Electron Configuration of Elements

in this video, we're gonna take a look at the concept of the electron configuration. Now, electron configurations represent the arrangement of electrons within shells and orbital's in order to connect the periodic table to the electron configuration of elements and ions. We have to have a redesign of the periodic table itself. Now here we're going to imagine the periodic table is being composed of different blocks. Now, the first two columns, which are these blue columns here. These blue boxes, which also includes this one here represents our S block. Our purple boxes here represent our D. Block are yellow ones here represent our P block, and then finally our two rows here represent together are F block. We're going to say here that we can further label this periodic table. We're going to say here that these first two boxes represent R, one S, orbital's. We're in the second row here. So this represents to us, these two third row, this is three s, four S, five S, six s and seven s. Now we're gonna go over to the D block after four S. We go down to three D. 45 d. and then six d. In the P Block, we're in the second row. This represents two p three p 4 p. All the way down to seven p. And then our two rows here. This first row was four F. And this bottom row here is five F. When it comes to each one of these letters of S. D. P and F. They can hold a maximum number of electrons S itself can hold up to two electrons, which explains that R. S. Block is composed of two slots. PerOT D. can hold up to 10 electrons. Which is why each row has (123) 456 78,910 P. can hold up to six electrons. 1 2 3456. An F. Can hold up to 14 in each row 123456789, 10, 11, 12, 13 14. Now, understanding how the periodic table itself is arranged within S. D. P. And F. Blocks. And how each of those blocks has different orbital's one S, two P four F etcetera. This is the key to understanding how to do the electron configuration of any element or ion. All you have to do is remember just reimagine the periodic table in this order and then apply it now that we've seen this periodic table. We can use it to answer this example question here, we're asked to find and write the full ground state electron configuration for the following element. So here we have to follow it. Find it in terms of silicon. Now here I say Z. Which represents the atomic number of silicon. Remember your atomic number is unique to an element. It gives us the number of protons for that element. And in terms of silicon, I've already labeled it on the periodic table. So here goes silicon here, all we have to do is count to it. Alright, so we gotta count to silicon. We're gonna go one s. And there are two elements that are one S. So one S. Two Then to s. 1 - two s. 2 two p. 123456. Next is three S. Which is 12. So three S. Two. And then we have to land on silicon. So that's it's in the three P. Row and its in slot 1 to 3 P. Two. So this would be the full ground state electron configuration for silicon. One S. 22 S. 22 P. 63 S. 23 P two. Again. As long as you can remember the order for the periodic table, just apply to find the electron configuration of any element or ion that's given to you now that you've seen how to just do a simple, basic ground state electron configuration of an element. Click onto the next video where we talk about the condensed electron configuration of elements.

So let's take a look at condensed electron configurations with condensed electron configurations. We start at the last noble gas Before the desired element. In this question, it says right, the condensed electron configuration for the following element here we have to write the condensed electron configuration of calcium here were given an atomic number of 20. Which would mean in terms of our periodic table, calcium would fall here. The last noble gas that I passed before I get to calcium is argon. Remember our noble gasses are in group eight. So here are a condensed electron configuration. Would start out with argon then we just have to count to calcium. Calcium is in the four S. Row and then it's in slot one to its in slot two. So that would be argon four as 2 condensed electron configurations are huge time savers when it comes to electron configurations configurations in general, that's because if I wanted to write the full ground state electron configuration of calcium, I'd have to start out with one S and count all the way to calcium, what would that look like? Well, that would mean that I'd have one s, 2, two, S 2, two, P six, three as two, three, P six. and then four asked to all of this here can be greatly shortened by just substituting an argon. It means the same exact thing. Again, when it comes to the full ground state electron configuration only give that if they're asking for specifically most of the time when it comes to electron configurations. Writing them in the condensed form because again, it saves a lot of time now that you've seen these two different types of electron configurations. Click on to the third video to take a look at our charged electron configurations and the rules that are applied when it comes to the electron configurations of these types of ions.

would charge electron configurations. Now with a cat ion, remember cat ion is a positive ion. We first remove electrons from the highest shell number, which is equal to our end value. Our end value represents the principle quantum number and deals with the atomic orbital size and energy. Furthermore, we're going to say that the print, the principle quantum number provides the shell number or energy level of the electron. For example, if we're looking at electrons within our one s orbital, because the number here is one, That means we're dealing with electrons in the first show. So n equals one electrons and two S or two P. We're dealing with electrons in the second shell because the number is two. So and it's too And then based on the pattern, we can see that here because these numbers are three, we're dealing with electrons in the third shell. So N equals three. Now here it says right the condensed electron configurations for the element and its ion. So here we're looking at vanadium. If we go back up to our electron or our periodic table, vanadium would fall right here. So we want the condensed electron configuration of vanadium. So the last noble gas we passed before we get to vanadium is argon. So we start out with argon. Then here we are at the four S. Row, so four S two and then we're here in the three D row and it's three D 123. So it'd be three D three. So coming back down here, we'd say that the electron configuration of vanadium is argon four S 2, three D 3. Now here, we need to figure out the electron configuration of vanadium three ions. So it's plus three plus three means that we've lost three electrons. Remember when you're losing negative electrons, you become more positive. If we have to lose three electrons are going to have to come from the highest shell number, which is what we set up above. So for looking at the electron configuration of neutral vanadium, we're not gonna look at argon because those electrons think of them as being locked away. So the only electrons we can lose are coming from either four s or three D. Because the number here is four. So that means n equals four. And because the number here is three, that means n equals three. So we need to lose three electrons. So the first electrons we lose would have to come from the highest shell number. So the first electrons would have to come from the four S orbital because its value is largest. So we have argon for us is now gone because we had to lose both those electrons. But remember we're not losing just two electrons, we're losing three electrons. So one more electron has to be lost and it'll be coming from the three D orbital here. So we'd have left three D 2. So again, remember when it comes to um charge electron configurations when it comes to cat ions were losing electrons from the highest shell number first then if you need to lose any additional ones, you take them away from the next orbital's next to it. So in this case three D. So we lost two from four S and then one from three D. Leaving us with at the end argon three D. Two as the electron configuration of vanadium three ion. So again, guys just apply the concepts that we've seen here to giving either the full electron configuration, the condensed electron configuration or are charged electron configuration of any element or ion. All of it starts with you remembering how to write the periodic table properly with its different S. Blocks and P blocks as well as orbital's such as two P and four F.

in this video we're gonna take a look at the electron configurations of specifically transition metals. Now remember we can reimagine the periodic table in terms of S. P. D. And F. Blocks doing this reimagining. We can then make a connection between electron configurations and the periodic table. Now when it comes to our transition metals they're located in two places. We have our transition metals that are within our D. Block. And then we have our transition metals which are in our F. Block. Now for the most part we're going to ignore the transition metals in the F. Block. That's because they have atomic numbers that are really large. And as a result they tend to have weird characteristics such as some of them being radioactive. They have odd electron configuration characteristics. So we're gonna tend to ignore those. Also a lot of the reactions that we're gonna see in organic chemistry that deal with transition metals are really reserved to the transition metals found within our D. Block. So we're gonna be concerned with only the transition metals in our D. Block. So as a result we're gonna say here that the transition metal elements or transition metals are the elements in the D. Block. We're gonna ignore F block. Our main group elements which are elements from groups 1 to 8. We coined them as being our group A elements transition metals, specifically the ones in the D. Block. We refer to them as group B elements. That's just some convention that came up with long ago. It doesn't really affect anything that we're going to see in terms of reactions. Just realize that main group elements, we label them as group A elements and transition metals as group B elements. Now with the transition metals we have our group three B, four B. All the way up to eight B, eight B is actually these three columns together And then we go to one B and to be here. So just realize that our transition metals are a little bit different now within a given period. Remember a period is a row. A transition metal fills its s orbital in the end quantum level followed by its D orbital. So, for example, if we're looking at vanadium, like we did previously remember vanadium electron configuration we had said was Argon four S 23 D three. So, with our typical D block transition metals, what do we do? We rolled out our four s orbital's first. So we wrote, we filled in our s orbital first before we moved on to our d orbital electrons. So that's just a characteristic of a typical transition metal. Also remember we said that when it comes to our different letters, S. P D and F, they can hold a certain number of electrons. That's because each one has a certain number of electron orbital's sore s sub level can hold up to two electrons, like we said, and it can hold up to two electrons because it has one electron orbital. So the electrons in that orbital one spins up one spins down. They have opposite spins. We said that R. P can hold up to six electrons. That's because it has three electron orbital's D can hold up to 10 because it had five electron orbital's and F can hold up to 14 because it has seven electron orbital's. These electron orbital's are important because later on we'll not only need to give the electron configuration of an element or ion but we may we may be asked to draw its electron orbital diagram. Also it's important to know because this justifies why are there two slots for S 10 slots for D. Six slots for P. And 14 for our F. So just remember when it comes to the transition metals we're gonna pay close attention to the transition metals within our D. Block because they more greatly relate to organic chemistry reactions. Then the transition metals we're gonna we would see in the F block also remember typical of transition metals. We fill up the s orbital. Then we move over to the D orbital now that we've laid the groundwork for some basic ideas of electron configurations and transition metals. I want you guys to click onto the next video and attempt to see how I approach this question. We're asked not only give the full electron configuration of titanium but also asked to do the electron orbital diagram for the element

Alright. So here it says, right, the full ground state electron configuration and electron orbital diagram for the following element. So we're looking at titanium here were given its atomic number is being 22. So if we come up here, I've already marked off where titanium is. It's right here. So all we have to do is count to it, starting with one S. So we'd say one S two, two S two, two P six. Which gives us all six of these slots, three S two, three P six for us to three D two. So that would be my full ground state electron configuration of titanium. Now we have to give the electron orbital diagram for the following element. Now, typically for electron orbital diagrams, we'd write the condensed electron configuration and then give the electron orbital diagram. So if we're talking about condensed electron configuration, we'd say that the last noble gas that I pass before I get to titanium is are gone. So for the condense we start out with are gone and then we'd say for us to three D two. So four asked to three D 2. Now we're gonna say for the so we'd write it here are gone for us to three D 2 for the electron orbital diagram. So our noble gas would fit within these brackets then remember an S sublevel has one electron orbital and then a D sub level has five electron orbital's. Remember that's coming from this section here. So that's why we have five boxes here Because they're part of three D. Alright, so we'd totally fill in for us. It has two electrons in it, one spins up, one spins down And then within 3D we have two electrons following hunt's rule, we first half fill. So we have two electrons. So up up and we stop because we don't have any more electrons. So this would represent the electron orbital diagram for titanium. So it's just what we've seen previously in terms of electron configurations. But now we're paying close attention to our transition metal. Now that you've seen this example, attempt to do the practice one on your own here. You just asked to write the condensed electron configuration for cobalt one ion. So remember the rules that we talked about in terms of charged electron configurations and apply to this example. I'll give you a huge hint it's best to write the electron configuration of the neutral form of cobalt first. Then that gives you an indication of what to do to write the electron configuration of its ionic form. Once you've done that, come back and see if your answer matches up with mine. Good luck guys.

in this video, we're gonna take a look at electron configuration exceptions. Now starting from the element chromium as the atomic number Z increases exceptions to electron configurations can be observed. Now, two major types of exceptions happens when we have electrons from our s orbital being promoted up to R. D orbital's. Now, when does this happen? Well, for the first one we're gonna say s orbital's can be promoted to create half filled orbital's with D four elements. So basically this exception happens when we have transition metals that are neutral and they end with D4 in their electron configurations. So for example, if we're looking at chromium on our periodic table initially we'll see it as are gone for S 23 D four if we're to fill out its electron orbital diagram remember year S sub level has one electron orbital & D has five. So for us to means we have two electrons one up, one down three D four, so we have four electrons and we have Phil first. Following the huns rule here, the three D orbital's would be even more stable if they were half filled unfortunately were one electron short to deal with this and become more stable. What happens is that one of our electrons from the four s orbital gets promoted up to this D orbital. So the correct configuration of chromium actually is Argon four S 1, three D 5. So now we only have one electron here And for our three D we still have our original four. Now with that additional electron that we took from the four s and promoted up. So our first exception happens when we have a neutral transition metal that ends with D four and its electron configuration, that configuration will not be correct. You'll have to take an electron from the four from the s orbital and promoted up to the D orbital, thereby going from a D four element to a D five element. Now, our second type of exception that arises is when we have s orbital's and they can be promoted to create totally filled orbital's with D nine elements. So in this case we still have transition metals that are neutral. But now, instead of ending with D four in their electron configurations, they end with D nine. So initially if we're looking at camper we'd see argon four S 23 D nine. So we have our four us orbital here And our 5 3 D orbital's here. So for us to three D 9. So huns really first half fill and then we come back around. We have to fill in fill in nine electrons. So there go our nine electrons. So D orbital's are most favorable and most stable when they're either half filled or totally filled. In this case where one electron short from having this orbital completely filled. So how do we get it completely filled? We take an electron from the s orbital and we promoted up to the D orbital. So the correct configuration for copper in actuality is are gone four S 1, three d 10. So there goes the one electron that I still have in for us, My three D orbital's. So now I have 10 electrons. So there goes my original nine And then we have our 10th electron there. So when it comes to transition metals that are neutral, if they end with D four or D nine, they tend to have an exception to the electron configuration, where we have the promotion of an electron from the S orbital to the D orbital, thereby making the D orbital either half filled or totally filled. Now there are more advanced explanations for these types of exceptions. But for the scope of this course we don't have to go into them. So if we're looking at our periodic table, remember we're looking at our D block transition metals when it comes to our D four elements. That's basically these two. So we have chromium and molybdenum. And then here these are D nines. They also have this exception. So again, just remember these six elements. They either fall into the D four or the D nine exceptions for the electron configurations. Now that we've seen these two examples will attempt in the next video to see if we can give the correct configuration to malignant. Um So click on the next video and see how I approach this example question

Alright. So here it says with an atomic number Z. A. 40 to illustrate the exception to the electron configuration of Melinda. So if we're looking at our periodic table, so hopefully you have one out. The last noble gas we pass before we get to malignant um is krypton And then we count to Molybdenum. So five s. 2 four D four. But we know that we can't have this electron configuration because it's a neutral transition metal Can't end with D four D 9. We're gonna have the promotion of an electron from the s orbital to the D orbital. So it's correct electron configuration is actually krypton five S one four D five. And in that way we make the D orbital's half filled which is more stable. So if we were to write out its electron orbital diagram here, we'd have krypton five S. And these will be four d. So five s. 1 And then 45 means that it's half filled and therefore more stable. Hopefully guys are able to attempt this even without my help remember, we have our six elements here that tend to fall into the D four or D nine category and therefore represent exceptions to our electron configuration. Now that we've seen this one, click onto the next video and see if you can attempt the next practice question. Remember when it comes to ions first write the electron configuration of the neutral form, then look at the ionic form, Come back and see if your answer matches up with mine

Alright. So here, right. The condensed electron configuration of the following ion. So we have to do it first for neutral silver. So if you're looking at your periodic table initially you would see it as Krypton five asked to 49. It is one of the exceptions. So we have the promotion of an electron from the s orbital to the D orbital. So now it's real electron configuration is krypton five S one 40 10 here though, we're dealing with silver one ion. So we've lost one electron. Remember we lose electrons from the highest end value Here. This electrons in the 5th show and these 10 are in the fourth show, We lose them from the highest shell number. So we lose it from the five S 1 orbital. So the electron configuration of silver one ion would just be Krypton four D 10. So you'd write it as krypton here, R5 S has no electrons in it. So don't write any electrons within it. And then four d 10. So up up up up up, then we come back around and do down. So that would represent the electron configurations of neutral silver silver ion and then its electron orbital diagram here, we can see that it wants to do this because doing this promotion of an electron from the s orbital to the D orbital gives us a completely filled in D orbital set. So remember D orbital are most stable when they're either half filled or completely filled. That explains this type of exception that arises. So hope you guys are able to follow along in terms of these exceptions. So keep in mind you're just adding this to the list of things that you're refreshing. In terms of electron configuration. We've gone over full electron configuration, condensed electron configuration of ions and of course, transition metals and their exceptions. So keep this in mind because later on it will be used as justification for some of the organic reactions. We'll see.

in this video, we're gonna take a look at the balance electrons of elements. Now remember the valence electrons represent the outer shell electrons that are involved in the formation of chemical bonds. Now for main group elements, the number of valence electrons is equal to their group number. So we have group one element such as sodium, it would have one valence electrons. A group four elements like silicon would have four valence electrons all the way up to Group eight. Like argon would have eight valence electrons. Now this is only true for your main group elements. So your group eight elements remember transition metals are group B elements and as a result they have to be treated differently when it comes to transition metals. We're gonna say the valence of a transition metal and any transition metal we're gonna claim as being metal M is equal to your s orbital electrons plus your d orbital electrons. If we take a look here at our periodic table we have a more abbreviated ending for each one of these transition metals. So for example, vanadium, vanadium is technically like we saw earlier are gone for us to three D. Three. We don't concern ourselves with the argon portion because again remember we're gonna say the valence electrons is equal to S plus D electrons and here are S and R. D electrons. So if we quickly look at all the abbreviated endings of all these transition metals we can figure out the number of valence electrons. So here we have scanned E um it's four S 23 D one. So that's two s orbital electrons plus one D orbital electrons. For a total of three valence electrons. All these elements have that in common. All of them have to s Orbital electrons with one D orbital electron titanium and its group below it. If you add them up, that's two plus two. So that's four valence electrons Then this would be five six. Now, tungsten and suburbia. Remember we said that they are not exceptions to the electron configuration. So they're just as two and D four but they still add up to six total valence electrons. This here would be 789, 10 Add all these up. You would see that they add up to 11 and this would just add up to 12. So remember when it comes to main group elements, we're going to say that the group number is equal to the valence electrons. But when it comes to transition metals it's different for transition metals. We look at the S and the D electrons and we add them together. Remember we're not we're only concerned with the D block transition metals. We're not gonna worry about the F block transition metals because those elements because of their high atomic mass is a lot of them are kind of weird and radioactive so we don't really worry about them when it comes to organic reactions. So again when it comes to transition metals just find the number of S and D electrons add them together and that equals your valence electron number. Now that we've basically seen how to figure out the valence electron number of transition metals, click onto the next video and see how do we approach this question? We were asked to figure out the total number of valence electrons for nickel plus one ion.

All right. So in this question, it says how many valence electrons are present in the following ion here were given nickel one ion. Its atomic number is 28. Now you can approach this two different ways. We can do it the easy way or we can do it the longer way Now. If we look up here, we see that nickel is right here And in its neutral form it would have 10 valence electrons Plus one means that we've lost one electron. So as a result, we should expect it to have nine valence electrons. So it's as simple as that. And that's our answer. But let's say your professor wants you to show how it only has nine valence electrons. Well in that case we'd have to write the electron configuration of the neutral nickel. Then find the electron configuration of the nickel ion. So, nickel as you can see up here, it's abbreviated electron configuration is for us to three D eight. Um It starts off with argon because that's the last noble gas we pass before we get to it four S 23 D eight. Now, nickel plus one ion means we've lost one electron. And remember that electron that we lose has to come from the highest end value here because this number is four. That means N equals four. So we're dealing with the fourth shell. And then here three means N equals three. We're dealing with the third show we're gonna lose the electron from the higher shell number, which means it's gonna come from the four S orbital. So the electron configuration of nickel ion would come out to be argon four S one, three D eight. And if we add up the number of s orbital electrons and d orbital electrons that will come out to one plus eight equals nine. So we could do it the quick way. We can just simply say that, oh it should have 10 valence electrons when it's neutral plus one means we've lost one of those valence electrons. So now it has nine. Or you could do it the longer way. What we have to give the full electron configuration of the nickel plus one ion? We give it condensed electron configuration here, look at how many S and D. Electrons it has, add them up and we get our final answer now that we've seen this example, attempt to do this practice one here. So for this one, I want us to provide the condensed electron configuration and the number of valence electrons for the following ion. In other words, I want you to go through the longer route of doing all this work to figure out the number of valence electrons. So it's good practice and it helps to justify why we have x number of valence electrons for iron three ion. So, good luck guys