 Hello everyone, once again I welcome you all to MSB lecture series on transformative chemistry. This is the 51st lecture in the series of 60. So another 9 are going to be left after this one. So let me try to complete the remaining topics and to give a completeness to this course Today let me start discussion on electron transfer processes and so far we learned about the substitution reactions in square planar complexes and then we moved on to substitution reactions in octahedral complexes and we looked into two mechanisms or two pathways that usually follow. One is dissociative pathway or SN1 mechanism and displacement or associative pathway that is SN2 mechanism and also we showed how important both the pathways depending upon the type of complex we are dealing with and also the type of ligands we are working with when we perform a substitution reaction and also I showed you how important the trans influence or trans effect in the substitution reactions of square planar complexes and when we come to substitution reactions in octahedral complexes again for first row transfer elements that is 3D elements SN1 mechanism plays a major role because of the smaller size whereas 4D and 5D metals have an option of showing both dissociative pathway and displacement or associative pathway and of course I also gave you examples to look into the stereochemical consequences in substitution reactions in octahedral complexes. Today let me start talking about electron transfer process this is one of the very important process this also we call it as redox process what is redox process let us look into it the simplest redox reactions involve only the transfer of electrons and that can be easily monitored by using isotopic traces for example look into the example I have given here we have taken 2 iron species hexasiniferate 3 minus and 4 minus and in both the cases iron or labeled one is 56 iron and another one is 59 iron when we perform or when we allow electron transfer by mixing these 2 what we get is 56 Fe CN64 minus and then we get 59 Fe CN63 minus so that means if you take MnO4 species and if you mix this one with unlabeled one here despite the rapid precipitation of BA MnO4 incorporation of label has occurred already and in this case precipitation is very rapid according to Henry Tobay who got Nobel Prize in 1983 for his extensive work on reaction mechanisms especially in electron transfer processes and redox processes what he said is any substitution reaction should be complete within a 30 seconds so then you may ask why we continue for 6 hours 7 hours 8 hours that depends on what kind of you know reagents we are using what kind of substrates we are using what kind of byproducts are formed for example byproducts are always if they in solution they can always tilt the equilibrium and go back to the back reaction so in that case what happens for the completeness of the reaction and all drive the reaction from left to right completely to get maximum product or maximum yield we prolong the reaction so let us continue talking about electron transfer processes so in the case of electron transfer between this tris bipyridine osmium 2 plus and 3 plus complexes you know the fact that both are optically active the rate of electron transfer can be measured studying the loss of activity it appears very easy here once we mix these two and see what would happen so optical activity changes so loss in optical activity can tell you or give a hint about the completeness of electron transfer process for example you take here this one plus here and we are taking minus here and then plus becomes minus and minus becomes plus after redox process is over or after electron transfer is over so according to Henry Tobay electron transfer process can be classified into two types one is outer spear mechanism and other one is inner spear mechanism so what is outer spear mechanism and what is inner spear mechanism in an outer spear mechanism electron transfer occurs without a covalent linkage being formed between the reactants that means we are bringing two species between which electron transfer has to take place simply you bring them together in the solution it happens electron transfer happens without actually bringing them with a bridging ligand so that means electron tunneling or electron hopping phenomena will come into picture but in inner spear mechanism electron transfer occurs via a covalently bound bridging ligand so in this case whenever we want to explain using inner spear mechanism or a redox process uses inner spear mechanistic pathway in that case what happens electrons would transfer or electron transfer occurs through a bridge formation so that means we should invariably have a ligand that is capable of bridging the two metal centers or it can exhibit bridging coordination mode in some cases kinetic data distinguish between outer and inner mechanisms but in many reactions rationalizing the data in terms of a mechanism is not straight forward so that means basically if we have a ligand in one of the substrates preferably the one which is low spin in that case what happens the moment we identify yes there is a ligand that is capable of bridging for example halides or hydroxide or any other bidentate ligand like pyrozyme yes you can tell probably when we perform redox process with this one it would prefer a inner mechanism because there is a bridging ligand but however rationalizing just in terms of mechanism is not always straight forward so let us look into inner spear mechanism in 1953 Henry Taube showed inner spear mechanism in the reaction of cobalt 3 plus and chromium 2 plus complexes in which the reduced form were substantially rabbi in which the reduced forms were substantially labile and the oxidized forms were substantially inert obviously the reduced species would have more electrons and less oxygen state as a result what happens it becomes labile it forms a high spin complex on the other hand that is the one getting oxidized would be having smaller size and then in that case what happens even a weak field ligand can form low spin complex so his predictions looks very reasonable here so let us look into the reaction given here we have taken pentamine chloro cobalt 2 plus low spin a non-labile complex because cobalt in plus 3 state the bond distances are very small and they have stronger bonds low spin complex and then we are taking high spin complex hexo aqua chromium 2 plus now once the redox process is over the moment you see now substrate molecules you should be able to tell which pathway this is going to follow when electron is transferred from one species to the other one so now this is the high spin now after the reaction cobalt hexa aqua cobalt compound forms a high spin and then the chromium 3 plus now chromium 2 plus has become chromium 3 plus this is non-labile so that means all the chromium 3 produced was in the form of this one you can see here so that means the bridging group has shifted from cobalt to chromium half redox process is over and tracer experiment showed that all the chloro ligands in this one in this compound is originated from this only so that means no other source of chloride was brought into the reaction that chloride has originally from cobalt has migrated from cobalt to chromium ion so this is about inner sphere mechanism you can see here visualize the bridging intermediate or transit state here so of course you know that chloride of establishing a covalent bond it has still three pairs of electrons and then a pair of electron can be readily donated to this one and then one water can come out and then it can establish a bridging like this prior to electron transfer so since cobalt 3 could not have lost chloride before reduction and chromium could not have gained chloride after oxidation the transferred chloride must have been bonded to both metal centers during the reaction this is what is very important the intermediate species is consistent with these observations so then I have put different colors to emphasize about the nature of the bridging complex or transient species or encounter complex here chloride is transferred between metal centers but such transfer is often observed of course it is quite common to see this kind of inner sphere mechanism as I mentioned when we have a bridging coordination more capable ligand on either of them let us look into another example here here I have taken the Sino complexes the intermediate also I have shown here is stable enough to be precipitated as the barium 2 plus salt similar to what we saw about BA MNO4 so here also one can trap the intermediate by precipitating it out as a barium 2 plus salt so is slowly hydrolyzed to Fe6 3 minus and then COCN5 aqua 2 minus without transfer of the bridging ligand so in this case what happens the bridging ligand from the parent compound is retained on the same it is not transferred unlike the chloro situation in case of cobalt we saw in the previous case so that means now it is very clear as long as we have a bridging ligand yes you can anticipate or speculate inner sphere mechanism for electron transfer processes and what are the ligands that can promote this kind of reaction they are all halides and OH minus CN minus CN minus also can act as a bridging ligand and NCS and pyrozine of course pyrozine is something like this something like this and of course 4 4 dash biperidine not 2 2 dash biperidine it is 4 4 dash biperidine so now we can look into the steps involved in inner sphere mechanism 3 steps are involved in the electron transfer process involving inner sphere mechanism first one is bridge formation and then electron transfer and then bridge cleavage these they are the important steps involved in a inner sphere mechanism as far as electron transfer process is concerned so now let us look into the different techniques that we have at our disposal to study electron transfer processes to determine the rate of that particular process of course now we have plenty of spectroscopic and analytical instruments at our disposal no matter what the rate is and how fast or how slow a reaction happens we can be able to determine the rate constant if the reaction is very fast and the rate process occurs between 10 to the power of minus 14 to 10 to the power of minus 8 or 9 one can go for femtosecond and picosecond flash photolysis or if it is in the range of 10 to the rise minus 9 to 10 to the power of 4 we have several options nanosecond flash photolysis or we can go for pulse radiolysis or we can even go for NMR spectroscopy provided we have NMR active nuclei in that species and again it is not paramagnetic even of course paramagnetic should not be a problem now however if we have diamagnetic species looking into NMR signals would be rather easy because we get short chemical shifts signals and also wing can also go for EPR spectroscopy if you have unpaired electrons and then if the range is between 10 to the power of 4 to 10 then we can go for conveniently stop flow or it is very slow we can go for spectrophotometry or even isotope separation so we have ample experimental techniques readily available for studying electron transfer process let us look into the steps involved in the sphere inner sphere mechanism as I mentioned the bridge formation electron transfer and bridge cleavage so based on that one three reactions I have given here first you can see here two species are coming into the picture and that is a corresponding back reaction rate also I have mentioned on top and bottom of these arrows and now the first step is formation of this bridge with the elimination of one of the water molecule present on chromium next step is electron transfer when the electron transfer happens it is already happened you can see here cobalt 3 has become here cobalt 3 has become cobalt 2 here and chromium 2 has become chromium 3 that means electron transfer has occurred here so here bridge formation is completed and here electron transfer and here bridge cleavage bridge cleavage is there so that means these are the very very important steps we come across in a typical redox process going through inner sphere mechanism so most inner sphere processes follow second order kinetics overall and the data interpretation is very simple so you should remember just three steps bridge formation electron transfer and bridge cleavage in the reaction between iron hexasino ferrate 3 minus and pentasino cobalt at 3 minus the rate determining step is the breaking of the bridge and is common for the electron transfers to be rate determining steps that means the always rate determining step is breaking of the bridge that is the rate determining step it is obvious that step is very slow where we have an opportunity to learn or study or measure the rate of that reaction for bridge formation to be rate determining step in case if the bridge formation is the deciding factor is a slow step or it has to be rate determining step the substitution required to form the bridge must be slower than the electron transfer in a substitution reaction if the bridge formation is going to be the rate determining step in that case the substitution required to form the bridge must be slower than electron transfer this is not so in case of this system what we saw in the first case cobalt chloride chromium system substitution in this one high spin is very rapid and the RDS that means rate determining step is electron transfer here however if chromium hexa aqua chromium 2 plus is replaced by vanadium hexa aqua vanadium 2 plus so this is a D3 system vanadium is 3D3 4S2 D3 system then the rate constant of reduction is similar to that of water exchange this is true for reaction between hexa aqua vanadium 2 plus and pentamine bromocobalt 2 plus or pentasino azide cobalt 3 minus indicating that the bridging group has little effect on the rate and that the rate determining step is ligand substitution required for bridge formation this which is very similar to water exchange reaction for this reaction you see here I have given X a general one so where X can be considered different anionic ligands for this reaction for this particular redox reaction with a range of ligands X the rate determining step is again electron transfer and the rate of reaction depend on X the increase in rate along the series if you go along the as halogen series increases or down the group correlates with increased ability of the halides to act as bridge so the tendency to act as bridging increases as we go from top to the bottom in the halogen series for both hydroxyl group and bromide K is similar but for water K is very small is also pH dependent so this is consistent with water not being the bridging species at all but rather hydroxyl group it is available in solution varying pH that means this one especially when we have this hydroxyl group it is pH sensitive here I have listed data you can see whatever I said you can see here the rate is increasing and in case of azide and OH is little because here this is a pH dependent whereas in this case water is not a bidentate ligand or water cannot bridge two metal centers and also through a mo diagram while discussing about bonding concepts I showed you why water despite having two lone pairs on oxygen it does not act as a bridging ligand and it can at most acts as a terminal ligand is because the second lone pair is little deeply buried and you cannot really activate to make water as a bridging ligand unless one OH bond is broken then it becomes hydroxyl group so as a result what happens you can see the influence of non bridging nature of water resulted in low rate constant here so that means now we have understood that as the ability to bridge for a ligand increases the rate of redox process also increases. So now with this let us look into outer spear mechanism when both reactants in a redox reaction are kinetically inert that means both are say low spin complexes electron transfer must take place by a tunneling or outer spear mechanism so that means here we do not have any ligand to establish link between two metal centers of course they are in solution but what happens the electron should go from one to other one so reducing species getting oxidized and oxidizing species is get reduced between them without establishing a covalent linkage electron should be transferred if that happens what would happen we call it as outer spear pathway or outer spear mechanism for such electron process reactions. Now if we look into this reaction here look very similar except for labeling on iron isotopic labeling this is 56 iron and of course oxygen states are different hexasino ferrate 3 minus with 56 and 59 hexasino ferrate is 4 minus now what we are getting for this one delta G equals 0 so but activation energy is needed to overcome electrostatic repulsion between ions of like charge to stretch or shorten bonds so that they are equivalent in the transition state. So that means if you take one labial complex and one inert complex it is not very easy to perform substitution reaction because of difference in the bond lens of metal to ligand so that means in that context activation energy is very essential for what activation energy is needed it is needed to overcome electrostatic repulsion between ions of like charge to stretch or shorten bonds so that they are equivalent in the transition state. For example if I take something like this so they are not stable so they keep coming and also of course they are magnetic you know that they try to interact so something like this if I assume the bond distances are very bond distances metal to ligand bond lengths are very different in that case what happens before electron transfer process happens what happens both of them should have a optimum or very moderate metal to ligand bond distance so that they look alike in the transition state so that electron can hop from one to other one for this process activation energy is needed. That means we need to stretch the bond where it is short in case of low spin complex in case of high spin complex we have to shorten the bond this is also requires the alteration of solvent sphere around each complex because when this happens what happens in the second coordination sphere solvent molecules also have to re-adjust their position so that means in order to do this work also we need activation energy. In a self-exchange reaction the left and right hand sides of the equation are identical so when the electron transfer and no net chemical reaction takes place so that means when we look into this reaction it appears that there is no chemical reaction of course there is no chemical reaction whatever the species we had we also have a right hand side but what happens this loses electron and this gains electron and vice versa so that means there is no net chemical reaction happens. So next now look into the rate constant given for some outer sphere redox reactions at 298 Kelvin in aqueous solution I have given here look into this carefully and look into the rate also we can get lot of information from this data given here for example if we just look into these two iron compounds here and of course you can see both of them are optically active and one way as I mentioned the loss of optical activity can tell you about the process and also to what extent it had happened also can give you and how quickly that happened also can give you the rate there is a way to measure so nevertheless the rate is in this order and when we consider osmium more or less in the same order they look almost alike when we go for this one phenanthraline this is a value here so there is a drastic change in the rate and then when we look into hexaco iron compound 2 plus and 3 plus you can see here the rate drops and then when you look into again ethylene complex here and 2 plus and 3 plus again here 10 to the power of minus 4 and then when you go for this one it is even slow to 10 to the power of minus 6 and once again when we go for two different type of species so one we have bipyridine other one is cyanone one is cationic complex one is anionic complex you can see the rate here and then when you go for again phenanthraline in one case and a cationic species other one is anionic species rate is this much and when you go for one iridium and one iron rate is this much is here so that means now let us try to understand the rate process in this one and try to get some important information to conclude about outer sphere mechanism and how ligands play a major role in these reactions. So let me continue little bit more discussion about this rate data I have provided here in my next lecture until then have an excellent time reading and I thank you very much for your kind of attention.