 I welcome you once again to MSP lecture series on advanced transfer metallic chemistry. In my previous lecture I had initiated discussion on AT electron rule and I hope I made you familiar with this method of electron counting and using both ionic method as well as covalent or neutral method. Let me continue from where I had stopped by giving more interesting examples here. So now I have given several complexes in this table and I hope you should be able to do it. Let me do one or two examples to make you familiar once again and let me start with covalent method. So manganese has 7 electrons and then chlorine is giving 1 electron and then carbon monoxide are giving 10 electrons so it is 18, fantastic, 18 electrons are there. So now if you take ferrocene here it is 8 electrons and then 10 electrons because covalent method it is 18 electrons so it obeys and if you take rhenium, rhenium comes below manganese so we have in the same group. So we have again here 7 electrons and then we have 10 electrons and then we have PF 3, 2 electrons and 1 positive charge is there so you should subtract 1 so it becomes 18 electron. So 18 electron so now let us look into HFE CO 4 minus and here 8 electrons, 1 electron is coming here from H and then again 8 electrons are coming and negative charge is there add plus 1 so we have 18 electron. When in covalent method when you are considering a cation depending upon how much charge is there that many electrons have to be subtracted when you are considering anion depending upon how what is the charge that many electrons if one negative charge is there one electron has to be added only in case of positive charge electrons have to be subtracted. And if you continue that one of course once when you go for ionic method it is already accounted there is no need to add or subtract an electron due to the charge present on that complex. Now let us consider that iridium complex so iridium 9 and then we have CO 2 and 3 triphenyl phosphine are there 6 and 1 chloride is there 1 and then NO is there and 3 and then 1 positive charge is there. So now if you count this one if you have given 3 electrons so there is a problem so basically if this goes what you have is it becomes 20 electrons so it is not true so that means here it is giving only one electron. So now you get some confusion here let us try to address that one when you go to ionic method I deliberately put this one here so just remember this. Now let us go to Mn 2 CO 10 in case of Mn 2 CO 10 we have 14 electrons and then we have 20 electrons equals 34 you need a metal metal bond you need a metal metal bond here this is a 17 electron system and now of course this already I showed you so this is no issues this also needs a metal metal bond a 13 electron species in the same way ethylene will be giving 2 electrons and rhenium we have 7 electrons and this is 1 electron and this is 4 electrons and this is 4 electrons and in this case it charges 3 so let us do it now let us consider this one as 6 in this case 6 covalent method sinets are giving 8 electrons and plus you add 3 negative charges are there so what we have is here 17 electrons let us verify that one here because 6 electrons are there from metal and 3 charges there and when you look into ACAC and again here 7 electrons are there and ACAC will give 3 electrons each so we have 9 electrons are there so then you can have 16 electrons here and now if you consider this one here Fe 16 electrons are there and 18 electrons this is 34 electrons here and here axlate axlate can give you 4 electrons so you can consider 3 electrons here 5 plus 3 into 39 plus 3 so it will be 17 electrons so let us now this is how it is covalent method now let us look into it now you can see here yes this is fine and here again no issues ionic method and covalent method both account for 80 electrons here in this case also no issues 80 electrons are there in this one also no issues and here if you see in ionic method when you go 9 electrons are coming it should have been not ionic method ionic method should be here and covalent method should be here if you just look into this one this will give you precisely if it is 9 electrons and then 2 electrons from here 6 electrons are there 1 1 and 1 so it gives you precisely so it cancels 18 electrons so this is correct so here in this should have been here and this should have been here so 18 electron is there so correction has to be made so it is not 3 it is 1 electron now you go to this one again this is a 18 electron species here because 1 covalent bond will be there 7 plus 10 and 1 metal metal bond will be there so it will become 18 electron species here and then in this one also metal metal bond is added here so it becomes 18 electron species and here no issues ok and here it also it is 18 electron species ok. And then if you consider here it is a 17 electron species it is a 17 electron species you can see and here it is 16 electron species and here it is 34 electron species and here it it is 12 electron species. Vanadium has 5 electrons, 6 electrons are coming from this one and plus negative charge it becomes 14 electron species. So, it is 17 is incorrect you should remember of course, oxalate when you write here, this is something here and of course, electrons will also be coming from this one you have to consider that you should be able to write in this fashion and sometime of course, when is coming. So, these 2 electrons will be coming here. So, it becomes essentially what happens 2 plus 2 4 plus 4 8 electron species it becomes. Now, another interesting thing is you can also use this 18 electron rule to identify an unknown transmittal in a given compound provided we say that the compounds obey 18 electron rule. If I say this compound obey 18 electron rule and you have to identify the metal I have given here what you should do is first you write 8 electron and you start subtracting the number of electrons coming from the ligand and also add or subtract depending upon what kind of charge you have then you will end up with electrons that electrons you have to show it to add to a d orbital and then count what is the metal atom or ion in this case. So, here 18 electrons are there and 14 electrons you subtract 4 4 plus 1 charge you subtract and then you have 5. So, this 3 D 2 4 S 2 vanadium. So, vanadium is in plus 1 state in this case. So, you can simply write it and now you go to this one. So, you have a negative charge. So, write 18 electron subtract 1 electron for charge subtract 1 electron 12 electron for metal you will be left with 5. So, again it is vanadium here, vanadium here it is a cationic complex with more electrons it is anionic complex 1 electron less in both the cases. That means, vanadium carbonyl you can make V CO 7 plus or V CO 6 minus both will obey 8 electron rule and here if the metal is not given and also the charge is also not given then you may have several options. For example, you take 18 electron and subtract 10 electrons you will be left with 8 electrons 8 electrons you can do it and here we do not know what is the charge here and if you assume this is a neutral compound such as Fe CO 5 then this is 0 then iron is 0 valent state and if it is in plus 1 then it will be cobalt and if it is minus 1 this will be manganese. So, that means both the entities are unknown then arriving at exact structure is difficult at least one of the entity should be defined here. So, otherwise what happens one should be able to write all possibilities precisely. Sometime in some challenging exams they may ask you then you should not end up with a question is wrong or anything you should make an attempt to identify all these species sometime they give this type of questions deliberately. Similarly, what you can do is you can work out electron counting for using both ionic method as covalent method if we have any problems let us know I shall tell you more details about those things and of course one the thing is here we have an allyl group is there. So, allyl group you should remember as a neutral ligand it is a 3 electron donor as a anionic ligand it is a 4 electron donor that you should know and of course here it is a cationic and if you consider here we have 8 electrons are there and 8 electrons are there 2 charges giving 2. So, it becomes 18 for example NEA 2 Fe CO 4 is an example and similarly you can see here 1 is 5 electron donor 1 is a 3 electron donor 5 plus 3 8 plus 2 10 10 plus 8 18 using covalent method in ionic method. So, this will be 6 electron this will be 4 and iron is 6 electron. So, 6 plus 4 10 10 plus 6 plus 2 18 you should be able to write in the same fashion now this will become 8 electron complex because you have if you just look into this one charge is there. So, 6 6 plus 12. So, 12 plus 17 19 19 minus 1 it will be 18 electron species. So, now this is stable why cobaltosin is unstable and cobaltosin cation is stable is for the same reason. So, that means 80 electron count can also tell you about relative stability of different complexes and here if you have negative charge you have to identify m here if it obeys 80 electron rule. Similarly, you can look into this one try to work out these examples here. Now I have given an interesting question here CP2WH2 on exposure to UV light loses a molecule of hydrogen and undergoes rearrangement reaction to give a dinuclear complex this is very important with tungsten containing 18 electron in its valence cell that means when it undergoes dimerization tungsten obeys 80 electron rule or it satisfies 80 electron rule that means it has 80 electron in its valence cell rather structure give oxygen state and of tungsten and the x plane. So, you have to do it now first thing you should remember is we are considering this molecule here on UV light what happens it loses a molecule of hydrogen. So, you take CP2WH2 you shine UV light minus H2 comes out and we will be left with this fragment here. Now this fragment undergoes dimeration if undergoes dimeration as of now let us see how many electrons are in this one. So, if you take it is like a tungsten oscene this we have is 12 electrons plus 4 electrons. So, 16 electrons are there if 16 electrons are there you should not think that there is a two metal metal bonds are between tungsten in this case the compound may not be stabilized because you do not have any bridging ligands to stabilize two tungsten units just in absence of any bridging ligand probably this dimeric structure having two double bonds may not be stabilized. So, then you should look for alternate methods what are alternate methods you have to use some of your chemistry knowledge to see what would happen to CP groups already present on it whether something else happen to CP groups. Let me write something like this. So, yes is there any activation of this one CH bond activation if CH bond activation happens what would happen one hydrogen can come out in that case and other one may remain something like this. So, now let me show you what would exactly happens to this one here can you see here CH bond activation happens when the CH bond activation happens one of the CH bond on one of the CP groups is activated and then the tungsten is added oxidatively to the CH bond and then same thing is to on either side that means we have a fragment one is eta phi one is eta 1 and then this will add up that means at the end what happens both of them are eta 6 in case of both of them but here it acts as eta 1 and in this case it acts as an eta 1. So, additional eta 6 eta 6 you have eta 1 and then two electrons are there. So, now let us try to do write electron count for this one it becomes very clear here for example if you take this tungsten here and if you see tungsten we have charges let us count here. So, this is also negative this also negative and this also negative. So, in covalent method it should be 6 electrons plus 5 plus 5 these two and then this carbon is giving one electron and this hydrogen is giving one electron. So, then it becomes 18 electron that means that is true in case of both also here also same thing. So, it is a symmetric molecule with centrosymmetry and now let us look into ionic method. Ionic method if you consider if you see 1, 2, 3, 4 so all are anionic ligands. So, tungsten should be in plus 4 state if it is in plus 4 state we will be left with only two electrons and then this is 6 electron donor and this is 6 electron donor and now this is H minus 2 electron donor and now this is carbon minus charge it is a 2 electron donor. So, it is 18 electron species. So, you should be able to write here 18 electron species and now the structure is drawn and oxygen state is plus 4 we know plus 4 and state of tungsten is 4 plus or one can also write like this. So, this is how you can write and you can also explain very easily. Sometime it may appear like challenging, but if you just start thinking about this one what are the possibilities and if you have some coupling reactions in your mind already know about coupling reaction where CH activation all those things takes place you can apply that one and you should be able to write conveniently the structure of this molecule here. And next is what evidence is needed to prove this kind of reagent. So, that means basically you start with a deuterated one and then when you shine UV light D2 will come out and of course once D2 come out if you monitor that reaction using 1 H and M R you should be able to see disappearance of these two indicates that these two goes and then we have a series and actual CH activation takes place and this hydrogen is not coming from here before the formation of this kind of bond through CH activation this is also eliminated and it is not coming back into reaction scheme again. So, now some more electron counting to make you familiar here I have shown here you can see this 70 electron species and you have to put a metal metal bond here and in this case the number is missing here and if we say that there is metal metal bond between chromium and then give the number of nitrosilic and present and one should be able to write in this fashion yes NO is 4 here you can count and you can tell and then in this one 5 are coming from CP and 10 electrons from D system nickel 3D8 4S2 and then it is giving 3 electron it becomes 8 electron species and similarly if you go for this one it is again 8 electron species. So, keep calculating for molecules you come across by studying coordination chemistry so that you become familiar and you will not do any mistakes and if you take this one to be able to write again and if you see 4 nitrosil groups are there and you can confine 2 nitrosil group completely contributing to one chromium and here it is C5 we take 5 electrons and here we have 6 11 so 6 plus 6 12 so 12 plus 6 18 and there is one matter matter bond so that means here what you have is a 70 electron species because this one is giving 5 electrons and then this has 6 electrons and this is 3 and this is 3 6 electrons we have 17 electrons are there you need one chromium to chromium matter metal bond so that explains why we have a bond here. Now we have another interesting ligand here we have an hydrocyclic carbene that is also a neutral ligand and it is a 2 electron donor now ruthenium is there and ruthenium is attached to chlorides we can go directly to ionic method we know that ruthenium is in plus 2 oxygen state ruthenium is 2 plus so that we have 6 electrons are there coming from ruthenium and then 4 electrons are coming from chloride ions and then substituted pyridines we are getting 4 electrons and then CO is giving 2 electrons and then n hydrocyclic is giving 2 electrons so we have 18 electrons. Now look into more of course this I already showed you so it is a linear geometry is there in this one it is a 3 electron donor so if you take 5 plus 10 plus 3 18 electron species and now if you consider this one the bent one is there in this case what you can do is you can take iridium is 9 and 4 and CO 2 and CL 1 and charge minus 1 and then how many we are plus plus 1 electron donor bent so it becomes 16 electron species so you should be able to write so you know when it is a bent it is a 1 electron donor when it is a linear it is 3 electron donor so that is also verified in this example. Now I have given a system here and in this one you have to identify the metal species just try to work out about this one in my next lecture I shall tell you about how to identify in this kind of interesting 18 electron puzzles the metal ion.