Hello everyone, once again I welcome you all to MSP lecture series on transformative chemistry. This is going to be the 52nd lecture and this is going to be the last lecture as far as inorganic reaction mechanisms are concerned. So once after completing this one, I shall move on to another important topic about interpretation of data when you make a compound. That means spectroscopic methods and how to use spectroscopic methods to characterize coordination compounds and organometallic compounds and also when you perform a reaction when we get the product where the product is in pure form if not if I if we get mixture of products in what ratio we got mixture of products and of course for separation you have to use the separation techniques and purification methods to get ultra pure product for analytical and spectral analysis. So those things we shall discuss but let me continue from where I had stopped to conclude inorganic reaction mechanism topic. So continue with again electron transfer processes. Before I conclude my previous lecture I showed you a table of data. Let me come go back to the data again. You can see here the second order rate constants K for some outer sphere redox reactions carried out at 298 Kelvin in the aqueous solution are listed here. The rates of outer sphere self exchange reactions vary considerably. Fastest reactions listed here in the chart are between two low spin complexes which differ only in the presence of an extra non-bonding electron in the T2G orbital of the complex containing the lower axis state metal. So that means the difference is only one extra electron in T2G because one is in plus 2 state other one is plus 3 state other than that both are low spin complexes because they are under the influence of a strong field ligand for example if you take bipyridine phenanthaline or ethylene diamine something like that. Such low spin complex pairs have similar bond lengths in their ground state. This is very important when we consider very similar low spin pair of complexes for performing outer sphere redox transfer process. So they will be having very similar bond lengths in their ground states. If we consider iron to nitrogen bond distances in the first case here it is 197 and 197 for 2 plus here it is 197 p-commeter and whereas in this case it is 196 p-commeter. That means very marginal difference is there and I would say there is no difference at all if we consider estimated standard deviation even if it is plus or minus 2 so both of them look alike. So such pairs where low spin complex pairs where bond lengths are very similar what happens the rate process can be very rapid because you have to spend less activation energy to prepare these two species having very similar bond parameters to do electron transfer process. So in a pair of low and high spin complexes if we consider bond lengths vary significantly in that case what happens we have to arrive at the transients or encounter state where we have to bring them one where we have the longer bond length in case of high spin complex and in case of low spin complex we have shorter bond length where the bond length is shorter we have to stretch it and where the bond length is shorter we have to contract it when the short one is there we have to elongate it when long is there we have to make it smaller. So in that case what happens for example if we consider is typical cases of octahedral complexes where we have say 2g 6 eg 2 1 case and another one is t 2g so here let me write here this is 5 and then whereas here we have 6 and eg 0 so that means here if you consider we are considering say cobalt 2 plus and we are considering cobalt 3 plus. So in this case what happens if you consider hexamine cobalt 2 plus hexamine cobalt 2 plus and hexamine cobalt 3 plus one is a high spin complex one is a low spin complex and here if you look into the cobalt 2 nitrogen bond distance it is 211 here whereas here 196. You can see the difference is remarkable so that means about 15 p-commuter difference is there so that means here alteration is very essential in order to transfer electron to complete the redox process and as a result you can see here the typical case here you can see the rate. The implications of differing bond lengths and spin states on the mechanism of self exchange reactions can be used using Frank Condom principle so that means when we talk about outer sphere mechanism Frank Condom principle has to be kept in mind and I shall tell you what is Frank Condom principle and how one can bring that concept in case of outer sphere mechanism. So reactants must approach closely for the electron to migrate from reductant to oxidant this is very very important here because we are not involving a bridging species. So however when we perform this kind of electron tunneling or electron transfer the Frank Condom principle imposes restriction what is that condition that is imposed by Frank Condom principle in outer sphere mechanism. So during electron transfer nuclei are essentially stationary so electron transfer between this one cobalt 2 plus cobalt 3 plus taking place it is not easy so that means you can see how to modify now it is not very according to Frank Condom principle between these two spaces having a remarkable difference in bond parameters it is not very easy it can only occur. So in between such cases such species involving outer sphere mechanism electron transfer can only occur with vibrationally excited states with identical structures. So we have to look for the excited states in case of both the species to have relatively similar bond parameters bond lengths. This reductant oxidant pair is called the encounter or precursor complex that means when we have dissimilarities in the bond lengths of two species between which electron transfer has to occur we have to prepare them to have a moderate or very similar bond distances in both the species this we have to prepare for eventual electron transfer and such intermediate a transfer state we generate are essentially called as encounter or precursor complex same terminology we are also using in case of water exchange reactions if you recall a molecular electronic transition is much faster than a molecular vibration the greater the change in bond lengths required to reach the encounter complex the slower the rate of electron transfer. So data here given in this table shows that the self exchange reaction between hexamine cobalt 2 plus having T2G5 EG2 electronic configuration and hexamine cobalt 3 plus having T2G6 EG0 electronic configuration is relatively slow while that between tris bipredyl iron 2 plus and tris bipredyl iron 3 plus is very fast because you can draw conclusion simply by looking into bond parameters. So bond parameters are similar in case of iron 2 iron 3 species whereas in case of cobalt 2 and cobalt 3 places with amine ligands bond parameters are distinctly different this data shows that the self exchange between hexamine cobalt 2 plus and hexamine cobalt 3 plus or between tris ethylene diamine cobalt 2 plus and tris ethylene diamine cobalt 3 plus is relatively low in both the cases you can see here this is also relatively slow here. So self exchange between tris phenanthraline cobalt 2 plus and cobalt 3 plus is much faster relatively compared to that one if you go here. So that means why this is different is much faster than the same between hexamine and tris ethylene diamine complexes although all processes are between high spin and low spin complexes if you consider all the 3 cases here in all these cases we have cobalt 2 plus has T2G5, EG2 and low spin complexes has T2G6 and EG and they have in all 3 cases distinctly different but whereas in case of phenanthraline ligand why the rate is little faster now we have to analyze this is consistent with the ability of phenanthral ligands to use pi orbitals to facilitate the intermolecular migration of electron from one species or one ligand to another and phenanthraline complexes tend to exhibit a forced rates of self exchange processes. So here the pi bonding nature in phenanthraline is responsible for enhancing the rate of electron transfer. All self exchange reactions involve cationic species in aqueous solutions the rates of these reactions typically not affected by the nature of concentration of the anion present in solution. So that means as long as we have cationic species what happens rate is not affected is independent of concentration of the anion present in solution. So the concentration of anion species present in the solution has little influence on the rate as long as we have another counter cationic species in the reaction sphere. So now let us look into the intermediate or encounter species that is needed for this self exchange reaction you can see here cobalt 2 plus is shown here and cobalt 3 plus is shown here again cobalt 3 plus and cobalt 2 plus are there. Can you see any difference between these two species? So one is T2G5EG2 this is T2G5EG2 and this is T2G6EG0 cobalt 3 plus. So can you see any difference when you look into the diagrams I have shown here of course here you can see bonds are little shorter here compared to this one this is obvious and now once the process is over now this one is elongated and this is shorter. So this is the overall redox reaction but it has to go through this intermediate here this is electron transfer for that one this process has to generate the encounter species or precursor species something like this you can see here this indicates this transient species here or transient state in which you can see both of them have similar bond parameters this is where energy is needed activation energy to prepare both the complexes to look alike so that electron transfer can be completed once electron transfer is completed they will revert back to this one this is how you can represent pictorially the transient state where both the species reductant and oxidant have similar bond distances. The outer sphere mechanism when the reactants have different bond lengths vibrationally excited states with equal bond lengths must be formed in order to allow electron transfer to occur this you should bear in mind. So on the other hand the rate of electron transfer between anions in aqueous solution generally depend on the cation and its concentration as I mentioned it has less influence less affected by the concentration of anionic species but what is important is the cation and its concentration. For example in a typical self exchange reaction if you consider between hexos ion of rate 3 minus and hexos ion of rate 4 minus with K plus as the counter ion proceeds along a pathway that is catalyzed by the K plus ions that means here in such self exchange reactions cations play a major role and they catalyze and they make this or facilitate this process through their catalytic activity. So interestingly it has been shown through experiments that in anionic complexes with K plus you add 18 crown 6 ether or krypton 2 2 2 so that potassium cation can be encapsulated. Once potassium cation is encapsulated with this kind of you know multi dented ligands what happens its activity is arrested its mobility is arrested when you arrest the mobility of cation while performing a redox process by trapping or encapsulating using 18 crown ether or krypton then what happens now the rate of electron transfer will be independent of cation mechanism. So that means here it has a tremendous influence on the rate rate probably decreases. So the rate constant often quoted is of the order of 10 to the power of 4 decimator q per mole per second whereas the value of K determined for the cation independent pathways only 2.4 into 10 tries to 2 so that means you can see now when potassium the cation was encapsulated and its mobility is arrested once then it is no longer can catalyze redox process in that case the rate constant is 100 times smaller. So this indicates the significance of cation and its mobility in order to make the reaction more facile we should not think of trapping the cation using encapsulating crown ether or krypton that is needed to stabilize and crystallize the molecule that is a different thing whereas in this process one has to bear in mind how a cation influences rate of the reaction and also what would happen if the cation is free or cation is encapsulated. The significant result indicates that one has to be cautious while interpreting of rate constant data for electron transfer reaction between complexes that means one has to analytically think and evaluate every species present in a particular reaction before arriving at any conclusion is what the lesson these points are telling us. The accepted and ideal method of testing for an outer sphere mechanism is to apply Marcus Hush theory so one more theory comes into picture which relates kinetic and thermodynamic data for two self exchange reactions with data for the cross reaction between the self exchange partners that means we can do two perform reaction and look into the cross reaction between those two independent reactions to arrive at the information about outer sphere mechanism that is about Marcus Hush theory. So for example let us look into first reaction here we are doing here self exchange here both of them are not labeled here both the species are not labeled we are doing this reaction and we are also performing another reaction where both the species are labeled now we are doing the reaction that is required for redox process one is labeled one is not labeled and this is called as this is let us call self exchange one self exchange 2 and self exchange 3 this is called cross reaction. So now for each self exchange reaction this is 0 so the Gibbs energy of activation then can be given for this reaction in using this following expression here so that means here towards Gibbs energy of activation four terms contribute which are those so of course the temperature T is there in Kelvin you are familiar that R is molar gas constant and K prime what we given is Boltzmann constant and H is Planck's constant Z in the term refers to effective collision frequency in solution this is in the order of 10 to the power of 11. So that means the contributions in this equation arise as follows if you look into the first term del W of G hash is the energy associated with bringing the reductant and oxidant together and includes the work done to counter electrostatic repulsion that means when you are bringing two cations you have to compensate the repulsion that try to keep away these two ions that means work has to be done to counter electrostatic repulsion this also included in this one besides bringing these two species together in the solvent seconds you know coordination sphere work is also done to counter electrostatic repulsion that would try to keep these two ions away from each other and del O G is the energy associated with changes with bond distances the long distance is there in the high spin complex that has to be shortened and the low spin complex it has to be elongated so work done to bring to optimum bond parameters is represented by this term and then delta S G arises from rearrangement within solvent spheres this will be referring to solvent contribution in this process and then RT ln term accounts for the loss of translation and rotational energy on formation of the encounter complex the energy lost during this formation of encounter complex where we have the optimum bond parameters so that is represented by RT ln so it is possible to calculate these terms to estimate this one for a self exchange reaction so the rate constant K for the self exchange can be calculated and verified by experimental data as we have now experimental techniques available for different rate for any given redox process the term represent this one K equals this term is there of course K that transmission coefficient which is always equal to 1 and Z is effective collision frequency in the order of 10 to the prof 11 now the rate and thermodynamic data are given by K X and delta G X for self exchange reaction 1 and K Y and delta G Y for self exchange reaction 2 and for the cross reaction corresponding rate is K Z and thermodynamic data is delta G Z so equilibrium constant is K Z equilibrium constant is K Z and standard Gibbs energy of reaction for this cross reaction is delta G not Z here so now we have defined all the terms involved in this equation so this Marcus Huss equation is given by expression this expression here and of course F Z is defined by a relationship something like this and in logarithmic form it can be written as often F equals 1 so log F becomes 0 and this term may be neglected so if you neglect this term then it would eventually end up in this format here log K Z equals rate of cross reaction equals 0.5 log K X plus 0.5 log K Y plus 0.5 log K so you should remember I had given in different color this is the equilibrium constant this is rate constant and then similarly if you call this one as equation 3 so values of K X K Y and K Z and K Z rate and equilibrium can be obtained experimentally or K X and K Y can be obtained theoretically and K Z is determined by E cell if K Z is calculated from this value agrees with the experimental value this provides a strong evidence that the cross reaction proceeds by an outer sphere mechanism so this is all about Marcus Huss equation and its utility in cross examining a reaction to confirm that yes whether it follows an outer sphere mechanism or not so with this I am almost completed discussion on inorganic reaction mechanism few other things I would have brought into picture as I mentioned because of time constraint and I have several other things to discuss in next 8 lectures I am stopping at this lecture so now quickly for a couple of minutes let us look into some problems and I leave those problems for you to solve with enough information I have given in my these five six lectures on inorganic reaction mechanism if I give a couple of problems you should be able to solve it without any problem. So now the first problem is that the octahedral complex undergoes substitution reaction as shown below optical one where I have used you should remember this is an optical active compound and Z is the entering ligand and X is the leaving ligand and then we get a compound like that assuming that stereochemical changes for both SN1 and SN2 pathways that means you have to consider dissociative mechanism as well as associative mechanism and there is a equal possibilities there so that means 50 percent for dissociative and 50 percent for associative pathway prove that the following data holds good so keeping the probability of undergoing dissociative pathway and associative pathway is 50 percent each so this data is given now we have to look into the validity of data by solving this problem for this what you should do is you take this one take this compound this is cis compound of course otherwise it is not optical active it is a trans compound it is not optical active we have to consider only cis here and then you go for first you know dissociative pathway in the dissociative pathway what you should do is you should look for square pyramidal geometry and trigonal bipyramidal geometry as I have shown and then look into the percentage what would happen to here optical activity retention is there resummitation is there or you get a trans compound and then go for associative pathway there you have to think of pentagonal bipyramidal geometry and how many intermediates we can think of when ligand is approaching from different directions and then you should be able to solve and verify the data given here and similar problem I also given here for a simple cis compound here and it is a cis compound now it is not an optical active compound now the data also I have given I have to see here the consequences now one more reaction the reaction of NiCO4 with L L equals Pph3 or POPH3 occur at the same rate resulting in substitution product NiCO3L comment on the reaction mechanism so read the question carefully for example if I have not heard of this question you think that you cannot answer but if you read couple of times you will understand the reaction what is the content in it and what is the question so that you should be able to answer the reaction of nickel tetra carbonyl with L where L is triphenyl phosphine or triphenyl phosphate occur at the same rate resulting in substituted product NiCO3L comment on the so statement is already made that whether you displace carbon monoxide one carbon monoxide with triphenyl phosphine or triphenyl phosphate the rate is same so rate is same means what happens here it is independent of entering ligand it is independent of entering ligand and rate determining step is the dissociation of one ligand so that is equally in both the cases that means it follows dissociative pathway and SN1 mechanism so it is so simple the answer is there in the question itself so one has to read one should not read between the lines one should read the question properly so that answer is there answer will pop out immediately one more interesting question the formation of cadmium complexes with bromide exhibits the successive equilibrium constants k1 equals 1.56 k2 equals 0.54 k3.06 and 0.37 suggest an explanation of why k4 is larger than k3 why if the formation is very facile for the last one and what is a hint aqueous medium is used that means if I am say if I am adding bromine here you should remember the hint aqueous medium in the aqueous medium if you take cadmium it will be hexa aqua cadmium 2 plus if it is hexa aqua cadmium 2 plus is there you take to begin with H2O 6 2 plus plus add bromide what would happen what you get is cadmium bromide H2O 5 times now it is plus here okay so plus H2O comes out so this is given so next one is similarly so this will give you next another this one plus H2O of course BR is added so next it will give one more BR is added BR 3 H2O 3 times so minus plus H2O so now this one just look into here this is where the catch you add another bromide what would happen 2 minus plus 3 H2O so 3 H2O means here entropy increases dramatically so the reason is increase in entropy here in this one so that is the reason there is a market difference is there so you should be able to see these things and what you should remember is this is the one to begin with yes I am performing bromide substitution on hexa aqua compound and again when you go for halides halides are you are not going to make hexa bromo so here only tetra bromo species is stable beyond that it cannot go so that means when the last bromide is coming the fourth one is coming three equivalents of water is coming out so that means entropy is increasing okay so that is the reason so this is how you can explain so let me stop here like that many interesting problems are there in Siver and Adkins book and also in CE housecraft and AG sharpie book and many other books please look into it and solve these problems and enjoy okay enjoy learning if you learn chemistry through problems chemistry is not a problem have an excellent time and thank you for your kind attention


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