okay folks we get to start with atomic properties of transition meal and our focus is going to be electron configuration why well because to understand transition metals and their ability to be reactive or exhibit all these different colors for example we need to know how many electrons occupi their orbital so electron configuration is essentially basically a way for us to predict chemical behavior and properties because it's going to help us describe how those electrons are distributed in those atomic orbs let's go ahead and see how we do that well first with the main group element we mainly will use the off bile princip right and the off bile principle would tell us that oh okay we're going to be filling our orbitals how do we propose from s to p and then go across from each period D here P orbitals and then we have the F orbital here we're going to continue to fill based on that offb principle and the arrangement of electrons are going to influence how atoms for bonds and of course interact with one another now to write the electron configuration we're going to have to use a specific notation that we have four electron configurations and we'll also talk about the condensed form which is what we'll use mainly for K12 so first the electron configuration is always going to be written with a principal quantum number and normally we denote that as n okay we're also going to show us the shape of the actual orbital wind or called the suble this is s but we can have f p d and f orbitals and also it's going to tell us the number of electrons within that particular U orbital suev okay you can also write the condensed formula that I was telling about which mainly has you noble gas configuration um in parenthesis and then whatever the rest of the electron configuration would be for that particular ele okay and then finally you can write orbital diagrams just so you can actually see how the electrons are going to be paired or unpaired in each orbital so let's say we're going to focus on the D orbital there going to be five different shapes for the D orbital okay um and these are going to be our D orbitals here and if each D orbital had one electron we can say that we have um five unpaired electrons here for example and those unpaired electrons have the ability to form potentially Bond now as we go higher we're going to learn more about the details of this potential um colors and magnetism that can actually happen with electron configuration and we'll get that in multiple videos later now let's focus specific on the transition metal portion of electron configuration so mainly we're going to be using the S and D orbitals we also can use the F orbital so for the F orbital meaning that you're going to have seven different orbital shapes okay and you can have a maximum of 14 electrons versus the D orbital which has five different orbital shapes and a maximum of 10 electrons and the S orbital which has one orbital shape and a maximum of two electrons okay also there's going to be different electron configurations for the D Block um and also period six and seven so we kind of have General formulas to help us understand understand really how to write any electron configuration for any of their transitions because they don't again follow the same kind of rules for everything else so for example if you have a d block conf configuration meaning periods 405 so here are the formulas with periods four and five the formula is going to be your noble gas in Brackets NS SAR meaning the um period for to 7 s which is going to be your orbital um with two electrons filled in it n minus one which is going to be your period numbers 4 to 7 minus one and then orbital D raised to a power of X and that X can be anything between 1 and 10 right because we're talking about the D orbital so it has up to 10 electrons again the formula is going to be your noble gas in Brackets ns2 n minus one D raised to the power of X so for example if you're looking at Scandium then you would have your noble gas which is going to be your argon you would have 4s2 because it's period four so my n will equal to 4 n minus 1 is going to be 4 - 1 which is 3 and then the D electron is going to Simply Be um one so it's going to be 4s2 here and then 3d1 so you look at the periodic table you also can use um the amount of electrons that we see for Scandium as well which is 21 electrons argon has 18 electrons 19 20 21 so this is going to be the electron configuration for a period or transition let's try with a period six or seven so let's use pum so half meum is going to be in Period six again half meum is here here6 that means it's going to be 6 S2 but the first we need our Noble Gaff our Noble Gaff here is going to be M we going to have 6s2 okay 6 minus 2 is four so that's going to be 4 F14 and then DX is going to be how many um electrons are in the D orbital for haum that's going to be D two okay so that is our of course I forgot my um actual number here so let me go ahead and raise this this should actually be 5 D2 because right here you go ahead and erase that go ahead and make sure you make this large change here what out this is actually n -1 DX okay so we were in for half we were in Period 6 that's how we got 6 S2 and then 6 - 2 is four that's we got the four here F14 and then n minus one is 6 - one which is five that's why we got the five here D2 I apologize I did forget about the act one name also did forget about the N minus one for this one too so it's actually n minus one here as well n -1 and D 2 power 10 okay I do apologize for that should I rewrite it perf okay folks so that's just a couple of examples of standi the cium and then for the exceptions you basically be going through but instead of filling in um for chromium and M midium for example instead of filling in um the entire s orbital you actually would have your period here s raised to^ 1 n minus one D to the^ 5 okay so this is one of the exceptions to make sure that we form stable compounds more exceptions copper and our silver other exceptions are b as well so you just want to follow those particular exceptions as you're writing the electron configuration it does make it a little bit easier when you see transition metals when you don't see transition metals you'll just for um basically follow the offb principle or follow the periodic table like you normally would let's get into some more examples um so that we can understand that first when it comes to this you just want to make sure you're locating things on the periodic table and then you want to just make sure that um one of the biggest things that comes along with transition metals is make sure high like this is that when it's time to lose electrons okay you're always going to lose lose those NS electrons first you're going to lose NS electrons first okay then you'll lose your D orbital electrons and then you'll lose your F orbital electrons okay so that means when I'm going through let me go ahead and get new highlighter color these NS um two electrons here these ns2 electrons here this ns1 this ns1 and this ns1 all these electrons are going to be removed first when you're thinking of ions okay so again when you're thinking of any kind of ions okay make sure I'm wri this very clear resol it it's so when you're thinking of ions you're going to lose the NF electrons first okay and then followed by D orbitals and of course the F orbitals so now the real goal is to put all this into perspective with some examples let's get to it okay so for these examples the goal is just to write condensed electron configurations for the following so we have a here um dronium we have v3+ we have M o3+ and we have re4 plus okay so the first thing I want to do is just locate these in periodic table I have Mar Z chronium here I have my vadian here of course this is going to be period four five 6 7 all right just so little you people remember what this little periodic table is telling us that information um let's go ahead and find the other ones billium boom M then r e boom okay so now that we located everything like that then we're kind of ask ourselves okay we know that we have our general d block we know we have our six and seven period um electron configurations and some of our exceptions and we have to ask ourselves if any of these kind of fit that molde okay so first things first we have our vadium and our zirconium here our ZR are both going to be in periods four and five okay um we also have our M Mo which is also in period five but one thing we have to talk about here is it's in the same group as chromium and chromium was one of our exceptions so because um MO is in the same group want to make sure that we are treating it just like we would chromium okay folks so just keep that in mind let's move forward now we have our re which is going to be in Period six so we definitely need um to use all three of our formulas I will first just write down all my formulas just to make sure that I'm all in the right and have my head space in the right area so again you can just stop and use flash cards like in the previous information you could use electron configuration have all three um your general or period six and seven and then your exceptions and then use that as you're going along with practice okay okay so period four and five like we see for these three here are going to be our noble gas ns2 n minus one D raised to the^ of X but we do have an exception um that we're going to be using for m o which is going to be very similar to chromium where we have our noble gas ns1 um n minus one D raised to^ 5 okay folks so that's what we have there now for ours uh period six group here which is re we're going to be using our noble gas ns2 n minus 2 F14 and then n minus one DX okay folks that's what we're going to be using um now let's go ahead and get into our other one okay so this one I'm going to use for youron name Ting let's start with our ZR we see that ZR is going to be in period five here so we can use our formula to basically get the electron configuration okay so we're using this one here our noble gas is definitely going to be our pyton of course you want to have a periodic table already out for yourself as well okay since we are in period five the N is going to be five S2 n minus one is four so that's going to be 4 d and then where is that actually placed in the D Block so this is going to be one here 2 3 4 five 6 7 8 9 and 10 so the chronium is going to be two so our answer would be uh KR in Brackets 5s2 4d2 okay let's move on to the next one B which is going to be the ion v3+ so any time I see ions I first go ahead and write um the by itself or badian by itself and then I go ahead and take away three electrons starting with those NS electrons so badium by itself is going to be um our r G we can see here and of course get your periodic table this is in period four so it's going to be 4s2 we can see in following this pattern 4 minus 1 is going to be three this is 3 D and of course um how many where is vadium in the D Block is going to be three so it's going to be 3 D3 okay that's what normal vadium looks like then I go back and I tell myself okay um vadium now has to lose three electrons so first these electrons are going to be gone that's two and then one from here there's two left over so the ionic form of AUM 3+ is going to be argon 3d2 so basically just keep everything that's left over next let's do number c which is Molly badum so uh first we're going to Geto by itself again we're in period five so that's going to be our pipon 5 S1 so again we're using the exception here remember folks 5 S1 now 5 - 1 is going to be four so we're going to have our 4 d and then we have to count all B these 1 2 3 4 here and instead of counting up here in D Block we're going to use our exception which means that we're going to have 4 D5 so this would be for our Mo now we have to go further and get it for m o3+ is what's the question asked so first we have to get rid of our s groups which is going to be one electron and then we need two more from here so it's going to be three left over I rewrite everything KR 4 D3 okay finally we're going to go ahead and do the last one which is re4 plus so again I'm going to get re no exceptions I just have to use the period 6 and7 formula so this is period 6 so I'm going to have 6s2 and then 6 - 2 is 4 so that's going to be 4 F4 okay next I'm going to do n minus one which is going to be five and then and um DX so where is r e 1 2 3 4 5 so 5 D5 okay now the next step is going to be removing however many electrons they want they want re4 plus so I'm going to remove four electrons first starting with the S so that's two here and then two from here which is going to leave me three so I'm going to rewrite what I have here oh and again I just made a mistake as well I forgot my noble gas here sometimes I'm just going fast and I forgot my noble gas the noble gas here is going to be Z Xenon so I have my x e 6 s e I'm tricking x e okay 4 F14 5 D3 as our final answer okay folks sorry about the little mishap at the end now don't forget to have yourself a flash card moment so that you can move through um this work a little bit easier and it's easier for you to review when you are doing some homework problems