 We will now consider in this lecture a few aspects of metal carbonyl compounds, the origin of the 18 electron rule and how some substitution reactions can be carried out with metal carbonyls. As I had mentioned in the lecture on metal carbonyls, carbonyl compounds are like alkanes in organic industry. In many of the compounds where carbon monoxide is the only ligand, you find that there are 18 electron complexes. These are the only complexes that can be synthesized and they are found in a stable situation. The only exception was the vanadium hexacarbonyl which was a black paramagnetic solid. We noted that that was really an exception and it could be readily reduced to give you VCO6 minus which was a diamagnetic species and a very stable compound. Although you can generate anionic species, cations as in the case of although you can generate anionic species, we noted that cationic complexes as in Werner complexes are not readily synthesized and they form very weak bonds to carbon monoxide. They do form compounds with carbon monoxide, but the bonds are generally weaker. The bond order to carbon monoxide or carbon monoxide like ligands are in general more than 1. You do have a partial bond order of greater than 1 and a maximum of 2. We will go through this in today's lecture also. So, we consider the fact that the best explanation for explaining all these factors that we have observed with metal carbonyls is to use what is called the DCD model or the Dewar Duncanson chat model or Dewar chat Duncanson model. These are the names of the three people Dewar, Chat and Duncanson who initially gave this give and take sharing of electrons. The ligand gives electrons to the metal and the metal in turn gives electrons back to the ligand. This give and take phenomenon is synergistic, a phenomenon which we will consider today and this leads to a very stable situation in metal carbonyl chemistry. The key factors which stabilize the metal carbonyl bond was a pi acidity of the carbon monoxide. Since, we have covered this before, we will just quickly go through this. The pi acidity or the accepting ability of carbon monoxide from the pi star orbital of carbon monoxide leads to stabilization of negative oxidation states. So, metals in negative oxidation states can be supported because you can pump an electron density into the anti-bonding orbitals of carbon monoxide. This results in a very strong bond between carbon monoxide and the metal. We also found that the carbon monoxide can form bridges between two metal atoms and even without bridging metals can form metal metal bonds as in two instances which we looked at MN2CO10 and CO2CO8. They are very readily formed metal metal bonds and they are quite stable. Let us for the sake of completion look at how the orbitals in chromium are evolved in CRCO6 and we will look at what is called this partial bond order and the origin of the 18 electron rule. So, to talk about first we will talk about the origin of the 18 electron rule. In order to look at the bonding in CRCO6, we first generate six orbitals which are suitable for interacting with the carbon monoxide lone pairs. So, here are the six orbitals on chromium and the valence bond formalism is the most convenient one in order to understand this bonding picture. So, octahedral set of ligands can be accommodated in six orbitals which are formed by D2SP3 hybridization. So, D2SP3 hybridization is what is needed in order to accommodate six ligands in an octahedral space. That is exactly what we have done here. Here are those six hybrid orbitals which have the shape which is given here in this on my right side. On the left side I have given you an approximate or a crude energy level diagram which leaves out three D orbitals in a lower lying set of orbitals, lower lying orbitals which are Dxz, Dyz and Dxy. These are called the T2g set of orbitals in an octahedral field. When we do that, we can interact them with six carbon monoxide. In carbon monoxide you have six lone pairs which are pointed towards the metal. That is what is called as a three sigma or in some books it might be given as five sigma. If the two core electrons are also counted, the core electron on carbon and the two core electron on oxygen, if they are counted then this becomes five sigma. But it is a sigma orbital on carbon monoxide. In addition to the sigma orbital which is at a slightly higher energy level, we have this is a sigma which is at a higher level. We should remember that carbon monoxide has got two sets of pi orbitals. These pi orbitals if you have six of them, there will be six into two, twelve pairs of electrons on carbon monoxide which are bonding orbitals. In general, we do not consider the interaction of this pi bonding orbital with the metal because they are low lying in energy. Secondly, their overlap is small because if you remember carbon monoxide has a smaller low bonding orbital on carbon and a larger low on the oxygen. So, these are the orbitals which are reasonably non interacting. So, we will transfer them to the molecular orbital in the same energy level. The six lone pairs are now stabilized by interacting with the six D 2 S P 3 hybrids which are available on the metal. So, the empty orbitals are pushed up in energy. So, the empty orbitals are pushed up in energy shown by this arrow here. The filled orbitals are stabilized which is shown by the arrow on my left side. So, all the filled orbitals are shown in blue. So, the blue are filled orbitals. They are filled with two electrons each. If you recollect the metal has got three D orbitals which are in fact stabilized by interaction with the empty orbitals on carbon monoxide. If there are twelve bonding pi pairs on the carbon monoxide, there must be twelve anti-bonding orbitals also. Fortunately for us only three of those have the right symmetry to interact with the T 2 G set. So, this is the T 2 G set and only three of these orbitals there are three on the metal and only three of this twelve pi star orbitals have the right symmetry. They have the T 2 G symmetry. These also should have T 2 G symmetry and have the right symmetry to interact. So, they are pushed up in energy. So, these are the pi star on carbon monoxide which have been pushed up and they are interacting with the filled orbitals on the metal. The filled orbitals on the metal are shown in dark blue. So, these are also dark blue and they are filled. So, all the blue orbitals are filled and the red orbitals are empty. Now, if you count the electrons around the metal which are mostly on the metal. So, you will realize that there are six into two twelve electrons which are coming from carbon monoxide and three into two which are coming from the metal. So, you have a total of eighteen electrons. You have total of eighteen electrons on CRCO6 and you can see that this is a typical energy level diagram that would stabilize the chromium hexacarbonyl system extremely well because the bonding orbitals are all stabilized and the anti-bonding orbitals are reasonably higher in energy and this gap is significantly large. So, this gap is significantly large and that leads to a very stable situation. Now, let us take a look at what would happen in another molecule and that molecule is tetrahedral system that is also nickel tetracarbonyl. Once again we found that nickel tetracarbonyl has got a complement of eighteen electrons around itself. Now, let us look at why although the coordination geometry is different. Although the tetrahedral geometry is adopted by this molecule we still have a set of eighteen electrons. You will remember that in order to generate vacant orbitals on the metal you on a tetrahedral coordination sphere you need to use the s and the three p's. You need to use s and three p's and if you have sp3 hybrid orbitals and the five d's are lower in energy you will notice that the two d orbitals which we have labeled as having e symmetry. These are orbitals which are having e symmetry and these two orbitals are slightly different from the t2 set of orbitals. Now, we do not have the g symbol because we do not have center of symmetry in this particular molecule. So, we have a t2 set and a e set and the t2 set remains non-interacting and it is in the same level as the d orbitals. Now, if you have two e orbitals and these e orbitals are capable of interacting with the carbon monoxide in a pi fashion. These two orbitals are stabilized. So, out of the four into two eight pi star orbitals on carbon monoxide only two of them can be stabilized by interaction with the d orbitals having e symmetry. So, you will notice that if you fill up these orbitals now we have four into two eight electrons here. You have four electrons here. So, that makes a total of 12 electrons and six electrons here because of the three orbitals here we have total of six plus four plus eight and that again makes up for 18 valence electrons around the metal. Now, we believe that because the carbon monoxide has pi orbitals these pi orbitals are not interacting significantly it is this 18 valence electrons that are around the metal. So, that is how you get 18 valence electrons both for the nickel case and in the chromium case. So, if you work it out for any other geometry also you will end up with a total of 18 valence electrons. So, now let us take a look at the pi bond order in these molecules. If you look at the previous case at the chromium hexacarbonyl system the pi bond number of pi bonds formed where because of this three orbitals which are having this t 2 g symmetry this t 2 g symmetry has got six electrons and the six electrons are the ones which are having pi interactions with the chromium hexacarbonyl. Although there are six ligands there are only three pi bonds that is coming from these three orbitals which are located here this d 3 system. So, you can see that the maximum pi bond order that can be there for chromium hexacarbonyl is only 0.5 the maximum pi bond order in a hexacarbonyl chromium is only 0.5 double bond order so the pi bond order is not fully six, but it is 3. Similarly, if you look at N i C O 4 you realize that although you have four carbon monoxide only two orbitals on the metal are stabilized and there are four electrons which are interacting in a pi fashion. So, these four electrons are interacting in a pi fashion and these eight electrons are working in a sigma fashion. So, the sigma bond order between carbon monoxide and the metal is one because two electrons are shared in that orbital. So, you can say that the bond order is one whereas, in the pi bond order it is only 0.5 once again you notice that the bond order that is present the pi bond order that is present between the ligand and the metal is not one, but 0.5. Now, we can let us analyze this a little bit and see why is it that it is only 0.5 and we will do that after briefly looking at another complex which is P T C L 4 2 minus. We will come back to this let us first look at the partial bond order. If you look at the bonding in detail between a metal and carbon monoxide we notice that there are two interactions which are present. One is a carbon monoxide interacting in a sigma fashion it gives a pair of electrons into this orbital which is there on the metal and two it is interacting in a pi fashion with T 2 G set of orbitals and one of them is pictured here with the pi star orbital. So, this is a sigma interaction and this is a pi interaction. So, here is the metal and here is the carbon monoxide. So, we can write the sigma interaction the pi interaction this fashion, but notice that both the sigma and the pi interactions are being shared. So, in the case of the sigma orbital you notice that the same orbital has to be shared with the ligand in the transposition, but in the case of sigma orbitals we generated 6 equivalent orbitals on the metal using d 2 2 d orbitals and the s and the 3 p orbitals. So, there are sufficient number of empty orbitals on the metal. On the other hand the pi bond order has to be only shared with the d set of orbitals and we have only 3 orbitals which are labeled as T 2 G. So, it is a T 2 G has got only 3 orbitals only 3 orbitals are available. So, only 3 orbitals are available on the metal for pi bond order whereas, 6 orbitals are available for the sigma bonding. So, because of this the pi bond order can only be 3. So, this has got significant importance in the case of complexes. The ground state structures of the complexes and here I have pictured for you two examples. Two examples where we have carbon monoxide bonded to the metal atom. In this particular instance you are seeing a rhodium atom. This is a rhodium atom which is bonded to two carbon monoxide. One carbon monoxide is trans to a nitrogen and the nitrogen is a good donor ligand and it pumps in electron density to the rhodium and because of that the rhodium carbonyl bond which is present is quite strong because you have electron density flowing from that direction and it pumps in electron density into the carbon monoxide pi star orbitals. Because you have a pi system on the opposite side which is pumping in electron density. On the other hand this carbon monoxide has got a bond distance of 1.892 angstroms. This is got a longer metal carbon bond and this has got a shorter metal carbon bond 1.861. If you have a weak donor opposite the carbon monoxide if you have a weak donor this results in a longer bond. So, this pi interaction that is present in the opposite trans geometry is very often essential for influencing or determining the bond distance between metal and carbon monoxide. So, we have one other example here. Once again you have in this case they are both one is a pi donor. This is a chlorine which is a pi donor and this is a carbon monoxide which is a pi acceptor. So, we have a pi acceptor which is trans to a pi donor and the bond distance is significantly short is 1.832 angstroms. Whereas on the other hand you have one carbon monoxide which is this carbon monoxide which is trans to a sigma donor. So, because you only have a sigma donor and not a pi donor this distance is longer it is 1.84 angstroms. So, this 1.84 angstroms is longer than 1.832 which is because of this pi interaction which is present. So, let us go back now and see the continue with our discussion on the 18 electron rule The 18 electron rule as we discussed with nickel tetracharbonyl comes about because we have a stable set of 18 electrons in the metal based orbitals. In some instances we also know that there are complexes which have less than 18 electrons and they are also stable. The classic example is this tetrachloroplatinate dianion ptcl4 2 minus and this is a system which has got 8 electrons and these 8 electrons are in a stable system. Let us try to understand how this could have come about. Suppose you have 5 d orbitals and 1 s and 3 p orbitals on the platinum. You can make 4 orbitals which are available for interaction with the ligand and these 4 orbitals could be made from the d x squared minus y squared, the s and the p x and the p y orbitals. If you have 4 ligands along the x in the x y plane. So, if you assume that the 4 ligands on ptcl4 2 minus is present on the x y plane. If this is the x y plane if you have 4 chlorines which are present in the 4 corners of the x y plane that I have drawn for you here. Then the d x squared minus y squared s and the p x and the p y can combine together to form a d sp2 hybrid. This d sp2 hybrid will now be stabilized by interaction with 4 lone pairs on the chlorine which I pictured here. These 4 lone pairs will be stabilized by interaction with these empty orbitals. That leaves out the d z squared orbital d z squared orbital and what we had earlier written as the t 2 set. So, d x z d y z and d x y and the d z squared orbital will combine together and will form a set of d 4 d orbitals which are left unperturbed at the level of the d orbitals in platinum. So, this is approximately going to be present in the same energy level and we can fill in 4 into 2 8 electrons here. So, that will be 8 electrons here and 4 into 2 8 electrons which are stabilized on the chlorine. So, a total of 16 valence electrons are present on a square planar complex. Now, one might wonder why is this different from the 18 electron rule? Why is it that so many compounds are formed which are all 16 electron and they are in especially when they are square planar in nature. That comes about because of the difference in energy between the d orbitals and the s and the p orbitals. When the p orbital is significantly higher in energy then the tendency of the metal to use the p z and the d z squared to form an octahedral complex is less. It prefers to involve only the 3 d 4 s 4 p x and the 4 p y orbitals. This results in a complex there is only 4 hybrids namely d x squared minus y squared 4 s 4 p x and the 4 p y orbitals. This is suitable for interacting with 4 ligands which are present in the x y plane and it leads to a square planar complex. So, instead of forming a complex in which we have 6 ligands it prefers to have a square planar complex and be stabilized only in the 16 valence electron systems. So, once again the nomenclature of the notation that we have used these are metal d orbitals and they are filled and they had 8 electrons and we had 8 electrons from the chlorine which are also filled. You will also recollect that C L minus has got several 4 pairs of electrons around itself. Only 4 of them have been stabilized by interaction with the metal and the other 8 of them the other set the other set is not interacting with the metal atom and they remain unperturbed in energy. Although they can interact with this d orbitals we in general we keep them unperturbed in an approximate energy level diagram. So, let us now go to the 18 electron rule and the electron counting as we have just mentioned this 18 electron rule could turn into a 16 electron rule especially if there are square planar complexes. We should always remember that electrons are not tagged. What I mean by this is that once the molecule is formed you cannot distinguish between the electrons that are coming from the metal or the electrons that are coming from the ligand. They are all equally orbiting the atoms in the molecule and they cannot be identified with a specific atom. So, there are 2 methods which are used in order to count the electrons. This is just book keeping. In other words I want to account for all the electrons in the molecule. I want to see how many are associated with the metal and since we know that in octahedral complexes and in some systems we do have stabilization of the 18 electron species. We would like to do this electron counting in order to get a quick way of identifying stable molecules versus unstable molecules. There are 2 general ways by which one can do this electron counting. One is called the ionic method and the other is called the neutral method. As the name suggests in the neutral method all the ligands are considered as neutral species. Whereas in the ionic method an arbitrary decision is made based on electronegativity of the atom to associate the charge with the ligand or with the metal. So, we will encounter this. We will try to understand this with some examples, but first let us take a look at some of the ligands that are pictured here which are one electron ligands. So, if you have any alkyl group. So, here I have pictured several alkyl groups. The simplest of them is a methyl radical. So, the methyl radical if you have in the place where I have drawn the dash if you have r dashed m I have only written the r this part of the group. So, if you have an r dash then let us assume that this will give one electron from the alkyl group and one electron from the methyl. So, in the neutral method we consider these as one electron donors. Similarly, if I have a C L, a N O 2 or an N 3 or a bent nitrocell radical all of them are also one electron donors. So, here C L dot is an actual donor N O 2 dot, N 3 dot and N O dot. These are one electron donors again. Now, this is purely a matter of electronic book keeping and so one should not confuse this with the charge on the atom. Now, if you have C 3 H 5 which is this group here it can be attached to a metal at the C H 2 position and if only one carbon is attached to the metal then we call this also as a one electron donor. So, this is also one electron donor. On the other hand carbon monoxide which has a pair of electrons given giving is given to the metal and triphenyl phosphene again two electrons are there on the phosphorous these two electrons are given to the metal. So, these are two electron donors. Now, if you have a molecule like an ethylene complex where the metal is bonded to two carbons where the metal is bonded to two carbons in equidistant fashion. If this distance that is there between the metal and the two carbons is almost the same then we call it as a eta 2 ligand. This is a eta 2 ligand and it is a neutral species and in the neutral method we assume that it gives two electrons because the pi orbital which has which is there on the ethylene is donated to the metal. So, in N O this is again a system where we have or we could have some complications. If you have a molecule in which this bond angle is 180 degrees this bond angle is 180 degrees. Then this is a linear N O and in the neutral method it is considered as a three electron donor. So, you will notice that this is an unusual ligand which gives three electron a total of three electrons to the metal when you have a linear geometry. Again if you have C 3 H 5 this is a system where we have C 3 H 5 and if all three carbon atoms are interacting with the metal in an equidistant fashion then this also turns out to be. So, each one of these lines has got a hydrogen. So, this is also considered as a three electron donor. So, you will notice that if you use the neutral method you have to assume that the ligand is completely neutral and look at the number of electrons that it will donate in that particular state when it is in the neutral state. So, methyl radical will generate or will give one electron to the metal. Similarly, a chlorine radical will give one electron to the metal. So, the neutral method assumes that the ligand is in without charge and then it counts the number of electrons that can be given. In the ionic method there is a slight difference. We use our knowledge of electronegativity of different atoms and assign a charge based on the electronegativity. In most instances the charge is based on the electronegativity and here I have methyl radical the same species, same set of species is talked about here. We have for example, C H 3 is a methyl anion. So, it is C H 3 minus in this instance the same molecule that we talked about earlier M C H 3 is now talked about as M plus 1 interacting with C H 3 minus 1 or in other words C H 3 minus is interacting with an M plus. So, this charge that we write on top of the metal is purely a matter of formality it is not a real charge which is present on the metal. For electronic book keeping we assign such charges and this charge is what we call as a oxidation state. So, one should not mistake the oxidation state for an actual charge on the metal. Similarly, C L minus N O 2 minus N 3 minus are all ligands which will be negatively charged because they are more electronegative than the metal itself and they are all two electron donors. In all these instances we have two electrons being given by the ligand, two electrons are given by the ligand and the metal is supposedly in a positive oxidation state. Similarly, if you look at C 3 H 5 minus which is an allyl moiety now in this instance we will have to write it like this. If it is only interacting with one carbon it is C 3 H 5 minus and there is a negative charge on the allyl group and there is a positive charge on the metal. Once again this is considered as a minus charged ligand and a plus one charged metal atom. However, when we look at the ionic method also there are some species which do not give any which are not associated with any charge. Here is an example carbon monoxide and triphenyl phosphine and they give a pair of electrons to the metal, but nevertheless there is no charge associated with this ligand. It is a neutral species whichever method you consider the neutral method or the ionic method they are neutral ligands. Similar is a situation with C 2 H 4 the same ethylene complex will have a zero oxidation state metal and the ethylene is also considered as a neutral species. NO when it is present in a linear fashion. So, in other words if M N O is present with an angle of 180 degrees then we end up assigning a plus one charge because NO plus is a ligand which is isoelectronic with carbon monoxide and it gives a pair of electrons to the metal. So, the most convenient way of understanding a linear nitric oxide in the ionic method is to associate a positive charge. Let us just indicate that a little more clearly. So, we have a positive charge associated with the ligand and so in this instance the metal is negatively charged and the nitric oxide is positively charged. So, you have a negatively charged metal atom interacting with NO and that gives you the particular description that we are talking about. So, C 3 H 5 minus will now be an ally group, but if all three carbons are associated with the metal we can consider it in this fashion. We can consider it in this fashion and think of it as if the pi bond of the allyl system is also donating two electrons and the allyl group the C H 2 group is also giving two electrons. So, total of four electrons are given by the allyl group and a minus one charge is associated with the metal. So, the metal has a plus one charge and the allyl group has got a minus one charge. So, I think you should be able to figure out some of the other examples where you have an odd electron carbon pi system interacting with the metal. We usually have a negative charge and if you have a even number of pi electrons interacting with the metal then you have a neutral species and in the case of ethylene it would be two electrons. If you have butadiene there would be four electrons and so on. So, if you go back to CRCO 6 which I have pictured here one can understand how you have a stable system if you have 18 electrons and you have either in the neutral method or in the ionic method you have a total of 6 electrons being donated by the carbon monoxide resulting in CRCO 6. Now, I want to talk to you about how one can substitute one of the ligands on this molecule with another ligand. Now, if you remember we have a stable system of 18 electrons and the dark blue colored and the blue colored orbitals are all the filled orbitals. These are all the filled orbitals and the filled orbitals are contributing to the bonding situation and there are no orbitals in the anti-bonding orbitals which are destabilized and are pictured here. So, these are anti-bonding orbitals. These are anti-bonding orbitals and you can think of it as if they are sigma star and these are sigma bonding orbitals and these are pi bonding orbitals and these are the pi star orbitals on the metal complex. So, in CRCO 6 all the blue ones are filled blue orbitals are filled and if you promote through photochemical activation, if you move one of the electrons from the ground state to the higher energy state, then what will happen is you will be removing one electron. You will have only 5 electrons in the bonding orbital and you will be populating the anti-bonding sigma star orbital. Now, this is easy to understand you are removing electrons from the bonding orbital. So, the depletion of electron density would lead to loss of pi bond order. The anti-bonding sigma orbital is populated and so that will lead to a weakening of the sigma bond. So, it is easy to see that if I photochemically excite CRCO 6, I would be destabilizing the metal again bond. One of the bonds is would be easier to break one of the bonds in CRCO 6 because of this photochemical excitation. Now, I would like to point out just one other factor in the excited state. You have a low lying orbital which is empty and so you can add an electron to this photochemically excited species CRCO 6. You can photochemically excited and then it would be easier to reduce this excited state species, easier to reduce the excited state species and because you have added one electron in a higher energy orbital, it would be easy to oxidize the CRCO 6 also. So, photochemically excited molecules turn out to be easy to reduce and easy to oxidize. So, here is the actual reaction. You can photolyse CRCO 6 when it is dissolved in a acetonitrile. It will lead to a co-ordinatively unsaturated species CRCO 5, which will have a vacant orbital with no electrons in it and free carbon monoxide. This CRCO 5 can be in fact studied in a matrix in a very cold matrix. If you photolyse the CRCO 6, it would be possible to look at the depletion of the infrared band which corresponds to CRCO 6. So, these are bands which are depleted and some new bands corresponding to the CRCO 5 species would be formed in an intermediate fashion. So, one can in fact study these species. One can also trap the CRCO 5 with a ligand. If you do not have a cold matrix and if you carry out the reaction in solution, if you add pyridine which is a good donor ligand, one can in fact form a nice complex in which the nitrogen is coordinated to the chromium so you can have a substitution of one chromium with a pyridine ligand. This would be thermodynamically a more stable situation because you do have removal of one carbon monoxide which is a pi acceptor and replacement with a good sigma donor. So, that leads to a stable situation for the trans carbon monoxide, the carbon monoxide that is trans to the pyridine. So, we can also carry out the same deactivation of the, we can also carry out destabilization of one of the ligands using radicals. Here, I have pictured a reaction which is destabilization of this pyridine bond, manganese pyridine bond is destabilized by heating it in at 110 degrees. So, thermally one can break one of these bonds. In this particular instance, the weakest bond is the pyridine manganese bond and the pyridine manganese bond is broken when you heat it and if you have a good ligand like triphenyl phosphine, a good ligand like triphenyl phosphine then it attaches itself to the manganese in the vacant coordination site. So, this is one way in which we can activate by heating it to 110 degrees, but at the same time it is possible to do radical activation of this whole reaction. How does one carry out radical activation? Remember, you have an 18 electron system which has got a set of filled orbitals which are the stabilizing bonding orbitals and a vacant anti-bonding set of orbitals. If I add an electron, if I add one electron to this manganese system here, if I add one electron then it will become an unstable situation and it will tend to lose an electron, lose a ligand and form have a vacant coordination sphere. So, in general it has been found that the negatively charged 18 electron species that we have pictured here can be activated towards substitution by making them, by destabilizing them by adding an electron. So, this is equivalent to say M L 6 plus E gives you M L 6, it is a negatively charged radical species and this M L 6 negatively charged species undergoes substitution reactions very readily. So, it becomes M L 5 L dashed and then this species can get re oxidized back to M L 6 or rather it will become M L 5. Let us just work this out again. This M L 5 L dashed will lose an electron and form M L 5 L dashed. Essentially, we have converted an M L 6 to a substituted species M L 5 L dashed by adding an electron and subtracting an electron. So, you need only a very small amount. In fact, 1 by 300 equivalence of an electron can be added to the species and one normally carries this out by adding sodium in stabilized in naphthalene, sodium naphthalenide anion. That is a good reducing agent and if you have a very small amount of that you can carry out this substitution reaction. Now, the same reaction which was proceeding at 110 degrees is now going at now proceeding at room temperature. The same substitution reaction can be carried out at room temperature by adding a very small amount of an electron donor. So, radical activation of this reaction can also be carried out. It is easy to see that addition of an electron can be done by electrochemical activation, which means that at the surface of an electrode you pump in electrons, you add an electron to the metal complex and then at the other electrode, at the other electrode you accept the electrons from the substituted species. So, this will now give the electron back here. So, you can have a complete circuit and the electrochemically activated species undergo substitution reactions fairly fast. Now, we have seen how or why the electrochemically reduced species undergoes faster reaction because it has got electron in the anti-bonding orbital. I have already told you that one can always heat the metal complex sufficiently high in order to destabilize one of the ligands. This is the usual most convenient way it is done in the laboratory, but the caveat is that L should always be a weaker pi acceptor and one needs a high temperature typically boiling in toluene, refluxing in toluene is what is carried out, what is done. So, temperature of 110 degrees is used in order to break one of this metal carbonyl bonds. So, thermal dissociation is always possible and in some instances, in favorable instances it is possible to isolate or characterize this intermediate spectroscopically and trap the intermediate with a more stable or a more stronger ligand. So, here I have pictured one example where you can dissociate a weak ligand cyclooctene. Cyclooctene is a weak ligand and that is dissociated in the process when you heat it in this ruthenium complex. So, thermal dissociation is the thermal dissociation of metal ligand bond is commonly encountered, but when you have a metal metal bond it is usually the weaker bond that is present in the system. So, if you heat Fe 2 C 1 9, it is likely that it will dissociate to give you Fe C 0 5 and Fe C 0 4 and the metal metal bond is broken in this case and the Fe C 0 4 is co-ordinatively unsaturated. So, it will react with styrene to form a styrene complex containing 4 other carbon monoxides. So, this is the complex that you will get by heating Fe 2 C 1 9 in the presence of styrene you would end up with a complex form between the co-ordinatively unsaturated species which was generated which is Fe C 0 4 and that is formed by breaking the metal metal bond. One can also cleave oxidatively cleave a metal metal bond and that can be done by addition of a compound which will add on to the manganese in this case. So, M N 2 C O 10 is having a manganese-manganese bond and if you heat it this manganese-manganese bond will break and so you would end up with manganese radical you will end up with a manganese radical and that will readily react with bromine and form a nice single bond by utilizing these two radicals these two one electrons which are present one electron which is present in manganese and one electron which is present in bromine will form a nice manganese bromine bond. So, you will end up with this bond here. So, there is one last consideration that we should there is one last example that I would like to take up if you want to do a substitution reaction especially with ligands like carbon monoxide. One can treat the carbon monoxide containing compound with an oxidizing agent mild oxidizing agent like trimethylamine an oxide. Trimethylamine an oxide might be pictured as having M E 3 N plus and O minus. So, this form of trimethylamine an oxide can attack the carbon monoxide in such a way that you can form this intermediate where you have a oxygen carbon bond between the metal carbonyl and the trimethylamine species. This will now break very readily to dissociate to give you carbon dioxide and trimethylamine. So, this will give you trimethylamine carbon and carbon dioxide and then you will end up with a vacant coordination site on the metal. So, now you will have a metal without a ligand. So, oxidation chemical oxidation of the ligand is one convenient way by which we can remove carbon monoxide from the metal complex. Not always is this possible, but in many instances trimethylamine an oxide or pyridine an oxide are weak oxidizing agents that will decompose a carbon complex. If you use a very strong oxidizing agent like seric ammonium nitrate which is serium in the plus 4 oxidation state it is a very powerful one electron oxidant. So, if you use small even small amounts of seric ammonium nitrate they tend to oxidize the metal which is usually present in a low oxidation state. It tends to oxidize it to the maximum oxidation state and once the metal is completely oxidized it does not stabilize the metal carbonyl bond. We realize that this has to be in a low oxidation state in order to stabilize the carbon monoxide metal carbonyl interaction. So, if you oxidize it to a high oxidation state it forms a very unstable complex and the carbon monoxide leaves the metals coordination sphere and you will destroy the organometallic complex completely. In general it is known that seric ammonium nitrate is an oxidant which can be used to decompose any organometallic species that you want to decompose. So, if you look if you remember we made metal carbonyls by reducing metals in high oxidation state to the 0 oxidation state and then reacting them with carbon monoxide. So, this was the reaction which we used. So, it is not surprising that if you take metal in the 0 oxidation state you would be able to oxidize it completely to M n plus and also remove the carbon monoxide from the coordination sphere. So, based on what we have looked at so far in metal carbonyl chemistry one should be able to draw the structures of various metal carbonyls. As we noted earlier these can be poly nuclear species and the difficulty with poly nuclear species is the fact that you can have bridging carbonyls and terminal carbonyls. That is the only difficulty that we have, but nevertheless the number of carbonyls that are arranged around the metal is not a difficulty because you can use a 18 electron rule to figure out how many carbon monoxides should be there on the metal. One can also arrange the carbon monoxides in terms of the stretching frequencies or in terms of the strength of the metal carbon monoxide bond order because the more the negative charge on the metal the more is the electron density that is pumped into the pi star orbitals. That increases the pi bond order between the metal and carbon. So, the pi bond order is increased when you have negative charge pi bond order goes up negative charge goes up these two are positively correlated. One can also in the same fashion predict the bond distance changes in carbon monoxide complexes. You can also predict relative stability of carbon monoxide complexes because we noted that if you have a trans ligand which is a good pi donor or a good sigma donor then you will have stabilization of the metal carbonyl bond. So, these are some of the things that we have learnt from the metal carbonyl chemistry that we have considered so far. There are as I said different ways by which you can distinguish between these metal carbonyls using spectroscopy. Finally, I will say that there are more complex factors that we have not considered in these lectures specially triply bridging carbon monoxide and large metal clusters.