 In this lecture, we are going to talk about some ligands, which are based on phosphorus in the oxidation state 3. So, we talk about it as phosphorus 3 ligands. In many instances, organometallic chemistry is associated with a metal carbon bond and that is correct. But you also need some other ligands to support the metal carbon chemistry that is going on. So, in general there are a variety of ligands, which are useful in organometallic chemistry. The most common ligands of course, are the ones, which have the carbon metal bond and that can be carbon monoxide, which is what we have already discussed. We can also talk about carbene ligands and carbine ligands, which are all having a single carbon attached to the metal centre. On the other hand, there are several ancillary ligands, which also support organometallic chemistry. These do not have a metal carbon bond and today we are going to talk about one such ligand and that is the phosphorus 3 chemistry. Metal hydride chemistry, metal nitrosilts, metal dinitrogen chemistry and metal highlights are all important in organometallic chemistry. Although the chemistry may not happen at the metal carbon centre, these are good supporting ligands. So, I am calling it as those ligands that support organometallics. So, today we will talk about phosphorus 3 ligands. There are some distinct advantages to phosphorus 3 ligands and we will consider them one by one. First of all, it is a good sigma donating ligand because phosphorus is in the oxidation state 3. A pair of electrons on phosphorus can be donated to the metal and the phosphorus 3 ligands can be varied extensively contrary to carbon monoxide, where carbon monoxide cannot be changed. You can only change the oxygen to sulphur or to selenium, but in the case of P 3 ligands, you can change the R group, which is attached to the phosphorus extensively. So, you can react this molecule P x 3. Usually it is the halogen, which is usually if x stands for halogen. You can change the halogen to an R group using a reaction with a Grinard reagent and make P R 3 molecules. Now, these P R 3 molecules can also be varied. You can change the R group to R 1, R 2, R 3 and this synthesis is relatively easy. So, because it can be done in a stepwise fashion, the other great advantage of phosphorus 3 ligands is the fact that x 3 crystallography of especially P P H 3 complexes becomes very, very easy because of some super molecule interactions that are there in tri phenyl phosphine. So, whenever you have P P H 3 as an ancillary ligand, the organometallic complex is easy to crystallize. If it crystallizes as a single crystal, one can do the crystallography of this molecule and study the solid state structure very readily and extremely accurately. One can also carry out phosphorus 31 NMR spectroscopy. This is again a very useful technique because no other element in the molecule might interfere in the spectrum and only the ligand is visible in the NMR spectrum. So, this becomes an extremely useful tool because of non interference and only the active species is visible. Infrared spectroscopy is unfortunately of not much value in this instance because the metal phosphorous bonds metal phosphorous bonds these single bonds are possible. It is possible to observe them in the infrared spectrum in the very low energy regions like 300 to 400 centimeter minus 1 and because this region 300 to 400 is quite crowded. It is very often not useful as a tool to analyze the phosphorus 3 organometallic chemistry that goes on. Let me say a few words about 31 P NMR. It is an extremely useful technique, but unfortunately it is an empirical tool and when you call it an empirical tool it only means that it is based on prior evidence that you have and you can in fact use this empirical tool based on prior evidence. If you have a crystallographically characterized molecule one can use the 31 P NMR usefully because one assumes that the solid state structure and the solution NMR have to be related coordination to the metal usually shifts the ligand signals up field. Now, this is what happens the coordination of the metal shifts the ligand signals up field, but the 31 P signals can also be shifted down field by up to 30 p p m. The usefulness of this tool is significantly affected because of this empirical nature and the fact that the signals are in unusual regions with respect to the free ligands. There is another erratic behavior of phosphorus that you want signals and that is the fact that if you have a chelated molecule. So, let me write down chelate an example of a chelate molecule if you have this diphenyl phosphinoethane which is this ligand. Now, this ligand is now capable of coordinating to the metal through two phosphorus centers and so a chelate ring will be formed depending on the chelate ring. In this particular case what I have written is a five membered ring 1, 2, 3, 4 and 5 this five membered ring is formed and then the shift can be a plus 20 p p m. Whereas, if you have a single carbon then the shift is minus 20 p p m. So, because of this erratic behavior of the chemical shift sometimes it becomes difficult to interpret the NMR spectrum of this phosphorus metal complexes, but otherwise if you have prior knowledge about the way the chemical shift changes on chelation it turns out to be an extremely useful tool. Based on these based on the information that we have so far we can see that phosphorus three ligands can be readily synthesized and they can be synthesized in a variety of R groups having variety of R groups. Now, the question arises what kind of a ligand is this P R 3? The P R 3 group is it a good donor? Is it a good sigma donor or is it a pie donor? One can also ask the question is it a pie acceptor? If you look at the complexes formed by triphenyl or trialkyl phosphines then one can understand based on the structure and the spectroscopy that we do we can figure out what kind of a donor we are dealing with. So, let us take a look at a little bit of the chemistry that is there in this molecules. First, let me give you an example of how one can synthesize this triphenyl phosphine complexes and that will illustrate the chemistry that I am talking about. Let us take a molybdenum hexacarbonyl complex. This is a molybdenum hexacarbonyl complex. I treat that with two equivalents of triphenyl phosphine. This appears to be a general way of synthesizing phosphorus containing molecules. You can take any labile ligand in this particular case carbon monoxide turns out to be useful because carbon monoxide will escape into the reaction medium away from the reaction medium and you will be left with the complex. You would expect the treatment of MOCO6 with two equivalents of triphenyl phosphine will result in the formation of a trans complex or it can result in the formation of a cis complex. It turns out that exclusively the cis octahedral complex is formed. Only the cis octahedral complex is isolated from this reaction mixture. If you look at the complex itself you can very easily see that carbon monoxide has no great steric influence. It is only the phosphorus containing ligand which has got some steric influence. So, if you want to pack two ligands, two phosphorus containing ligands around the molybdenum it would be best to have the trans geometry because the two L groups are far away from each other. Surprisingly it is the cis ligand which is cis complex which is formed and the trans complex is not formed. So, there must be an electronic reason for this particular preference. Very often it is good to form a particular opinion by looking at a variety of complexes. So, this has been done in the case of phosphine complexes. Both trans and cis have been attempted. Preparations of these have been attempted and almost always the cis complex is what is observed in many metals and with many carbon monoxide geometries. One forms only the cis complex. Now, one can also make the Tris coordinated complex. The Tris tri alkyl phosphine complexes have been synthesized and here again you can have two particular geometries. One is called the Mer geometry where the three ligands are placed in a meridional plane. The three ligands are placed in a meridional plane. Then it is called a Mer complex and then you can also have a FAC complex and the FAC complex you have a facial geometry. So, one phase of the octahedron is occupied by the three ligands which are the phosphorus three ligands. So, between these two geometries you will notice that the FAC complex always has one L trans to a carbon monoxide. Always you have one carbon monoxide trans to an L group. So, all three L groups have got a trans carbon monoxide. You will notice that each of these L groups has got a trans carbon monoxide. Whereas, in the meridional geometry you do not have that particular situation. In one case which is this particular L group you do have a trans carbon monoxide. But the other two L groups do not have a carbon monoxide opposite them. Now, it turns out that you can distinguish these two species very easily using the carbon monoxide stretching frequency because that depends on the trans ligand. So, based on carbon monoxide stretching frequency you can find out that it is only the FAC isomer which is formed and the merisomer is not formed. So, we have two situations now both in the dye substituted case and in the trisubstituted case. We can see a clear preference for the ligand L to be trans to a carbon monoxide and this electronic preference needs to be explained in order for us to understand why exactly L prefers this geometry and what is the electronic reason for the same. So, let us now look at the traditional explanation which was available until the mid 1980s. People normally gave this explanation and we will discuss this first because it illustrates an important scientific principle. Whenever there is a argument or a discussion regarding a concept one achieves a better explanation at the end of the discussion. So, here is an explanation that was given originally people thought that P R 3 groups are pi acceptors and this pi accepting property arose from the 3 D orbital which is empty because the phosphorus is a element which has got the valence electrons in the 3 S and 3 P. So, remember 3 S and 3 P are the valence electrons and these are the ones which are having the 5 electrons which are there on phosphorus and the 3 D is usually empty. So, because the 3 D is empty people thought that the acceptor strength the acceptor strength of the P R 3 group comes from the 3 D orbitals and they have the right symmetry to overlap with metal orbitals. So, let us take a look at what we are discussing here. Here is the molybdenum center interacting with the phosphorus 3 molecule. The 3 molecules the 3 arms of the phosphorus where the R groups are present will be distributed just like ammonia in an umbrella like fashion. And the lone pair on the phosphorus is now pointed towards the metal atom it has got 2 electrons and this lone pair is going to be donated to the metal. So, this is the electron density flow in the sigma orbital that we are talking about and if the phosphorus atom has got a 3 D orbital it has got a 3 D orbital. Then the 3 D orbital can interact with the metals 3 D orbital and if this is empty and if this is filled. So, you have a filled metal orbital and you have an empty P orbital phosphorus orbital which is a 3 D system then you can have electron density flowing in this direction. So, this kind of an electron density flow will result in a pi acceptor behavior. So, people thought that phosphorus ligands are good pi acceptors because of this electron density flow going from the metal D orbital into the empty D orbital on the phosphorus. So, how does this help us to explain the geometrical preferences that we observed earlier. Because carbon monoxide is an excellent pi acceptor carbon monoxide is one of the best pi acceptors that we know because this is the best pi acceptor that we know. It is natural that carbon monoxide would like to have a poor pi acceptor in the transposition. Now, you might ask why is this situation why should it be in the transposition and that is because this D orbital that is there on the molybdenum center. This is the D orbital that is there on the molybdenum center I have drawn it slightly pushed away from the center. So, that you can understand the overlaps. You can see that the D orbital that is involved in this D pi D pi D pi interaction. This interaction that we are talking about is on the same axis as this carbon monoxide in the transposition. It is the same orbital which will have to donate electron density in both directions. It will have to donate electron density from the metal D to the phosphorus D. It will have to give electron density from the metal D to the pi star of the carbon monoxide. So, this is the interaction that we are talking about. There is a competition between the phosphorus empty D and the carbon monoxide pi star. This is the pi star and you can draw the second set of orbitals also. This would be shaded and this would be empty. So, this is the pi star orbital and the electron density has to flow into the pi star of the carbon monoxide. That is what usually one observes in all the complexes. So, because carbon monoxide is such a good pi acceptor, the phosphorus is a poorer pi acceptor. So, by putting three phosphorus ligands trans to carbon monoxide, the carbon monoxide ligands are kept happy because they will have enough electron density moving in this direction. Whereas, if you have the phosphorus ligands all in the meridional position or in the trans position, you will notice that the carbon monoxides also have to be trans. If the carbon monoxide is trans, one carbon monoxide competes with another carbon monoxide for the same D orbital and that is an unhappy situation for the metal. So, it does not prefer that situation and it likes to have a phosphorus, which is a poor pi acceptor opposite carbon monoxide, which is a strong pi acceptor. Now, how do we know that whatever we have said so far is true? If you look at carbon monoxide stretching frequencies in the fact complexes, here is a fact complex that we are talking about. This is a fact complex because the three ligands, the substituents are in the facial position of an octahedron and then the carbon monoxides are on the opposite phase. You can look at the symmetric stretching, this is the symmetric stretching and the anti symmetric stretching of the carbon monoxide. You will find that if you have a very good donor on the phase or on this phase for example, if you have a very good donor, then the carbon monoxide stretching frequency reduces quite significantly. You will notice that in MOCO6, that is in molybdenum hexa carbonyls, where carbon monoxide is trans to another carbon monoxide always. So, this is the observation for MOCO6. Then you have a stretching frequency of 2004 centimeter minus 1. Adding a ligand like di n, which has got three nitrogen donors in the facial position results in a very low stretching frequency for the carbon monoxide ligands. That is because electron density flows from the nitrogen to the metal and from the metal to the carbon monoxide very effectively. So, the stretching frequency goes down. The moment you start adding a ligand, which can compete with the carbon monoxide for electron density, then the frequency of the carbon monoxide slowly keeps going up. When you have tri ethyl phosphite for example, in this case it is p o e t thrice. If you have p o e t thrice, then the tri ethyl phosphite happens to be a better pi acceptor than tri ethyl phosphite. This is tri ethyl phosphine p e t 3. So, this is a poor pi acceptor. So, poor pi acceptor this is a good pi acceptor. So, if you have a good pi acceptor trans to the carbon monoxide, then the stretching frequency is higher. So, you can see that as the stretching frequency of the three carbonyl ligands go up, you can see that the electron density on the metal has gone down because the phosphorous atom is also competing for the electron density. So, that is the reason why you have this competition between the d pi d pi interaction between the phosphorous and the metal. The d pi pi star interaction on the carbon monoxide these two compete with one another. So, it is better to have the phosphorous ligands in the transposition. One can look at this graphically to understand it and appreciate it better. Here is an example where you have the three carbon monoxide ligands having two stretches and a symmetric stretch and an asymmetric stretch. Both of them are plotted in a series and you find that the metal carbonyl complex, the pure metal carbonyl complex which is marked here as m o c o 6 has got the highest stretching frequency, but you put good donors on the transposition. The best donor that is the nitrogen donor has a very low stretching frequency, but as you increase the pi accepting nature of the phosphorous ligand, the stretching frequency keeps increasing. So, the electron donating power of the trans ligand keeps increasing in this direction, keeps increasing in this direction and the trans carbon monoxide stretching frequency keeps increasing in the opposite direction. So, this is a very clear indication of the fact that the trans ligand competes for the same d orbital for d electron density for pi donations. So, this explanation this traditional explanation was quite satisfactory and it was used in the textbooks for a very long time, but in more recent textbooks you might find it that it is being disputed because people found by computational methods that 3 d orbitals are actually not available or accessible for the metal d orbitals. So, the metal d orbitals are in one particular energy level and the 3 d orbitals are much higher in energy. So, if one has to represent this graphically one can say that these are the metal d orbitals and the phosphorous d orbitals are much higher in energy. So, the interaction between these two orbitals will be very poor because the energy matching is important for forming bonding and anti bonding molecular orbitals. So, because of this controversy people have abandoned this explanation that the d orbitals on phosphorous are actually involved in d pi d pi pi interactions. So, this gap that is large as it resulted in abandoning of this explanation and one also notes the fact that you cannot measure the energy of empty orbitals. So, it is difficult to disprove or prove this particular point that we are talking about. As a result there was an alternative explanation that was built around what is called a sigma only theory. That means phosphorous ligands are only good sigma donors and there is no pi interaction that is present. So, this particular theory was able to explain a few of the observations that were present. Let us just take a look at some of the explanations that could be made using the sigma only theory or the no pi bond theory. The decrease in the new CO is proportional to the electron density on the metal atom. If the P y 3 ligand is a good donor, if P y 3 is a good donor then metal has got greater electron density. If the metal has greater electron density, CO has less stretching frequency because the pi star orbitals are populated to a better extent. If P y 3 is a weak or it is a weak sigma donor then the new CO is higher. This explanation also seem to be reasonably sufficient because the type of electron donation that we are talking about is an electron donation from the 3 S P hybrid on the phosphorous to the metal. If this is x and this is y, this is d x squared minus y squared accepts electron density from the phosphorous and it is the d x z on the phosphorous on the metal which is interacting with the pi star on the carbon monoxide. So, this type of an interaction would be sufficient to explain the type of electron density changes that are happening on the carbon monoxide and also subsequently the carbon monoxide stretching frequencies. One way in which people try to explain these changes is to look at the ionization potential of the phosphorous atom because the electron density that is being donated from the phosphorous is on the hybrid orbital the lone pair on the phosphorous. If you do ionization of the phosphene the electron is coming out from the phosphorous hybrid orbital. So, the extent to which the phosphorous is able to donate a pair of electrons to the metal must depend on the ionization potential. If the ionization potential is low then this is a good donor. If the ionization potential is very poor as in the case of P f 3 this would be a poor donor. So, as you increase ionization potential the donor ability of the phosphorous will come down. A second way to look at this is to also look at the ligand and the p k a which means the easiness with which you can remove a proton from the protonated phosphene. So, the extent to which you can give the electron density on the phosphorous to the proton is depend on this particular equation. This is obviously an equilibrium and if you have a very high value of p k a then that means that the proton is not easily dissociated. So, that means that this is a good donor of electron density. If you have a very small value as in the case of p p h 2 o m e then this is a poor electron donor. So, this can also be correlated with the stretching frequencies and it was indeed possible to analyze the type of changes that you have with electron density donation from the phosphene to the metal and the frequency changes that are there in fact complexes. So, this explains all the observed results from the infrared stretching frequencies and one could almost confidently say that there were no pi effects in the interaction of p x 3 with a metal atom. So, p x 3 is a sigma donor and there is no pi interaction at all. However, it is possible for us to explain the stretching frequencies, but not the bond distances. In fact, if you look at the bond distances that are there between phosphorous and the metal one can see that you would one can expect longer phosphorous metal bonds due to poor phosphorous to metal bonding in the sigma manifold. So, if you have electron donation from the phosphorous to the metal, if you have a poor sigma donor you should have longer bonds. If you have strong sigma donation you should expect stronger or shorter bonds, but this is what you expect, but what is observed is something else. Here is where the no pi bond theory fails. Let us take two examples. One example is the case of p p h 3 with c r c o 5. Here we have a monosubstituted carbon carbonyl complex. So, you have c r c o 6 where only one phosphorous ligand has been added. You can also compare it with the tri phenyl phosphite complex which is given here. We can now look at the phosphorous chromium bond distances that are there in these metal complexes. Suppose you have a p c r distance of 2.4 to 2 angstroms in the case of the tri phenyl phosphine complex. This is the instance where the p o p h complex will have less electron density to donate. So, you would expect a longer bond distance for this particular complex where you have a weak donor. So, p o p h 3 is a weak donor. This is a weak donor and this is a better donor. So, the better donor should have the shorter distance, but we find that this is exactly the opposite of what you would expect. You have a shorter bond distance for the tri phenyl phosphite complex. What is interesting is that you would expect for the same reasons that we have discussed. If you have good sigma donation, the trans carbonyl should have the longer bond distance, but this is also in the opposite direction. So, it is very clear that the no pi bond theory or the sigma bond only theory is not sufficient to explain all the data that we have in terms of crystallography. So, you expect something else other than the sigma bonds to explain these bonding interactions. One of course knows that p o p h 3 if it if because it is a better electron withdrawing group, this would form better pi bonds between the phosphorous and the chromium. If this forms better bonds between phosphorous and chromium, the pi bonding is there, then this is exactly what one would expect. The phosphorous chromium bonding would be short and the phosphorous chromium bonding in this case where there is poorer pi interaction would be longer. So, the pi bond theory is able to explain the bond distance changes in these cases, whereas the sigma bond only theory is not able to explain this changes. I will give you one more example where this has been conclusively shown and I take an example where the two phosphorous metal distances are in the same complex. This is always a good way to make comparisons because there are many many factors that go into a metal ligand bond distance. So, here is a case where you have p p h 2, which is a good electron donor compared to p c f 3. C f 3 is an electron withdrawing group. So, this would be a poor donor and better accepted. This is a good donor relative to this phosphorous relative to this phosphorous. One can see that this phosphorous is a better donor and if this is a better donor one would have expected based on sigma bond only theory. This should have a short distance whereas we find that this distance is longer. So, clearly there are some pi f x in the explain that need to be used in order to explain the short bond distance that is observed 2.17 for this phosphorous platinum bond and this bond distance is 2.24 angstroms. So, one can explain these bond distance changes using the pi f x because this is a better pi acceptor. You have multiple bonding between the phosphorous and the platinum. You have multiple bonding and so this bond distance is reduced from what you expect. A similar explanation can be offered for these complexes when you put two triphenyl phosphenes trans to each other. You are competing for the same d orbital electron density when you have a chloride trans to a triphenyl phosphene. Then you have a pi donor trans to the triphenyl phosphene and if there is a pi effect then the electron density can flow from the chlorine to the rhodium and from the rhodium to the triphenyl phosphene. So, because the pi bonding is reinforced the pi bond in the phosphorous trans to the chlorine is reinforced by electron density being donated from the chlorine to the rhodium and from the rhodium to the phosphorous. This pi bond is in fact making this bond distance shorter. It is in 2.22 angstroms about 0.1 angstroms less than what you expect for this triphenyl phosphene which is in this complex. So, you have a same complex two complexes that are being compared in both complexes. You compare the bond distances within them. We are not comparing the 2.24 here with the 2.32 here rather we compare it within the system within the platinum complex. You compare two phosphorous platinum distances and within the rhodium complex you compare this phosphorous rhodium distance with the rhodium phosphorous distance which is there along this direction. So, the pi bond theory is able to explain why the triphenyl phosphite has got a short distance 2.309 versus the triphenyl phosphene which is got the longer distance. That is because of the pi bond which is there between the phosphorous and chromium which is more effective in the case of the electron withdrawing phosphite which is present here. Now, if we cannot explain the pi bonding using the d orbitals how can we explain the pi bonding? To do this the theory the computational chemist came up with the idea that you can have what is called negative hyper conjugation. Or donation of metals filled d orbitals to the sigma star orbitals of the p x group. So, if you have a p x bond and the p x will have a sigma bond or a sigma orbital corresponding to the sigma bond and a sigma star orbital. If this sigma star orbital on the phosphorous is capable of accepting electron density then one can say that there is negative hyper conjugation from the metal to the phosphorous p x sigma star. Now, can these p x sigma star orbitals accept electron density? Can they accept electron density? If so what is the shape and symmetry? So, depending on the shape and symmetry you can expect them to behave as good pi orbitals. So, this is what we are going to see in the next section. Here I have for you the molecular orbitals of a hypothetical phosphine which is p h 3 p h 3 which is arranged according to the energy. If you look at the three sigma bonds of the p h sigma bonds they form a set of molecular orbitals bonding molecular orbitals. So, all of these are filled these are filled molecular orbitals these are filled molecular orbitals. Then we have the traditional lone pair which is sitting on the phosphorous which is pointed away from the three groups which are there on the phosphorous. So, this is if you want to talk about the metal interacting with the phosphorous then this would be the direction in which we are orienting the phosphorous ligand. So, the phosphorous has got a large lone pair which is sitting on the phase opposite the three hydrogens and it is pointed in such a way that can be now donated to the metal atom. That is your sigma bond between the phosphorous and the metal. Now, the pi acceptor orbitals are actually coming from the sigma star orbitals which is the sigma star orbitals of the p h bonds. The sigma star orbitals are primarily phosphorous p x and p y. If this is the z axis if this is my z axis then the p x and p y x and y this p x and p y orbitals are what you see here. This is my p x and this is the perpendicular p y direction. So, the p x and the p y orbitals are the empty orbitals. These are empty and they can accept electron density. So, this is how the phosphorous atom is able to accept electron density from the metal. From the metal electron density flows into the sigma star orbitals of the phosphorous x bond. The p x sigma star is capable of accepting electron density. It turns out that it has the same shape and the right symmetry to overlap with the metal orbital. So, phosphorous is a good pi acceptor not because of the empty d orbital, but because you have the sigma star orbital and that sigma star orbital has got greater contribution from the phosphorous greater contribution from the phosphorous. It is primarily the p orbitals which are present on phosphorous which contribute maximum to the sigma star orbital. So, it is able to accept electron density from the filled orbital of the metal. Arsenic x 3, arsenic y 3 and antimony y 3 groups should also be good as they are also less electronegative than r groups. So, they have a good option of accepting electron density into the sigma star orbitals. Now, one can also ask the question why is it that a means or n r 3 groups are not good pi acceptors. This is surprising because in the periodic table we find in the same group n r 3 is only a sigma donor. These are only sigma donors and not pi acceptors. One can in fact categorically say that these are not pi acceptors. So, why is it that they are good sigma donors and not pi acceptors? Nitrogen is in fact more electronegative than most other groups. In the case of ammonia for example, it is definitely more electronegative. Nitrogen is more electronegative than hydrogen. So, nitrogen has got good contribution to the bonding molecule orbital, but the sigma star has most of the contribution from hydrogen. I will show you the molecule orbitals of n h 3 as an example. Here you can see that the nitrogen contribution to the bonding molecule orbitals are significant. These are all filled now. You have filled molecule orbitals here and this is the lone pair. So, that again is a good donor orbital, but you can see that the sigma star orbitals where you have this 3 sigma star orbitals. These are the sigma star orbitals. You can see that the contribution of hydrogen is significantly greater compared to what you had in the case of tri phenyl or tri alkyl phosphines, where phosphorus because of its lesser electronegativity compared to nitrogen contributes more to the sigma star orbital. So, one can say that the electronegativity between a and h in h 3 is responsible for the sigma star orbital being capable of accepting electron density. If a is less electronegative, then a will be a good sigma star pi accepting ligand. So, in the case of phosphorus, arsenic and antimony, they are good sigma star orbitals capable of accepting electron density. Now, we can go back to this figure and see if the pi donation to the sigma star orbitals are capable of explaining the carbon monoxide stretching frequencies that you have observed. First, let us take a look at what we saw earlier. If you have OET groups, which are strongly electron withdrawing, we find that the frequency of the trans carbon monoxide is in fact higher. So, the frequency is in fact increasing in this direction. When the group was a phenyl group, then it was not as good a pi acceptor, but when we have OET, it is a very good pi acceptor. So, you find that relative to MOCO 6, which was listed here, the OET has got the frequency has got the trans carbon monoxide stretching frequencies closest to MOCO 6. So, P OET thrice is a very good pi accepting ligand. So, P OET thrice is a very good pi accepting ligand. So, let us now look, go forward and we see that suppose we had chlorine and phosphorus. In the previous diagram, we did not have the PCL 3 and the PF 3 ligands. Suppose, we add the PCL 3 and the PF 3 ligands. We find that these complexes have got stretching frequencies, which are even higher than what you expect for PR 3 groups. So, here we go MOCO 6 had this stretching frequency 2004 centimeter minus 1. In the case of PCL 3 and PF 3, the stretching frequencies are even higher. They are 2040 and 2090 centimeter minus 1. So, clearly you have a situation where the trans ligand is capable of attracting pi electron density as much as carbon monoxide. PCL 3 and PF 3 are able to compete with the trans carbon monoxide for pi electron density. So, the trans carbon monoxide stretching frequency has increased beyond what you have for MOCO 6. So, this explanation is again something that could not have been given by the sigma only theory. So, in fact compounds carbon monoxide is competing against the poor pi acceptor. That is why it is a situation, which leads to greater stability. So, fact compounds are more stable than mer isomers. It is also true that the carbon monoxide stretching frequency is changing in the way, in which it changes for a series of complexes based on the trans ligand. The trends are reversed in the case of PCL 3 and PF 3. So, when you have PCL 3 and PF 3, you have a situation where the trans carbon monoxide has greater stretching frequencies than MOCO 6, where the trans ligand is carbon monoxide. Now, based on these factors, Tolman in fact devised what is called Tolman's electronic parameter. Tolman's electronic parameter measures the extent to which the complex NiCO 3 L has two stretching frequencies for the system which is pictured here. So, the reason for choosing this nickel complex is the fact that you can readily make it by reacting NiCO 4. You can react NiCO 4 with any ligand and it will readily form NiCO 3 L. You can readily measure the infrared spectrum and you can see that as you substitute a good carbon monoxide with a phosphorous ligand, the stretching frequency of the trans carbon monoxide are lower than what you expect for NiCO 4. But as you substitute the trans ligand, you tend to have an increase in the stretching frequency. So, much so that in the case of PF 3, the stretching frequency is close to that of free carbon monoxide. In other words, there is very little pi star electron density on the carbon monoxide. When you substitute the ligand with L equals PF 3, there is very little electron density that flows into the carbon monoxide ligands, the pi star orbital. So, the stretching frequency is close to what you expect for free carbon monoxide. So, this is the reason why you end up with a good pi acceptor in the transposition is always competing for the carbon monoxide electron density. So, you have poor pi acceptors in the transpositions and if you have a good pi acceptor like PF 3, then the stretching frequency of carbon monoxide is close to the free carbon monoxide stretching frequency. So, pi f x alone may not be sufficient to explain many of the interactions that we have described. The charge induced on the metal also plays a role and this is something which we have not discussed in this lecture. But we will discuss it in a lecture on carbon monoxide and one should also remember that bonding is dependent on steric effects. This is again a factor which we will explain in a future lecture. Now, discuss so far what we are going to do is to look at a diagram which will tell us what we have discussed so far in a diagrammatic fashion. We first looked at phosphorous ligands and we looked at P 3 ligands and how it is easy to synthesize P 3 ligands. Because we can make this in a stepwise fashion, we are able to make P R 2 x, P R x 2 and P R 3. Because stepwise formation is possible, we can make ligands which have got three different ligands or three different R groups on the phosphorous. This is a very useful tool because if three R groups are present, the phosphorous becomes a chiral species. After we looked at synthesis of phosphorous 3 ligands, we also looked at how one can make phosphorous 3 complexes. This is done by a simple substitution reaction. Having looked at these complexes, we looked at some of the spectroscopic factors that are useful in these systems. One noted that the fact that phosphorous that you want NMR is extremely useful although it is an empirical tool. Although it is an empirical tool because phosphorous that you want signals are readily observed in these complexes, it is possible to use them effectively. Then we looked at phosphorous bonding to the metals. Initially people thought that there was a double bond connected due to the presence of D orbitals on the phosphorous. Then it was realized that it was not just the D orbitals on phosphorous. It is possible to have electron donation to the electron donation to the sigma star orbitals on the phosphorous due to negative hyper conjugation. The alternative suggestion that it is electron density is flowing into the P x sigma star is quite sufficient to explain all the factors which we have observed in the case of metal phosphorous bonding. Finally, we looked at characterization of phosphorous 3 ligands. We looked at the Tollman's electronic parameter. The Tollman's electronic parameter is extremely useful to understand the type of bonding that is there between the metal and the phosphorous ligand. In future classes, we will look at Tollman's cone angle and the buried volume concept which is also useful for explaining the phosphorous metal bonding. If you look at the range of complexes that are formed by phosphorous ligands interacting with metals, one finds that it is truly a remarkable range. One can use a variety of R groups and one can use a variety of metals and a truly amazing number of molecules can be made using phosphorous chemistry. This turns out to be a very rich field which is being actively pursued even today. This is finished.