 Hello everyone, once again it is my pleasure to welcome you all to MSP lecture series on Transstitial Chemistry. I am sure you are having excellent time reading and understanding coordination chemistry that I am discussing in my last 47 lectures. This is the 48th lecture in the series and another 12 are left. In another 12 I shall try to cover whatever the part that is so far not considered to ensure that it becomes a complete course as far as coordination complexes are concerned. With this let me start the discussion and talking about substitution reactions in square panor complexes. In my previous lecture I showed you demonstrated how one can explain trans effect and trans influence on substitution reactions in square panor complexes using three bonding or three trans theories. One is polarization theory holds good for neutral ligands where induced dipole plays a prominent role that is water and ammonia type ligands. And second one when we have anionic ligands kinetic electrostatic theory explains very nicely using a trigonal bipyramidal intermediate having coordination number 5 and next when we have back bonding ligands such as phosphines and carbon monoxide one can use comfortably by bonding theory where very nicely through back donation it generates void or less electron density in your position trans to the strong pi acceptor ligand and hence the nucleophilic attack would be made very facile. So using these three concepts when we perform substitution reaction we should be able to write products without any problem. So now let us look into one more example to make you familiar with this substitution reaction and also preparing different isomers possible for a combination of different type of ligands. Earlier I showed you having four different type of ligands on platinum that result in three isomers and how one can prepare three isomers very comfortably without any problem. Now let us look into another similar problem here by making use of trans effect phenomena in platinum two complexes of course most of the data at present is available on platinum two complexes because as I mentioned platinum two complexes can be readily made and they are stable and also the substitution process is relatively slow so that understanding the mechanistic aspects and pathways would be very easy that is the reason extensive studies have been carried out with platinum two complexes. Nevertheless when we come across this type of compounds with palladium or nickel or rhodium and iridium square pen are complexes even including gold three we should be able to conveniently use these aspects without any problem. So now let us look into the preparation of all possible isomers of this compound here and here in this case we have ammonia we have we have pyridine we have bromide and one chloride. So that means two anionic ligands are there and two neutral ligands are there one is ammonia one is pyridine. So let us see how we can make let us start with the trichloro amine platinum anionic complex to begin with. So now bromine bromide ligand is entering so of course here both of them are have transfect so that either we can use this one or this one does not matter let us put bromine here and then other position does not change. Now we put pyridine here when you put pyridine pyridine will come here you get a neutral complex this is one isomer. Let us look into second isomer. Let us start with a pyridine complex. If you ask me how this pyridine compound is formed of course you take tetrachloroplatinate add one equivalent of pyridine you end up getting this one so that is the reason it is obvious so that is the reason I did not mention about the preparation of this one and also preparation of this one. If you take tetrachloroplatinate add one equivalent of ammonia we can generate readily this one and if instead of ammonia if you add one equivalent of pyridine we can generate this one. It is obvious the moment you look into it here there is no need to worry about trans influence or anything since we are starting with a homo elliptic molecule. Now again I shall add bromide here again no changes we end up getting something like this but we have pyridine here and we have chloride here again this is anionic. Now I am adding ammonia of course now you should be able to tell where ammonia goes and again bromide will direct ammonia to a position trans to it and this is the second isomer. So now let us look into third isomer for third isomer let us start with cisplatin add pyridine when you add pyridine it will go here we are generating a cationic complex now. Now we shall add bromide so this is how the third isomer can be prepared you should remember it is very easy to start with an appropriate metal complex always it is ideal to start with tetramine platinum 2 plus or tetrachloroplatinate 2 minus or with corresponding palladium or nickel as well. So we can just look into more examples in text books and try to make yourself familiar preparing this kind of complexes. Let us look into another term called ligand nucleophilicity very very important when we talk about substitution reactions in square planar complexes what it says here in a substitution reaction the rate of substitution by y this is the entering ligand in a given platinum 2 complex depends on the entering group and the rate constant K2 increases in the order I defined 2 rate constants for a substitution reaction in square planar complexes considering both dissociative pathway as well as associative pathway I shall give that equation in a minute. So now the rate constant increases in this order you should see here this sequence is called nucleophilicity sequence for substitution reactions at square planar platinum 2 the ordering is consistent with soft nature of platinum 2 metal. So that means a term called nucleophilicity factor represented by npt npt subscript is shown by this is the one I was telling about we have K1 and K2 you just look into this one of course the moment you look into the rate equation and the parameter that we have included would tell you the nature of it for example here we are considering only the concentration of this one not concentration of the entering ligand whereas here concentration of the complex plus entering ligand both we are considering so this is K2. So now npt so nucleophilicity factor is given by minus dpt l3x concentration of the starting compound over time equals K1 into this one so here npt is given here log K2 over K2 prime or npt equals this should not be K2 prime this should be K1 or npt equals log K2 minus log K1 you just correct it it is not K2 prime it is K1 so that means log K2 minus log K1 this is for a neutral ligand you should remember this nucleophilicity factor that is shown in this term is for a neutral ligand Y if you consider Y as a solvent molecule such as methanol so in this reaction if you take for example di pyridyl dichloroplatinum trans compound and if you add Y what we are getting is a trans compound here and cl goes off so we are getting a compound of this type. So now the rate constant for this reaction is K2 prime so for Y equals if you consider this one and then npt would be 0 so you can go back and use this one you will come to know now for the various ligands I have given nucleophilicity factor here for chloride 3.04, NH3.07 and pyridine 3.19 and bromide 4.18 and this data is taken from Inarin chemistry book by CE Housecraft and EAG Sharpey. The nucleophilicity parameter describes the dependence of the rate of substitution in a squapen or platinum 2 complex and the nucleophilicity of the entering group so that means what it says is it essentially describes the dependence of the rate of substitution reaction in squapen or complexes on the nucleophilicity of the entering group that is very very important so nucleophilicity of the entering group is very very important when we talk about the rate of substitution reaction in squapen or complexes. Since this term utilizes two pathways one is dissociative pathway and another one is associative pathway in dissociative pathway what happens first the leaving group would leave and generate an intermediate having one coordination number less intermediate or a transition state for example if we are considering squapen or complex if say ml3x if x leaves then end up with ml3 plus x in that case we will be having a intermediate having coordination number 3 and this is less likely and we know that substitution reactions in squapen or complexes are SN2 type bimolecular nucleophilic SN2 as a result how to bring in dissociative pathway in a manner it should look like bimolecular process. So for that one we should remember one fact when we take squapen or complexes in donor solvents sometime what happens donor solvents can establish weak interaction in the axial position to generate a tetragonal distorted octahedral complex it is very similar to what we come across in John Taylor theory of course John Taylor distortion you cannot talk about case of d8 system the d8 system in the plane two ligands would fail to establish bond in case of d8 system whereas d9 and d4 in case of chromium 2 and copper 2 that is what happens but nevertheless for example if we take a metal complex in pyridine or in THF or something always you can anticipate some sort of interaction of solvent molecules in the axial position in that case what happens we will assume it has a pseudo octahedral geometry and then during the substitution reaction what one of the ligand goes then it appears like dissociative pathway very similar to what we come across in octahedral complexes with this assumption this mechanism is shown here for example let us have a tetragorally distorted squapen or complex having solvent molecules when the y enters okay naturally y will come here and it will be generating a squapen complex and then the leaving group would leave when the leaving group is leaving what would happen is you end up with a square pyramidal geometry so for example something is there and then it is gone then if you just see square pyramidal geometry we end up with square pyramidal geometry and then what happens in the first step so the rearrangement takes place and then we are having something like this and of course here it should be a because x is gone and then or x if we are taking x can also come here so that means here if you consider this path again x can depart before leaving this vacant site and this vacant site is there and orbital is empty on the metal in the first step the solvent can occupy that position prior to y and of course in case of this one y is added in the first step and then that means the slow step is in if you consider dissociative mechanism the slow step is elimination of the leaving group elimination of leaving group is the slow step and hence there is a rate determining step and addition of y once intermediate is generated in dissociative step addition would be very fast and immediately substitution will be completed so that is what this reaction says and of course solvent having a tendency to establish weak bond with the metallocent so the dissociation mechanism of substitution in square panel complexes through a square pyramid intermediate here square pyramid intermediate is there everywhere you can see here square pyramid intermediate so is it possible to have trigonal bipyramidal intermediate that is the question next one yes dissociative mechanism in square panel complexes with this kind of assumptions can also be explained using trigonal bipyramidal geometry so basically what happens let us say we have something like this y is entered here y is entered in the place of one of the solvent in the slow step what happens it will rearrange through the elimination of solvent to form a TBP intermediate TBP intermediate is there and then this TBP intermediate would rearrange again when solvent comes and it would have something like this and in the fast step or it can go from here and in the fast step what happens substitution is completed it will go back to this one but having y in place of x so this one the dissociation mechanism of substitution in square panel complexes having tetragonal structure we should remember so here we are assuming it has coordination number 6 pseudo octahedral geometry and then with this kind of mechanistic path you can convincingly explain substitution reaction in square panel complexes using dissociative mechanism involving an intermediate having trigonal bipyramidal geometry with this let me conclude substitution reactions in square panel complexes and let us move on to substitution and resumization in octahedral complexes and most of the mechanistic studies on substitution reactions in octahedral metal complexes are based on again Werner type complexes involving chromium 3 D3 system or cobalt 3 D6 system and these complexes are kinetically inert the reason why Werner carried out a systematic substitution reactions during those days by considering very inert complexes are because that time analytical instruments or spectroscopic instruments or spectrophotometry were not very handy in fact they were never heard of in that case he has to go for very very conventional methods to quantitatively understand the kinetics as a result what happens one has to go for slow process very slow processes and of course if you try to do substitution reactions with inert complexes the rates are very slow and hence understanding and determining the rate would be rather easy with this intention Werner of course Werner when he proposed the coordination theory you recall even electrons were not known electrons were discovered by JJ Thompson a couple of years later the in that context we can understand why he took inert complexes and carried out substitution reactions and also established almost in a convincing way and even now that is accepted that shows the clever experimental skills he had and analytical thinking he had to arrive at some of very important postulates and propose coordination theory if you consider rhodium 3 and iridium 3 complexes they are low spin and having D6 electronic configuration they also undergo slow substitution reactions one can of course lot of data is there on Wilkins G Wilkins himself has carried out lot of work since every inorganic reaction is unique when you perform substitution reaction every inorganic reaction is unique no universal mechanism is available to explain the substituted reaction in octahedral complexes of course that also we saw in case of square pressure complexes also because every reaction we perform with a little modification or little change because we have hundreds of ligands with different properties you saw while classifying the ligands and even the similar ligands when they are coming together as a bidentate ligand or tridentate ligand and they differ completely to the corresponding monodentate ligands so in that context understanding reaction mechanism in inorganic chemistry is not very easy and also one cannot have a universal mechanism like we come across in case of organic chemistry because we always revolve around one carbon center and with the tetrahedral geometry preferably so that is the reason substitution reactions are understand reaction mechanism is not very simple it is complicated nevertheless attempts are made during Varna's time itself to understand and rationalize some of these reactions in that context understanding substitution reactions in octahedral complexes is very very important let us continue discussing about that so now to begin with let us start with water exchange reactions water exchange reaction is the simplest one to understand and we have seen from D1 to D10 all metal ions have a tendency to form hexalqua complexes in that context what happens the comparison would be very easy and understanding the process of water exchange can give some clue about the substitution reaction we can carry out with other ligands in that context understanding and studying thoroughly water exchange reactions in octahedral complexes is very important so how this water exchange studies have been carried out simply by exchanging coordinated water with labeled isotropically labeled water like H2 17 what I have shown here has been studied for a wide range of octahedral complexes of the type this one hexa aqua complexes so cobalt 3 is unstable in solution cobalt 3 is unstable in solution and hence no exchange reaction has been carried out a typical water exchange reaction is shown here you take hexa aqua compound add labeled water and label water replaces one of that and you get water comes out now we have to study the substitution if you consider ascent P block metals we have ascent P block metals they also form octahedral complexes many of them and when we look into their properties are their substitution reaction the rate constant increases with increasing cationic radius again we go back to periodic trends periodic trends play major role in dictating the reactivity of main group elements ascent P block elements in that context if you just look into water exchange reaction among ascent P block metals the rate constant increases with increasing cationic radius for example if you consider cations of similar radii lithium plus magnesium 2 plus and gallium 3 plus they have more or less very similar radii so an increase in ionic charge slow down the substitution the increase in ionic charge slow down this that means when we are considering ionic radii similar then we have to focus on the charge is increasing here so that means charge has influence on the rate of substitution reactions when we consider D block metal di cations there is no correlation between rate constants and ionic size so when it comes to transfer elements periodic trends of very little usefulness and because we find some other facts play important role here so as I mentioned there is no correlation between rate constants and ionic size in case of D block metals but correlation is seen with respect to electronic configuration that is very important okay hexawatt components can be made as I mentioned from D1 to D10 electronic configuration as a result one can make an attempt to correlate the electronic configuration of the metal with the rate of water exchange so data whatever is available about D block you know tri cation species or M3 plus also support correlation between rate constant and electronic configuration so whether you take di cationic or tri cationic okay M2 plus ions or M3 plus ions what happens you can you can always make an attempt to correlate the rate constant with electronic configuration of that particular metal. Another term I am introducing here activation volumes what is activation volumes so activation volumes I already defined in my previous you know couple of lectures before when I initiated discussion on substitution reactions or reaction mechanism activation volumes for water exchange reactions for selected 3D metals I have shown here you can see here for the water exchange reaction here the change in the values from negative to positive indicate the change of mechanism from A to D that means when the mechanism is changed from associative pathway to dissociative pathway a negative to positive value is observed which suggests that bond making becomes less important on going from D3 to D8 system bond making is less important and bond breaking is what matters so that means for M3 plus ion if you just see here M3 plus ions are given in the last four of them volume data suggest associative mechanism data indicates 4D and 5D metals prefer associative mechanism which is consistent with the large size of metal ions which can easily accommodate the entering ligand before the leaving group departs from the metal center for the 3D series the first order rate constants K for water exchange reaction very greatly as all our high spin complexes you can see for example if you take chromium 2 plus copper 2 plus are kinetically very labile having this in the range rate in this range and chromium 3 D3 is kinetically inert you can see the rate is slow and when you consider manganese 2 plus D5 iron 2 plus D5 cobalt 2 plus D7 and nickel 2 plus D8 are kinetically very labile and hence the rate can be seen here when you consider vanadium 2 plus with D2 is considerably less labile then the later M2 plus ions so let me stop at this juncture and continue discussing more interesting aspects revolving around substitutes reactions of octahedral complexes in my next lecture until then have an excellent time and before I conclude I shall thank you for your kind attention.