 Although, the metal hydride implies that there is only a metal hydrogen bond and not a metal carbon bond, metal hydride chemistry is extremely important in organometallic chemistry. That is because in several organometallic complexes where there is a metal carbon bond, you might end up with some chemistry that generates a metal hydrogen bond. We will see how this can happen as time goes along. Let us first start with why they are so important? Why metal hydrogen complexes are so important? If we look at the industrial chemistry that is going on, you will notice that much of it is oriented towards fuels and metal hydrides and catalysts containing metal hydrogen bond are in fact a key component of the fuel industry. It is possible to make methanol from carbon monoxide and hydrogen and this also requires the use of a metal catalyst that contains a metal hydrogen bond. So, metal hydrides are important in the fuel industry. Secondly, we notice that in the case of manufacturing fine chemicals, one often uses metal hydrides as catalysts. This is true of bulk chemicals like fertilizers and plastics also. All of them end up using metal hydrides. Another important place where metal hydrides might be involved is in the generation of metals from metal ores. During the reduction of a metal ore from the oxide or the sulphide, it is possible that metal hydrides are intermediates. So, let us take a look at the chemistry and the synthesis of metal hydrides. If you look at the periodic table and look at all the metal hydride complexes, you will notice that there is a distinct gradation in the properties of metal hydrides that are formed. There are some metal hydrides which are formed by alkali and alkyne earth metals which are on the left side of the periodic table. These are marked in pink and you will notice that they are marked as saline. They are marked as salt like metal hydrides. These are salt like metal hydrides and the salt like metal hydrides implies that they will react with water and generate hydrogen. So, the type of chemistry that you have here is that of H minus and M plus. There are other metal hydrides or hydride compounds which are completely molecular and these are on the extreme right side of the periodic table. These are marked in yellow. So, these are molecular metal hydrides with the exception of aluminum which we can understand readily as being an intermediate metal hydride because you have both covalent character and also a little bit of ionic character. But other than that most of the hydrides on the right side of the periodic table are covalent metal hydrogen compounds such as methane, ammonia, water, phosphine and so on. So, these are extremely covalent and they will not ionize in water as H minus or H plus. On the other hand the group of transition metals which lie in between these two extremes are compounds which may not form sometimes only metal hydrogen compounds. There are some which form metallic hydrides which means that after the hydrogen compound is formed with just M H N. So, we are talking about species which form only M H N and these species are completely metallic. So, they are marked in green and that is what we have here on my left on the left side of the periodic table and these are marked as metallic. Now, what is left in white corresponding to ruthenium, ion, cobalt etcetera they are either complexes which form alloys or species which are not definite compositions and so they are left as unknown hydrides. So, practically the whole periodic table can all the elements in the periodic table can be converted into hydrides or hydrogen containing compounds and their chemistry has been extremely well studied. So, these are as I mentioned hydride complexes they are usually called hydrides and they are in striking contrast with several molecular hydrides like covalent hydrides like ammonia, methane etcetera. Metal hydrides are usually those compounds which have other ligands as well apart from hydrogen. The few complexes of transition metals with hydrogen and hydrogen only where there is only hydrogen as a ligand are few and rare, but nevertheless they are present and one example, one very interesting example is that of Re H 9 2 minus. You will notice that rhenium is now if hydrogen is in is acting as H minus then rhenium is in the plus 7 oxidation state. That is a maximum oxidation state that one can think of for a group 7 or a rhenium metal and it is interesting that it forms a poly hydride like that. The transition metal hydrides are considered hydrides because the electronegativity of the transition metal is less than that of the hydrogen. Now it is difficult to assign electronegativity to complexes because once you have a complex and you have several other ligands the electronegativity of the fragment has to be considered. In this particular instance if I write a metal carbonyl the metal carbonyl fragment might have an electronegativity that is different from that of the transition metal itself. So, we are talking about the ability of the fragment to pull electrons to itself when it is bonded to another element. So, if you look at T M C O N which is the species which is pictured here then the ability of the transition metal in this fragment to pull electrons to itself might be greater than that of hydrogen. So, if this is the case then we have a problem in the free state as a transition metal its electronegativity was less than that of hydrogen and so we called it a hydride meaning hydrogen will pull electrons to itself and in a complex containing T M H the hydrogen will behave as a hydride. But when it is in this situation where you have a carbon monoxide which has increased the electronegativity of the transition metal. Transition metal hydride might not be real hydride. So, if T M C O N has got an electronegativity greater than hydrogen heterolithic splitting of T M H might occur by which we mean that T M H might now behave as H plus and T M C O N minus the negative charge would be stabilized by the transition metal C O fragment the carbonyl fragment and so H might behave more like a proton. So, this will create some confusion because we call all metal hydrogen containing compounds metal hydrides which implies that it is a H minus whereas in the complex that we have synthesized which is H T M C O N H T M C O N the hydrogen is behaving as a typical proton. So, this is something that can lead to a confusion because the chemistry implies that hydrogen is H plus in the complex whereas the nomenclature we refer to it as a hydride. So, here is a example of a few compounds which are formed by hydrogen which are all metal carbonyl containing complexes and the situation that I just described to you is very easy to follow H C O C O 4 is a hydrid o complex of a cobalt carbonyl complex. Cobalt has got 4 carbon monoxides and it can stabilize the negative charge and as a result if you dissolve H C O C O 4 in water you will end up with C O C O 4 minus which is there on the right side and a proton would be generated in the aqueous medium which means that it is really an acid. Here I have listed for you the equilibrium constants this is the K a which is listed here and you will notice that the K a of H C O C O 4 is 10 to the power of minus 2 which is almost like that of sulfuric acid. So, you will have a large amount of protons in solution when you dissolve H C O C O 4 in water. So, P K a is actually the minus log 10 K a. So, P K a of H C O C O 4 would be equivalent to 2. So, the first compound would have a P K a of 2 and so it is a very strong acid. You will notice that when we replace one of the carbon monoxides with a triphenyl phosphine what we have done is we have reduced the capacity of the transition metal cobalt in this case to accept electron density and that in turn reduces the effectiveness of cobalt in pulling electrons towards itself and in turn the effectiveness with which it will release H plus in solution. So, what happens is the equilibrium constant as a smaller value 10 power minus 7 and it behaves as a weak acid. So, just by replacing carbon monoxide with triphenyl phosphine we have reduced the acidic nature of this metal hydride species and it behaves like acidic acid whereas, the H C O C O 4 was a strong acid. Now, this is a common phenomenon and you can experience this with many metal carbonyl complexes which have hydrides. The hydrogen will behave as an acid when you have only carbon monoxide, when you replace carbon monoxide with poorer pi acceptors then it will behave as a weaker acid. Now, let us proceed, let us take another example H 2 F E C O 4. This is an extremely strong acid, there are two hydrogens attached to the iron and it will ionize to give you H 3 O plus in solution, but when it ionizes it leaves behind H F E C O 4 minus the species which we have listed as a second compound here as the last compound in this table. So, H 2 F E C O 4 will ionize like a strong acid, but the second ionization constant just as in the case of oxalic acid, the second ionization constant turns out to be a very small number. In other words it does not dissociate very readily and you have a small k a and in turn the p k a would be close to 14. So, the p k a of this compound which is listed last in this table is going to be something like 14 which means it is a very weak acid. This is exactly what happens in organic acids also and I have written here confer oxalic acid because that is a this acid where the second ionization constant is much much lower or it ionizes to a lower extent compared to the first ionization. So, the chemistry of transition metal hydride is a bit confusing because it is something that we call as a hydride and it behaves like a acid in solution. So, Heber was the first person in fact he discovered it way back in the early 1900's and he made transition metal hydrides and metal hydrogen complexes. He in fact was studying the reaction of sodium hydroxide with iron pentacarbonyl and during the course of this reaction he made a compound which is Na H Fe C O 4. Now, you will notice that the essential reaction in this particular case is that sodium hydroxide the OH minus has reacted with carbon monoxide and formed an anionic compound where you have this anion and Na plus. So, this sodium salt is formed by any OH reacting with Fe C O 5. OH minus attacks one of the carbon monoxide and generates the sodium salt which can now decrease decompose because you have the elements of carbon dioxide in this compound. If you eliminate carbon dioxide what you end up with is a hydride complex H Fe C O 4 minus. Notice that OH minus was what attacked the carbon monoxide. So, the net charge on this compound would be minus 1 and after you eliminate the neutral carbon dioxide molecule you would be left with a hydride complex which will be negatively charged. So, this chemistry is in fact reminiscent of what happens in the water gas shift reaction. The water gas shift reaction which is a extremely important industrial process which is used to generate hydrogen from water using carbon monoxide as a reductant is something that can be written in this fashion C O plus H 2 O to give H 2 and C O 2. If you recollect the reaction of OH minus in water results in the conversion of the net conversion of carbon monoxide to C O 2. Fe C O OH minus the species that we generated by reacting a carbon monoxide on the ion coordination sphere with OH minus can. In fact, this is Fe C O 5 reacting with OH minus to give you the anionic salt. The anionic salt now decomposes to give you the hydride and if this hydride reacts with proton in water this is H plus in water then what would happen is it would lead to the generation of di hydrogen or hydrogen molecule and Fe C O 4 as a neutral species because we are reacting an anionic species with a cation proton and as a result we will eliminate a neutral water molecule and this results in the formation of Fe C O 4. Fe C O 4 is a 16 electron species and it will grab carbon monoxide from the atmosphere in the reaction medium and form Fe C O 5 again. So, you can write a catalytic cycle which involves Fe C O 5 and water and carbon monoxide. Fe C O 5 is a catalyst and it can convert carbon monoxide and water to carbon dioxide and di hydrogen. So, this is basically how the water gas shift reaction might be happening. So, let us take a look at some of the general methods by which we can synthesize insertion metal hydrides. The synthesis of metal carbonyl complexes involved taking metal in its higher oxidation state and reducing it with some reductant. In the case of metal hydrides you often encounter very similar chemistry you can start with a metal sulfide. So, you can start with M S or M 2 S depending on the oxidation state of the metal and you can reduce it with hydrogen in the presence of carbon monoxide and then you form very nice hydro carbonyl complexes. In this particular instance you liberate a molecule of H 2 S hydrogen sulfide and the reaction has to be carried out at a higher temperature 200 degree Celsius and 200 atmospheres of pressure of hydrogen and carbon monoxide and you liberate hydrogen sulfide. It is convenient to push the reaction to the right hand side of this equilibrium you have to add copper powder. So, that the hydrogen sulfide is completely adsorbed by the copper to form copper sulfide. So, this is a convenient way of making hydro carbonyl complexes. Hydrogen can also reduce oxides and this is what we refer to in the beginning of this lecture when we said that to form pure metals from ores one very often uses hydrogen gas. So, if you can take osmium tetroxide and reduce it with carbon monoxide and hydrogen one can generate a hydro osmium carbonyl complex and here pictured two complexes that are formed as a result of this reduction reaction you can form hydro osmium complexes both dinuclear and mononuclear species. Apart from oxides and sulfides one can also start with chlorides and in the case of chlorides very often early transition metal hydrides or saline hydrides can be used. I have shown for you two possibilities either the use of lithium aluminum hydride. Lithium aluminum hydride is Li Al H 4. So, this is essentially an aluminum hydride or sodium borohydride and that is B H 4 minus is a reductant. So, either B H 4 minus or L H 4 minus are used as reductants to reduce C P 2 Z R C L 2 and in this instance what we have here is a cyclopentadienyl compound where C P in fact stands for cyclopentadienyl anion. Two of them are there in the zirconium and the chlorides are reduced to hydrogen. So, this can be carried out with fairly easy laboratory methods you can convert the chloro compounds into the hydride. Interestingly, if you take cis platinum complex, cis chloro platinum complex and reduce it you end up with a trans platinum compound during the reduction. Now, we will talk about what happens with hydrogen and it is bonding in a few minutes, but this is one of the few instances where a ligand has only sigma bonding interactions. Whereas, most other ligands that we encounter in coordination chemistry and in organometallic chemistry there are pi interactions which are present. They are either very strong pi interactions or weak pi interactions, but nevertheless they are there. In the case of hydride of course, you do not have the possibility of having a pi type orbital on hydrogen and so only sigma interactions are present. Reduction with hydrogen gas of chloro compounds can be carried out in group 8 metals especially the ruthenium group. They are readily reduced with dihydrogen and we know this chemistry from Wilkinson and here we have generated a H-R-U-C-L tris-triphenyl phosphine complex. P actually stands for triphenyl phosphine and so if you reduce the tetra case triphenyl phosphine complex with hydrogen you end up with this very interesting very valuable ruthenium complex which is something that we will encounter later on when you study the chemistry of hydrides. So, interestingly up to this point I have talked to you about the use of hydrides and the use of dihydrogen for reducing. There is also this possibility of adding a proton. This should not surprise us because when we talked about metal carbonyls metal hydrides what are so called metal hydride carbonyl compounds released protons H plus. So, if you take a cobalt complex which I have written for you here and the net charge on this cobalt complex is there is a plus charge on this cobalt complex and this tetra coordinated phosphine complex of cobalt 1 can be protonated with H plus and it generates a 5 coordinate species which grabs the C L minus and it now forms octahedral complex of cobalt with a net charge of plus 1 with a net charge of plus 1. So, if you consider the hydrogen as a hydriding species then you will realize that we have we have changed the oxidation state of cobalt from plus 1 in this instance to plus 3 in this on the left hand side. But these are reactions which are easy to carry out in the laboratory and have been extensively studied in the recent past and I will come to that towards the end of this lecture. Now, it is also possible to take the sodium salt of carbonyl compounds in this particular instance I have pictured for you a sodium pentacarbonyl rhenium compound. So, this is actually R E CO 5 minus and it can be reacted with hydrochloric acid to generate H R E CO 5. So, if you remember your manganese chemistry rhenium is in the same group as manganese you can recollect that H M and CO 5 was behaving as an acid it will liberate H plus. And here is a reaction where we have used sodium salt of the carbonyl anion and protonated it in order to generate H R E CO 5. So, this chemistry should not surprises that we are using an acid to generate a hydriding so called hydriding complex. The hydriding complex in fact is formed in the reaction which requires an acid H plus. So, it is also possible for us to cleave metal metal bonds as I had mentioned metal metal bonds in carbonyl chemistry can be cleaved very easily with hydrogen. And in this example M N 2 CO 10 which is decacarbonyl complex of dimanganese it is reduced with H 2 that is di hydrogen to generate H M N CO 5. If you remember we made H R E CO 5 which is the analog and now we are making H M N CO 5 by cleaving the manganese manganese bond. And manganese 2 CO 10 is actually a combination of 2 CO 5 M N species. So, there are 5 carbon monoxide on each manganese and there is a M N single bond M N single bond M N. So, this is the single bond between 2 manganese atoms and if you wish you can write H 2 below that and see how heterolithic splitting or homolytic splitting of this 2 bonds can generate H M N CO 5 very easily. A similar reaction is pictured here with a cobalt analog of carbon monoxide complex. And here also the rhodium rhodium bond is broken and H R H CO complex is formed. So, the metal metal bond is easy to cleave and the formation of a metal hydride which is also forming a strong bond leads to the synthesis of these complexes. Having considered the synthesis of these complexes let us talk a little bit about how one can characterize these species. Metal hydrides are often characterized by carrying out infrared spectra on these species. They have characteristic stretches M H stretch around 2200 to 1900 centimeter minus 1. So, the C H stretch usually comes around 3000 centimeter minus 1 because the metal is heavier. You now have a shift to the lower wave numbers and you are going to 2200 centimeter minus 1. So, this stretching frequency is very diagnostic of the formation of a carbonyl species and is dependent on the other ligand the other ligands which are present on the metal species. Unfortunately this stretching frequency is occurring at a region where the carbon monoxide itself absorbs. If you remember free carbon monoxide is 2143 centimeter minus 1 and metal carbonyls range all the way from 2100 to 1900 centimeter minus 1. So, you have a conflict between the metal carbon metal hydride stretch and the CO stretch in a metal hydro carbonyl complex. But nevertheless it is easy to distinguish the two by a simple substitution of the hydrogen with the deuterium. If one does that because deuterium is heavier the reduced mass of the species changes and you have approximately a change of stretching frequency which is in the ballpark of 1.44 the ratio between new M H and new M D is 1.44. So, as a result if you substitute H M N C O 5 with deuterium D M N C O 5 will have a stretching frequency which is 1.1783. Divided by 1.44. So, this is this would be the range at which the stretching frequency occurs if you make the D M N C O 5 complex. So, it is easy to use infrared spectra as a diagnostic tool and very often metal hydrogen spect metal hydrogen frequencies are readily observed because it is in a region where other species do not vibrate. And except in the case of metal carbonyls then you would have to use some another diagnostic tool. So, here I have shown you two metal hydrides in one case it is a cyanogroup which is trans to the hydride and here is the metal hydrogen which would have a stretch of 2041 centimeter minus 1. If you have a nitro group then the metal hydrogen stretch goes up to 2242 centimeter minus 1. Now, what is going on here and why is the frequency changing drastically when you replace a nitro group with a cyanogroup. We know very well that a cyanide is a great pi acceptor and so the bonding between platinum and C N group here is extremely strong. Whereas, the nitro group is only a pi donor and it is a weak ligand. So, the strong ligand in fact weakens the trans bond. So, this bond is weak and this bond is much stronger. So, the metal hydrogen when it is strong will have a frequency it will require greater energy obviously to stretch that bond. You have a stretching frequency which is close to 200 centimeter minus 1 greater in the nitro complex. So, the metal hydrogen stretch can in fact be used as a tool to find out how is the ligand in the transposition interacting with the metal. So, the metal hydrogen stretch is a great useful tool for understanding the bonding. So, I call it the reporter ligand for the trans ligand. So, we can also use one can imagine that you can use NMR spectra very readily for understanding the chemistry in these cases. NMR is indeed very useful, but it can be difficult to interpret the NMR spectra that we get for several reasons. One of them is the fact that signals from hydrogen in M H complexes can be very broad. These hydrogen atoms are fluxional which means they do not stay in one position in the metal complex. When they are moving about the electronic environment changes and that makes the resonance very broad. So, you have a range of chemical shifts for this hydrogen which leads to a broad spectrum and you might even miss the metal hydrogen resonance. Usually, if you have a complex which is not so fluxional and you can observe it readily, then the stretching frequency ranges from minus 7 to minus 24 p p m, which means that they do not interfere with normal organic compounds. To look at metal hydrides, one in fact has to look at the negative region in the NMR spectrum. Now, the value of the chemical shift depends very much on two factors which is the D electron count on the transition metal. So, the D electron count on the transition metal can make a difference and the separation between the lowest energy level and the next highest energy level or the D D separation usually makes a difference in the chemical shift. So, it is difficult to predict exactly where you would get resonance for metal hydrides and the added complication is the fact that it can be broad. But nevertheless, there are several instances where it has been possible to use NMR spectra very effectively and that is because the coupling between a nucleus of the metal and the hydrogen can be observed and this when it is observed is a great indicator of covalency in the metal hydrogen bond. So, it is impossible for an ionic species where there is no overlap of electrons between in the hydride and the metal to show any coupling interaction. So, once we observe that there is a coupling between the metal and the hydrogen, then we are quite sure that there is covalent interaction for transferring the spin information from the metal to the hydrogen. So, rhodium for example, at 1, 0, 3 and platinum 195, these are atoms which have hydrides being formed and these hydrides have a very strong coupling with a metal atom. So, observation of spin-spin interactions clearly imply covalency in these compounds and I mentioned that there can be fluxional processes and this can be a boon and a bane. It is a boon when you can in fact observe them at say low temperature, you observe the chemical shift, then it is useful for identifying such fluxional properties occurring in solution. Structural difficulties for metal hydrides are significant when you look at the solid state structure of metal complexes. Very often they are not located in the x-ray diffraction pattern because the low electron density on the hydrogen, hydrogen has got only a one shell and it can at most accommodate two electrons. So, when it has two electrons, maximum of two electrons, you would be able to locate it in the diffraction pattern. It would give rise to a peak, but if the hydride behaves less like H minus and it behaves more like H plus, the position of the hydrogen atom is difficult to identify because there is very little electron density around the nucleus, the hydrogen nucleus. To add to this complication, you can have a large electron density on the metal. The metals electron density can be very large and the hydrogen electron density is relatively small. So, one can say that the electron density at the hydrogen is literally swallowed up by the metals electron density. So, one way people have solved this problem to understand the bonding between a hydrogen and a metal is to use neutron diffraction. Only problem with neutron diffraction is the fact that you need very large crystals in order to get this, but this is being this problem has been solved by better and better technology. It is now possible to do neutron diffraction fairly easily. One is able to locate the position of the nucleus without difficulty. This leads to accurate structures for metal hydrides. The problems of disorder and fluxional behavior are solved to a greater end and greater extent. Let us move on. Let us take a look at a typical metal hydrido complex. Here I have pictured for you a metal hydrido complex, which is in fact a ruthenium complex. It has been studied recently for water soluble polymerization. This is very interesting because people have been looking at how one can use green solvents for various reactions. If you want to do polymerization of styrene, which is usually done in organic solvents, if you can do it in water with the help of a metal complex, it would be a great advantage. This is a recent paper, which has been published, which suggests that ruthenium hydride complex can be an excellent catalyst for water soluble or aqueous polymerization of styrene. Here is a complex. I will show you this structure in three dimensions now. This picture now shows you how a hydrogen is present right on the ruthenium. The hydrogen is present on the ruthenium. Let me show this to you. Right here is the hydrogen and that hydrogen is connected to the ruthenium, which is the metal center right here. That has got a biprodyle which is complex to it and also a benzene ring, which is coordinated to it. We will talk about the chemistry of these compounds a little later, but let us take a look at the three dimension structure in order to appreciate the system little better. So, this is the complex in 3D and you can in order to simplify the orientation. I have removed the carbon atoms, which were attached to the benzene ring here. The benzene ring, which is pictured here had other carbon monocarbony other methyl groups. This benzene ring had methyl groups and I have removed it, so that you can see how the hydrogen is linked to the ruthenium and a simple biprodyle ligand. So, there are two nitrogens, which are marked in blue here and these are interacting with the ruthenium and you have this hydrogen. You can see that this is just as good as any other simple ligand like a chlorine or a bromine or a triphenyl phosphine interacting with the metal center. So, this is a hydridocomplex which has been characterized quite well. So, let us get back to our presentation here. We have the distance between ruthenium and hydrogen is around 1.54 angstroms and it is exactly what you expect for a single bond. As I mentioned, hydrogen is the only element which will interact through only a sigma bonded interaction because there is only an s orbital and no pi interactions are possible. So, let us now move on to some other complications, which occur with hydrogen. It was discovered fairly recently about 20 or 30 years ago close to 25 to 30 years that it is possible to form not just hydride complexes, but one can also form dihydrogen complexes. So, these are compounds in which the metal is interacting with H 2. So, instead of simple hydrogen as H minus, now you have a whole molecule of dihydrogen interacting with the metal. So, this was discovered when a substitution reaction was carried out between cyclohebter triene. So, this is a ligand cyclohebter triene. I have listed for you here. This is cyclohebter triene. It has got three double bonds and these three double bonds can interact with molybdenum. As a result, you would have 18 electron complex formed between three carbonyls and cyclohebter triene. It is possible to do a simple substitution reaction by replacing the cyclohebter triene with triphenyl phosphine. During this displacement or substitution, what happens is you are left with a vacant coordination side because this is now a 16 electron species. Only 16 valence electrons are present around the coordination sphere and you have one vacant coordination sphere. If this reaction is carried out in the presence of dihydrogen or H 2, it was found that dihydrogen complexes are formed in this reaction. So, dihydrogen complexes are formed when you do the substitution reaction. So, this was reported recently by Cubas and one can read about it in the accounts of chemical research paper in 1988. Now, what is exactly going on is that a bis triphenyl triphenyl triisopropyl complex of this molybdenum has got a vacant coordination side. It cannot accommodate an extra phosphine, but instead a small molecule like dihydrogen can form a complex. So, because of the bulky nature of this trisisopropyl phosphine, it results in a situation where the molybdenum would rather interact with dihydrogen than another molecule of trisisopropyl phosphine. Now, let us take a look at the metal complex. The molybdenum complex could not be isolated and characterized, but the corresponding tungsten compound was characterized crystallographically. These are compounds which have to be carried out, which have to be studied using neutron diffraction. I have pictured for you the complex that was formed with tungsten. I will show you a three-dimensional version of the same molecule, but here is a nice drawing of the same molecule. You will notice that instead of a single atom, you have a dihydrogen or an H 2, which is pictured here. You have a dihydrogen, two hydrogens here. So, it is this dihydrogen, which is sitting in the vacant site, which is formed when you have three carbon monoxide and two trisisopropyl phosphine complexes. Let us take a look at the three-dimensional structure in order to appreciate the system a little better. I have it pictured here. You can see two hydrogen atoms. Once again, the hydrogen atom is shown for you in white color. You can see the two hydrogens, which are marked here. One of them is at a distance of 2.126 angstroms. That is the one on the right side. The two hydrogens are pictured here. The one on the right side is at 2.126 angstroms. The one on the left side, which is slightly higher in level, is at 1.7, 1.898 angstroms. So, there is a slight asymmetry in the way in which the dihydrogen is interacting with tungsten, but nevertheless it is a full molecule of hydrogen, which is interacting with tungsten. How do we know that this is not two atoms of hydrogen interacting independently? First of all, we realize that the distance is not symmetric and it is not a dihydride. But, we also notice that the distance between the two hydrogens is in fact zero. So, the distance between 0.755 angstroms, which means they are much closer than what you would expect for a dihydrogen species. So, here is an instance where the distance between the two nuclei are so close that you have to think of it as a dihydrogen interacting with the metal complex and not two hydrogen atoms. So, let us take a look now at back to the presentation. So, how do we understand the bonding in these systems? When you have a simple hydrogen interacting with a metal atom, you can think of it as m interacting with a metal complex. So, the one is orbital on the hydrogen. Whereas, if you have dihydrogen, then you have two hydrogen atoms here. You have two hydrogen atoms here and in between the two hydrogen atoms is a little bit of extra electron density that is collected because of the formation of a molecular orbital. In this molecular orbital, this extra electron density is stored and this can now donate this electron density to a vacant orbital on the tungsten. So, the vacant orbital on tungsten is pictured here and into this vacant orbital, you can donate this electron density on the dihydrogen. On the other hand, you can also think of a situation where there is a sigma star orbital on the dihydrogen. This is your sigma orbital. The sigma star orbital on the dihydrogen will have two lobes of opposite sign and this can now interact very easily with the empty with the filled d orbital. This is a filled d orbital on filled d orbital on tungsten and this can now pump in electron density into the dihydrogen. So, dihydrogen chemistry is significantly different from hydrogen complexes where you have m h bonds, hydride compounds and in these cases, you do have this is a pi type interaction. Now, there is happening between the filled d orbital on tungsten and the pi star or the sigma star orbital on the dihydrogen. So, here is the pumping in electron density into the sigma star, but this is a pi type interaction because if you rotate it around this axis, you will break the bond. So, this is a pi type interaction and this is a sigma type interaction and you have weakening of the hydrogen, the two hydrogens and that bond would be elongated. This bond would now be elongated, dihydrogen bond would be elongated from what you would expect for free dihydrogen. So, dihydrogen chemistry is slightly different from hydride chemistry, but what is interesting is that in several of these complexes, it has been shown to be in equilibrium with dihydrites, which means dihydride can convert to dihydrogen complex. So, which means that you have this h 2 going to m h 2. So, this is the equilibrium that we are talking about, which is pictured here. You have two distinct and this bond is now being broken. There is no bond here between the two hydrogens and you can have this equilibrium very readily, which means that on the right hand side, I have two h minuses interacting with the tungsten and on my left side, I have the dihydrogen molecule interacting with the tungsten. Now, to complicate matters, there are species where three hydrogens are present and from the neutron diffraction data, it has been shown that two hydrogens are close to each other. Let me illustrate this with this particular structure, which is shown here. The two hydrogens, which are close to each other are the ones on the top part of this metal hydride system. The one on the lower side is the one on the is a hydridor complex m h. So, it is as if you have m interacting with two hydrogens and then it is also interacting with a hydride. So, that is what is happening here. Surprisingly, the distance between these two are much much longer at 1.748 angstroms and the distance between the dihydrogen compound, the dihydrogen ligand is much closer and it is reminiscent of what you would have if h 2 was interacting with the metal. So, here is another instance, which another complex of ruthenium, which also has dihydrogen interacting with the metal. You can see how you can have only dihydrogen or dihydrogen and hydrogen or the species in equilibrium. So, the proof for existence of a dihydrogen complex is quite difficult because it is difficult to distinguish between the two species on the basis of NMR spectra. Surprisingly, X-ray evidence and neutron data needs to be used. Neutron data is more reliable and it has been shown that the distance between two hydrogen atoms can be little bit longer than what you have in the free hydrogen. That is 74 picometers in free hydrogen and 84 picometers in the complex species. So, this has allowed the study of some very complicated chemistry with metal complexes and dihydrogen and hydrides. As I mentioned earlier, the systems are in equilibrium sometimes, which complicates matters. But in many instances, there is a range of hydrogen-hydrogen distances that are observed. So, these are in picometers. These distances in picometers and you go from 74 picometers for h 2 species all the way to 130 picometers for a dihydrate complex. So, you have elongation of the hydrogen-hydrogen bond as you pump in more and more electron density into the anti-bonding orbital of the dihydrogen. One way is to look for the hydrogen-hydrogen coupling, but unfortunately that would not be visible. So, HD coupling is what one usually looks for and it turns out that if you have a dihydrogen compound, if you have a dihydrogen compound, you should be able to observe HD coupling and this is in the range of 33 hertz. For free dihydrogen, if you substitute one hydrogen with a deuterium, then the coupling constant is around 43 hertz. So, the reduction coupling constant is indicative of the fact that you have reduction in the bonding between the two hydrogen atoms. So, you have a lower coupling constant and lower covalency and this is because of the bonding description that we just went through which suggests that the anti-bonding orbitals in dihydrogen are populated. So, MH 2 complexes and MH 2 complexes are interesting species which can be studied and you can use HD coupling to study them and you can also use what is called a relaxation time measurement in order to study them. This is beyond the scope of today's lecture, but it is easy to measure relaxation times and distinguish between when a species is present as you can find out when a species is present as a dihydride and when a species is present as an MH 2 complex. So, when the hydrogen atoms are close by, the relaxation rate is around 4 to 100 milliseconds. If they are far away, the relaxation rate is between 150 and 400 milliseconds, but these measurements are more time consuming and they are they can be quite complicated. So, the relaxation time measurements have also been studied by Halpern and he has pointed out that there are problems with even that technique and distance is not the only factor relaxation rates also depend on how freely the dihydrogen rotates. So, let us take a look at what factors favor H 2 complexes and what factors will favor the dihydride complex. If you have electron withdrawing ancillary ligands like carbon monoxide trans to the ligand then you would favor H 2 complex. If you have positively charged system then once again you would form the hydrogen complex and if you have a lot of electron density on the metal you can pump it in and form break it and form dihydrides and usually this is formed for the second row elements. You can have the if you have less electron rich first transition metal series metal then you can form the dihydrogen complex more easily. If you have a 4 D and 5 D transition elements it become easier to do oxidative reaction and transfer electron density into the pi star orbital. So, let us stop at this particular stage when we have discussed dihydrogen and hydrogen chemistry and we will try to summarize what we have just learnt. So, the key points to remember in this section is that hydride complexes are in fact species where an m h bond is there, but this term itself that it is a hydride is misleading because you have the ionization of an m h complex many times as h plus and m minus and the difficulty in assigning oxidation states arises from this particular problem. The ambivalent nature of hydrogen in hydrides is something that we have covered and later on in a future lecture we will look at some of the chemistry that is behind the metal dihydrogen and metal hydrides. So, with this we conclude today's discussion on metal hydrides.