helium which is simply 1 s2. But the other noble gases will be ns2 np6. So if you see this, this corresponds to the column or the uh group of elements in the periodic table. So elements in the same column, they share the same or similar chemical reactivity. oxygen, sulfur, selenium, they have similar configurations, NS2, NP4, and they behave in a similar chemical way since they have the same number of electrons. Well, what I meant to say is same number of veence electron. Now, it turns out that there's a way you could find out which column it corresponds to. For example, if you have Ns2 NP4, simply add the exponents 2 + 4 that gives you six. So this corresponds to the elements in group 6A. That's oxygen, sulfur, selenium. those to calc if you add two and five you get seven which corresponds to group 7A which would be the halogens like chlorine, bromine, iodine, florine. NS1 corresponds to group 1 A which is the alkaline metals. NS2 corresponds to group 2 A which is the alkaline earth metals, magnesium, calcium, strontium, things like that. So let's say if you have a few elements such as manganese, aluminum, germanmanium, phosphorus and xenon. Which element is or corresponds to the group that's ns2 np1. ns2 np1. If you add the exponents 2 + 1 is three. So this is going to be group 3a. So if you don't see like a d somewhere, it's probably not going to be a transition metal. So, which element is found in group 3 A or which element has three valence electrons? It's not going to be xenon that has eight valence electrons. It's not phosphorus. It has five. Phosphorus is in group 5 A and it's not germanmanium because that's in group 4 A. The answer is aluminum. It has three veence electrons and it's in group 3A. So, let's try another example. Let's say if you want to find the element that corresponds to the group NS2 and D7. So if it ends in D, you know it's a transition metal. So let's say the options are venadium, manganese, Fe, CO, and Ni. So 2 + 7 is nine. So it has to be group nine which is basically the ninth column. So this is going to be cobalt. Cobalt has the configuration 4S2 3d7. If you count it starting from the left, it's in the ninth column. So it's in group nine of the periodic table. Now the last topic of interest is determining if a set of four quat numbers if it's allowed or if it's unallowed. So let's say if N is 3, L is 2, ML is negative 1 and MS is plus a half. Are these four quantum numbers are they allowed or unallowed? Is there any error with these numbers? Now there's some things you got to remember. L is always less than or equal to N minus one. So if L is equal to or greater than 3, it's not going to work. So L is less than N, which is okay. ML has to be between L and L. So if L is 2, ML has to be between -2 and 2, which in this example it's okay. And MS can only be a fraction. So there's nothing wrong with these four quantum numbers. Now what about this one? Let's say if n is 4 and if l is 4 and this is -2 and if that's negative half is there anything wrong here. l has to be less than n. L cannot equal n. So it's wrong. Now let's say if n is 5, l is 2, ml is -3 and ms is plus a half. Is there anything wrong? So, first compare N and L. If L is less than N, then it's good. Now look at ML and L. If L is 2, ML has to be between -2 and two. -3 is outside of that range. So this doesn't work. So let's say if this is four, that's one zero. Let's say negative a half. So L is less than N which is good. And if L is one, ML has to be between -1 and one. So zero is within that range. So that's okay. And MS can be a fraction. So that works. Here let's try another one. Let's say if N is three, L is 2, ML is -2, and MS is one. Is it okay or is there an error? So let's compare L and N. L is less than N. So that's okay. And if L is 2, ML could be -2 to 2, which -2 is in that range. So that's okay. The only issue we have here is that MS is not a fraction. MS can only be plus a half or minus one over two. So this doesn't work. So that is basically it for this video. We've covered almost every topic that you'll see um in relation to electron configuration and quantum numbers. So, thanks for watching and have a great day.