Electron Configuration - Quick Review!

In this video we will focus on the

electron

ic

configuration

and how to write the orbital diagrams of the elements, so let's start with the element sulfur if we 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

electron

s. Atoms have the same number of protons and electrons, that's why they are neutral ions, they have an unequal number of protons and electrons, so because we are dealing with a sulfur atom sulfur has 16 electrons now how can we write the electron

configuration

of the ground state for the element sulfur?

This is what you can do in the first energy level, we only have one s sublevel, in the second energy level you have two sublevels 2 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 that far with sulfur now the s sublevel can hold up to 2 electrons p can hold up to 6 and d can hold a maximum of 10 electrons so we will start with 1s so it will be 1s2 and then after 1s, the next sublevel is 2s, so 2s2 and then after 2s it will be 2p, then 3s, now p can hold up to 6.

So 2p6 now we have a total of 10. 2 plus 2 plus 6 is 10 and we need to get to 16. So let's continue after 2p we have 3, so it will be 3s2, so we have a total of 12. So we need four more, now three p could hold up to six, but we don't need three p six, otherwise the total will be eighteen. We want to stop at three p four so that the total is equal to the number of electrons in this sulfur atom. This is the electronic configuration of the ground state of sulfur. Now, how can we represent it using orbital diagrams?

So if you want to write the orbital diagram for sulfur, this is what you can do. So we'll draw pictures. This is for the 1s orbital and then the 2s the s sublevel only has one p orbital it has three orbitals so this is 2p and then 3s and then 3p so one s is completely filled there are two arrows each of these arrows will have opposite spins, so in each 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 must add one arrow at a time with parallel spins, so one, two, three, after they are completely unpaired, the fourth must be paired but with opposite spin, so you have to add it one at a time according to hund's rule, this is how can write the orbital diagram for the element sulfur.

Now, how can you represent the ground state electronic configuration using noble gas notation? So if you go to the periodic table, look for sulfur that has atomic number 16 and choose a noble gas that has an atomic number less than 16 but as close to 16 as possible. There are only two noble gases that have an atomic number less than 16. That's helium, which has an atomic number of two, and neon, which has an atomic number of 10, but neon is closer to sulfur, so let's write neon now. Neon has an atomic number of 10, which is equivalent to 1s2 2s2 2p6, that is the configuration of 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 below to be 3s2 3p4, so you can also Write the electron configuration for sulfur.

So, 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 electron configuration for sulfur. We know it's um 1s2 2s2 2p6 3s2 and 3p4 and the reason why it is like that. It is because sulfur has a total of 16 electrons now how many electrons does the sulfur ion have to find the number of electrons is equal to the atomic number minus the charge the atomic number of sulfur is 16 and since an atom has zero charge then a sulfur atom has 16 electrons but for an ion it is different as we said before the ions have an unequal number of protons and electrons in this case the charge is negative two which is 16 plus two so the sulfur ion has 18 electrons to write the electron configuration for the sulfur we need add two to the 3p4 so the configuration will be 1s2 2s2 2p6 3s2 and three p six as you can see if you add all the exponents you should get eighteen two two and six is ​​ten two and six is ​​eight ten and eight gives you eighteen, this is how you can write the electron configuration of an anion which is a negatively charged ion, whatever the charge is just add that number to the electron configuration so if you have a charge minus two add two electrons if it's a minus three add three and so on for practice, go ahead and write down the ground state electron configuration for the nitrogen and the nitride ion, so feel free to pause the video while you work on these examples so we have 1s 2s 3s 2p 3p 3d For this example, I don't think we need more than this.

Remember that s can hold up to two electrons, p can hold up to six, and d can hold a maximum of ten, so let's start with nitrogen. Nitrogen has an atomic number of seven, so we need to get the exponents up to seven, so it's going to be one plus two and then two plus two, so we have a total of four. Seven minus four is three, so we need three more instead of going to 2p6, we'll stop at 2p3 because 2 plus 2 plus 3 is 7. and that's the electron configuration of nitrogen now, if you want to write it using noble gas notation, We can't use neon because it is an atomic number ten, we have to use helium because it is less than seven. now helium, which has an atomic number of two, will replace s2, so it will be helium 2s2 2p3, so that is the electronic configuration for nitrogen used in noble gas notation.

Now what about the nitride ion? What is the electron configuration? So let's calculate the The number of electrons in the nitride ion will be the atomic number minus the charge, so we must add 3 to the number of electrons so that the nitride ion has 10 electrons, so we must write the configuration until we reach 10 So it's Now it's going to be 1s2 2s2 instead of writing 2p3, let's add 3, so we'll 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 nitride ion now how can we write the orbital diagram for nitrogen atom?

How would you do it? This is for the 1s orbital. The 2s and p orbital have three orbitals, so 1s and 2s are. full but the 2p orbital is not full it is half full and according to hund's rule we should add it one at a time with parallel sprints so according to hundred you don't want to do this that is not the right way to fill a 2p orbital. empty or what am I 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 empty according to hund's rule you want to add it one at a time Electrons prefer to have their own orbitals.

If there is an empty orbital, they prefer not to share it. They prefer to have their own orbitals because whenever you have two similar charges next to each other, they repel each other, so when two electrons share an orbital. orbital is a less stable situation so to speak opposite charges attract so this configuration is less stable than this configuration because the repulsion of electrons is minimized so according to hund's rule make sure of adding them one at a time before pairing electrons in a single orbital. Now here is another question for you related to elemental nitrogen: is nitrogen paramagnetic or diamagnetic?

Sometimes you may see a question like that on a test and what you need to know is that a paramagnetic substance is one that contains unpaired electrons and is therefore attracted to an external magnetic field, which is why nitrogen is paramagnetic, it contains three unpaired electrons. Substance that is diamagnetic is weakly repelled by a magnetic field and contains no unpaired electrons, so the nitrite ion has no unpaired electrons, it is 2p6 at the end, so it is completely paired. then the nitride will be considered diamagnetic while the nitrogen atom is paramagnetic. Now sometimes you may be asked how many s electrons there are in this element or how many p electrons.

If you get a question like that, just write down 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, 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 is, electrons in the outermost energy level or in this case the second energy level, so Nitrogen has 5 valence electrons, seven minus five equals two. so these are the core electrons, it has two core electrons.

Now what about the sulfur? We already wrote the configuration for sulfur. How many p electrons and s electrons does sulfur have and how many valence electrons and central electrons does it contain? So let's write the configuration so I know it's uh 1s2 2s2 2p6 3s2 3p6, I mean, not 3p6, I went too far, 3p4, so the number of electrons p is based on 2p6 and vp4, so you can add 6 and 4, that means sulfur has 10 p electrons. Now, how many s electrons do they have? You see 1s2 2s2 3s2 2 plus 2 plus 2 3 times is the same as 2 times 3, which is 6.

So sulfur has 6 s electrons. Now, how many valence electrons does it have? Remember that valence electrons are defined as the outermost electrons. Energy level those are the electrons that participate in chemical reactions, so the highest energy level is not the first energy level, it is not the second energy level, but in the case of sulfur it is the third energy level , so only these numbers have a 3 in front of them. so it is 3s23p4 so sulfur has 6 valence electrons and you can also 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 chalcogens like sulfur, selenium, oxygen, their configuration always ends in ns2 and p4, they have six valence electrons to find the nucleus electrons, you can count everything else, two plus two plus six is ​​ten or if you have a very large atom where the configuration is very long, simply subtract 16 times the valence electrons and you will get the nucleus electrons 16 minus 6 is 10, which gives you this number. Now sulfur, is it paramagnetic or diamagnetic answer? Realize that everything before three p four will be filled completely, so to answer the question, you only need to look at the last part, the three p four, so this will be three p one two three four.

Notice we have two unpaired electrons, so since we have unpaired electrons, it's paramagnetic now as a question for you. If we have two unpaired electrons, how many are paired? We know the total is 16. So if two are paired, I mean if two are unpaired, then 16 minus two, the other 14 must be paired, so sulfur has two unpaired electrons. 14 paired electrons. Let's move on to another element, let's try aluminum, what is the electronic configuration of aluminum and aluminum plus three cations? Feel free to pause the video and solve this example so that aluminum has an atomic number of 13.

So, following the same trend that we know we have. the 1s orbital then 2s 2p 3s and 3p then it will 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 will stop at 3p1 now what about the aluminum ion plus 3? 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 three, so it's 13 minus three, so aluminum. plus three ions have ten electrons, so we need to write the configuration down to 10.

So we need to stop at 2p6, so for the most part just remove these three electrons and that will give you the electron configuration for aluminum plus three . cation, then it is 1s2 2s2 2p6 that is the electronic configuration of the ground state. If you ever wonder what the difference is between the ground state and the excited state, the ground state is simply the state where electrons have their lowest possible energy levels if an electron absorbs energy. you can jump to a higher energy level and it is in excited state 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 jump to the force level, so this could still representaluminum, but it is no longer the electronic configuration of the ground state, this would be an excited state, it is still aluminum because the total still adds up to thirteen, two, two and six is ​​ten plus two and one is still thirteen simply represents an aluminum atom in the excited state, so if you ever see an electron configuration where there is a jump, if you are missing 3p, if it goes from 3p to 4s or if it jumps to a high level, you know it is an excited state.

Now, how would you write the ground state electronic configuration of aluminum used in noble 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, so here it goes. will be neon and neon will replace everything up to 2p6 and then we'll write what's next, so neon 3s2 3p1, that's the configuration that uses noble gas notation for aluminum. Now, how can you write the orbital energy level diagram for aluminum? is not the orbital diagram we made before, but the orbital energy level for aluminum, so sometimes you can see a graph like this: 1s 2s 3s and above 2s we have 2p and also 3p, the 1s orbital is more close to the nucleus, so it is going to have the lowest energy, the three p sublevel is furthest from the nucleus, so it has the highest energy compared to all the sublevels that are listed here now according to the off principle of the rod, we must first fill the lower energy levels with electrons before filling the higher energy levels, so we must fill level 1 first and then level 2 once we reach 2p.

We have three orbitals of equal energy, so in this case Hund's rule applies, which states that whenever you add electrons to degenerate orbitals, that is, orbitals of equal energy, you must 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 in the same orbital, you want to add the second in the empty orbital but with a parallel spin, so if the first arrow goes up, the second arrow should also go up and this will be 3 4 5 6. now that we are done with the 2p orbital, we need to go up to the next orbital 3 so that it is filled and then 3p1 now this substance aluminum is paramagnetic or diamagnetic since it has one electron not Since it is paramagnetic now, how many s electrons does aluminum have and how many p electrons does it contain?

So aluminum has six s electrons one s 1s2 2s2 and 3s2 in terms of p electrons, it has 7 2p6 and 3p1 now how many valence electrons does aluminum have? has and how many core electrons it contains, then the valence electrons are the electrons at the highest energy level, in this case the third energy level, so we have three s two three p one two plus one is three, so aluminum contains three valence electrons if you look at the periodic table aluminum is in group 3a and group 3a elements contain 3 valence electrons now the core electrons will be 13 minus the 3 valence electrons which is 10 which correlates to this part 1s2 2s2 2p6 now how many even and non-even how many 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 that we are going to use. cobalt as in our example, cobalt has an atomic number of 27. so go ahead and write down the ground state electron configuration for cobalt and also the noble gas notation so that we have 1s 2s 3s 4s and then 2p 3p 4p 3d and 4d, as we mentioned before. s can have a maximum of two electrons it has one p orbital it has three orbitals that can hold up to six electrons d it has five orbitals that can hold a maximum of 10 electrons so let's start with 1s after 1s it will be 2s 2p and then 3s after the orbital 3s is 3p, then 4s and then after 4s we have 3d and 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. The goal is to get to 27. So it's 3s2, which is 12 3p6 12 plus 6 is 18. 4s2, which is 20. So we need seven more to get to 27, so we'll stop at 3d7, which is the ground state electron . setup for cobalt now what is it? What is the ground state configuration used in noble gas notation? What 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 is equivalent to everything up to 3p6. For neon, it is 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'll write everything that's left to be argon 4s2 3d7, so that's the electron configuration of cobalt using the 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 2 4 6 8 s electrons in the cobalt atom.

Now, how many p electrons are there? So it's two, p, six and three p six six and six is ​​twelve so there are twelve 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 paramagnetic or diamagnetic and how many pairs and no pair electrons has so to answer this, let's write the orbital energy level, so I'm going to rewrite the electrical configuration, it was 1s2 2s2 2p6 3s2 3p6 4s2 3d7, so we will have the level 1s 2s 3s and 4s and then 2s we have 2p 3p and after 4s we have 3d the d sublevel has 5 orbitals so 1s2 is completely filled and then 2s so we have 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 hund's rule we need to add one electron at a time with parallel spins so 3d 3 4 5 6 7. so cobalt has a total of 27 electrons as an atom and as you can see it has three unpaired electrons so if 3 are unpaired then 27 minus 3 the other 24 are paired electrons and you can count them this is 2 4 6 8 10 12 14 16 18 20 22 24 now is it paramagnetic or diamagnetic as it contains electrons not mated?

In general, cobalt is now a paramagnetic substance. that you know the electron configuration of cobalt write the electron configuration of cobalt plus two ions and of cobalt plus three ions now be careful when working with transition metals because the way you remove electrons from transition metals may differ from that of representative ones elements so first let's calculate the number of electrons the cobalt plus two ion has 25 electrons to acquire a charge of plus two it must lose two electrons for cobol plus three it lost three electrons so it has 24 electrons now we know the configuration for the The cobalt atom is 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? we remove it from the 4s sublevel the general rule is that first you want to remove the electrons from the highest energy level now you don't want to remove it from the 3d you want to remove it from the 4s because 4 has a higher number than 3. So generally speaking remove the electrons from the fourth energy level before removing them from the third energy level, so we will remove two electrons from the 4s2 sublevel, so sometimes it will be 1s2 2s2 2p6 3s2 3p6 4s0.

They won't even write four zeros, but I'm going to write it four zeros and then three d7s. Now that 4 is empty, you can ignore 4. You don't have to write it if you don't want to, but just for yourself. I can see what happened. I'll leave it there now for the cobalt plus three. We need to take one more electron from the cobalt plus two ions so that there are no more electrons to remove from the forest level, so the best you can do is remove it. the 3d sublevel, so it will be 1s2 2s2 2p6 3s2 3p6 and then 3d6, this is how you can write the configuration for a transition metal cation.

There is one more topic we will need to cover for electronic configurations and that is exceptions. The exceptions you should be aware of are chromium and under chromium, molybdenum and then on the right copper, silver and gold, we will cover these two exceptions: chromium and copper, they are very common in a test typical chemistry, so if you see them. We know these are exceptions, so let's start with chromium which has an atomic number of 24. So go ahead and write down the ground state configuration of chromium, so starting with 1 it will be 1s2 2s2 2p6, so now we have a total of 10 and then after 2p6 comes 3s, so it will be 3s2 and then after level 3s it is 3p then 4s, so 3p6 4s2, so at this point we have a total of 20.

So we need four more after 4s, we have the sublevel 3d , so 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 that argon covers everything up to 3p6. So it's argon 3p, I mean argon 4s2 and then. 3d4, now it looks like 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 just for force level ii 3d4, notice what happens now Initially, it looks as if the force level is completely filled and the 3d level is not, so notice that we have this empty orbital, this electron does not want to share the force orbital with the other electron because of the electron repulsion, there is a lack of stability, so this electron sees the empty orbital and wants to jump to this orbital now it can jump because the force level and the 3d sublevel are not far apart in terms of energy, they are very close to each other, for so the amount of energy it takes to jump from the full s level to the 3d level is offset by the gain in stability when you reduce the repulsion of the electron by having its own orbital, so it is more stable for this electron to jump to a level of higher energy and being unmated rather than staying mated in strength. level, so you do it and this will be the orbital energy level diagram, it will look like this, then we will have 4s1 3d5, so this is not the correct configuration, it will be argon 4s135, so all you have to do is is to take an s electrode and add it to the 3d sublevel and the same goes for copper.

You can do the same now. How many unpaired electrons do you see in chromium? So be careful if you don't know. the exception you will choose four, but the answer is not 4. Chromium has 6 unpaired electrons, so it is very paramagnetic, so to speak, so if you are asked which element has the most unpaired electrons, it is probably be chromium if it is not an element if it is not an internal transition element because it can have up to seven unpaired electrons, maybe even more, but if it does not have any elements like lanthanides and actinides, mostly chromium or molybdenum, they are let's go to have a very high number of unpaired electrons now if chromium has six unpaired electrons, how many paired electrons does it have then all you have to do is 24-6 so that it has 18 paired electrons now let's write the configuration for copper but using Notation noble gas, so copper has an atomic number of 29 and we're going to start with argon.

We know that argon contains 18 electrons and is all up to 3p6, so after 3p6 it is 4s and 3d, so 4s2 would give us a total of 20. I need 9 more, so 4s2 3d9 now we know this is an exception, so the actual answer will be argon 4s1 10. So looking at the 4s and the 3d sublevel for the first one, initially the forest is completely filled and the 3d sublevel. The level is not, but for some reason this electron wants to jump to the 3d sublevel and it just works that way, so that's something you'll have to know for the test, so for copper it won't be 4s2 3d9, it will be strength 1 3d 10. for some reason the 3d subtier prefers to be paired rather than the forest subtier, so be sure to keep that in mind for copper and the same goes for the elements under copper, silver and gold. s-electron is going to jump into the 3D sublevel, so that's it for this video, thanks for watching and have a great day.