 Hello guys. Good evening. Can you hear me? Yes, sir. What we did last class? I think we did with the Werner's coordination theory, right? Yes, sir. We wrote drawbacks of Werner's theory. That was the last thing. So primary secondary equivalency we have discussed. We also discussed the number of ionizable ions. Okay. So, right, so we have few drawbacks in Werner's coordination theory. That's why we require another kind of theory for bonding in coordination. Right. And then we got balance bond theory. This balance bond theory is very much similar that we have done in chemical bonding if you remember. Okay. All those hybridization and all will discuss it here. So the next theory is balance bond theory. One second just to give me. Okay. So Werner coordination theory, just one question we'll see and then we'll move on to the next theory in coordination compound that is Werner's bond theory. Okay. So look at this question. You're on mute. Yes. So formula and moles of AGCL precipitate per mole of the compound with excess AGNO3. Right. So formula is given here. The first one is PdCl2.4NH3. NiCl2.6H2O. PdCl4.2HCl. COCl3.4NH3. PdCl2.2NH3. These are the molecules given to the primary and secondary valency and what is the formula of this molecule because there's nothing like, you know, it is not written in the square bracket or something like that. So how many ligands are present in the complex part that you need to find out plus the primary and secondary valency for all these compounds you need to find out if the number of moles of AGCL precipitated for the per moles of this is 2 for the first one. Then 2, then 0, then 1, and then 0. This is the data given by now. Secondary valency and the formula. One second guys. Okay. So the first one. Yeah. You see the number of moles of AGCL precipitate. Right. AGCL that precipitate, it depends upon the number of ionizable chlorine atom there. Right. So moles of AGCL precipitate is 2. It means there are 2 chlorine atom which is present outside the complex, the coordination complex. Right. And it's the formula for the first one. If you write down the formula should be this palladium is the metal NH34 and then CL2 outside. Okay. So from here, if you find out the primary valency, what is the primary valency here? The primary valency is 2, the oxidation state and secondary valency is 4. That is the coordination number. Number of ionizable chlorine atom is 2. Second one if you see, we have 2 AGCL precipitated. Again, it means this 2 chlorine atom presents outside the complex and hence the possible formula is this according to the given data dot CL2. Okay. Not dot CL2. Okay. CL2. So from here again, the primary valency is 2 and secondary valency again is the coordination number 6. Okay. So the third one, you see, there is no precipitation here. It means all the molecules ions are present inside the complex. So we'll have PT. So basically we can write this as H2PT CL6. This way also you can write H2PT CL6. So primary valency here is, is the oxidation number. So primary valency is again we don't have no condition at CL is there. What should be the primary valency here and secondary valency? Secondary valency to 6. What is the primary valency? I think we should write it down this way. The correct representation here is this, no correct representation here. The third one, chlorine is the donor from HCL. So we can write down this as PTCL4 dot 2 HCL like this. There's no charge on this complex. That's why the primary valency is zero here. Secondary valency is 6. The HCL itself is one again. Our chlorine is the donor atom there. So I want it ionized. It is there in the complex. See, I'll write it down this formula based on this data. If it is outside if I write no H2 then it will ionize also. It will be H2PT CL6. Yes, H2PT CL6 if you're right. That won't satisfy the value which is given over here. It says zero AGCL. See actually primary valency is also we consider the number of ionizable chlorine atom because of that only will have the charge on the complex. If you have, like this you see, CL2 present outside. You see the primary valency is the number of chlorine which is present outside the complex here. Here all chlorine will write down inside. But oxidation state if you check, this you can also write down this way. I am not writing down the primary valency with respect to the oxidation state here because if you write down this one H2PT CL6 then the charge on platinum you are getting as plus 2 minus 2 minus 6 so plus 4 you are getting over here. But plus 4 is not the primary valency. Actually what happens this out of 6 CL, 4 CL satisfies here the primary valency of this. But this AGCL that precipitate here this is only possible when the chlorine is present outside the complex. And we also know the primary valency is nothing but the ionizable. Chlorine atom or any other atom which is present. So here you see here ionization is possible H plus and this PT CL6 2 minus we get. But chlorine does not comes out like this. So like this you can write but this is the most appropriate way to represent the molecular formula of this. Since this satisfies that there is no precipitation there is no ionization in this complex we can see. Here still we have ionization. That's why this way you can write but in some book they have written this way also. As far as AGCL precipitation is concerned this also holds true because this chlorine won't go out into the solution. This chlorine also won't go out into the solution. So both way you can represent but primary valency here since there is no ionization of chlorine atom we are considering this to be 0. Sir what the primary infallency isn't dependent on the number. Sorry. I actually hit a mistake. Sorry. Okay fourth one is right fourth one. What is the primary valency you are getting in fourth one? Only one should be outside right. So four NH3 or one CL should be inside right. So five you know is the secondary valency and the primary valency would be? Sir three. Sir one. I'll write on fourth one. Yes correct. So since one chlorine atom gets precipitated so formula should be CO NH3 for CL2 and CL. This is the possible formula we can write because the only one precipitation is there okay. So it's a primary valency we have three CL no huh. Secondary valency is six four plus two six. Primary valency is what? Only one chlorine atom that goes out okay and minus two will have here. Primary valency would be three. So that's when the previous one why can't we say that the primary valency is four? It was in one second I'll check that. I think PT NH3 twice and CL2 in this case what we are getting. We are getting primary valency as zero obviously zero here. There's zero here and secondary valency as four. Only secondary valency. Here we have primary valency one. You get here primary valency four. This is five. See actually one thing you have to you must take care here. This two chlorine atom which is present inside this satisfies both valency right primary as well as secondary right. So secondary valency if I've written six right. So this we are already considering here. So we won't consider these two chlorine atom if you write the primary valency here. The same thing I'm doing here only. So one thing this difference here is what usually if you see the last one is the condition theory we have this because that primary valency corresponds to the oxidation state of the metal. If you find out the oxidation state we are getting it as plus three. But primary valency in this compound is the valency which considers the negative ion which you know which satisfies only primary valency. Secondary valency we are not considering here. So this two chlorine atom is satisfies both primary as well as secondary. So while calculating the primary valency will count only those atoms which are present outside the complex. So here also you see if you write down this way or this way obviously there is no chlorine present outside the complex hence primary valency. Primary valency is two primary valency is two. Here there is nothing outside no chlorine outside primary valency. This chlorine why we are not considering primary valency again I'm repeating because this satisfies both primary as well as secondary. So here on the basis of this data we consider only those ions which satisfies only primary valency that is one. Yes. The last class you gave that same fourth example and your primary valency is three. Yes. We are talking about this compound fourth one. Yes. So the question is what? Question is the primary valency. So primary valency we consider only see it's basically the language that is written. Okay. If you because if you look at the coordinates one or coordination theory the primary valency is corresponds to the oxidation state of the matter. So what is the oxidation state of cobalt here? It is plus three right? Is it? Yes. According to this the primary valency should be three but that is a maximum value out of three two chlorine atoms satisfies the secondary valency also that we are considering here in six. That's why in this case based on the data given we only consider those atom chlorine atom here in this case which is present outside the complex according to that only we'll write down the why this we are not considering because this contributes into the secondary valency as well. That's why we are not considering this only this one. Okay. So this kind of questions easy questions you will get you easily understand that what is the possible structure of the molecule in Wernher's coordination. So one kind of question is this another kind of question. They can give you the formula and they can ask you the like the primary valency or secondary valency is given. They can ask you the number of ionizable chlorine atom. Okay. Like suppose in this one because there are many different kinds of question possible in this one you see if they give you this formula and says the primary valence the secondary valency is four for this one. Secondary valency is four. What is the number of ionizable? What is the number of ions we get in the solution? This kind of questions they can pray. Okay. This is another type. They can also ask you how many you know moles of AgCl precipitate when this molecule with secondary valency four reacts with AgNO3. This kind of question is other way here and there they can change the question. This kind of question they can ask on the basis of you know the molecular formula they can also ask you the order of molar conductivity again that depends upon the number of ions. Any of these kind of questions possible here in Wernher's coordination theory other than this I know there is not there's no possibility of getting any other kind of questions right. This is the first bonding theory which obviously fails reason and the drawback I have given you already in the last class. So to overcome this you've got the new theory the other theory of bonding that we call it as valence bond theory. Valence bond theory again we have you know postulates and then you know this one that postulates we can find out the bonding of coordination compound all these things we are going to say very important topic this one is write down the another theory of bonding of coordination complex that is valence bond theory BBT in short write down this theory mainly deals with deals with the geometry geometry and and the magnetic property magnetic property of the complex property of the complex okay the main postulates are the first point write down the main postulates the first point the central metal atom CMA the central metal atom loses a definite number of electrons to form ion a central metal atom loses a definite number of electrons to form ion this is the valency of the metal again I'm repeating the central metal atom loses a required number of electrons to form ions definite number of electrons to form ions this number is the valency of the is the valency of the of the metal next one depending on the coordination number the central metal atom and when I say central metal atom write down CMA okay depending upon the coordination number the CMA has equal number of vacant CMA has equal number of vacant SPD orbitals these orbitals forms the CMA has equal number of vacant SPD orbitals these orbitals forms hybrid orbitals forms hybrid orbitals together right and that thing is same like the number of atomic orbital combines equal number of hybrid orbital forms okay that is same but it is very much similar to the you know yeah a little bit similar to the valence bond theory in the simple simple compound okay depending on the coordination number the central metal atom has equal number of vacant SPD and D orbitals these orbitals forms hybrid orbitals together this is same till here now the difference here is the third point the difference is in case of strong ligand in case of strong ligand there may be some rearrangement there may be some rearrangement of electrons in the in the atomic orbitals against against the and soon right one thing here I said in case of strong ligand okay so what is a strong and weak ligand we haven't discussed yet okay this will understand when we do CFT that is crystal field theory okay there I will tell you how to find out basically you have to memorize that which ligand is a strong and big and not that difficult if you solve some questions you will memorize it easily but for this one I'll let you know that this particular ligand is weak ligand or strong ligand okay so this is the rearrangement against the next fourth one the ligand and and the metal bonds and and metal bonded with bonded with coordinate bond which contains which has considerable amount of polarity considerable amount of means the bond is polar now the last point here is the same thing if there's unpaired electron if the complex contains if the complex contains unpaired electron then it is paramagnetic otherwise dynamic electrons are paired so third point is the new thing we have here otherwise you know all other things right depending upon the coordination number we can have the we can we can have the understanding of you know hybridization of the complex and hence we can understand the geometry of the complex so this table all of you have drawn all of you have written this yes okay this table you draw one side one side we have coordination number CN stands for coordination number and then we have geometry and then we have hybridization coordination number could be two three four five six geometry is just a second space here four and then five if it is two then the geometry is linear hybridization is sp if it is three then it is trigonal planar hybridization is sp2 if it is three then two different geometries are possible according to the hybridization if it is tetrahedral hybridization is sp3 if it is a square planar is the sp2 for coordination number five coordination number five two different geometries are possible trigonal bioparameter or the square pyramid a hybridization is we can have either dsp3 or sp3d keep this in mind it is not like trigonal bipyramidal is dsp3 and square pyramidal is sp3d it could be anything okay here we have this thing tetrahedral sp3 the square planar is dsp4 if it is six then we can have a square octahedral or square bipyramidal both are the same thing only octahedral or square bipyramidal the hybridization is d2sp3 or sp3d2 yes am i audible yes okay no no no see guys whether the geometry is tdp or square pyramidal it depends upon the molecule okay it has nothing to do with the geometry do with the hybridization here okay don't consider this as tbp is dsp3 or square pyramidal is sp3d okay it's not like that one more thing here this d2sp3 and sp3d2 is what see here if you write down d2sp3 means in this how this hybrid orbital forms this hybrid orbital forms by the combination of two 3d orbital plus one 4s orbital and three 4p orbital right these are the numbers we have sp dsp3 okay so one 3d two 3d one 4s and three 4p so here you see we have 4s 4p orbital and along with this we have two inertial orbitals 3d orbital is there 4d is not there so there is inertial orbitals present here when you write down d2sp3 d2sp3 habit orbital forms when 3d combines with 4s and 4p so this is inertial orbitals hence we call it as low spin complex this kind of orbital it forms low spin complex because inertial orbital is involved with this low spin complex but on the other hand if you write down d2sp3 sorry sp3d2 this means this means we have 4s plus 4t plus 4d at this kind of combination is there it means it is high spin complex okay this is the term we use high spin complex just a second I wish just a second okay one more thing low spin complex is generally forms like low spin complex generally forms with strong ligand in case of strong ligand low spin complex generally forms high spin complex forms when we have weak ligand what is a strong and weak ligand we will discuss that okay just to write it down here we will see that what is a strong ligand and what is weak ligand okay this is again I will go back yes copy this wait I will just wait low spin means when lower d orbital involves then it is low spin complex just that term we use when the inertial d orbital involved is low spin complex when outer shell 4s 4t 4t involves it is high spin complex so these are the postulates of balance bond theory based on this we can find out the geometry plus the magnetic nature right whether the molecule is paramagnetic diamagnetic if it is paramagnetic and how many unfit electrons all those informations we can get one by one we will see one more very important thing is here all these discussion that we are doing out of all these coordination number 6 is the most important one coordination number 6 is the most important one mostly they ask question from this one coordination number 6 okay but we will discuss all the you know coordination number like 4 5 and 6 we will discuss so first of all examples will lie down here example we are taking for coordination number 4 first one see this question the question is we are n i c l 4 2 negative is found to be paramagnetic is found to be paramagnetic explain hybridization and hybridization now for you for your information all helite ions are helite ions are weak ligands that you should know X minus are weak ligand must remember this generally a strong ligands are those in which the carbon is a donor atom then the ligand are said to be strong again in general these two things you must keep in mind but once you solve some questions you'll understand that which ligand is strong and which ligand is weak got it so how do we do this question you see first of all what do you do you find out the oxidation state of the metal which is here plus 2 so nickel has how many electrons 28 if you draw the electronic configuration argon 4 s 2 3d 8 okay for ni 2 plus because the metal ion is ni 2 plus this would be argon with 3d 8 so if I draw the orbital diagram here this is 3d for example this is 4s and this is 4d 1 3 4 and 5 so 3d has 8 electrons so 1 2 3 4 5 6 7 8 okay and and this is 4s 4s and this is 4p and 4d is not required 4d is this side not required because coordination number is 4 only we require only 4 orbitals so that we can get 4 hybrid orbitals in which the ligands 4 ligands can donate its loan pay right so what happens this 4 orbitals empty orbitals goes under the hybridization before the bonding and it forms 4s p 3 hybridized hybrid orbitals in which the 4 chlorine atom donates its electron pair right the ligand donates its electron pair and makes a coordinate bond with the metal right obviously if you look at this the hybridization is Sv 3 geometry is tetrahedral right there are two unpaired electrons two unpaired electrons and hence the molecule is paramagnetic which is also given in the question yes cl minus is a weak ligand so there is no pairing against the hands to not suppose if this is a strong ligand like cl obviously is a weak ligand then there is no pairing against the hands to if it is a strong then this electron pair this jumps over here and makes this orbital vacant in that case then this orbital also will take part in the reaction and 1 2 and 2 orbitals will take from this possible from this p sub shell then the configuration will be odd the hybridization would be d s v 2 in that case square planar geometry ok this depends upon the nature of the ligand that you should know whether the ligand is being not a strong these two terms in mind ok this helps you a lot in solving this kind of questions now you see the second question here we have a molecule say m i c o 4 carbonyl group is diamagnetic magnetic explain hybridization and geometry tell me ramsa c o the carbonyl group c o yes d s v 2 square planar ok c o is a strong field ligand that you should know right carbonyl group is a strong field ligand so there will be pairing against the hands to right there's no zero oxidation instead of nickel is zero here hence n i is argon 4 s 2 3 d 8 so this is 3 d orbital that is 4 s 4 d since it is coordination number 4 we require only 4 vacant orbit 3 d 8 1 2 3 4 5 6 7 8 ok and 4 s will have to 4 s will have to then what happens you see all of you are getting d s v 2 there will be pairing against the answer so these two electron it jumps here into the inner orbital of 3 d makes this 4 s orbital vacant ok so after this rearrangement the orbital diagram is this trance in each of these orbitals this is vacant this is vacant 4 s and 4 p so this goes under hybridization and it forms again sp3 hybridized orbit this is sp3 hybridized orbital so hybridization for this complex is sp3 all like so it is diamagnetic diamagnetic geometry is tetrahedral this is what the answer how did you get d sp2 oxidation state is zero no the oxidation state is zero try this one n i c n 4 and cyanide is also a strong ligand n i c n 4 2 minus is the complex find its magnetic property geometry and hybridization is it d sp2 you are getting yeah it is d sp2 it's not sp3 yes because here the oxidation state is plus 2 so whatever you have done in the previous case that is applicable here right so this would be if I write down that electronic configuration of n i 2 plus it is argon 4 s 0 and 3d a orbital diagram is this 0 electrons really cross this c n is strong ligand then there would be pairing against the unskilled and this electron which is unfair it jumps back into this orbital and makes the electron pair and hence this orbital becomes empty here so here after this rearrangement the orbital diagram looks like this 1 2 3 4 each of these orbitals will have 2 2 electron 1 year of orbital is vacant 4 s and it is 40 so it is 3d 4 s and 4t so 3d orbital 4 s and 4t 2 orbital here this goes under hybridization and it forms 1 2 3 4 d sp2 hybridized hybrid orbital so hybridization is d sp2 geometry is a square planar magnetic properties is diamagnetic is three things is three things we can understand if I ask you one more question they ask in this particular kind of geometry that which d orbital is involved in this because we know we have d 5d orbitals we have d x y d y z d z x d x y y square and d z square right so out of this 5 which d orbitals is involved in this geometry do you answer this sorry so can you repeat I said we have 5 d orbitals right in 3d sub shell we have 5 d orbitals out of the 5 which d orbitals involves in this geometry because only one d orbital is involved no only one d is there so which the which the orbital is out of the 5 have to first of all 3d is a sub shell guys okay 3d is a sub shell I'm talking about orbital so within 3d will have 5 orbitals like what 3d should be 3d x y 3d y z 3d z x 3d x square y square 3d z square okay so which d orbital is this one 5th one okay why it is the x square y square yeah just guessing okay fine by the way your answer is correct the answer is the x square y square there is no you know you know defined logic for this that this is why this is happening okay this is factual actually but I'll discuss one small thing here probably with that you can understand that which d orbital is involved right obviously it is not you know a given method in the book but yes from this particular thing if you do not know if you forget in the exam then you can think like this you will get the answer okay what d orbital or which d orbital is involved no it does matter it does matter because the orbital has its own orientation around the nucleus that's why it does matter see actually what happens a square planar geometry is what with geometry you can understand it is not that difficult also all of you will get it the square planar geometry is something like this once again the square planar geometry is this correct suppose this is the square planar geometry we have and metal is present somewhere here in the center of the square and how do we get the square planar geometry when the ligands are attached at these corners yes or no so to metal to get attached with this metal at these corners of the square it has to approach the metal from these directions I'll show you what direction it is so if the metal has to form in the no ligand has to form d you know the square planar complex and this must have to approach the metal in these from these directions along these lines we can say or along these directions must have to approach this way so suppose all the four ligands are coming from this direction and it is trying to approach the metal for the bonding it's trying to donate its electron to the metal from these directions all these directions the ligands are coming correct so obviously the orbital and this metal has all the d orbitals obviously the orbitals which are present along these lines that particular orbital will have the maximum probability to involve in bonding yes or no are you getting me the angle or the you know like the line through along which the ligand is approaching the metal along that particular line whatever orbital is present if it is right if it is whatever orbital is present that particular orbital will have the maximum probability to involve in the bonding because the ligand is approaching towards that particular orbit now you see the another thing if you consider this line obviously the square diagonal it by 6 at 90 degree it's the 90 degree so what I am trying to say here that I can assume that this axis is suppose this I am assuming as x axis and this I am assuming as y axis why because the angle is 90 degree and perpendicular to this one suppose we have a z axis which is not required here in this case but is required in the other cases right so obviously we know this dx square y square orbiters are are along this axis the other axial orbital it is present along this axis which orbital is this dx square y square is it all these things I have discussed in chemical bonding already right this orbital is what this orbital is dx square y square that's why in this square planar complex the orbital that is involved is dx square y square because the ligand is approaching head on to this orbital did you get this yes right yes now what is a square bi pyramidal we have these four ligands and along with these four ligands we have two more one is from this side and other one is from this side yes or no that is only square bi pyramidal so if I ask you which d orbitals are involved in a square planar in square bi pyramidal geometry what is your answer x square y square then dx square y square and dz square yes very good so with this kind of you know you know the pictorial diagram that we have discussed you can have the idea that which orbital is involved in a given geometry ok d orbital there was many times these questions in need exam even in j also they have asked this question ok so right on this you know this thing that we have discussed here the geometry here and the d orbitals involved geometry and d orbital so if the geometry is a square planar the d orbital is dx square y square if the geometry is a square octahedral or square bi pyramidal the orbitals are dx square y square and dz square if it is a square pyramidal square pyramidal it is only dx square y square if it is tbp trigonal bi pyramidal it is only dz this is the orbital d orbitals involved in sir yeah tell me so so can we say that the pz orbital is left out in the 4p orbitals in which one the 4p orbitals out of the 3 4p orbitals pz will be left out pz no we cannot say whether it is pz or px or py ok mostly pz you know we don't consider because pz normally forms sigma bond first that's why we always assume inter-nuclear axis along the other axis right but yes they won't ask you which orbital is left behind in p sub ship whether it is px py pz because that is not in our control you cannot you know say that so this is also important the coordination number 4 we have discussed you see coordination number 5 same kind of thing you have to do nothing you know difference here but yes we'll do one or two one two examples so that we can understand it's important also so we have a question on molecule that is feco5 feco5 ok and information with this is feco5 is found to have found to have 0 dipole moment mu net is 0 for this 0 dipole moment find its hybridization and geometry try this one sir what is that coordination number is equal to 5 on top coordination number is 5 of fe sir yes this is the complex coordination number is always defined for metal what is the charge of the complex nothing 0 moment wound up but yes with dipole moment see you try this aditya once ok you will understand why this dipole moment thing is given ok try once to do it ok sir what is the geometry 60 square by pyramindal c o is a strong ligand correct yes what is the geometry pratham you are getting see this obviously the observation state is 0 and iron has 26 electron so the electronic configuration is argon 4s 2 3d 6 ok fine so anahita is getting trigonal bipedal ravikiran also trigonal bipedal fine ok so 4s 2 3d 6 so 1 2 4 5 6 and 4s 2 but since it is a strong filled ligand see you arrangement against the hunts 2 right so this electron this 2 electron will jump into the inner shell this comes over here this comes over here and this also will jump here after this rearrangement the orbital diagram is this 5 is the coordination number so we required 5 vacant orbital so one is this d then s and then b the hybridization is dsp3 dsp3 is the hybridization geometry like I said it can be tbp or it can be square bipyramidal which one is the correct geometry here that depends upon the information given in that question now you see the square sorry pyramidal not bipyramidal okay so if you look at the Bose geometry here tbp is this this is tbp and the square pyramidal is in the square pyramidal what happens this ligand all these ligands you see because the bond has considerable amount of polarity because of this polarity we have dipole moment right so because of these two dipole moment we have a net dipole moment in this direction net dipole moment in this direction and these two cancels out but because of this we have some mu net in the molecules mu net should not be equals to 0 for a square pyramidal geometry but if it is trigonal bipyramidal this dipole moment will get cancelled by this dipole moment and this dipole moment will get cancelled by this dipole moment right so mu net is 0 for tbp trigonal bipyramidal since in the question it is given mu net is equals to 0 hence the geometry for this particular complex is trigonal bipyramidal it is not a square pyramidal yes correct understood no doubt in this yes right so some some information will be given based on that you can find out the geometry of the molecule like you see the another one NICN5 3 minus and in this one it is given that it has two types of NICN bond length two type of NICN bond length out of which four bond length are same are equal and one is different find out hybridization magnetic property and geometry of the molecule dsp3 answer for this question I'm writing it down if you have any doubt you can tell me I will do this the hybridization will be dsp3 and it is diamagnetic it is diamagnetic and the geometry is a square pyramidal because four bond lengths are same so square pyramidal geometry any doubt in this finished all of you next you see coordination number six coordination number six the first one is this is the most important one we have co cobalt NH3 and plus three on the complex hybridization and geometry for this one it is d2sp3 NH3 is a strong ligand NH3 is a strong so cobalt plus three oxidation state and its configuration is argon 3d6 okay so 3d6 the geometry is the orbital diagram is this and NH3 is a strong ligand so they will be pairing against the against the unstable and it gives and this six orbitals will go into hybridization and forms d2sp3 hybridized orbital okay geometry is octahedral and all the electrons are paired so it is diamagnetic and since we have d2sp3 hybridization so it is it is low spin complex you see low spin complex forms because of the rearrangement against the hunts rule and rearrangement against the hunts rule is possible when the ligand is a strong that's why whenever low spin complex forms we always have strong ligand okay right down with the hybridization whether it is low spin or high spin it is the nature of ligand we can identify or we can understand sir how do we determine if it's octahedral or square bipyramidal same thing anyone you write it is correct sir so in the actual molecule will both structures exist simultaneously at the same time structure is same two names are there octahedral is same as the square bipyramidal same thing yes sir I'm sorry okay for a square planar also you can say low spin high spin but generally we don't you know assign this kind of you know terms with square planar geometry generally with four coordination number but yes if you have to pick between the two dsp2 is the low spin complex but sp3 we don't say whether it is low or high right but d2sp3 is obviously low spin complex if you talk about sp3d and dsp3 if you have to choose then dsp3 is the low spin complex sp3d is the high spin complex but mostly we talk about these two terms in coordination number six yes sir okay so one is then one exception we have here right NS3 I told you it is a strong ligand but in some cases NS3 behaves as a weak field ligand only one or one case we have here the metals like nickel okay so this one you have to memorize because these are the exceptions you have to keep this in mind for example you see if you write down this complex could you find out the hybridization and geometry here and I NS3 6 2 plus in this one first of all you write down one note and then you can try one note you write down with most of the metal with most of the metal NS3 behaves as strong ligand behaves as strong ligand but with some metal like nickel it behaves as a weak field ligand this is one of the drawback we can say or exception of VBT reliance bond theory right couldn't repeat the last point yeah with most of the central metal iron or atom NS3 behaves as a strong ligand but with some metal iron like nickel here it is ni 2 plus it behaves as a weak field ligand it behaves as a weak field ligand this is the drawback of VBT try this one now hybridization geometry and magnetic property sp3d2 yes sp3d2 you are getting octahedral complex how many unpaired electrons are there how many unpaired electrons 2 unpaired electron that's right you see there is no bearing of electrons ok so here in this complex NS3 is behaving as a weak field ligand ok which usually does not happen NS3 usually a strong ligand but here it is behaving as a weak ligand ok so this example you must remember ok I'll quickly write down this ni 2 plus is the charge we have oxidation state and its electronic configuration is nickel has 28 electron so it is argon 4s0 3d8 and it is 4s and then it is 4p ok so this bearing is not possible here is not taking place and hence this 6 orbital goes into hybridization sp3d2 and it forms sp3d2 hybridized orbital so geometry is octahedral or square pyramidal bipyramidal paramagnetic with 2 unpaired electron magnetic with 2 unpaired electron you can also think of one thing here like if you try to understand this if you do not memorize also you can do this easily see if you think of chalo NS3 is a strong ligand so bearing against the hunts 2 so what you'll do you'll shift this electron here right you will have only one vacant orbital then if you try to write down the hybridization of it then dsp4 is not possible dsp3d this kind of hybridization is not possible right that's why this orbital won't take part in hybridization at all right and hence it should not be vacant that's why the pairing against the hunts 2 is not possible hence it is sp3d2 ok and this also they can also ask the magnetic moment of the complex ok magnetic moment magnetic moment is calculated by this formula which we have also done last year in atomic structure root in that n into n plus 2 bm for magneton where n is the number of unpaired electron you substitute n value 2 here 2 into 2 plus 2 that is root in that 8 and that value as 2.83 bm so this value you must remember sometimes they'll give you the the value of magnetic moment in the question right so you should know suppose for this question for this question the magnetic moment is given this complex has 2.83 magnetic moment so first of all what you do you just find out the number of unpaired electron with this value what you need to do you need to equate this with root in that n into n plus 2 right this should be equals to 2.83 but sometimes what happens this is become this becomes very complex to solve this you know it is square and then solve the trick it's difficult to do so for that you should memorize this value when the magnetic moment is 2.83 the number of unpaired electron is 2 however you can you know roughly you can put some value of n and you can find out maximum 2 3 unpaired electrons they're not more than that in most of the cases but what point I'm trying to make if you memorize these values like 2.83 is a magnetic moment when the number of unpaired electron is 2 then this helps you a lot in solving the question so once you know this value of n you know how many electrons you have to left you know unpaired and then you can think of the hybridization of the molecule similarly if n value is 3 like this one if mu value is suppose the magnetic moment mu s we are getting 3.287 Bm 2.83 n value is 2 3.287 n value is 3 we have unpaired electron is 3 okay with 3 you'll get root in a 15 so root in a 15 you see it is somewhere in between 3 and 4 this kind of you know observations helps you a lot in resolving this kind of questions okay so with nickel it behaves as weak free ligand NH3 okay one last question we'll see here try this one CR NH3 6 3 plus sir what is chromium configuration chromium configuration electronic configuration yes sir 4S1 3d5 you should know it argon 4S1 3d5 okay see that other things magnetic property is it die I think triple you have written should be D2 right yes I was getting D3 as D2 D3 yes are you getting D3 preparing will happen sir no it's not it won't happen sir sorry so we're pairing happen no no no sir even though it's a strong field again this is what the two things we have discussed actually here you know in the question if they ask usually they give the magnetic moment here new value that is 3.28 this is the magnetic moment we observe then they'll ask you what is the hybridization you know because you should know that this chromium also this is not happening and one more thing in coordination complex D3 you don't consider D3 you know what we say hybridization is not possible we are not getting D3 hybridization here but this question was actually when they ask you they'll give you the magnetic moment value with this you will have the idea that how many unpaired electron so here also my point is here also it is behaving as weak field ligand so these are the few exceptions we have of BBT drawbacks of BBT right so answer for this question I'm not doing this answer is it is D2 SP3 low spin complex number of unpaired electron is 3 and it is paramagnetic how did you get diamagnetic aditya yes correct right so these two examples you must remember because it is not you know because these are exceptions you must take care of especially that nickel this one at least you can understand that D3 hybridization is not possible here so we won't think of the pairing of electron actually the reason is NH3 is behaving as weak field ligand okay these are the drawbacks of BBT okay that's why and the major drawbacks is what but again you know it could not explain the color of the complex and what color the complex shows this also could not explain my valence bond here that's why we got the new theory after this which we call it as crystal field theory with crystal field theory we can hide out the color of the complex obviously we cannot do theoretically because we need to know the wavelength of each color right and then only we can say okay this is the wavelength we are getting so this is the color of the complex so color thing you have to memorize okay but yes there is the CFT crystal field theory gives us the concept that how do we understand or how do we know that what color of the complex should be how do we calculate the lambda value there right that we get from the CFT crystal field theory so next write down the major drawback one note you write down after this the major drawback for BBT is it could not explain the color of the complex could not explain the color of the complex sir here it is the lowest spin complex or high spin complex this one low spin only sir yes whenever d2 you write first d2 sp3 then this see this s and p belongs to fourth cell right 4s 4t orbital is there no but this one is 3d so lower spin complex is when the inner shell d orbital is involved in the hybridization okay okay high spin complex when we have spt 4s 4t 4t high spin okay sir and is it possible to have unpaired electrons when it's like when some pairing takes place but still unpaired electrons are remaining that is also possible like all the electrons won't get paired that see everything depends upon the characteristics of the molecule like you see the one that you did this one what you did I think you did this one in this one you'll get three unpaired electron initially and then what you did you paired like this right and you say there's one unpaired electron yeah okay this is what you did right but it will not do here because with this the number of unpaired electrons you are getting one and this does not set it well but it is possible in some complex that there are five electrons unpaired electrons out of five you will get paired and accordingly we'll get mu value in the complex possible but those questions what happens either they'll give you mu value then only you can find out that whether the electrons are getting paired or not or if five unpaired electrons are there then how many electrons are getting paired depending upon the mu value which should be given in that question okay yes yeah next write down crystal field theory the crystal field theory I'll give you a brief idea of it because if I dictate you the theory it takes a lot of time so there are so many things in this actually what happens first of all in this theory there is one assumption and what is the assumption assumption is that the ligands and the metal iron be considered as a point charge okay so assumption is what ligand and metal iron are considered as as point charge okay now it is point charged so there will be what there will be you know this thing electrostatic attraction ligand is negatively charged metal metal iron is positively charged so we will have electrostatic attraction so the entire theory is based upon this particular assumption that ligand and metal will have electrostatic attraction but now what happens we have metals and d orbitals of metals are there right and ligands are trying to approach the metal right with its lone pair of electron and there are electrons present in d sub deal orbitals of the metal already right so when this ligands approach the metal and then suppose we have certain electrons present in some of the orbitals of metals like this some arrangement we have like randomly I have drawn this okay so when this ligand tries to donate the electron pair right these electrons repels the coming electron from the ligand okay repels in the sense what the orbital which consists of which has electron present into this the d orbitals of the metal in which the electrons are present those orbitals and the coming electron we have repulsion between the two so it is possible that some of the orbital has one electron two electron or some of the orbitals are vacant also right so obviously the orbitals which has electron they experience more repulsion and the orbitals which has no electron they experience less repulsion right so what we can conclude from this that when this ligands tries to donates the electrons into the orbital of the metal right there will be uneven repulsion between the orbital and the electrons that comes from the metal that comes from the ligand right because of this uneven repulsion uneven repulsion means some of the orbitals experience very high repulsion and some of the orbitals experience less repulsion right because of the presence or absence of electron because of this uneven repulsion this orbital splits into two sets of orbital mainly two sets depending upon the geometry we can have different number of orbitals right it splits into two sets of orbitals which we call it as splitting of orbital and this is splitting we call it as crystal field splitting right in this splitting what happens some of the orbital goes on to the higher energy level and some of the orbital comes down to the lower energy level by maintaining I'm sorry by maintaining the average energy of the complex average energy will be maintained this will increase energy some positive value of delta O and some negative value of delta O so that average energy will be maintained like this that splitting takes place ok so this is what actually happens in the coordination complex ok and this is crystal field theory so to some of all this wait a second to some of all this what we can say that crystal field theory is based upon the electrostatic attraction between the ligand and the metal ion why electrostatic attraction because the ligand and the metal ion are the point charge ok so when the electrons from the ligands approach the d orbital right because of the electron already present in the d orbital will have uneven repulsion in the d orbital and hence the d orbital to minimize the repulsion it splits into two parts ok or two sets of orbital or more than two also possible depending upon the geometry it is splits into two parts this is splitting we call it as crystal field splitting in this splitting some of the orbitals goes on to the higher energy level some of the orbitals goes down to the lower energy level in order to maintain the average energy of the orbital this is crystal field splitting did you understand the basics of it yes got it so all these things are given in ncrt properly they have written in ncrt I would request all of you after the class at least ten fifteen minutes to give to go through the entire theory of crystal field theory because if I dictate it will take a lot of time because I am not dictating this theory whatever I said now so I want you to go through once ok I am not dictating all these things so the thing is next when this orbital splits right this orbital splits into groups actually when we have octahedral complex because this kind of splitting we observe in a different different geometry so we'll have a splitting in octahedral complex we'll have a splitting in square pyramidal complex we'll have a splitting in tetrahedral complex right all these three splitting we'll see understood so first of all you write down just basics right down here according to this theory according to this theory the bonding in the coordination complex according to this theory the bonding in the coordination complex splits sorry is completely electrostatic is completely electrostatic again I am repeating according to this theory the bonding in the coordination complex is completely electrostatic and we assume and we assume the ligand and the metal ion as a point charge the complex sorry and we assume the second line of the first line from the beginning I'll repeat from the beginning according to this theory the bonding in the complex is completely electrostatic in which the ligand and the metal ion are considered as point charge in which the ligand and the metal ion are considered as point charge explain the complex is regarded as the combination of the complex is regarded as the combination of CM a that is central metal atom complex is regarded as the combination of CMA surrounded by the ligands. In the entire theory, I'm cutting short to it because it takes a lot of time. A small line after this you add, because of the interaction between the ligands and the metal ion, because of the interaction between the ligands and the metal ions, the d orbital experience, uneven repulsion, the d orbital experience, uneven repulsion, it splits into, and it splits into, it splits into two sets of different orbitals having different N and G, and it splits into two sets of different orbital having. So these sets that we get, that is the splitting of orbital, it happens in this way that all non-exial orbital, for example, dxy, dyz, dzx, all these non-exial orbitals, it goes into one set, okay, and this set we call it as t2g orbitals, these are non-exial orbitals. Exial orbitals are called eg orbitals, for example, dx square, y square, and dz square, these are axial orbitals, right, this t2g and eg, you don't have to think about it, these are spectroscopic terms, we call it as spectroscopic terms, just a group of orbitals, it's said to be this and this, not required this, but when we distribute orbitals, then this grouping, this terms helps us to understand the distribution of electron, that's why we have given this up, right, not much important, fine, so this splitting of orbitals, right, it, there are certain factors which affects the splitting of orbitals, okay, like we have, you know, a position of transition metal, oxidation state, and other things, right, there are various factors which affects the splitting of orbitals, so how splitting takes place in different, different complex, whether it is, you know, octahedral, square panel, tetrahedral, that we'll discuss a bit later, first we'll see that what are the factors which affects the splitting of orbitals, right, right on the magnitude, explain, the magnitude of splitting, the magnitude of splitting depends on the following factor, depends on the following factors, these factors are not important, only two things, you know, you have to keep in mind here, otherwise it is not at all important, okay, the first one here, the first point is a bit important and the last one, okay, just you need to know the, you know, the basic result of it, right on the position of the first factor is position of transition metal in the periodic table, in the periodic table, the magnitude of splitting increases as we go down the group, the magnitude of splitting increases as we go down the group, for example, if you compare the splitting in PTCL4 to PTCL4 2 minus and PDCL4 2 minus, okay, as we go down the group, the magnitude of splitting increases, so platinum produces more splitting than palladium, okay, the second factor we have oxidation state of metal, oxidation state of metal as oxidation state of CMA, central metal atom increases, splitting increases because more positive charge, more will be the attraction and hence more will be the repulsion in the d orbital, more will be the splitting, okay, so easily you can compare this FeCN6 3 minus and FeCN6 4 minus, obviously in the first one the repulsion will be one more, in fact two more, if the charge on the central metal atom, if the charge on the central metal atom increases, splitting increases, oh this is again, it's the same thing, charge on the central metal atom is same, if it is same, then the metal with higher number of D electrons, higher number of D electrons will have lesser splitting, so if you consider this CO2 plus, suppose you have a complex in which you have CO2 plus oxidation state and Ni2 plus oxidation state, so in CO2 plus the splitting would be more, because if you look at the configuration of this, it is argon 3D7, argon 3D7 and for nickel it is argon 3D8, so more electrons lesser will be the repulsion, the most important one is the nature of ligands, okay this is what I was talking about, strong ligands produce more repulsion and weak ligands produce lesser repulsion, so magnitude of repulsion actually gives us the idea about the nature of ligands, right down into this, the ligands which affect the ligands which affects only a small degree of crystal field splitting, the ligands which affects only a small degree of crystal field splitting are called weak field ligands, ligands which affects only a small degree of crystal field splitting are called weak field ligands and those which affect a large splitting are called a strong field and those who affects a large splitting are called strong fielding and next slide. When these ligands are arranged in order of the magnitude of their crystal filled splitting of the magnitude of the crystal filled splitting will get a series, will get a series which is called a spectrochemical series, a series which is called a spectrochemical series. So a spectrochemical series right now, this is a spectrochemical series from I-2 EtOH these are weak ligands. These are moderate and all these from NH3 to this these are a strong ligand in general. Actually water if you see water is present in between whatever ligand is present out right side of water Strong ligands in general and including water whatever ligand is present left to the water molecule these are weak ligands. So whether the ligand is weak or strong that idea we get from the splitting in the orbital. So this is the spectrochemical series and the nature of the ligand weak or strong okay. So if the two complexes given and the ligands are different then depending upon the nature of the ligand weak is strong or moderate you can easily compare the splitting in the complex okay. Now you write down crystal filled splitting in octahedral complex. Three complex we have to discuss octahedral complex, a square planar and tetrahedral. So octahedral complex will discuss in detail everything is same only in other two complex. So we'll just see in other two complex we'll just see the splitting of orbitals or the orbitals splits. So we need to first of all you know we need to first of all identify that each two orbitals will have the maximum repulsion like which two orbitals will have the maximum repulsion because the complex is octahedral so one ligand comes from the top another one comes from the bottom this is the third one then fourth one fifth and then sixth right in all these direction the ligand approach the middle. Metal is present here in the center all these are ligands one so we can easily you know consider the axis along these lines we can consider the axis because the angle is 90 degree one x y and z. Obviously along this line we have dx square y square orbital dx square y square orbital present along this line right and along this line we have dz square orbital. So the two orbital set of orbital dg and dg t2g is all non axial orbital dxy dyz and dzx and t2g easy easy has two orbitals which is dx square y square and dz square orbital. So when this ligand approach the metal to form an octahedral complex okay so eg orbital will have the more repulsion and this will go to the higher energy level and t2g will come down to the lower energy level by maintaining the average energy of the of the complex okay. So what is how the splitting takes place I'll show you the diagram for that you have to draw this this is for tetrahedral complex sorry octahedral complex see this what happens here this y axis is the energy and this x axis is the distance decreases between the metal and ligand like if you're going left to right here means ligands are coming closer to the metal okay. So this x axis you write down distance decreases between the metal and ligands as the distance decreases you see that repulsion starts increasing because the ligands are approaching the metal the energy of this orbital starts increasing is goes to the value here this step one energy goes to this value then further the electrons the ligands comes closer and the orbital splits into two part one is this other one is this you see this one this one is eg and this one is t2g by maintaining the average energy average energy right this you let it be very central it's not required simply average energy level is here. Now this value is 0.6 delta O it's not 0 or not it is O O stands for octahedral okay it's basically the unit of unit of energy in this complex octahedral okay this equals to a fixed value of energy for a given complex this value will be different for different different okay those energy we don't have to calculate right it's not required 0.6 delta O below it is minus 0.4 delta O right total energy is delta O that's how the splitting takes place. Pop it on this first sir why is it minus 0.4 delta O because the energy is decreasing from that point so the minus energy decreases it's just a sign to mention you can see yes sir can we say that the energy of the d orbitals in step one raises up to 0.2 delta naught or delta O? 0.2 delta O no no it's not 0.2 some is this plus this is delta only plus minus sign is just represent the energy is increasing for this and decreasing for this total sum is delta only okay yes sir yes sir right nothing in step one as the ligands are coming closer to the metal slowly the energy of d orbital is increasing and it reaches the maximum value and then it's it splits into the two parts step one is the excited state excited state we cannot say excited state because in that excited state generally when electron jumps into the higher region if the transition of electron is not there but because of the interaction of the electron between d orbital and the ligand overall the energy of d orbital is increasing goes to a maximum value sir and we have assumed the ligands are approaching the metal atom like along the axis is it only an assumption or is it actually we are assuming octahedral complex no to form an octahedral complex obviously ligand has to approach this way otherwise it won't form so then the axis can be like oriented in any other way also right see this is actually this way I have you know we explain things this way so that you can memorize it you can understand but actually what happens when suppose the ligands are approaching from you know from the other side you know from the every side towards the metal then obviously the ligand also repel each other no there will be some crowding and then some don't pair with don't wear the ligands are coming the lone pair will also repel each other then the ligands will arrange themselves in such a way so that the repulsion will be minimum and then actually the geometry of the complex we get we get right we are understanding this in a different way what I'm telling you that to form an octahedral complex the ligand has to approach this way so that you can understand this that these orbitals are involved and why see if I tell you this thing no that ligands approach the metal to minimize the crowding and repulsion of the electron pair you know it will arrange this way and forms the octahedral complex like that also I can explain but then when I tell you that this orbital these two orbitals goes to the higher energy level then you will ask me sir why these two orbitals goes to the higher energy to clear this thing we are understanding it this way because if octahedral complex is forming the ligand has to approach along these lines and along these lines we have these two orbitals present that's why in these two orbitals the maximum repulsion so we are trying to understand this in a different way but the actual thing is ligands are randomly approaching the metal and whatever the you know arrangement of ligands in the space which provides the minimum energy to the complex and maximum stability that will be the geometry of the complex all of you on this yes okay so this is the splitting of orbitals now you see if you want to distribute the electron into this one from which orbital you have to start the distribution of the lower energy on the lower energy orbit yes that's right because in atomic orbital we also did the same thing lower energy orbital gets filled first right by n plus L rule so here also if you want to distribute the electron in this because this is the actual case we have right now if you want to distribute the electron this three orbital will get filled first and then the electron goes into e g orbital so first t2g and e g that's one thing okay very very carefully listen to me now whether this distribution of electron follows hansel or not that depends upon the nature of the ligand if the ligand is a strong then this is splitting will be high that difference of energy will be more between the two orbital and if the difference of energy is more than the distribution of electron does not follow hansel right we'll have one two three four five six seven eight nine ten like that if the ligand is strong energy gap is high if energy gap is more is less than the distribution of electron follows hansel in that case it will be one two three four five six seven eight nine ten like that okay so this is the distribution of electron first thing is that right how do we find out the crystal field splitting energy because obviously the two sets of orbital will have an energy difference between the two sets of orbital so this energy difference also we can calculate by a formula and that formula is called CFSE crystal field splitting energy and it is given by minus 0.4 times x plus 0.6 times y delta o again it is oh it is not not okay delta o is transfer x what is x x is the number of electrons number of electrons 0.4 is associated with which orbital 0.4 is associated with t2g orbital right t2g 0.6 is associated with e g orbital you see this x is the number of electrons present in number of electrons in t2g orbitals why is the number of electrons in e g orbitals okay so distribution of electron if you know you know the x y value that x y value you can substitute here and you can find out CFSE in terms of delta clear so yeah so we always have 0.6 delta o and minus 0.4 delta o right for when it splits I didn't get you come again so like we'll always have 0.6 delta o for e g and minus 0.4 for t2g right even for square cleaner also it's the same thing this is the splitting and all these are experimental all the you know orbital aspects so yeah so but then how do we determine whether the splitting is large or less even though the fraction is the same see when the or see for this delta o value no it depends upon the complex it is not fixed okay I'll give you one example if you want this delta o I'll come to this give me some time I'll come to this otherwise everything will mess up okay so you see here I was talking about this CFSE so once you know the number of electrons in the two orbitals you have the value of x and y and then you can find out CFSE in terms of delta o delta o for a given complex will have a fixed value it's not like this value is universal like gravitational constant right delta o will be differ for will differ for different different no complex it has different value for different complex okay it's not constant for all the complex okay but we do not calculate those values okay even you don't have to calculate CFSE normally hardly you have to calculate this that is not but even they ask you also this in the option also they'll give delta only you just need to know this x and y that is okay but for your understanding I'll discuss one example okay so the point is how to distribute the electrons like I said the distribution of electrons depends upon whether the ligand is strong or not okay so in case of a strong ligand the you know the gap of the energy between the two orbitals e g and t to g will be more and in that case we do not follow hunts rule when the ligand is strong when the ligand is weak will follow hunts rule and will distribute the electron as we do in atomic orbit okay so heading you write down here distribution of electron first one in case of in case of weak ligand write down simply in case of weak ligand will follow hunts rule will follow hunts rule so the distribution will be according to hunts rule so first the electron will go into t to g because it has the lower energy then the electron will go into e g because we are following hunts rule then again electron will go into t to g and again the electron will go into e g right so first three electron will go into t to g orbital fourth and fifth electron goes into EG orbital. Sixth, seventh and eighth electron goes into T2G orbital, ninth and tenth electron goes into EG orbital. Right? So suppose the ligand is weak and we have D5 configuration. Suppose you need to find out CFSE, CFSE for D5 configuration. Right? So D5 configuration is still here. What is the X value? Could you tell me? What is the X value? Three. X value is three. Y value is two. Right? Yes sir. So this X and Y value is substituted here. Three. Y value is two. You substitute it here. You'll find out CFSE with this formula and that is minus 0.4 times 3 plus 0.6 times 2 delta O and this is equals to zero crystal building splitting energy. Is it zero? Yes. Right? CFSE is zero for this one. Similarly, for any configuration you can find out depending upon the ligand, the nature of the ligand. Suppose this D8 configuration we have, then the number of, the value of X is what? The value of X is six and Y is two. Substitutes X and Y here, you'll get CFSE in terms of delta O. Did you get this all of you? Yes sir. Yes sir. Yeah, fine. Now when the ligand is a strong, second case, in case of strong ligand, the energy gap between the two orbitals, EG and T2G will be more and it does not follow, distribution of electron does not follow henceforth and hence the distribution of electron will be like this. We have T2G and then the electron will go into EG. So first six electrons will go into T2G orbital. One, two, three, four, five, six, then seven, eight, nine. So CFSE is equals to, suppose we have an example CFSE for D8 configuration. So for this case, X value is six and Y value is two. CFSE is equals to minus 0.4 times six plus 0.6 times two delta O. So it is minus 1.2 times delta O is the energy here. One more thing we have here that we call it as average pairing energy. Okay, note you write down this. All these small, small information, they have asked questions on this. That's why I told you in the beginning that in this chapter, you'll get many different types of questions. Okay, that is possible they can frame. All these small, small information are important, right? Like you see, average pairing energy. Because of this only, the electron does not follow Hansel over there. Average pairing energy. It is represented by P, capital P, write down. It is the energy required, it is the energy required for the pairing of electrons in the same orbital. It is the energy required for the pairing of electrons in the same orbital. Okay, so in case of weak ligand, what happens? The delta O, the energy gap that you have between the T2G and EG orbital is lesser than the pairing energy. And that's why pairing takes place over there, right? And follows Hansel. But when the ligand is strong, when the ligand is strong, the relation is this. Delta O is greater than P, average pairing energy. And that's why it does not follow Hansel, okay? This relation also they have asked many times in the exam. Sir, could you explain this again? Sorry? Could you explain this one again? This one. You know, I think it's just a definition. See, when you want to pair up an electron in the same orbital, some energy must be required because the electron has to be filled into that particular orbital. That energy, we call it as average pairing energy for the pairing of electrons. So when this gap, the T2G and EG orbital, if this gap is lesser than this case, lesser than the average pairing energy, then the pairing takes place and it follows Hansel. If this gap is lesser than this energy. But when the gap is more, then the pairing energy is lesser than the gap of the two orbitals, two energies orbital like T2G and EG. So in that case, the pairing of electron is difficult. That's why the distribution of electron does not follow Hansel. So, but if more pairing energy is required, then, okay, one second. No, more pairing energy is not required. Average pairing energy is fixed for the complex. Now, depending upon the ligand, we can have different, different Delta O, the gap of the two orbitals. If that gap is less than the pairing energy, then pairing takes place. If that gap is more, then the pairing won't take place. No, average energy level is the energy of the orbital, average energy of all the orbitals. Pairing energy is for electrons in the same orbital. Now, one more thing you see, the drawback in the previous one is mainly the color we cannot find out for the complex. This calculation is difficult and they won't ask you to calculate the one that I'm going to talk right now. You see, if you want to find out the energy of the complex, then this is the, because suppose here is the electron, right? Here the electron, one orbital, one electron is there, two electron is there, two electron is there. It is present into this orbital. Then what you do, you just, you know, provide some energy with the help of light on this complex, right? Electrons takes up this energy and converts and jumps to the higher energy level, provides energy so that this transmission takes place, right? So this energy we can find out by CFSE, right? And this energy is equals to what? You can write Delta E is equals to Hc by lambda per mole if you do, then Na you have to multiply here, divide here. This energy gap is nothing but CFSE and that is nothing but CFSE is equals to Na times Hc by lambda. All these value, you know, you can find out the wavelength, like, which is getting absorbed in this process. And that wavelength corresponds to one particular color, that will be the color of this. You know, the complementary of that color will be the color reflected by the complex. So with this method, we can find out the color of the complex. So that drawback has been resolved, but for us, we are, we cannot find out these values, right? Suppose by any, how suppose any means you can find out lambda, but how do you memorize that this lambda corresponds to this particular color? Are you getting one point? You cannot memorize wavelength for all the colors, no? That's why they generally do not ask this question in the exam. Sometimes they have asked, in that case, you cannot do anything, but you have to know that this ion in this oxidation state will show this color that you have to memorize. But how to find out the color of the complex method is this. Did you understand it? So the energy difference is the CFSE. Yes. So then what is Delta naught? CFSE? Delta O is that only. See, if I write down CFSE, so in terms of Delta O, the total is Delta O only. It is a split into this, so 0.6 plus 0.4. Delta O only. So Delta O is this gap and this is equals to, like, this is a unit. It's, it's, you know, see, I'll take one example. Suppose I have a complex. This you will understand. For this complex, usually this kind of example is not given in most of the book, but I have, I have, you know, taken one, one example for this. TI H2O 6, 3 plus. This is the complex we have. For this, if you find out Delta O, that would be equals to the energy E per mole calculation is there. So this would be equals to NA times HC by lambda. So what wavelength it absorbs that we know by the experiment, right? So when you substitute all the value, I'm giving you this, you know, this fact here. NA value, you know, 6.022 into 10 to the power 23. H value, you know, 6.6 into 10 to the power minus 34 joule per second. 3 into 10 to the power 8 meter per second. This divided by the wavelength that this complex absorbs is 20,300 into 10 to the power 2 meter inverse. No, the value is and this is the experimental value. Okay. When you solve this, you'll get Delta O equals to 241.97 into 103 joule per mole, which is approximately 242 kilo joule per mole. Do you see this? For this complex, this Delta O is nothing but the value of energy, which is this. And that gap is nothing but here. And that is what I'm talking about here. You see, this gap is the energy, right? This energy is Delta O, right? That is the splitting thing we have Delta O, which is equals to hc, any times hc by lambda because for mole calculation. Did you understand it? So that CFSE is not Delta O, right? So CFSE is not Delta O, right? If SE is the energy associated with the distribution of electron because x and y values, oh, at that time I said TFCFSE. No, it's not. It's Delta O actually. The energy gap is Delta. CFSE is the energy associated with when you distribute electron in different orbitals. In that case, there will be some exchange in energy. That is CFSE. Okay, sir. What is that signify that exchange energy and what? Which one? So that's CFSE. CFSE, you see CFSE formula is what? Minus 0.4 times x plus 0.6 y is equals integrate out, right? x and y is what? x and y is the number of electrons. So with different value of x and y, you'll get different value of CFSE, correct? Okay. CFSE is what? CFSE is because you are distributing the electron and when the electron goes from one orbital to another orbital or you put one electron into one orbital, there will be some exchange in energy, right? That energy signifies by CFSE. Okay, sir. Because x and y are the number of electrons, right? So x and y changes, CFSE also changes. That is associated with the number of electrons present in the orbit. Yes. Okay, so this thing Delta O is, you can say it is a type, kind of unit for each complex. Okay, Delta O, I have given you for one, okay, but it will be different for the other, you know, complex, but you don't have to calculate this. They won't ask, I have never seen any question in which they ask you to calculate Delta O. If they ask, they'll ask you to calculate in terms of Delta O. So you don't have to worry about it, but yes, to understand the concept, you should know this. Like this, we can find out energy, wavelength, and then Delta O, corresponding to that, the complementary color is the color of the process. Clear understood? Yes. Fine, we'll take a break now. After the break, this is actually the major thing we have done. Only two things we have to discuss. And this much the distribution of electron and everything, we don't have to discuss for a squared, linear, and tetrahedral. Just we need to see how the orbitals splits in the list of this, in these complexes, squared, linear, and tetrahedral. Which orbitals goes onto the higher energy level, and which orbitals comes down to the lower energy level that only we need to see, that is a diagram we need to understand first. Nothing much like this distribution of electron and all. This has the same concept. We don't have to do that. They won't ask also. Understood? Yes. Yeah, okay. Fine, take a break now. We'll resume the session at six or seven ten. Okay, 20 minutes. Seven ten will resume. Take a break guys.