 Hello everyone. It is a pleasure to welcome you all once again to MSP lecture series on Transmittal Chemistry. Since my last lecture that is in my 45th lecture I started the topic inorganic reaction mechanisms with brief introduction. I also initiated discussion on substitution reactions in square planar complexes. So let me continue from where I had stopped. So transmittal complexes having D8 electronic configuration and we can consider rhodium, iridium, palladium and platinum in plus 2 state, rhodium and iridium in plus 1 state and also gold in plus 3 state. So they all prefer square planar geometry with an exception of nickel 2 complexes. They have a tendency to adopt both tetrahedral and square planar geometries even with sometimes strong field ligands. However, majority of kinetic work on square planar systems has been carried out on platinum 2 complexes because the rate of ligand substitution is conveniently slow so that the mechanism can be established with much easy without going to very fancy fast reaction rates systems and other things. Although substitution reaction data is available for palladium 2 as well as gold 3 complexes which also show similarities they are not always follow the same mechanism that is followed by these metals such as rhodium, iridium or platinum. So let us look into those differences as we progress with this topic. The extensive mechanistic studies on platinum 2 complexes and observed negative values of delta S and delta V suggest that the nucleophilic substitution reactions in square planar complexes proceed by associative mechanisms. Once again I tell you the extensive data we have on mechanistic studies on substitution reactions in platinum 2 complexes and also observed negative values of delta S and delta V suggest that most of the nucleophilic substitution reactions that we perform on square planar complexes proceed by associative mechanism. For example if you look into the rate constant for the displacement of chloride ligand by water in tetrachloroplatinate or you can consider here Cl3 and NH3 having anionic ligand or you have diamino dichloro or having something like this. If you consider any of these 4 complexes for substitution of chloride by water they suggest associative mechanism. The dissociative pathway is expected to show a significant dependence on the charge on the complex. Nevertheless there are ways to explain substitution reaction in case of platinum complexes or DI system or square planar complexes using dissociative pathway by looking into some donor solvent interaction which make weak interactions on the axial position to have a tetragonal geometry. So let us look into those things later. Now let us consider a typical reaction here square planar complex of platinum 2 that undergo substitution. Here again X is the leaving group and Y is the entering group and we have this one and it is forming here. So the rate law can be written this one in this format where we are considering two terms here K1 and K2. So the typical reaction of this type are studied under pseudo first order conditions in the presence of an excess of Y and also solvent S. The two terms are there you should recall. So the reason is the solvent interference in the substitution reactions. For example if Y is in large excess of course solvent is always in large excess. The change of concentration of Y with time is almost 0. So that means after certain time the Y concentration does not change. Similarly solvent concentration also does not change with time because the minimum change in the solvent can be ignored. So it more or less remains same. So that means with respect to time the change observed is 0 and then what happens the observed rate constant can be simply written as in this format. That means as well as K observed equals K1 plus K2 into the concentration of Y. So K1 and K2 can be evaluated by carrying out a series of reactions with various concentrations of Y. I showed you two terms K1 and K2 they can be evaluated by carrying out a series of reactions with various concentrations of Y. For example let us look into these plots here. The experiment shows that the term K1 becomes dominant if the reaction is carried out in polar solvents and its contribution diminishes in a polar solvents. So this indicates solvent participation can be represented by modifying the equation in this form you can see here in this form. So you can see the experiments shows that the term K1 becomes dominant if the reaction is carried out in polar solvents and its contribution diminishes in a polar solvents. You should remember that this indicates solvent participation can be represented by modifying the equation even as E2 can see here. So for example here we are taking K1 into Ptl3x concentration of this one and then concentration of this one also. Whereas in this case when the solvent is there you can also add the term solvent here in this case. So this is how the influence of solvent can also be added to the rate equation. Since SC is in vast excess that means solvent SC solvent is in large excess its concentration is unaffected during the reaction. Literally there is no change and hence it can be considered of first order condition by comparing equation 1 and 2 what we can get is this is the one. Simply by comparing we get K1 equals K3 into the concentration of solvent. If the solvent is a potential ligand such as water it competes with Y the entering group in the rate determining step of the reaction. So that X can be replaced by Y or S to begin with. Yes X is the living group substitution of S by Y takes place in a first step that is non rate determining step. So therefore two pathways by which the reaction probably proceed are represented in this one. For example this is a typical reaction here so that we can see first initially before X is leaving S comes here and then it establishes this intermediate and then this intermediate it would react with Y to give the product in the first step. That means the decrease in the rate constants with sterically bulky ligands Y or L is an indicative of associative pathway favoring both the terms K1 and K2. So that means the steric factors when we are considering we have to consider the ligands already present on the metal such as L and also the steric bulk of the entering ligand Y both have to be considered to understand about the nature of the pathways. But here very clearly one can tell about the decrease in the rate constant with sterically bulky ligands indicates the associative pathway. That means when associative pathway is there that favors both the terms K1 and K2. In the majority of substitution reactions of Fui-Penner Platinum II complexes yes stereochemistry is retained you should remember in majority of substitution reactions of Fui-Penner complexes stereochemistry is retained the entering group occupies the leaving group position. The entering group occupies the leaving group position and associative or IA mechanism involves a 5 coordinate intermediate or transient state and since the energy difference between the 5 coordinate geometry is small one would expect rearrangement of 5 coordinate spaces to be facile unless for example it is sterically hindered that means here IA or it is lifetime is too short like IA you have to consider. So once again I tell you in a substitution reaction on Sui-Penner complexes the stereochemistry is retained the entering group occupies the position left by the leaving group and associative mechanism or interchange associative mechanism involves a 5 coordinate intermediate or transient state IA transient state in A intermediate. Since the energy difference between different 5 coordinate geometry is small especially in IA one would expect rearrangement of the 5 coordinate spaces to be very facile for example this is the one I am telling you very facile unless for example it is sterically hindered A or IA does not matter or it is lifetime is too short in that case it is IA. The stereochemical retention can be envisaged as shown in this figure here you can see in this figure why does intermediate or the transient state species has a trigonal bipyramidal geometry. So that means now the question is if associative mechanism is followed exclusively with square penner complexes during the substitution reaction and it increases the coordination number by 1 in the intermediate that is going from 4 to 5 in that case the preferred geometry is trigonal bipyramidal. You can see here it is added now it is revert back to square penner geometry of x leaves that means here it indicates substitution is completed. Now the question is why it adopts trigonal bipyramidal geometry why not square pyramidal geometry square pyramidal geometry it can come and attack something like this and then this can go on it can revert back to square penner geometry so that is the thing. Now this is the time appropriate time to bring the concept called trans effect trans effect is very very important when we talk about the substitution reactions of square penner complexes perhaps you are wondering why I did not bring the term trans effect or why I did not introduce trans effect prior to the discussion on square penner complexes of course I was waiting for this juncture to introduce trans effect the choice of leaving group in a square penner complex is determined by the nature of the ligand trans to it this is the trans effect and is kinetic in origin that means in other words I can also explain the ability of a ligand to labelize a group trans to itself so that the entering ligand can occupy that position is called trans effect that means the labelizing effect of a group trans to itself during a substitution reaction to make way for the entering ligand is called trans effect. Let us look into the preparation of well-known complexes of cis and trans platinum that means PTCL2NH3 twice it is a square penner complex you all know and it shows geometric chisomerism cis and trans so that means how cis and trans complexes can be prepared using trans effect so how trans effect operates in substitution reactions of square penner complexes we can look into it just with the beginning of the understanding of the formation of cis and trans diamino dichloroplatinum complexes so look into it here we started with here tetrachloroplatinate here you should remember of course we are adding ammonia and by spectrochemical series also you know that the moment a strong field ligand such as ammonia comes chloride relatively weaker ligand it leaves yes it goes and now we have three chlorides are there and one ammonia is there in this one and next continue adding another molecule of ammonia so that replaces another one and then we get a cis compound here so that means we are getting a cis compound means either is replacing this one or this one not this one so we are getting the cis product here in contrast we started with tetramine platinum 2 complex so now in order to make a chloroamine complex we have to add Cl minus when we are adding Cl minus obviously one of the ammonia goes does not matter which one goes we end up getting a 3 ammonia and chloride comes next when we add NH3 this NH3 adds to a position trans to the chloride here so that means that results in the formation of trans compound here so trans compound is obtained that means tetrachloroplatinate if you take and add in the sequential order ammonia that leads to the formation of a cis product here on the other hand if you take tetramine platinum 2 complex add chloride in 2 steps that leads the formation of trans compound so this was the beginning of the reactions that led to the understanding and explanation of trans effect and hence the substitution reactions in square penar complexes and in case of octahedral complexes this trans effect may or may not influence the substitution but whereas in case of square penar complexes this trans effect influences a large and it has tremendous influence on the rate of the reaction and also in understanding the mechanism of the reaction so the general order of the trans effect that means the ability of ligands to direct trans substitution spans a factor about 10 to power of 6 in the rates that means their influence is remarkable if you can see to what extent the influence is represented by this order that indicates to what extent this trans effect has an influence on the substitution reactions in square penar complexes the trans series is given here for example if just look into this one you should be able to see how a ligand can be replaced with another ligand can be clearly seen so this is the trans series and one can also extend it adding some more but these are the important ligands we come across when we perform substitution reactions in square penar complexes so now with this information now let us look into the classification of ligands in my previous lectures when I was talking about classification of ligands by donor atoms so I tried to convince you about the type of ligands we have and also how they can be classified that means classification of ligands one can do into three categories and their characteristic properties have been thoroughly explained in my previous lecture so in order to understand the trans influence of these ligands one has to look into different trans theories that means the three class of ligands I classified and and told you are the pure sigma donor ligands and sigma donor and pi donor ligands and sigma donor and pi acceptor ligands that means we do not have a very general trans theory to explain the influence of these ligands on substitution reactions but we have to go for different theories depending upon the nature of the ligands we are using that means since we have three categories of ligands we also should have three categories of trans theories that means we cannot have a general theory that can explain the influence of all class of ligands okay so we have to conveniently three theories to understand the trans effect of these three class of ligands what I will do is in my next lecture I shall elaborate on trans effects and what are the trans theories we have at our disposal and what kind of ligands will follow which theory all those things I shall tell you in more detail so until then have excellent time in reading chemistry thank you for your kind attention