 Hello everyone, I welcome you once again to MSB lecture series on Advanced Transmittal Chemistry. In my previous lecture, after giving introduction to 80 electron rule, I showed you several examples with interesting properties and how you can assess and count and show whether they have 80 electron or not in their valence shell and also whether they satisfy effective atomic number or not. Today let me continue with few more interesting examples to make you familiar with the electron counting process. I have shown here a structure where 4 atoms are hidden behind this orange spears and you can see the rest everything is clear. There are quite a few things one can give as hint. If no hints are given and from this one you can see through metal-metal bond you should be able to identify the metal despite the total valence electrons present in this cluster is not given. So to begin with let me not give the total valence electrons present in this cluster but however I say that this cluster satisfies 80 electron rule then what is the metal atom present in this cluster. So it is very simple except for metal atom rest is known and if you simply start using neutral method or covalent method you should be arrive at the answer to identify the metal atom here. So let us begin. Here let us start with covalent method. According to covalent method C5H5 is giving 5 electrons. So let us begin with one metal center 5 and we have 2 carbon monoxide per metal here because we have 4 carbon monoxide are there totally 8 electron donors but each one should have 2 electron share in that case. So this one now we have a total of 7 electrons are there and if you believe that this satisfies 18 electron rule then 18 minus 7 equals 11 now 11 and if it is satisfying 18 electron rule let us detect 3 electrons for each one then it will be 8. So then metal should have 8 electrons if the metal should have 8 electrons first we should put into S2 N plus 1 S S2 and remaining Nd now we can put D easily. So that means the metal ion we are talking about should have of course if it is a 3D series it should be 3D6 and 4S2 or if it is 4D it should be other way round okay it is 4D6, 6S, 5S2 and it goes on does not matter. So this is the one and if this is the one means 8 electrons in the balance shell means it goes to the iron group okay if it is 3D it should be iron, if it is 4D it should be ruthenium, it should be 5D it should be osmium, if it is 3D iron 4D ruthenium and 5D osmium. On the other hand so we can check whether the metal atom is iron or not if it is 3D. On the other hand if I say this cluster has a total of 60 electrons identify the metal center. So in this case what we should do is we have to take 7, 7 we have each one 7 is there 7 into 4 28, 28 goes then we have 32, 32 by 4 equals 8 so we are also coming here only. So that means already here the total number of valence electrons present in this cluster is also be known given 60 in that case it is 8 so in both ways you should be able to tell. But provided if we say that this cluster satisfies 80 electron rule okay let us look into it. Iron atom you can see this bonds are there one bond here and two bond here another bond here. So each one is coming so that means each one has 15 electrons and in order to satisfy the electron they establish three bonds and you recall when I give this example here dimer in this case we established triple bond between them because here only two iron atoms are since we have four iron atoms are there will be establishing single bond each because each iron atom is making three bonds with neighboring iron atoms which are arranged in a tetrahedral fashion here. You can see this is arranged in tetrahedral fashion. So this is how 80 electron problems can be solved almost like mathematical puzzles. Then let us go to one more example of course here you just recall MnCO5 if you make that exists as a dimer having this kind of composition. The reason why will be answered by this example here just look into this one. We have four carbonyls are there eight are there and of course this is 3d54s27 and then this is linear three electrons are there so this is 18 electron. Since here in case of carbon monoxide we have a NO and instead of two electron donor like CO it is a three electron donor now it satisfies 18 electron rule there is a reason it does not undergo dimerization. So this is an 80 electron complex. Now let us look into this one here we have 10 electrons are coming from 5 ammonia molecules by covalent method cobalt has 9 electrons and then n is bent so one electron it is giving and two charges are there subtract to so you get 18 so it satisfies 18 electron rule. And now let us look into this molecule here again go with covalent method so 8 electrons are coming from ruthenium and then one is coming from chlorine and then two or two each coming from two triphenyl phosphine and one is linear three electrons are coming one is bent two electrons are coming so we have a total of 18 electron and you go to ionic method same thing happens here also so eventually you can use both the methods and you can decide in some cases like here even if the mode of coordination of NO is not given you should be able to tell of electron count that one is linear one has to be bent so that 80 electron rule is satisfied. Now I have given another similar example here in the compound shown below identify the first row transfer metal that means we have to identify which 3D metal is there in place of M here in this case and then here the total valence electrons are giving and of course when total valence electrons are giving you have to keep in mind that you should not count metal metal bond. Metal metal bond will not give you extra electrons you have to manage with the electrons by taking from neighboring atom so that at a given time each metal center satisfy 18 electron rule that is what I told you I am again stressing upon that one you should remember when we add metal metal bond will be managing within the electrons present in the valence shell we are not taking extra electrons from anywhere. So now if this cluster contains 48 electrons we have to identify the metal and now NO let us assume 4 NOs are there 4 NOs will be giving 12 electrons and then we have 3 CP groups are there they are giving 15 electrons so it becomes 27 and what we have is 48 48 minus 27 equals 21 because 3 metal centers are there equals 7 if I put S2 here I have to put D5 so it is D5 S2 D5 S2 means it has to be manganese. Now if I ask you to state the oxidation state of each metal it is bit tricky here and NO comes as NO plus so that basically what happens whatever the electrons they are giving for CP to make it CP minus so each one is assume each one will be Mn will be in plus one state so 4 NOs are there 4 NOs are there one will give one electron here another one give another electron here and then the 2 NOs will give 2 electrons to one of the manganese as a result what happens it becomes Mn 0 this becomes Mn 0 and then it becomes Mn minus 1. So now we know that you can do ionic method to identify the metal here you should remember if you consider CP to become CP minus manganese has to give one electron it is given here so all manganese are in plus one state because each one is making a bond with CP group and now 4 NOs are there out of 4 NOs 2 NOs have to give 2 electrons to one of the metal centers in that case let us assume 2 NOs are giving here NO NO and here 1 NO is giving one electron and 1 NO is giving one electron so it becomes 0 0 and it becomes minus 1 so this is how you can also identify the oxygen state by doing a proper counting and through the electron counting the way I described using both ionic method as well as covalent method you can see here I showed whatever I described is here here so 2 metals have 0 oxygen state and whereas 1 has minus 1 oxygen state. Now another interesting one is there R H F E C O 5 C 7 H 8 and what is C 7 H 8 also I have shown here cyclo heptatriene. Now if a system like this is there you should assume that there is a metal metal bond and now one metal metal bond is there but what is interesting is this group is sharing its electrons with both rhodium as well as iron depending upon how many electrons are needed for each one so it is very interesting structure I should show you here so this is how first you should write. You should ask me why I have put only 2 carbon monoxide and rhodium and 3 carbon monoxide and iron you should know that iron has less number of electrons compared to rhodium in its valence shell to begin with because iron has 3d6 4s2 whereas rhodium has 4d7 5s2 so already it has excess electrons it needs very few electrons to satisfy its act as a result you have to add more ligands to those which have less electrons in their valence shell. Now if you start counting here by neutral method 4 electrons are there and 9 4d7 5s2 9 and here we have 6 electrons plus 6 we have 9 electrons are there and here 13 electrons are there so 13 electrons are there one bond is coming let us say from covalent bond then we have to see how many electrons are needed now 8 electrons are there here so 8 plus 4 14 electrons are there. So now let us put it here in such a way that it can bind to both the rhodium and iron atoms now so what does it show now you can see 1 is eta 4 1 is eta 3 now eta 4 means 4 electrons are coming so 13 plus 4 17 and 1 is coming from this bond and here 3 are there 14 plus 3 17 so another is coming from this one so now you can see very nicely so that means in order to satisfy 18 electron rule in this dimeric in this heterodimetallic complex cycloheptate rein binds in this fashion having eta 4 towards rhodium and eta 3 towards iron this explains very nicely and you can see now both the metal atoms are obeying 18 electron rule and this whatever I showed with star this represents metal to metal bond so now I have one more interesting molecule is there assuming this complex as a neutral one and obeying 18 electron draw the structure right the best possible structure for ruthenium C8 H8 2 that means cycloheptate rein here we are talking about if you are taking about cycloheptate rein and both of them should satisfy and of course ruthenium has 8 electrons are there and ruthenium has 8 electrons what you need is now 10 electrons are needed if 10 electrons are needed how one can write so you can 1 2 3 4 5 6 7 8 are there if 8 are there let me see whether I can put bonds like this and also let me write one more 1 2 3 4 5 6 7 something like this if I put something like this if I take this one and if I take this one it becomes sort of 3 electrons it recall it is like a benzene and it is ethylene so now I let me try to put in this fashion and this will be eta 6 and this will be eta 4 so that means this is giving 6 electrons this is giving 4 electrons we have already have 8 electrons are there and 6 electrons are there and 4 electrons are there so it satisfies 18 electron that means you should remember you may ask me why it should be eta 8 and eta 2 if you see is a structure of cyclotetra trine it is very difficult to take all the 4 to ruthenium because of the orientation of this molecule here so it is very difficult as a result what happens you cannot bind all 4 double bonds to single metal but this is the most appropriate metal prefers to bind having eta 6 and eta 4 coordination so that means one should give some thought and analytically one should try to analyze how one can write a structure without much stress on the molecule but it satisfies 80 electron rule this completes the discussion about electron counting and 18 electron rule and effective atomic number I believe now if any metal complex is given you should be able to solve it without any difficulty now let us move on to another interesting topic called metal-metal multiple bonding and I shall tell you in this one the type of metal-metal bonds we come across among coordination compounds as well as organometallic compounds and also I shall tell you more about quadrupole and quintuple bonding that means according to the convention we cannot have more than 3 bonds between two metal centers but due to the painstaking work from Carton's group we came to know that yes metals can have more than 4 bonds or even 5 bonds between two metal centers that we call it as quadrupole bonding and quintuple bonding and if these bonds are there whether we can use molecular theory to explain this bonding is what interesting I shall tell you in detail to make you very familiar with identifying some of this multiple bonds between two metal centers when you come across in coordination chemistry let us start discussion what is you know metal-metal multiple bonding so metal complexes with metal-metal multiple bonds have been known for decades even while doing electron count also I showed you the existence of metal-metal double bond triple bond but bonds formed by the overlap of valence diarbates of trans metal ions received greatest attention after the discovery of quadrupole metal metal bonds by F. A. Carton in 1960s Carton's group at MIT earlier he was a professor at MIT and later he moved to Texas A&M so during that time he was interested in Rehenium chemistry for which he started working on Rehenium salts such as K2RECl4 and one of the Russian scientists earlier that means in 1950s had characterized this compound and composition and he called this one as K2RECl4 salt. Carton was purchasing Rehenium salts from one mining company called Coors in Colorado in USA where the metal was obtained as a side product while mining molybdenum one such sample having K2RECl4 obtained from Coors group was found to have several crystals suitable for single crystal x-ray structure determination as a result what happened Carton performed structure and he determined the structure of the so-called K2RECl4 to his surprise and his group surprise the sample was shown to have a stoichiometry of KRECl4 but not K2RECl4 what it was thought earlier and it was found to be a di anion with 2 K2RECl4 moieties connected together with a metal-metal bond so then curiosity was there to think about that one before looking into those things while working with Rehenium salts Carton has solved the structure of this trimeric molecule Re3Cl12 and he found they have double bonds between the metals in this fashion so he had prior knowledge of this one however Russian group predicted this to be having this kind of composition but later when they found they found it as a di anion connected through these Rehenium atoms this is how the structure was found after establishing through single crystal x-ray analysis it was thought to be mono anionic and then the dimer was di anionic so the di anion was found to have d4 symmetry you can see from what I showed with eclipsed tetragonal Cl4RERECl4 planes something like this they are eclipsed and with a very short Rehenium Rehenium distance of 2.2 anionic units to support the short Rehenium Rehenium bond and the eclipse structure the chlorine atoms were found to be bent away from bond with an angle of one so that means they were not exactly like this what happens in order to prevent this kind of they these four went up little bit and then they went down like this in this fashion so something like this establishing a bonding of 104 degrees between the metals something like this so clearly the eclipse structure appeared to be due to the electronic effects but if you consider the steric consideration staggered would have been much more stable so a graduate student of cotton named Charles Harris presented the theoretical arguments for a quadruple bond between Rehenium atoms so that means now they went for bonding concepts how to explain this bonding or short bond between the Rehenium atoms using some bonding concepts so this is how it was appearing as I mentioned they went up little bit these four to minimize repulsion you should remember the fact that each chlorine has three lone pairs are there and they will be having some impulsive forces to keep them away from this one in the eclipsed configuration they were slightly pulled up and they were slightly pulled down in the structure. Now let us look into the orbital interactions the orbital interactions between the Rehenium centers included a sigma bond found by overlapping of dz square orbitals from two Rehenium atoms and then there were two degenerate pi bonds having the similar or same energy due to the overlapping of dxz and dyz orbits of two Rehenium atoms and a delta bond was established through weak overlapping of dxy from two metal atoms Rehenium atoms in particular positive overlap of the adjacent dxy orbits requires that the tetragonal or recl four planes adopt eclipsed structure. So that means I will show in a couple of minutes or maybe in my next lecture why this square planar units should have an eclipsed configuration not the staggered one. So dx square square orbital remains involved with ReCl bond formation and hence it does not participate in Rehenium-Rehenium bonding. Since Rehenium 3 plus ion has d4 electronic configuration if you consider four electrons from each d orbital we have a total of eight electrons are there these eight electrons would be occupied to in in a fashion such as sigma 2 pi 4 and delta 2 so that bond order will be 4. So I shall give you a clear m mod diagram and explain step by step to make you familiar with you know understanding this quadruple bonding concept and also predicting the nature of the metal-metal bond in molecules where we have different electronic configurations until then have an excellent time reading chemistry.