 Welcome to MSP lecture series on Advanced Transmittal Chemistry. In my previous lecture, I had initiated discussion. In my previous lecture, I had initiated discussion on metal metal multiple bonding and also I gave you some background to metal metal multiple bonding by the work carried out by Carton's group especially while working with rhenium salts. Let us continue from where I had stopped. Before we try to establish a bonding concept such as molecular orbital theory to understand the nature of these bonds, let us try to classify the metrometer bonds we come across among coordination compounds and organometallic compounds. In general, there are three class of metrometer bonding we come across in coordination chemistry as well as in organometallic chemistry. They are covalent bond first one is covalent bond so here they are electron precise bonds mm bonds counts as one electron from each metal center. So most common type of metrometer bonding we use while electron counting and also to know whether a given complex satisfies 18 electron rule or not especially when we have more than one metal center in a complex. So that means this is a covalent bond and then we have another type of bonding called dative bonding. So one metal uses a lone pair from the field d orbital to coordinate to an empty orbital on a second more unsaturated metal for simple electron count purpose can be considered as covalent bond. But here what happens one metal is electron rich other metal is electron poor and has vacant orbital so that the electrons can be given very similar to what we come across in case of acid base interaction, Lewis acid and Lewis base interaction that is called dative bond. For example, if I take a metal center having 18 electrons and other metal having 14 electrons so it can give and that can also become 16 electron complex if not 18 electron. Now we have the last one very rare one that is called symmetry metal metal bonding what does it mean? So weak metal-metal interactions caused by molecular orbital symmetry interactions of field empty metal-metal bonding and or anti-bonding orbitals seen in case of d8 metals but they are less common in nature. Now let us come back to RE2Cl8 diann here you can just imagine this molecule in this fashion or in this fashion and here this XY plane is there and your principal axis is z axis in which two rhenium atoms are establishing a material-metal bond and now we have four chlorides here and four chlorides are here and here for all practical purpose let us consider this as z axis here and then this is XY plane and then accordingly we have to consider the d orbitals for the overlapping to establish various bonds to accommodate these electrons present on metal to arrive at different bonds between two metal centers. So here we have a quadruple bond that is four bonds between two rhenium atoms and we have to show convincingly using a mo diagram. Let us say if you consider this is the z axis if it is the z axis what happens we have x square and y square is there and then little angle we have XY at 45 in the same plane and then we have d x z and then d y z is there and d z square will be in this fashion. So that means now if two rhenium atoms are in z axis let us say they can overlap something like this they can overlap something like this this is d z square so this overlapping you can call it as sigma bond and now if you talk about pi bond let us assume this is d x z and this is another d x z this is another d x z from 2 and if they overlap this is pi overlapping this is one x z and then this one one is y z so this will degenerate so that accounts for 2 pi bonds and then this one is d z square accounts for 1 sigma bond. Now we have XY is like this now XY is like this if the XY is like this now they do not have any option other than overlapping like this this kind of overlapping we do not come across in main group chemistry or in organic chemistry when we talk about pi bonds because this is not pi bond pi bonds will be something like this this is d x z and this is d y z and then this is d z square and now if you look into the orientation XY is like this so they overlap sideways so this is XY so this is responsible this is called delta bond the higher energy energy of this one is very high and the lowest one is d z square and the next one will be this 2 d x z and d y z and then we have d x y now assume like this it is there this is the z axis and one d z square is there another d z square is there and now if you ask me why we are using only 4 orbitals such as d z square d x y d x z and d y z why not d x minus y square if you just look into this ml 4 square planar complex formation and if you recall again valence bond theory in valence bond theory we are using sp d sp 2 hybrid orbitals if it is d sp 2 hybrid orbitals the d orbital is d x square minus y square that means d x square minus y square is already participated in making metal to ligand bond as a result that is not available for metal metal bonding that is the reason we are not considering so other orbitals that are left unutilized while making metal to ligand bond are d x square minus y square is used so d x y is not used d x z is not used d y z is not used and also d z square that is the reason we are using these 4 unutilized bonds to establish bonds between 2 metal centers. Now so you bring this Cartesian coordinates so that can give you precisely the orientation of different orbitals once again you take this one here another plane so another plane so now we put now this is what we are considering 2 are in this plane so you can see now how sigma bond is established between 2 d z square which are along z axis so now with this one let us continue forming other bonds so one bonding molecular orbital is formed one anti-bonding is formed here and now this is sigma bonding and this is sigma start bonding if 2 electrons are there one electron here one electron here they come here and one sigma bond will be established. Next we have d x z and d y z they degenerate here so they overlap like this and they overlap like this to generate bonding and anti-bonding molecular orbitals having 5 symmetry so 2 are there so capacity is 4 electrons now consider d x y and this d x y will be forming delta and this will also be delta but this is sigma bonding and this is delta anti-bonding this is delta bonding whereas this one is delta anti-bonding so this is how you can explain quadruple bonding here if you have 4 electrons are there 1 2 3 4 so here 2 electrons are there here 4 electrons are there here 2 electrons are there 8 electrons are there here we do not have any electrons so bond r would be c 8 minus 0 by 4 so that is 8 minus 0 by 2 so that is 4 bonds so that explains the delta bonding or quadruple bond between 2 rehinium atoms so now to make it clear again I am showing you so this is sigma bond this is one of the pi bond and this is the second pi bond and then this is the delta bond with d x y to sideways overlapping okay and now that leads to this sigma molecular orbital and this one and this one to pi bonds in this fashion at orthogonal to each other and then we have the delta bond so that means we have a sigma bond 2 pi bonds and one delta bond they are responsible for making bonding molecular orbitals and remaining will go in the same sequence delta pi and sigma they will be anti-bonding orbitals so this is how using in a very simple way one can explain the multiple bonding between 2 metals in coordination chemistry to even further simplification I have used these cartoons you can see this is sigma bond head on collision and this is one of the pi bonds okay now this is orthogonal another pi bond and now we have delta bond I hope it is very clear now how to write bonding molecular orbitals and anti-bonding orbitals are to show the magnitude of bonding among coordination compounds if there are matrimatial bonds so now I have shown here what would happen if you have more than 4 electrons so that means d 1 d 1 we put 2 electrons here that results in a single bond and d 2 what happens we will be having 2 electrons here and 2 electrons so bond order will be 2 so it will be a double bond if you have triple bond we will be having 2 plus 2 plus 2 here 6 electrons 3 from each so we have a triple bond and if you have 4 electrons from that means d 4 system we will be having a maximum bond order of 4 that is called quadruple bonding and then the 5th one will be if you have one more electron so that goes here a pair so that means 2 4 2 2 10 now the bond order will decrease so it will become a triple bond similarly if you go with the d 6 electrons then it will become double bond and d 7 up to here completely filled you have a single bond if d 8 system both anti-bonding and anti-bonding are completely filled so bond order is 0 no net bonding results in case of d 8 system in this fashion so this can you know give an idea about and the nature of the bond through simple electron counting here simple electron counting and placing them here what you should remember is you should remember one sigma 2 pi delta delta star pi star sigma star that is it and d z square is responsible for sigma bond and d x z and d y z are responsible for 2 pi bonds and d x y is responsible for delta bond and in a similar fashion so we can see the corresponding anti-bonding molecular orbitals now let us look into some examples here so 4 electrons as I mentioned you can start putting here 2 2 first they are singly occupied and then you have they are doubly occupied 6 and then yes okay so this explains now 8 electrons are there in the bonding so bond order is 4 now let us look into this molecule here first you have to identify tantalum nothing doing identify tantalum metal here okay tantalum means where it comes you should know vanadium neobium tantalum that means it has d 3 s 2 d 3 s 2 means 5 electrons are there if 5 electrons are there and you have to identify anonyclic ends so we have 1 2 3 4 5 6 6 are there it has a symmetric structure so that means each one would contribute 3 electron so that the titanium is in 2 plus state d 3 s 2 d 3 s 2 loses 3 electrons and then we have a d 2 system here if d 2 system is there you can go back to the previous diagram and see whether we have a double bond between 2 tantalum atoms we have a double bond between 2 tantalum atoms and then here the tantalum tantalum bond is 2.68 angstrom units now let us go to another example here so now we look into this one each molybdenum has 2 carbonyl groups and 1 Cp group is there and of course here molybdenum is let us consider d 4 s 2 d 4 s 2 6 electrons are there and now 1 is negative so molybdenum should be in plus 1 state as a result what happened this is a d 6 becomes d 5 if d 5 is there you should recall the previous diagram I showed you so d 5 will result in how many electrons it is d 8 so if you put here 2 electrons 4 electrons here and then 2 electrons here and 2 electrons here so this is sigma 2 pi and then delta delta star so 10 2 4 6 8 minus 2 8 minus 2 6 by 2 so there should be a triple bond yes there should be a triple bond okay so triple bond is there so okay so now you go to this one here so here we have try anionic ligands are there so molybdenum in the same molybdenum d 4 s 2 but molybdenum is in plus 3 3 plus 3 state means what we have is out of 6 3 d 3 system so d 3 system how to write so here so here 6 electrons are there from 1 d 3 here and another d 3 here so what we have is 1 sigma 2 pi so it should be 3 triple bond so triple bond is there so this is how you should be able to identify the magnitude of matrimatial bond and simply by electron counting and then adding those electrons to already establish the m o diagram that I have given to you in the order sigma 2 pi delta delta star 2 pi star and then sigma star again so here molybdenum distance is 2.17 and strike units and it has a triple bond this is almost identical with re2 cl8 di anionic compound so d 4 system this is again d 4 system and what is interesting is despite ch3 groups are very bulky still it prefers eclipsed conformation so that it can establish delta bond what happens the other one turns this way what happens your x y won't be having in the same fashion it will be something like this they have to be something like this both the x y if it turns something like this there is no overlapping so in order to have this one facing each other exactly like this for sideways overlapping they have to be eclipsed that is the reason despite it war comes kinetically stable staggered conformation and prefers eclipsed of course in order to minimize what happens it lifts these four orbits little up and this one down so that it can have a quadruple bond okay so now this is very similar to what I showed for molybdenum again you should anticipate a triple bond between these two chromium atoms here chromium chromium distance is 2.27 and strike unit now let us look into this example here in this example again we have this pyridine is there on each one neutral ligand and we have a bisphosphine is there bridging two metallocentrous neobium vanadium neobium tantalum and we have again a d5 system 1 2 3 4 5 6 are there that means you have to get rid of three electrons for each one so it is d3 s2 is that let us assume d5 to remove three electrons it becomes d2 so d2 system in d2 system you will be having a double bond between these two okay so that means in this one it has a double bond between two neobium atoms it is very simple it is much simpler than doing 18 electron rule but do not mix up 18 electron rule with this way of electron counting there are two different entities you should remember you should not even make an attempt to make it satisfy or obey 18 electron rule because early metals or metal ions having less than 6 electrons in their d orbital which are available for bonding you cannot have even with octahedral geometry you cannot satisfy electron rule so 18 electron rule is a different concept and then looking into the nature of the metal metal bond in this kind of molecule is a different concept please do not try to mix up even if somebody ask a question they are referring to different concepts okay one need not have to worry about that one but there is no harm in verifying whether a given molecule the magnitude of the bond metal metal bond can be explained using both the methods we may have some examples but you cannot generalize for all complexes or coordination compounds both the methods now let us look into another interesting molecule here we have osmium compound we have four acetates are there and also we have a osmium chloride bond on each one so that means here you should know that acetate is a mono anionic one two three four mono ionic are there and two negative are there so that means basically six anionic ligands are there each one should go to osmium so they pull out three electrons from each osmium metal osmium is a d6 s2 so out of eight electron d8 electrons you have to take out three three means you will be osmium is three plus this is a d5 system is a d5 system we will be having a triple bond okay how d5 system will be triple bond again two pairs here sigma pi two pi and delta we have two electrons and then delta star we have two electrons so 10 electrons are accommodated and now what we have is a triple bond so eight minus two six by two equals three so you should have a triple bond so easy so you can see now triple bond bond order is three so this is how no matter which metal ion is given you can also say how many bonds are there whether the bonding is possible or not all those things you can analyze simply using this a very general mo diagram so now I have another interesting molecule here I have shown the molecular formula I have given here and how to write the structure yes you can take the compound here so this is how the ligand is described if you see here the valence is not satisfied this should be negative charge this should be negative charge and this is phosphonium positive charge so that means two negative charge and one positive charge is there and then this can also one can write something like this so it should be something like this so now this looks almost isoelectronic with acetate that means we have four of them are there how the four of them will be forming bond something like this one two three and four so that means just one I write so then another one should be something like this I can write another one I can write something like this other one something like this no four are there so four are there and this is a let us assume we have six electrons are there one two three four anionic are there to each so it will be a d four system if it is a d four system very similar to rhenium case you can anticipate a quadruple bond between them okay you can see yeah anionic ligand and then resonance structure when you write yeah this is how the structure you can write and you can see yes this molecule has a quadruple bond so four anionic ligands and two chromium atoms are there and chromium is in plus two state once it is in plus two state it is a d four system obviously it shows quadruple bonding let me stop here another interesting question that comes to your mind is whether we can have more than four bonds between two metal centers yes we can also have five bonds that is called quintuple bond so what are the conditions for a metal complex to have a quintuple bond that is discussed with a couple of examples in my next class until that okay enjoy multiple bonding concept reading and understanding