 Today, we will discuss the concept of isolobal analogy. The problem that we have in organometallic chemistry or even inorganic chemistry is a fact that we do not have the same systematization that is available in organic chemistry. So, a concept of isolobal analogy was introduced by professor old Hoffman and this analogy allows one to understand a variety of organometallic compounds structures and sometimes the reactivity. Primarily the structures are explained extremely well using the isolobal analogy. So, let us just develop this concept in the following few slides. If you look at organic structures you notice that there are different functional groups. These functional groups can be transformed in a fairly systematic manner in a wide range of compounds. So, you could think of for example, a ketone which is present in an organic structure which can be present in a very different scenario in a cyclic molecule and it would behave very similar to the ketone which is present here in this linear acyclic form and these groups are replaceable in a systematic fashion also. So, if I have two hydrogens which are present on this carbon each one of them can be replaced by any R group which is a univalent radical. Similarly, a C H 2 group can be replaced by an oxygen and so on. So, this type of systematization is unfortunately not available in in organic chemistry whereas, in organic chemistry you have a systematic replacement that can easily explain wide range of compounds and it would be nice if you can do this for transition metal chemistry and organometallic chemistry. So, that is really the goal and we will see how far we can succeed in this particular venture. There are indeed common fragments. If you look at organometallic compounds in general you find that there are common fragments like the cyclopentadienyl fragment which is highlighted for you or for example, the iron cyclopentadienyl group that is pictured here in yellow and you can have these groups such as M N C O 3 or M C O 3 a metal with 3 3 carbonyl groups. So, these are recurring fragments and so if one can in a systematic way replace them in different molecules just like we can do it in organic fragments it would be nice. Now, how did we form the organic fragments? If you really want to go back to how one can think of fragments you are normally looking at the C 1 fragment and the saturated variety at that. So, that is methane and we systematically remove hydrogens one by one in order to generate a variety of molecules or fragments which can be used as groups that can be plugged in in a molecule. So, you have the alkyl group which is the methyl radical here and you can also have the methylene group and the methane group and so on. So, let us take a look at the inorganic fragment and see if we can do such a plucking of groups so that we can arrive at different fragments. So, let us take hexacarbonyl chromium and that is like a typical octahedral fragment that is available in organic chemistry and organometallic variety of the inorganic icon an octahedral metal complex. Let us remove one by one and then we get these various fragments of a 5 coordinated metal atom 4 coordinated metal atom and so on. When it comes to 3 and 2 or even 4 we tend to have or we can have a variety of geometries, but nevertheless these fragments can they be used as a fragments that we encountered in organic chemistry. Can we have a kind of a mapping of one to one and that is the question that we are asking really. So, in order to do that let us take a look at the electronic structure of these molecules because that would give us a key or that would give us a way to systematically understand these molecules. So, as we had described earlier in octahedral fragment which is pictured here we have to define our axis. So, we will use the usual axis that we the convention the z axis will be the one which is going up and down on the screen and the x axis is the one which is going from left to right and the y axis is the axis that is projecting into the screen. So, you have for you the 3 axis defined and in these axis if you have a bunch of orbitals valence orbitals in the case of transition metal systems. We normally talk about N D and the empty N plus 1 S and N plus 1 P which are pictured here the energy ordering is approximately in this fashion. So, now if you want to form an octahedral complex we have to choose the right set of atomic orbitals that will form a combination of molecular orbitals with the 6 ligands which are coming in the plus x minus x plus y minus y and the plus z minus z directions. So, if you want to form a set of molecular orbitals we normally end up taking the D 2 S P 3 we take 2 D orbitals 1 S and 3 P orbitals. So, that we can combine them and form a combination of molecular orbitals which will be pointed along the x y and z directions. We are in fact mixing 2 methodologies the valence bond formalism and the molecular orbital formalism here. In the molecular orbital formalism we normally do not talk about hybridization of the D 2 S P 3, but if you look carefully at the ligand group orbitals that have the right symmetry to interact with 6 ligands coming in the x y and z directions. The molecular orbitals are in fact those which are formed by mixing the same set of orbitals which we have used in the valence bond formalism. So, this type of mixing of these 2 theories gives us a easy way of explaining the fragment molecular orbitals that we are going to obtain. So, here I have for you pictured the D 2 S P 3 hybrid which is oriented in the right way to interact with ligands coming along the Cartesian coordinates. So, the 3 orbitals that are left behind that are not going to interact with any orbitals on the along the axis because they are pointed in between the axis are the x z y z and the x y orbitals. These are in the crystal field formalism we encountered or even in the molecular orbital formalism we would call them the T 2 G set. So, the situation is akin to what we encountered in organic chemistry when we in fact hybridize S and 3 P orbitals on the carbon to generate 4 hybrid orbitals which we call as a S P 3 hybrid. So, these hybrid are now pointed along the 4 vertices of the tetrahedron and you can you are familiar with this and each one of these lobes would look in a in this fashion. So, each one of these lobes which are formed by combining the S and the 3 P's would look approximately like that and they are pointed along the vertices of tetrahedron. So, let us now look go back to the metal fragment and if you look at it you get exactly a similar picture you have a set of orbitals which are pointed along the x y and z axis. If a metal complex is formed by combining 6 ligands each with a pair of electrons and they come towards the metal atom. We are hypothetically using a metal which neither has d electrons or no valence electrons basically. So, we have the set of t 2 g orbitals which are empty and the 6 ligand group orbitals which we formed metal group orbitals which are suitable for interacting with the ligand group orbitals are mixed and they would have an average energy corresponding to the weighted average of the d 2 S P 3 orbitals. So, here we can interact the ligand group orbitals which presumably are filled and are usually at a lower energy level. So, they would combine together or mixed to form molecular orbitals. So, these are our molecular orbitals. So, these are our molecular orbitals which are formed where you have a set of bonding orbitals. This is our bonding group and this is our anti-bonding orbitals. In our approximate picture we are not worried about the individual levels or energy levels, but approximately you would have a bonding and an anti-bonding group. This would be essentially the non-bonding group. Now, let us take a closer look at this picture. We have the ligand which has come in with 12 electrons 6 into 12 electrons and these are filled nicely in the bonding set of orbitals. Now, if you have in addition to the electrons that are there in the ligand group, suppose I have a maximum of 6 electrons on the metal. So, I have 6 electrons which are sitting here and these 6 electrons would fill up only the T 2 G set of T 2 G set which is in the middle and it is a non-bonding group. You will notice that we had 6 into 12 electrons here and 6 electrons here. So, we have a total of 18 electrons and this is roughly the origin of the 18 electron rule. When you have more than 18 electrons you tend to form complexes where you are populating the anti-bonding orbitals that are located here. Since you populate anti-bonding orbitals, you will have weaker metal ligand bonds. Essentially, what we are saying is these 6 electrons which are present on the metal go into non-bonding orbitals do not contribute really to the bonding situation here. Even if you have less than 6 electrons, so if you have instead of 18, if you have 16 or 14, you would still have a reasonably good bonding situation between the metal and the ligand. Now, let us proceed further. Let us take a look at what would happen if you had to remove one of these ligands. But before we do so, we must emphasize the fact that this energy gap between the non-bonding and the anti-bonding group of orbitals is prohibitive compared to the energy gap that is existing here. As a result, usually one does not fill in these anti-bonding orbitals at all. In most cases, you would stop at 18. So, although 18 electron rule is touted as this most stable configuration, you could have less than 18 electrons and still have stable configurations without difficulty. So, if you remove one of the electrons, one of the ligands that have come in with a pair of electrons. These ligands are, let us say you remove these ligands by just pulling them out from the top and the plus z direction. You just remove one electron. Now, what would happen is the following. Let us say we prepared the metal in order to interact with six ligands. So, one of these orbitals, metal group orbitals which are suitable for interacting with the ligand is now left without a appropriate combination to form the bonding and the anti-bonding orbitals. So, it would be approximately at the level where it had mixed. So, this is approximately the d 2 weighted average of the d 2 and sp 3 orbitals. So, it would be at a slightly higher energy compared to the t 2 g set. So, you have one electron here and what we are going to find is that this leaves an empty orbital which is present on the metal atom in the place where the ligand should have been. So, you have an M L 5 complex and you have an empty orbital which is present on the metal. Let us go further and think about the combination, think about what would happen if you fill in some electrons into the metal. If you have the approximate number of electrons, let us consider M N C o 5. Manganese pentacarbonyl is a fragment that is formed when you break M N 2 C o 10. If you break M N 2 C o 10, if you split it in half, you would end up with M N C o 5 species which would have exactly the same structure as what we have drawn here. This fragment M N C o 5 fragment would have 7 d electrons. You remember that the electronic configuration of manganese which is usually written as d 5 s 2 and in the free state would in fact in a chemically significant environment would switch over to d 7. You would end up with the arrangement of filling in the t 2 g set with 6 electrons and the empty orbital that was available for the ligand had it been there is now sitting in this one orbital which is at a slightly higher energy level. Notice that when you form M N 2 C o 10, we would have simply combined this one electron that is present on the M N C o 5 fragment with the other M N C o 5 fragment to form M N 2 C o 10. Now, we have these 7 electrons and you have a single electron on the manganese fragment. Consider the situation which you have when you remove one ligand from methane. If you remove one ligand one hydrogen atom from methane, you end up with a C H 3 radical and the C H 3 radical has a S P 3 hybrid present on the carbon and the S P 3 hybrid has a single electron present in it. That single electron which is present on the methyl group can be compared with the M N C o 5 fragment which also has a single electron. Notice some more similarities which are present between the two groups. What you have on the on the right hand side is the methyl group and the methyl group has got a single lobe and the manganese M N C o pentacarbonyl is also got a single lobe which is a S P 2 S P 3 D 2 hybrid now, but the forms are reasonably similar. Now, these are schematic diagrams. What about the actual diagrams? We should let us take a look at them also. Before we do that, we should tell ourselves that these are not iso structural fragments. The manganese has got 5 groups, the carbon has got only 3. Neither are they isoelectronic. The carbon has got only 7 electrons now and the manganese has got 7 valence electrons. There are more electrons on the manganese. So, the carbon and the manganese cannot be considered as isoelectronic fragments. Nevertheless, the total valence electrons that are present if you consider the carbon and the manganese, they appear to be similar. What we have forgotten is that there are 5 bonding electrons which we have omitted and here we have 6 bonding electrons which we have counted. So, the total number of electrons on manganese should really be 17 and this one should be 7. So, we are not talking about isoelectronic fragments. The comparison of the frontier orbital tells us now that the two have a similar form. They have a similar lobe and if you look at the results of the computation, you really find that there is a large lobe on the manganese which as if you think of the fragment as being oriented in this fashion. Here, you have contributions from all the carbonyls also to a small extent. So, the highest occupied molecular orbital which was having the single electron. In fact, looks as if it has got a large lobe which is pointed away from the MNCO5 fragment, but it is very similar to what you have on the methyl group. Which also had a single lobe, only difference seems to be the small contributions from the carbon monoxide as well. So, if you look at the methyl radical, if you look at the methyl radical, we can ask the question since the lobes are similar, would they have similar reactivity. Here, we have the methyl group. What does it do? It combines with another methyl radical and forms ethane. A very similar situation happens with MN2CO10. You take two MNCO5 fragments, both of them have got one electron each on the highest occupied molecular orbitals. They can form a manganese bond. Here, on the organic side, you form a carbon-carbon bond. There, the two similarities seem to be fairly striking. Now, we can also ask the question, can you combine the methyl instead of combining with another methyl? Can it combine with a methyl? Combine with the methyl. So, the answer is yes. We have this familiar molecule MNCO5ME, which we used for looking at insertion reactions. So, methyl pentacarbonyl manganese is a perfectly stable molecule, which is formed by combining these two radicals. The organic fragment, which is on your right and the metal fragment on the left. You can combine these two, just like two methyl radicals would combine and form ME MNCO5. So, you can see that there is a lot of similarity between the two in terms of reactivity, in terms of the behavior. We can in fact go ahead and look at them as isolobal fragments, which have similar reactivity. Now, if we do this further, let us push this analogy a little further. Let us go back to our drawing board and look at the fragments that would be formed, if you remove two ligands from the octahedral complex. So, here is the bunch of orbitals, which are generated for the six metal group orbitals, as if six ligands are going to come in. But now, if we remove two of them, if we bring only four ligands and these four ligands, we are going to bring in along these axes. So, we do not bring in one, which is away from you on the y axis and one on the right side of the x axis. So, the plus x axis and the minus y axis do not have ligands now. So, what will happen is that the four orbitals, initially there were six orbitals and these are pictured here. Then we looked at the fragment, which would be formed when we removed one ligand and that is a fragment here. We are left with an orbital, which did not have a suitable combination. Now, we are having two orbitals along the minus y axis and the plus x axis. They do not have a suitable partner to interact with. As a result, there will be four ligand group orbitals interacting with the four metal group orbitals, which are pictured here. But two metal orbitals are left without partner to interact with and form molecular orbitals. So, again they will be left in the state, where they have no bonding or anti-bonding. They will be similar to the non-bonding orbitals, which are formed by the T 2 G set. So, here we have the familiar T 2 G set, which is left in the non-bonding state. We had six metal group orbitals, out of which four of them have now bonding orbitals and anti-bonding orbitals, but two are still left in a non-bonded state. Now, before we proceed further, the two orbitals, which are in the non-bonded state have in fact, they are in fact, equi energetic, if they are hybridized together. But in the molecular orbital formalism, you will note that even in the case of water, where we are used to writing two fragments, two m o's or two lone pairs, sorry, two lone pairs on the oxygen, which are identical. We call them the rabbitier lone pairs. These lone pairs are called rabbitier lone pairs. They are from the V S valence formalism. These are the valence orbitals, which have two pairs of electrons, which are pointed in the tetrahedral direction. But if you look at the m o formalism, we will end up with a sigma pair of electrons. Let me write it for convenience in this fashion. I write it on the plane of the screen, so that I have two hydrogens in the plane of the screen. The third orbital, which is called the sigma lone pair, is present on the plane of the screen as well. Now, perpendicular to the plane of the screen, I will have another lone pair, which is going in and out of the plane of the screen. Now, I would have p-type orbital or a pi-type lone pair. Here are two types of lone pairs. One is a sigma-type lone pair and another is a pi-type lone pair. The molecular orbital formalism, in fact, distinguishes these two electron pairs. It suggests that there would be two different energy levels for them. One would be lower than the other. In fact, if you do electron spectroscopy on a water molecule, you would find out that in fact, there are two different energy levels for the non-bonded lone pairs. Although the valence bond formalism, which we have used to derive the rabbitier lone pairs on water are very common and very familiar to us, the actual situation in an isolated water molecule is that there are two different lone pairs. One is a sigma lone pair and the other is a pi lone pair. So, a very similar situation happens when you look at the two other rabbitier valence orbitals, which are present in the M L 4 system. The M L 4 system has got a rabbitier group of orbitals, which have to be split into a sigma and a pi set of orbitals. So, these are the sigma and pi set of orbitals and that is the reason why they have different energies and these different energies come out. They are very slightly different in energy, but both are empty. Now, you see that the M L 4 fragment should have some analogy with the fragment, which is present in the groups that we formed after removing hydrogens from methane. So, let us take a look at the Mendeleevs abacus and see what would be similar to the methylene group that is formed by removing two hydrogens on the methane. You remove two carbon monoxides on the M L 6 fragment and if you want to have two electrons on the two orbitals, which are the valence orbitals. Now, for the iron carbonyl fragment you would need eight valence electrons and we know that the metal, which has got eight valence electrons is actually iron tetra carbonyl and you would have iron has got the eight configuration. So, the T 2 G set would be filled with six electrons and one electron would be available for the two molecular orbitals, which are present and which have the symmetry of A 1 and B 2. So, you have the same and or a similar situation for the C H 2, the methylene fragment, except that the energy of the sigma lone pair is lower in the case of the C H 2 fragment. Whereas, in the F E C O 4 fragment you have the pi type the P type orbital, which is lower in energy. Nevertheless, because these two orbitals are reasonably close in energy, you end up with the electronic configuration that is very similar for the two fragments. So, the two lobes, which we derived for C H 2 and F E C O 4 in the valence bond formalism would have two rabbitier type lone pairs or two rabbitier type orbitals, which are filled with one electron each and in the molecular orbital formalism, we would have to split them as sigma and pi type and they have the symmetry of A 1 and B 2 and this is the way in which they are filled with one electron each. So, these two fragments also appear to be similar and the C H 2, the methylene fragment is now considered to be isolobal with the ion tetra carbonyl. So, notice the symbol that Professor Hoffman has suggested for these two groups. They are, it is a double headed arrow with an orbital lobe below the double headed arrow. So, this is a double headed arrow, which we have and the orbital lobe is attached to the lower half of the double headed arrow, suggesting that the two fragments have got similar lobes in the valence orbitals of course. So, let us take a look at the ion tetra carbonyl species and the type of molecular orbitals they have. Here I have shown for you the valence orbitals, which are available for Fe C O 4 and this now has got a very large lobe, which is the sigma lobe, which corresponds to the sigma type interaction on the ion and that is pointed exactly midway between the two carbon monoxide units, which are like this, one coming towards you and the other going away from you and two carbon monoxides are up and down on the plane of the screen and this lobe is in the plane of the screen also, but it is a large lobe, which has got only one phase and so it is a sigma type orbital. This is a sigma type orbital whereas, the other orbital, which we have has got a nodal plane, which is passing in between the along the two carbon monoxides, which are along the plus z and the minus z axis. So, you have a nodal plane passing along the x z plane and so you have two lobes, one is shaded green and the other is shaded red, so this one is red. So, that is the those are the two m o's that are present on the Fe C O 4 fragment and so now, this is very similar to the P orbital that would have been present on the carbon and the S P 2 hybrid, which would be present on the carbon in a sigma type orbital. So, the Fe C O 4 fragment and the methylene fragment are very similar. Now, let us take a look at the reactivity. If you take two methylenes, they would combine together, they would combine together to form ethylene. Now, it turns out that ethylene is quite stable, a similar situation happens when you take the iron tetra carbonyl, it can form di-iron octa carbonyl, but unfortunately this species, which has the iron-iron double bond is not as stable as the one, which has a carbon-carbon double bond. Now, you know that in most systems, when you have a heavy metal or a heavy atom other than the first atom, they do not form strong multiple bonds. So, also the iron-iron bond is weaker and it is not as stable as ethylene itself, but nevertheless this is a reactive intermediate that is formed in the reaction of Fe 2 C O 9, which is a stable molecule. It can in fact react with a methylene, a carbene fragment and that is what is pictured for you in the center of the screen. You have a carbene complex of Fe C O 4, that is a simplest carbene complex that you can form with a C H 2 group and iron tetra carbonyl unit. So, there are differences between the two, although the two are supposedly iso-lobal, you do have some similarities. You can combine them together to form, you can combine the organic fragment and the inorganic fragment together to form iron tetra carbonyl complex, which is complex to C H 2, which is a prototypical carbene. The Fe 2 C O 8, the unsaturated molecule is more reactive than ethylene itself. Now, it is possible for us to have apart from this reactivity, it is possible to have more fragments, which can be compared and I have for you in this in this in the screen a series of molecules, which can be compared to the organic fragment. Suppose, you take three carbines and put them together, you would form a cyclopropane, that is a saturated cyclopropane that we are talking about. Now, if you replace one of the methylene units with an Fe C O 4 unit, then you end up with iron tetra carbonyl olefin complex. Notice, that although it is written in this fashion, what we are really talking about is the olefin complex of iron tetra carbonyl. So, this is the complex, this is the way we normally write iron tetra carbonyl species, where each of these lines is kept with a carbon monoxide and you we normally write it to the center of the carbon carbon bond. But, if you wrote it as if the two carbons are directly bonded to the iron, which in fact is a way in which they are interacting, then you would end up with a form where a cyclopropane has been formed by an isolable replacement. So, there is yet another way in which we can go from the inorganic side. Let us take Fe 2 C O 8 and treat it with a carbene, then we end up with a carbene complex, where a methylene group is bridging a Fe 2 C O 8 unit. And lo and behold, this is exactly the same thing that we have done, except this is exactly the same type of a replacement we have done. Now, instead of replacing two instead of replacing one Fe C O 4 unit, we have replaced two Fe C O 4 unit from the cyclopropane. So, from cyclopropane we have added two iron tetra carbonyl units and we have obtained the olefin complex or the methylene carbene complex. Now, we have a replacement of all the three carbons on cyclopropane with Fe C O 4 unit and then you would obtain Fe 3 C O 12. Unfortunately, Fe 3 C O 12 has got this structure, where there are bridged carbon monoxide. So, it looks different, but however, remember you can just replace iron with ruthenium and ruthenium with osmium and if you do that, you get O S 3 C O 12. O S 3 C O 12 is extremely similar to the cyclopropane unit. So, we have gone from one metal cyclopropane. This is an organic cyclopropane. From that, we have gone to one metal cyclopropane to a two metal cyclopropane and then we have gone to a completely inorganic cyclopropane system. So, the analogy can even be extended further. You can replace a methylene group with a simple replace two hydrogens on the methylene group with another molecule. So, here I have for you tin atom, which is in fact bridging two Fe 2 C O 8 units on either side. So, imagine this to be a CH 2 group initially and then the two hydrogens have been replaced with a ethylene molecule, the inorganic part of the ethylene molecule. So, you end up with Fe 2 C O 8 unit interacting with this carbon, the central carbon and since carbon can be replaced with tin silicon and then subsequently tin and then replaced by lead, you have here an inorganic analogue of the spiro compound, which we would have formed if we just combined two cyclopropanes together. So, this is the spiro compound that we are the organic analogue and this is the inorganic spiro compound that is formed by replacing the central carbon, which we had here with a tin atom. So, now let us go on to the M L 3 fragment. The M L 3 fragment could be formed by simply removing three ligands from the M L 6 fragment that we are familiar with now. So, let us remove them and let us take them out from plus z plus x and the minus y axis and so, we would have the fragment that is pictured for you on the top here and this fragment can be rotated in such a way that it has a C 3 axis, which is now pointed along the z z plus z of the Cartesian framework. So, this is the, so we wanted to be pointed in such a way that you have a C 3 axis passing through the plus z axis. Usually, the plus z axis is considered the principal axis and so, we rotated. So, that we have a proper orientation of the M L 3 fragment. Now, if we have three electrons present in them, we would have a C H fragment something equivalent to a C H fragment. So, here is the organic analogue and I have a hydrogen here. So, this hydrogen has got three lobes, which are pointed along the places where the other three hydrogens would have been in methane and so, these two turn out to be looking very similar except for one striking difference. The methane fragment, the C H fragment, which was formed from methane has got three lobes and it has got a single hydrogen, which is pointed downward. Whereas, the three carbonyl fragment that we have here has got three legs and three orbitals pointed upward, but the three lobes seem to be looking very similar. Let us now see what would happen if we, if we want to form the analogue fragment for the carbine. So, this is C H or the C R unit, which has got three electrons. So, we want one electron for each one of these orbitals and if you remember, the T 2 G set has to be filled with six electrons and we want three electrons on the three valence orbitals, which are available for the M L 3 fragment. If you have to fill up with nine electrons, it has to be a D 9 system. So, a D 9 M L 3 would be tricobalt octa, tricarbonyl cobalt. So, C O C O 3 is the right fragment, which would have three electrons available for interacting with other fragments. So, just as we explained in the previous instance, we had the inversion of the symmetric orbital and the unsymmetrical orbital. Here also, the degenerate set of orbitals, which we have on the carbon framework will be higher in energy. We have two P orbitals and this is the hybridized orbital, which we have at a slightly lower energy. Similarly, here also you would have to split them into a A and a E set of orbitals and these orbitals would have the form that I have shown here for you. Here is the form of the metal orbitals. You will, if you were orienting the C O C O 3 unit, such that one of these carbon monoxide fall on the X Z plane, then the nature of these orbitals would look somewhat like the P orbitals on the cobalt. So, this is what you would have a symmetric orbital, which is pointed along the Z axis and P Y and P X set of orbitals. So, a very similar situation happens in the C H also, the C H fragment. So, the D 9 M L 3 fragment and the C R group turn out to be very similar. So, now let us take a look at the actual molecular orbitals, which have been computed for tri carbonyl cobalt. Here are the three orbitals, which are present. The carbon monoxide, the three carbon monoxides have been oriented in a slightly tilted fashion. So, that you can see the lobes little better. There are two P type orbitals on the cobalt and they are marked lomo and homo, but in fact the three orbitals are filled in this fashion. It just depends on the electron count that we have used for doing the calculation. So, here I have for you the lowest energy orbital has got a single lobe, which is symmetric and it is pointed along the Z axis. That is rotated in such a way, so that you can see it fully. It has got only one phase in the whole side that is pointing towards you. You have two P type orbitals, where you have red lobe and the green lobe and a red lobe. So, these two, whether they are oriented in such a way that you have a nodal plane. So, they are degenerate and they have pi type symmetry for interaction with other fragments. So, just like you have in the case of carbon, the carbine unit you have a system, which has got sigma and pi type interactions. Now, let us take a look at tetrahedrene, which would be the fragment that is formed by combining four carbine units C H fragments and that is tetrahedrene. You can replace each one of these fragments with a tri carbonyl cobalt unit. That is the inorganic analog of C H. So, you have, if you form cyclopropenyl unit, cyclopropenyl compound and that can be complexed with cobalt tri carbonyl. So, COCO 3 is a pianostool complex, which has been inverted. So, the pianostool is in the bottom now and the COCO 3 units are on the top. I can replace two of them. So, this is my alkyne complex. So, this is my alkyne complex, where I have CO 2. So, where I have a CO 2 CO 6 unit interacting with an alkyne and the axis of the alkyne is perpendicular to the axis of the cobalt cobalt bond. So, here is an alkyne complex. Here is a cyclopropenyl complex, which is a three carbon fragment that is interacting with the cobalt. Now, we have made it in an inverted stool structure and you can also have three cobalt atoms. It is a tricobalt non carbonyl complex, which is capped with the carbine. So, here you have three cobalt units, which are capped with the carbine and the finally the completely inorganic analog of tetrahedrine is the IR 4 CO 12 unit that is pictured here. So, you can see very clearly that each one of these fragments, each one of these carbine fragments, which are present in tetrahedrine has already been replaced in organometallic chemistry and the equivalent structures have been made and characterized. So, it is possible to understand some of these complex structures as simple combinations of iso lobel fragments. So, we have seen that the CH 3 fragment can be equated to a D 7 M L 5. You can equate the D 8 M L 4 with the CH 2 and the D 9 M L 3 with the CH fragment. So, each one of these is an iso lobel replacement. Now, if you removed other if you generate other geometries, you would end up with different structures and these also have iso lobel equivalence. Now, we are going to look at at least two examples, where we can have such replacements. Now, imagine the scenario, when we have two ligands in the trans plus z and minus z axis, which are removed simultaneously. You will recall that octahedral complex is sometimes destabilized and it becomes a square planar system and that is exactly what we are talking about here. Let us remove two of the ligands in the trans position and so, what would happen is the two ligands, which are involved, which is the P z and the D z squared. The D z squared are not going to interact and so, instead of you of having a T 2 G set, instead of having the simple T 2 G set, you have in addition the D z squared, which is not interacting. Similarly, the P z, which was mixing with the lower energy s and forming a combination of metal or group orbitals, you are only having four group orbitals and the P z is left non interacting. Now, let us take a look at what would happen, if you removed one ligand along the plus x axis. So, notice that you have in this picture, first of all the modified set of orbitals, the D z squared has been relegated to the level of the T 2 G set. So, you have four orbitals, which are available for the metal at the lower energy level. I have only one orbital, which has been removed or which has no suitable partner ligand partner and so, which is left non interacting at the ligand or the metal group orbital level. So, that gives you an empty orbital and here I have, if I have a D 8 system as I have in the case of platinum, I would end up filling up all these orbitals. So, that so much so, that I have only one empty orbital and that is at the higher energy level. So, P T C L 3 minus for example, is a fragment that is formed by removing one ligand, which is present along the plus x axis and that leaves behind a hybridized orbital, a hybridized orbital, which is present on the metal and it is pointed along the plus x axis along the plus x axis. You have a lobe, which is pointed in this direction and this empty lobe will make it look as if it is a C H 3 plus. So, now, we have a as in the case of a C R C O 5, where we have five ligands and one empty orbital and you have P T C L 3 minus, which is formed from the 4 M L 4 fragment. You have a species, which has got an empty orbital very similar to C H 3 plus. So, that will also have an empty orbital. So, you have the possibility for forming other geometries and other fragments, which are isolobal. So, we have seen here what would happen if you formed from the M L 4 fragment, M L 3 fragment in a T shaped geometry. Now, let us take a look at what would happen if you removed two ligands along the minus y axis and the plus x axis. In this particular case, you remove two ligands, you end up with two orbitals, which are hybridized and left empty. The two orbitals are the metal are hybridized and left empty. I have a total of 4 metal orbitals, which have not interacted at all. As a result, if I take a D 10 system, I would fill up the 8 electrons here, 8 valence electrons here. I would have two electrons, one here and one here. So, that looks similar to what we had for a C H 2 fragment. So, M L 2 system on a D 10 metal would look like F E C O 4, except that now we will have a different geometry on the metal. It will have a V shaped geometry and it would similar to my carbene, which is a C H 2 diradical. So, C H 2 diradical is iso lobel with iron pentacarp, iron tetracarbonyl and the iron tetracarbonyl is iso lobel with my N I L 2 or N I P R 3 2 unit. So, much of this lecture that I have given is described beautifully by Professor Hoffman himself in his Nobel lecture, which has been reproduced in Angovatashemi International Edition and that reference is given here for you to go through in detail. Now, what we can do is to extend this analogy and also talk about iso electronic systems. If you have a C H, the C H group can be converted into a nitrogen, as if the proton has been pushed into the nucleus of the carbon. In a similar fashion, a B H 2 unit can be thought of as a C H unit and that in turn can be converted into a nitrogen. So, we can make replacements of C H groups in by nitrogen and even a B H 2 unit with a C H and a nitrogen. We talked about sandwich complexes earlier and here is a iron sandwich, which has on one side cyclopentadienyl unit and on the other side of P 5 minus. A P 5 minus is iso electronic with C 5 H 5 minus. So, C 5 H 5 minus is identical to a P 5 minus. This is similar to a P 5 minus group. So, notice that each C H unit has been replaced with a phosphorus and these molecules have been well characterized and structurally have been well characterized both spectroscopically and structurally and in a solid state. Both of these molecules, which I am showing for you on the screen, have well defined crystal structures. So, this is an interesting system where you have an iron 2 plus with an inorganic sandwich bread on one side and an organic sandwich on the other side. So, the titanium molecule of course is completely inorganic. Let us move on and this is the synthesis of the titanium sandwich molecule. This was recently published in 2002 in a science paper and it has got the titanium sandwich between 2 P 5 units. So, you can see very clearly that this has got a 2 it is a 2 minus charge. It is a dianion and it is stabilized by 18 crown 6 coordinated to a potassium plus and you have a stable system being formed. So, here is a crystal structure of the molecule with a bond distances indicating that you have a very stable structure and something that can be well characterized both spectroscopically and crystallographically. So, with this we end this discussion about iso lobel analogies. What we find is that iso lobel and isoelectronic replacements can be done and these replacements allow you to make a wide range of molecules with very interesting properties and understand the type of interactions that are present between the molecules especially the inorganic fragments. Because of the type of lobes valence electrons and the lobes that they have in the valence shell and using a combination of valence bond theory and molecular orbital theory you can readily explain the type of molecular orbitals that are available for these fragments to interact with other fragments.