 Hello everyone, good evening. Hello Paras, how are you? Where are others, Paras? Where is Aditya? Okay, guys, please type in your name. All of you who is present in the session. Hello Shreya, Samyukta, how are you doing? Okay guys, tell me what have we done last class? I think nomenclature we finished, no? Ardra, right. Nomenclature we finished last class, correct. Did we finish Werner's Coordination Theory? Bonding in Coordination Compound? Okay, Werner's we did. We have seen the examples also, no? That MNCO5L example we did, right? Dimerization, oxidizing, agent and reducer reducing agent, correct. So, right. Okay, so hello everyone once again. So, we will start today with the Werner's Bond Theory. Okay, Werner's Bond Theory of Coordination Compound. See, it is exactly, you know, the similar thing we have that we have chemical bonding. Okay, so in chemical bonding, we have different different theories for molecule because one particular molecule, you know, does not, like, from one theory, all kind of molecules, the bonding of all kind of molecules was not defined. Okay, VBT we have also done, is it? Can we do the last, what we did the last example? Is there any example I discussed? Last example? Did we see the postulates of VBT? Yes or no? NiCO4, like hybridization, bonding and all, right? Okay, so we have seen that, means NiCl4, NiCO4 we have done, right? Correct, correct. So, like, okay, then we will start from this, another example, which is this one, you see this example, we have NiCN4, NiCN4 2 minus. Okay, what we need to find out for this, again the same thing, hybridization, geometry, magnetic property, property, number of, number of unpaired electron, number of unpaired electron. Can you tell me this answer of these questions? CN4 minus, CN, sorry CN minus is a strong ligand. Okay, CN minus, this information I am giving you, it is a strong ligand. This roman range is at the, right? Tell me, hybridization, geometry, magnetic property is a number of unpaired electron. Then tell me, anyone? Okay, see I am doing this. First of all, can you tell me the oxidation state of NiCl? CN is minus, minus 4 and minus 2 this side. So, oxidation state of NiCl is plus 2, right? Oxidation state of NiCl is plus 2. And the electronic configuration of NiCl is what? NiCl is 28. So, it is argon 4S2 3D8, right? So, we have Ni2 plus in the complex. So, for Ni2 plus, the configuration is argon 4S0 3D8. So, the orbital diagram of this will be 3D, then 4S, then 4P, okay? 3D has 8 electrons, that is 1, 2, 3, 4, 5, 6, 7 and 8. And this is the thing we have, right? But we know this is a strong ligand. So, in case of a strong ligand, what happens? Pairing takes place against the Hans rule, right? This electron jumps over here and will have the pairing of electron here. So, the orbital diagram will be like this, okay? Now, the coordination number of this molecule is what? Coordination number is 4. So, we require 4 atomic orbitals, which takes part in hybridization, correct? So, the 4 atomic orbitals are 1D, 1S and 2P, right? So, 1D, 1S and 2P forms hybrid orbital and this hybrid orbital is 4 hybrid orbital. Hybridization is DSP2, right? In this 4 hybrid orbital, CN minus ligand donates its electron pair, right? So, first of all, the hybridization is DSP2. Now, when hybridization is DSP2, geometry is a square planar. Magnetic properties, since all electrons are paired, so it is diamagnetic. And number of unpaired electron is 0, is it clear? So, this is the hybridization we have here. One kind of question that they ask, like in suppose this one, when the hybridization is DSP2, what D orbital is involved? Means out of 5 D orbitals, which D orbital, which D orbital involves in this hybridization, which is DSP2? This is also one of the question that they ask. This question we have done already, everything we have already got. But this kind of question also they ask, which D orbital involves into this? So, first of all, you have to memorize this thing, okay? But logically also we can understand. All of you have written the last thing, you know? You see, first of all, you write down one side geometry. This side we have geometry and here we have D orbital. In which geometry, which D orbital is involved, okay? So, if the geometry is square planar, if the geometry is square planar, the D orbital involved is DX square, Y square. If the geometry is trigonal bipyramidal, the D orbital involved is DZ square. If the geometry is octahedral, if the geometry is octahedral, D orbital involves here, or DX square, Y square and DZ square. If the geometry is square pyramidal, it is DX square, Y square. These four things you have to memorize. In which hybridization, what D orbital is involved, okay? So, we can understand this also. Let's discuss this. But eventually, you know, you have to memorize this. This four thing you have to keep in mind. But with this, you can, you know, it will be easy for you to memorize if you understand this one. You see, okay? Now you see the center we have metal, right? We have ligand present here. One, two, three, four, five and six. You know this structure, right? Square bipyramidal. Because square is the base we have here, right? And one ligand on the top and other one on the bottom that forms an square bipyramidal structure. Okay? If you imagine this, you can easily understand. You see, this looks like a pyramidal with a square base. Pyramidal with a square base. You have to imagine this, okay? Looks like a pyramidal with a square base. Okay? So it is a square by pyramidal structure. Okay? One ligand on the top, one on the bottom and four in the plane of the square. Okay? Now you see this. We know the square diagonal bisects each other at 90 degree, correct? Now how this square bipyramidal structure forms when the ligand approach the metal atom by this direction. One from this direction, one from this direction, one from this direction, fourth one is this, fifth one is this and sixth one is this. Then only the square bipyramidal structure forms. Okay? Now when the ligand is approaching the metal atom by all these directions, the orbital which is along this line will involve in the bonding. What I said, the orbital which is along this line, this line, this line, will have the maximum probability to involve in the bonding, correct? Now you see, since the bond here between this diagonal is 90 degree, so we can imagine here easily that this is 90, this is also perpendicular to this plane. Suppose if this is x axis, this we can imagine as y and this we can imagine as z because bond angle is 90 degree everywhere. So we can easily imagine these lines along x, y and z axis. So obviously the orbital which is along these lines will have the maximum probability to get involved into the bonding. And what is the orbital which is along this x and y axis? This is one orbital? This orbital is what? This orbital is d x square y square which is along the x, y axis. And along this line z axis, we have d z square. That's why we say in octahedral complex, d x square and d z square orbitals are involved. How many of you understood this CLR? Clear? So the thing is, that's why octahedral complex has these two d orbitals involved. Now the point is, if you have a square pyramidal, square pyramidal means what? We have a square and only one side we have ligand present. Either this side or this side. Then only it is pyramidal. See, when you write the structure like this, square and this ligand one we have here. It is a square pyramidal. When we have both ligand like this, it is a square bipyramidal. Both sides we have ligand. When we have a square pyramidal structure, this d z square orbital is not involved because from this side we do not have any ligand. That is not present. So this ligand we have, but this side we do not have ligand. So we cannot assume that this is along z axis. But we can assume that these two along x and y axis and that's why square pyramidal has this orbital involved d x square minus y square. Now we have a square planar. A square planar, how do we get? This is square bipyramidal. One ligand you remove. A square pyramidal. This one also you remove. A square planar. A square planar is nothing but a square and the ligand is approaching the metal from this direction. We can assume this as x and y axis. So the orbital is d x square y square. Tbp. Triagonal bipyramidal is what? Triagonal bipyramidal is this. Instead of this square bipyramidal, instead of square as a base, in triangle bipyramidal triangle is the base here and one ligand is from this direction another one is from this direction. Now you see this bond angle is not 90 degree. It is not 90 degree. It is 120. So we cannot assume any axis along these lines. This is in between the x y axis. It is perpendicular to the plane of this. So we can anytime assume this as the z axis and that is why the ligand which is involved in this bond may have the maximum probability to get d z square into the bonding. That's why Tbp has d z square orbital involved. Let me know. We'll move on to the next question. Any doubt? Correct. Right. You see. So for coordination number 4, we have discussed three examples. And one kind of question is that where you have to find out geometry, hybridization, magnetic properties and all. Another thing, which d orbital is involved according to the geometry of the molecule. So this is again one important thing. Next one you see, we have the next example of molecules in which the coordination number is 5. Same thing you have to do. We'll see some more examples. The next question is FeCo5. FeCo5 is found to have bound to have zero dipole moment bound to have zero dipole moment explain its hybridization geometry. One more question I'll write down this side. NiCN5 3 minus is found to be diamagnetic with two different CN bond length bond length out of which four NiCN bond length are same and one is different. These two questions done. First one both are strong ligand. NiCN4 CN5 3 minus is diamagnetic with two different. Square pyramidal. And the first one is DSP. What is the first one? It is DSP3. DSP3 DSP3 is the first one. Triagonal bipyramidal. Why not square pyramidal? Correct. Your answer is correct so I'm not going to explain these things again because the same thing you have to do. The answer will be hybridization is DSP3 DSP3. And with DSP3 hybridization both kind of geometry is possible. Either we can have square pyramidal with coordination number 5 or we can have trigonal bipyramidal both possible. But which one of these geometry will have that depends upon the information that is given in the question. So square pyramidal if you see square pyramidal will not have zero dipole moment. You see this bond, this bond, this bond, this bond. This four bond cancels out the dipole moment here. But because of this bond will have some dipole moment in the molecule. So for this molecule mu does not equals to zero. But when it is trigonal pyramidal bipyramidal right trigonal bipyramidal so these two cancels out right and this two also cancel out because these two the resultant will be in this direction and this will cancel out this one. So for this molecule mu net will be what? mu net will be zero. And that is what the information given in the question is found to have zero dipole moment and that is why this geometry is not possible. This is correct geometry. Is there any unpaired electron in this molecule? Can you tell me? Number of unpaired electron no unpaired electron zero and that is why the molecule is diamagnetic. Square pyramidal is not possible trigonal bipyramidal is possible. Okay similarly this question also the hybridization will be what? Hybridization will be DSP3 okay. But here you see what is given found to have found to be diamagnetic with two different NICN bond length. You see in this molecule if you have trigonal bipyramidal structure it has two different NICN bond length right it also has two different NICN bond length but three bond length are same two are different from the other three. Here the information is given what? Four NICN bond length are same and one NICN bond length is different from the other NICN bond length which is only possible in this case. These four NICN bond length are same and this one is different that is why for this one the geometry will be square pyramidal right diamagnetic so zero unpaired electron okay hybridization in this we have already done. Okay so based on the information given in the question we can find out the geometry of this. Next one you see we will see two examples of coordination number 6. Coordination number 6 molecules which has coordination number 6 is most important in this chapter okay. So first question is CO NH3 6 3 plus. What is the number of number of unpaired electron hybridization geometry magnetic properties one more question we will have FeF6 FeF6 3 minus same thing hybridization geometry number of unpaired electron tell me quickly. NH3 is strong correct NH1 is paramagnetic D2SP3 first one you are getting right first one is D2SP3 is it para or diamagnetic D2SP3 hybridization is correct it's diamagnetic this is also right zero unpaired electron and geometry is octahedral see whenever the ligand is strong okay generally the nature of the molecule will be dia will have one or two reflections that is possible but generally the since the pairing against the hunts rule takes place so generally the molecule will be what diamagnetic okay what about this one next one you tell me FeF6 3 minus done hybridization is what tell me what is the hybridization it is SP3 D2 correct paramagnetic number of unpaired electron you tell me geometry is octahedral F minus is a weak ligand first of all not D2SP3 try this let me know if you are not able to do this I will explain this answer will be SP3D2 because F minus is a weak ligand halide ions are the weak ligand 5 unpaired correct we have 5 unpaired electrons samikth I got it SP3D2 yeah wait see first of all the oxidation state of ion is what plus 3 so Fe oxidation state is argon 4S2 3D6 for Fe plus 3 this will be argon 4S0 3D5 F minus is a weak ligand so pairing against hunts rule is not possible now orbital diagram is what we have 3D 4S 4P 4D 3D has 5 electrons so 1 2, 3, 4, 5 now the pairing of electron is not possible here because the ligand is weak correct and the coordination number is 6 so for that for this we require 6 atomic orbitals so 1 this is 4S this is 4P and this is 4D so 1 and 3 4 so 4 plus 2 6 orbital takes part in hybridization and hence it forms SP3 D2 hybridization so hybridization is SP3D2 5 unpaired electrons paramagnetic the matrix of the hydrogen now one more thing you see D2 SP3 means what this D2 SP3 means this two inner D orbital 1S and 3P is involved suppose if this orbitals are involved then the hybridization will be D2 SP3 D2 SP3 so when inner D because it is 3D orbital 4S4P4D so when inner D orbital involves in the complex we call it as lowest spin complex lowest spin complex outer D orbital involved high spin complex low spin complex high spin complex low spin complex means what the ligand is strong that also we can withdraw the conclusion with strong ligand low spin complex high spin complex the ligand is weak all these things we can draw got it? as in this VBT again there are few drawbacks or limitations so limitations of of VBT ok the first you know drawback or limitation is what it does not explain explain the color of the complex the color of the complex second point it could not explain could not explain why magnetic moment magnetic moment changes with temperature magnetic moment can be given as we have a formula you must have done it last year in chemical bonding or periodic properties N into N plus 2 this is magnetic moment where N is the number of unpaired electron write it down N is the number of unpaired electron so for all diamagnetic molecules the magnetic moment will be 0 for paramagnetic molecules the number of unpaired electron you substitute here you will get the magnetic moment of that substance ok one more example is important here like for this one you see NH3 6 this molecule ok for this molecule you see what happens in this this is a strong ligand ok I am explaining this you just see this this is a strong ligand ok and 6 is the density here so coordination number is also 6 ok so oxidation state of Ni is also plus 3 ok you see Ni the electronic configuration is argon 4S2 3D8 so Ni3 plus Ni2 it is not 3 plus it is Ni2 plus here so for Ni2 plus the configuration is argon 4S2 sorry 4S0 3D8 8 orbiter diagram you see 1, 2, 3, 4 then we have 4S 4P and 4D ok so this 3D has 8 electrons so 1 2, 3 4, 5 6, 7 8 ok so we require what we require 6 atomic orbitals to form hybrid orbital so when the ligand is strong so ideally what should happen this electrons should jump over here pairing takes place against the Hunts rule and then this involves in the hybridization and we will get the hybrid molecule right but in this molecule what happens pairing does not take place here this too will be as it is and 1S3P and 2D orbitals involves in the hybridization and it forms SP 3 D2 hybridization this is not possible here ok however this ligand is a strong ligand then also pairing does not take place right so what we say NS3 being a strong ligand with most of the metal it like it behaves as strong ligand and pairing takes place against the Hunts rule but with Ni2 plus this Ni2 plus actually this NS3 is behaving as a weak ligand is a weak ligand because what happens we cannot have hybridization like DSP3D that is not possible inner D and outer D at the same time cannot involve into the hybridization ok so either all the inner D or all the outer D orbital involves so in this case one of the postulates of VBT fails which says that NS3 being a strong ligand or any strong ligands in case of strong ligands the pairing takes place against the Hunts rule which is not true in this case and that is why this is one of the drawback of VBT so you must keep this in mind Ni2 plus metal if you have then NS3 is behaving as a weak ligand any kind of more oriented ligand with coordination number 6 here in this molecule this kind of shifting of electron against the Hunts rule is not possible because at the same time inner D and outer D orbital cannot involve in hybridization that is not possible either inner D or outer D both is not possible so this is the hybridization we have magnetic property is paramagnetic 2 unpaired electron geometries octahedral so this one you must remember because this kind of question you may get where the things does not follow the general pattern general rule that we have so you must keep this in mind all of you understood can you move on correct the next theory because again we have few drawbacks into this so we have got a new theory another new theory of bonding and that is crystal filled crystal filled theory so in this write down first few points and then we will discuss what this theory is all about write down according to this theory point wise write down according to this theory the bonding in coordination compound the bonding in coordination compound is completely electrostatic okay is completely electrostatic next point ligands are ligands are considered as as point charge considered as point charge in case of ions in bracket you write down point charge in bracket in case of ions or dipole in case of neutral molecule neutral molecule so ligands are either point charge or dipole if it is neutral then dipole if it is any anion any ion or anion right then it is points charge that is what we consider okay next point this theory this theory this theory mainly focus mainly focus on the repulsion repulsive you write down not repulsion repulsive force repulsive force between the lone pair and the d electrons d electrons means what the electron present in d orbital of the metal right so we are mainly considering the repulsive force between the lone pair of ligand and d electrons of metal this is for metal d electrons of metals so what happens because of this repulsive force write down because of this repulsive force because of this repulsive force repulsive force the degeneracy degeneracy of d orbital d orbital disturbs disturbs and it splits and it splits splits into two set of orbitals two set of orbitals having different energy having different energy right this is the crystal field theory we have okay the bonding is the bonding in coordination compound is completely electrostatic okay so we are now considering as either point charge or dipole depending upon whether it is charged or neutral theory mainly focused on the repulsive force between lone pair and the d electrons of metal right because of this repulsive force the degeneracy of the orbital disturbs and it splits into two set of orbitals having different energy what is this degeneracy of orbital you know that what is degeneracy of orbitals okay degeneracy of orbital are the orbitals which has equal energy right like suppose if I write down the 5 d orbitals like this this means what all the d orbitals are at the same energy level so this is degenerate orbitals degenerate orbitals equal energy orbital equal energy orbitals are degenerate orbitals correct so what happens you see like I said ligands ligands approach the metal correct metal has d orbitals right so these d orbitals the electrons present in these orbitals and the lone pair of ligands obviously when it try to you know approach the metal will have repulsion between the orbitals and the lone pair of electrons correct now the d orbital which is directly towards this ligand right or the lone pair of ligand will have the more repulsion right so it is as simple as that suppose one of suppose the electron lone pair of electron is going like this and one d orbital suppose it is in this direction directly towards the you know lone pair of ligand and other one is different direction like this ok the one which have which is directly you know towards the lone pair of ligand will experience the more repulsion right because this head on to this you know lone pair of electrons the one which is directly towards this lone pair of electron will experience the more repulsion in comparison to the other orbitals and more repulsants repulsions leads to the more energy of the orbital basically what happens because of this repulsion the energy of this orbital increases ok it keeps on increasing as these electrons are going closer to this metal right when it goes when it goes closer to this metal the repulsion increases and with this repulsion the energy of the orbital also increases and that is why one set of orbitals will have different energy another set of orbitals we have different energy one one set of orbitals will have higher energy the other set of orbitals we have will have lower energy right so that is why the orbital which has higher energy will go to the higher energy level and the orbital which has lower energy will come down to the lower energy level and this is leading takes place in such a way that average energy of the orbital will remain same, right? This is the energy of the orbital, one of few of these orbitals goes to the higher energy level, few goes to the lower energy level. This splitting takes place in such a way that average energy will be same, right? So this splitting of these d orbitals because of the lone pair of ligands, we call it as crystal filled splitting, correct? We call it as crystal filled splitting. Did you understand this, tell me? Yes or no? What is the splitting you understood? The orbital which has more repulsion towards the lone pair of ligand will experience more repulsion and hence goes to the higher energy level, right? If this repulsion is same everywhere towards all these orbitals, right? Then the repulsion will be same, repulsion is same, it means they have the same energy, okay? This is what crystal filled splitting. Now the next point, next point, this is it, next point is what? If this ligand is strong, correct? If this ligand is strong, then it provides more repulsion to these orbitals in comparison to the weak ligand. So more repulsion means more splitting, more splitting means more energy difference, correct? This energy difference or this splitting, like I said, we call it as crystal filled splitting and the energy difference between these two set of orbitals, we call it as crystal filled splitting energy, CFSE, right? Strong ligands produce more splitting, weak ligands produce less splitting, correct? More splitting means more crystal filled splitting energy, less splitting means less crystal filled splitting energy, right? Third point, now when you have different different ligands, correct? So with each and every ligand, if you try to find out this crystal filled splitting energy, so according to the energy that we get according to each of these ligands, so we'll have a series of ligands according to their repulsion to the d orbital of metal or we can also say the crystal filled splitting energy. So according to this, the crystal filled splitting energy that we get with respect to each ligand will arrange those ligands in a series and that series, we call it as a spectrochemical series, right? What I said, first of all, this splitting of orbitals, then second point, what I said is strong ligands, more splitting, weak ligands, less splitting, third point what I said, if we arrange those ligands according to their crystal filled splitting energy, we'll get a series from high to low, right? We'll have few ligands which produce more repulsion, most or more energy. We have few ligands which produce less repulsion, less energy. So we'll get a series of this crystal filled splitting energy with respect to the ligand. This series, we call it as the spectrochemical series, correct? Two, three terms you understood? Yes. Depends. We'll discuss that Shreya, why it is 2 plus 3, it depends upon the geometry of the orbital, geometry of the complex. So we'll see that in what geometry, what kind of splitting takes place, okay? That we'll discuss in, just now we'll discuss, okay? So the thing is here, when we say this ligand is strong, this ligand is weak, it is because of the splitting that they provide. According to that only we are saying that this particular ligand is weak, this particular ligand is strong, correct? So that all those ligands that we discussed, that Cl minus is a weak ligand, Cn minus is a strong ligand, CO is a strong ligand, NS3 is a strong ligand, that we can only say with respect to this splitting of orbitals, okay? So coming back to the next point here, like I said, the magnitude of splitting depends upon different, different factors, okay? And with splitting only, we'll understood that whether the ligand is weak or is strong. So first of all, we'll see what all factors we have that provides that affects the magnitude of splitting, okay? So heading you write down, the magnitude of splitting, splitting depends on, depends on following factors, following factors. The first one you write down, position of, position of transition metal, metal in periodic table, position of transition metal in periodic table, write down the magnitude of splitting, okay? I'll write down in this mod, this means magnitude of splitting, mod of splitting means magnitude of splitting. The magnitude of splitting increases as we go, as we go down the group. All these factors you have to memorize, as we go down the group, the magnitude of splitting increases. You see one example I'll give you here, suppose we have two compounds like PtCl4, 2 minus and PdCl4, 2 minus. So first of all, you see the magnitude of splitting increases as we go down the group. So if you see this group, it is present, we'll have platinum first and then we have palladium in the private table, right? So magnitude of splitting increases as we go down the group, oh sorry. So we'll have, okay, just one thing you see, platinum will have plus 2 oxidation state, right? And palladium we have plus 2. So in the private table if you see, the platinum and palladium if you compare, platinum is in the next period of palladium. So Pt2 plus will produce more splitting here in Pd2 plus. Means in these two compounds, because the ligands are same, obviously the ligands must be same. We cannot change the metal and the ligand both at the same time. Either you change metal or you change this ligand, okay? So since Pt is present in the next period of, next period, that's why we are going down the group and hence the platinum will produce here more splitting. Aditya is asking, sir, can we finish the class by 8.30 without break? Without break if you want, I can finish by 8.45. Is it fine? Let me know with all of you, right? So here in this platinum Pt2 plus will have more splitting than this, okay? Now the next factor we have, if all of you are agree with it, without break 8.45, I am fine with it. Next is, write down oxidation state of the metal, write down as oxidation state of central metal atom increases, as oxidation state of the central metal atom increases, splitting increases, splitting increases, okay? For example, you see FeCN6 3 minus, two examples I'll give you, FeCN6 3 minus, FeCN6 4 minus. Obviously, here we have more oxidation state plus 3 plus 2, so splitting will be more over here, okay? The reason behind this is what? More positive charge on this metal, okay? More will be the attraction of the lone pair of electron. So the lone pair of electron is strongly attracted towards the metal if the positive charge increases. And more strongly attracted, more will be the repulsion of the metal and orbital, the lone pair and orbital. And that's why we'll have more splitting, right? Next, write down the third one is the nature of the ligand. Third one, the nature of the ligand. Write down to this, write down the ligand which affect, the ligand which affect or produce, produce right and that would be better. The ligand which produce only a small degree of crystal field splitting, just a second, I'm coming, just a second, right down the ligand which produce only a small degree of crystal field splitting are called weak field ligands. The ligand which produce only a small degree of crystal field splitting are called weak field ligands. And those which produce and those which produce a large degree of splitting, large degree of splitting are called strong field ligands, are called strong field ligands, are called strong, large, sorry, strong field ligands, strong field ligands or strong ligands, same thing, okay? Next, let me write down when these ligands are arranged in order, next line, when these ligands are arranged in order, in order of the magnitude of there, when these ligands are arranged in order of the magnitude of their crystal field splitting, magnitude of their crystal field splitting will get a series, will get a series which is known as, will get a series which is known as spectrochemical series, I'll write down this, will get a series which is known as, which is known as spectrochemical series, spectro, so this is spectrochemical series is actually based on the splitting that the each and every ligands produce, correct? So, we write down this spectrochemical series, this order you write down, I minus minimum, then Br minus, then S minus 2, then SCN minus, CL minus, NO3 minus, N3 minus, F minus, OH minus, C2O4, 2 minus which is almost equals to H2O, NCS minus, CH3, CN, Pyridyl, NH3, ethylene diamine, bipyridyl, then we have phenyl, then we have NO2 minus, triphenylphosphonium phosphine, PPH3, CN minus, which is almost equals to CO, this is, this is what? Spectrochemical series, now all these things you don't have to memorize, okay? This series you don't have to memorize, in general what you have to keep in mind, that oxygen atom donor if you have, right? Oxygen atom donor except this NO2, oxygen atom donor in general it is a weak field ligand except this NO2, okay? So, what you have to keep in mind, in this series, in this series you see H2O, right? H2O is present in bit grain and whatever ligands which is stronger than H2O are strong ligands, strong ligands and like left side of H2O the ligands are weak ligands, including H2O also, H2O also we consider as weak ligands, correct? So, in general what you have to keep in mind that halide donor, halide ion or halide then oxygen if it's the donor atom in the molecule, then nitrogen donor and then carbon donor. Like I said whenever nitrogen and carbon atom are the donor atom these are strong ligands, right? Halide ion is weak, oxygen to some extent it is also weak ligand but if you have to compare the, you know splitting between the two molecules they may give you different ligands into that. So, like this you can compare halide, oxygen, nitrogen, carbon. If the carbon is the donor atom there will have the maximum splitting, okay? That is why it's important but in general whenever carbon and nitrogen atom are the donor atom ligands are strong, fine? Yes, tell me now we are going to discuss how is splitting takes place in a given complex according to the geometry of the complex. So, write down the heading next, splitting of d orbital of octahedral complex. Octahedral is again the most important one, octahedral complex. Now you tell me one thing I wanted to, you know ask you here. Let me, you know, draw the geometry first of octahedral complex. Octahedral complex is this, we'll have metal here at the center, ligand at every corner of the square along with one ligand, this side one ligand, this side. This is the metal. Now can you tell me we have three or, you know, we have five orbitals here like we have five orbitals, dx, y, dy, z, dzx, dx square, y square, dz square. Okay, fine. No issue, you can resume the video wherever you have stopped it, right? From there only, fine. So, can you tell me what are the orbitals we have here which experience most repulsion? Out of the five d orbitals, the orbitals which experience most repulsion here, can you tell me the orbitals? See the point is here, the orbitals which is, you know, which is along the, you know, direction of the ligand, like from which direction the ligand is coming towards the metal. If the orbital falls in that exactly the same path, then that will experience the more repulsion. Suppose, if like I already discussed, if the ligand is approaching the metal like this, if one orbital is like this, correct, then this will experience more repulsion than the other orbital which is in this direction or in another direction, correct? So, ligand if it is head on to one of the orbital, that will experience maximum repulsion. Now one thing here, can you tell me along this line, right? Or this one, because ligand is trying to approach like this, then only octahedral complex forms along this direction only the ligands are approaching, then only tetrahedral complex forms, sorry octahedral complex forms. So, what are the orbitals we have along this direction, this direction and this direction, can you tell me? Tell me, what are the orbitals we have? See, this is what the two diagonal we have this and this is the, this line, okay? This is what, we can assume this as x axis, this as y axis and this is z axis, z axis because the mochil angle is what, 90 degree. So, obviously we know along x and y axis, we have dx square y square orbital present, dx square y square. This is dx square y square orbital and this is what, dz square orbital we have, right? So, out of the five orbital, we can say that dx square y square and dz square orbital experience more repulsion than the other three. Other three orbitals are what, dx y, dy z and d zx, right? That's why these two orbitals, the energy increases in comparison to these two orbitals and we'll have a set of these two orbital and we'll have a set of these three orbitals, two different sets we get, okay? So, these two orbitals, we call it as eg orbitals. It is again an spectroscopic term, eg is an spectroscopic term and this we call it as t2g orbital, right? So, what is this eg and t2g? These are spectroscopic terms. We just use this, there's no, you know, there's nothing in this we need to understand. I'll explain it. Eg means these two orbitals, t2g means these three orbitals, that's what the thing is, okay? This is spectroscopic term, this is spectroscopic term. See, what I said, since the ligand is approaching in this direction, right? Which is, in other words, can I say it is approaching along the axis, x, y and z axis? Can you tell me that? Can I say this that in octahedral complex, ligands approaches the metal along the axis, x, y and z axis? Can I say this? Yes or no? Tell me fast. So, all those orbitals, d orbitals, which are present along the axis or the orbitals which are axial orbital will definitely have the more repulsion because the ligands are head on towards those orbitals. Since the ligands are approaching towards the metal along the x, y and z axis, so obviously axial orbitals are those orbitals which are present along the axis, right? So, axial orbitals out of the 5d orbitals, we have two axial orbitals, dx square, y square and dz square because these two along the axis, this is along x, y and this is along z axis, but the other three orbitals are non-axial orbital. This you already know because these three orbitals, dx, y, dy, z and dz, x present in between the axis. This is present in between x, y, in between y, z in between z, x. So, obviously since the ligands are approaching the metal along the direction of the, along the axis itself, that is why the orbital which is along the axis or axial orbital will definitely experience more repulsion than the non-axial orbital. Now, you understood this, clear? All of you, tell me. All of you understood? So, what happens here? These two set of orbitals which are eg orbitals and these are T2g orbitals. So, eg orbital will experience more repulsion and goes to the higher energy level, right down this point. Eg orbitals, eg orbitals experience more repulsion. Eg orbitals experience more repulsion. In bracket, you write down since these are axial orbitals. Eg orbital experience more repulsion since these are axial orbital, bracket close and goes to the higher energy level and goes to the higher energy level. Next line, T2g orbitals in bracket, you write down non-axial orbital. T2g orbitals experience lesser repulsion, lesser repulsion and hence experience lesser repulsion and hence goes to the lower energy state and goes to the lower energy state. One note you write down here is splitting of orbitals, one note you write down takes place in such a way, splitting of orbitals takes place in such a way that the average energy that the average energy of the orbitals remains same. Average energy of the orbitals remains same. Average energy won't change. One set of orbital will go to the higher energy level. Another set of orbital will go to the lower energy level. Average energy won't change. So orbital, like energy diagram if I draw here, next you write down the heading, energy diagram. See if you understood this, next we have to do with tetrahedral complexion is square planar also. So that is also exactly same. We just need to see what all orbitals are directly towards the ligand. That's what we need to understand. So energy diagram, I'll draw it first and then I'll explain. Okay, you also draw it. X axis is decreasing distance, decreasing distance between metal and ligand and this is the energy profile, energy axis. What happens here you see, this is the energy of orbitals. These are d orbitals, all these are d orbitals. So energy in free ion, when there is no attraction towards ligands, okay, when it is free, when you see when you're going from left to right, the distance between the metal and ligand is decreasing, correct? Means the ligands are coming closer to the metal. So when the ligands are coming closer to the metal, the repulsion will keep on increasing, right? Repulsion will keep on increasing. You see when repulsion increases, the energy also increases and will go to the average energy state, average energy. This is the average energy we have. Overall energy increases and this will be the average energy. After this further ligands goes closer to the metal, then one of one set of orbitals experience more repulsion and another set experience lesser repulsion, right? According to whether it is axial or non-axial, okay? And then splitting takes place, right? One set of orbital goes to the higher energy level, another goes to the lower energy level, right? This orbital here is EG orbital because EG orbitals are axial, more repulsion, more energy. This set of orbitals are T 2G orbital, lower energy. This energy difference here, this energy difference here is represented by this energy difference. Here it is represented by del O. It is not zero, it is del O. Del O is the energy difference. This O stands for octahedron, okay? Experimentally, this energy is found to be 0.6 del O and this is found to be minus 0.4 del O, 0.6, 0.4 del O, okay? This is the energy diagram of an octahedral complex. Is it clear? The important point here is that you should know EG will have the higher energy state and T 2G will have the lower energy state, okay? Now you see this kind of distribution is taking place, right? This kind of distribution, sorry, splitting is taking place, right? After this, what happens? If you try to distribute the electrons, correct? Now it is very logical. You try to understand here carefully. Since the orbital has different, different energy, what we had earlier, all the five orbitals has same energy, correct? Like this, then this distribution of electron follows Hernst's rule 1, 2, 3, 4, 5 and then pairing. And why this kind of distribution we had? Because all orbitals have the same energy, correct? You must have heard about N plus L rule. What it says? The orbital which has lower energy will have the electron first. The electron goes into the lower energy orbital first, right? So the thing is here, if the orbitals have same energy, then the distribution of electron follows Hernst's rule. If the energy difference will be there, then the first electron goes into lower energy orbital and then into higher energy orbital, correct? So in this, suppose if you have to distribute electron in these five orbitals, correct? Then the electron starts filling into this T2G orbital first and then it goes into EG orbital. Is it clear? Is it clear? Tell me. First electron goes into T2G and then electron goes into EG because T2G is the lower energy orbital, EG is the higher energy orbital, right? So distribution of electrons here, just that's what we are trying to understand now, how distribution of electrons takes place. And according to the number of electron present into these orbitals, we can also find out crystal-filled, splitting energy. For that, we have a formula that we'll see, Babme. We'll see that later, correct? So first of all, you see how distribution of electrons takes place in this complex. So write down the heading, distribution of electrons, right? Since this distribution of electron depends upon, right down, it depends upon the nature of the ligand, depends upon the nature of the ligand. See, what happens before going into this? I'll tell you one thing. If the ligand is strong, then what happens? This splitting will be more, correct? If the ligand is strong, splitting will be more and in that case, the distribution of electron does not follow Hans' rule because the energy difference will be more. So first we'll have 1, 2, 3, 4, 5, 6 and then here, 7, 8, 9, 10. But if the ligand is weak, the energy difference is less over here and hence the distribution of electron overall follows Hans' rule. That's what we did in the last VBT theory. What we did in VBT, that when the ligand is is strong, right? The distribution of electron, the arrangement of electron does not follow Hans' rule. That's what we did. Pairing of electron takes place against the Hans' rule. Why? Because when the ligand is strong, the energy difference will be high and hence first here pairings takes place and then here pairing takes place. That's what it happens. But this kind of thing is not possible when ligand is weak because in that case, the energy difference is not that high, okay? The two cases we have here in terms of ligand, if we say, in case of weak ligand, right? On first point here depends upon the nature of ligand, in case of weak ligand. Right now into this, in this case, the energy difference, in this case, the energy difference between T2G and EG orbital, in this case, the energy difference between T2G and EG orbital is not high. T2G and EG orbital is not high and hence, and hence the distribution of electron and hence the distribution of electron is done according to Hans' rule. It's done according to Hans' rule. Distribution of electron is done according to Hans' rule. Okay? So you see what happens? First we have T2G, then electron goes into EG, then again the electron goes into T2G, then again the electron goes into EG. Why electron goes into T2G first? Why not EG? Can you tell me? Why the electron goes into T2G first? Why not EG? What is this more attraction? Why I don't know? You see what happens? When in, see why electron goes into 1S first and then 2S and then 2P and then 3S? Why this order we have? Why electron the configuration is this? Can you tell me? We draw the configuration, no? 1S2, 2S2, 2Pc, 3S2, 3Pc like that. So why electron is going into 1S first? Why not it goes into 2S, then 2P and then 3S like that? You don't know this? Yes or no, tell me otherwise I'll explain this. Energy levels are different, different energy orbital. See what happens here, let's take few minutes here. See we have different, different orbitals. 1S we have, we have 2S, 2P, 3S, 3P, 3D all these orbitals we have. Poly-exclusion principle, correct, N plus L rule. So what happens here? We'll calculate this N plus L value of these orbital. Right? As N plus L value increases, as N plus L value increases, energy of the orbital increases. Right? Energy increases. Electrons always goes into lower energy orbital first, then next higher, then next higher and then next higher. So N plus L value, N plus L value of 1S is what? N value is 1, S value is 0 for, L value is 0 for S. So for this, N plus L value is 1. 1 plus 0 is equals to 1. For this, N plus value is 2 plus 0 is equals to 2. For this, N plus L value is 2 plus 1 is equals to 3. Right? N value is the principal quantum number like 1, 2, 2, 3, 3, 3. All these are N. L value for S, P, D, F orbital are like this, 0, 1, 2, 3. This you should know. For S sub shell, L value is 0. For S sub shell, L value is 0. N value is 2. N value is 1. For 2P, N value is 2. L value is 1. So 3. You see 1, 2, and 3 we are getting. Lower value of N plus L, lower will be the energy of that orbital. That's why the electron goes into this first, then here, then here, then here and so on. Understood this? The point I am trying to make is what? If the orbital has lower energy, then electron goes into that orbital first, then next higher, and then next higher like that. Is it clear now? Yes, correct. Now if you go back to the topic that we were discussing, in this we have to distribute the electron. The T2G orbital is at the lower energy state. EG is the higher energy state. So obviously T2G will, the electron goes into T2G first and then EG first. So in case of weak ligand, the energy difference is not that large. The first electron goes into this, then this, then pairing takes place. So if you have T2G orbital, so 1, 2, 3, first 3 electron goes into T2G orbitals. 1, 2, and 3. First 3 electrons goes over here, 1, 2, 3. And since this difference is not that large, so pairing won't take place against the Hund's rule because we are assuming weak ligand. So 1, 2, 3, then 4th, 5th electron goes into EG, then 6th, 7th, 8th goes into T2G, 9th goes into EG, 10th goes into EG. Is it clear now? Yes or no? Thank you, Hamsini, you are back. What will you do? The portion that I did already. Now the next one you see. Second case, if the ligand is strong, in case of strong ligand, down, in case of strong ligand, down, if the ligand is strong, if the ligand is strong, the energy difference, if the ligand is strong, the energy difference between T2G and EG orbital, the energy difference between T2G and EG orbital is high and it does not allow and it does not allow right, right, Hamsini, and it does not allow the electrons to follow Hund's rule and it does not allow electron to follow Hund's rule and it does not allow electron to follow Hund's rule. Now you tell me, first electron goes into which orbital? T2G or EG? This, Hamsini, this emoji is not the Baywala emoji, it's slapping. You know that? Right, Paras and Likith regarding this Emojiwala comment. Yeah, that's what, that's valid logic. Aditya, he's doing wave. Anyways, coming, coming back to the point, coming back to the point. Okay, so in which orbital the electron goes first? Tell me, Shreya, Samyukta, why EG? Why EG are the rightest? See, electron always goes into lower energy orbital first. Okay, you see first of all here, when the octahedral complex we have, so whether the ligand is weak or strong, the lower energy orbital is this only T2G. So it's the other thing that if the ligand is strong, the energy difference will be high. But in that case also, the lower energy is T2G orbital. Right, so in this case what happens, like whatever the ligand is, whether it is weak or strong, first electron always goes into T2G orbital. A difference comes when if the ligand is weak, then it follows Hund's rule. But if the ligand is strong, it does not follow Hund's rule. What happens in this, you see, first electron goes into T2G and then it goes into EG orbital. This is what the possibility we have. So first electron 1, 2, 3 goes into EG. Fourth, fifth, sixth again goes into EG because it does not follow Hund's rule. Then seventh, eight, ninth and tenth. This is the distribution of electrons. Is it clear? Got it? Okay. Yes, all of you understood? Just a second. Okay. One note you write down here. In case of weak ligand, write down, in case of weak ligand, in case of weak ligand, just a second, just a second, somebody is there. In case of weak ligand, the delta O, weak ligand, this delta O is less than P. What is this P? I'll tell you. First you write down. When this ligand is strong, this delta O is greater than P. This P you write down, it is, this P you write down, it is the average pairing energy. It is the average pairing energy. It is the average pairing energy, which is the energy required to pair, it is the energy required to pair the electrons in the same orbital. It is the energy required to pair the electrons in the same orbital. Okay. average energy required. Yeah, just we are talking about only for octahedral now. Okay, but these formulas are same like this P thing will be same, but this, you know, the splitting of orbitals is only for octahedral for tetrahedral, it will be different that we'll discuss for square plan it will be different. But this formula or this P definition that I'm giving you, it is same for all all kind of complex whether it is octahedral tetrahedral or square plan it correct. So, you see, again, there is a like this this question that they have asked sometimes in the exam, like in neat exam, they have asked this question or any other bits or other exam, they have asked this question, what is the relation of Delta O and P for week or, you know, strongly again. Okay, so this is again important. The formula of CFSE write down crystal filled splitting energy, crystal filled splitting energy. In short, we write it as CFSE. Okay, the formula of CFSE write down crystal filled splitting energy is equals to minus 0.4 x plus 0.6 y into Delta O. Okay, where x you write down, where x is the number of electrons present in T2g orbital, x is the number of electrons in T2g orbital. And why is the number of electrons in Eg orbital? Why is the number of electrons in Eg orbital? See, P is nothing but see what is P, we have an orbital here, suppose this orbital you have, and it has one electron into it. So, when you try to pair up this electron, so you have to add one more electron into it. Okay, so for that, he required some energy. That energy is pairing energy. It is the average pairing energy, which is the energy required to pair two electrons in the same orbital, in the same orbital, if you want to pair up the electron. So for that, we required some energy that energy is pairing energy. You don't have any numericals into it. They won't ask you this numericals. Yes, yes. In splitting, like if you have orbitals like this, in that also you can say this orbital and this orbital. Here the pairing energy is this. Here the pairing energy is this. That is what we're talking about. Done this one. You see this one table I'll draw here. This side we have metal ion configuration. Here I'll write down metal ion configuration. Then we have, suppose if the ligand is strong, strong ligand, then CFSE, CFSE in terms of delta O, then we have weak ligand, then we have CFSE delta O. So you see the configuration of metal can be D1. It can be D2. It can be D3, D4, D5, D6, D7, D8, D9 and D10. This is the configuration. Draw the table first. Now you see what you need to do here. If the ligand is strong, we have two set of orbital, T2G and EG. So for D1, the configuration is what? T2G has one electron and EG will have zero electron. For D2 configuration, T2G will have two electrons and EG will have zero electrons. For D3, T2G will have three electrons and EG will have again zero electron. But when the configuration is D4, T2G will have what? Three. If the ligand is strong, then what happens? The pairing will not follow according to Hohenstuhl, right? So 1, 2, 3 will have still four. EG will be zero. T2G, five. EG will be zero. T2G, six. EG will be zero. After this, all these EG will have one, two, three and four. And the number of electron present in T2G orbital will be six. T2G will be six. T2G will be six. T2G will be six. This is the distribution of electron in case of strong ligand. Did you understand this distribution of electron? Now you know the number of electrons present in T2G and EG, you know, orbital according to the configuration of the metal. So we can find out CFSE, the formula we have already. Use that formula, find out this CFSE. Similarly, this we can do for weak ligand, right? But this I will let it be for you to do at home, right? This is the homework for you. I will just write down the configuration here, right? So you see here T2G will be one again. EG will be zero. T2G, two. EG will be zero. Again, T2G, three. EG will be zero. But here what happens next? T2G, three. EG, one. Like this, you feel this? Tell me the CFSE for this one. Just you write down the distribution of electron here. Calculate this. This part you do at home. Understood? Tell me the value of CFSE. For weak ligands, you let it be. This you do it at home. Just you tell me this value and then distribution of electrons here. That is it. What is the value of this one? For D1 configuration, you see the formula is what? Minus 0.4 into x. x value is what? One plus 0.6 into zero. Delta O I have written here. So no need to write here again. That will be what? Minus 0.4. This is what we get. So first we'll get minus 0.4 and then we'll get minus 0.8. Then we'll get minus 1.2 and then it will be minus 1.6 minus 2.0 minus 2.4 minus 1.8 minus 1.2 minus 0.6 and this will be zero. Here the distribution will be what? T2G3, EG2 according to one-stool, T2G4, EG2, T2G5, EG2, T2G6, EG3, sorry EG2, then EG3 and then EG4, T2G6 and T2G6. This is what we have. It's okay. I'm sure. So this you do it at home. This you calculate and you do it right. Next one we see right on crystal field splitting in tetrahedral complex. Crystal field splitting in tetrahedral complex. Tetrahedral complex. Okay. Yeah, okay. No issue. You are going for family party, family function liquid. Say tetrahedral complex is this. All these are ligands 1, 2, 3 and sorry fourth one is this. Correct? So you see, so this you see since along this line the bond angle is 109 degree 28 minute. So we cannot assume any axis along this line. Axis, you know, can be like this. Suppose this is x, this one is y and this one is suppose z, x, y and z. Okay. This is again. So point is what the ligand is approaching here the metal in between the axis. It is not along the axis like in octahedral complex what we have discussed that ligand is approaching the metal along the axis right. So here in this case what happens is the metal and ligand are you know the bond of metal and ligands in between the axis. So obviously the non axial orbital will have the more repulsion right. Non axial orbital orbital will have the more repulsion. Correct? So in this case what happens t2g orbital will have the higher energy than e g orbital just reverse of the first case understood this. So it is exactly reverse to that of octahedral complex. Okay. So write down few points into this then we will see the energy diagram. In this in tetrahedral geometry in tetrahedral geometry the ligands are approaching the metal. The ligands are approaching the metal in between the axis in between the axis and hence and hence the non axial orbital and hence the non axial orbital will have more repulsion and hence the non axial orbital will have the more repulsion right and it goes to it goes to higher energy state. It goes to higher energy state. Okay. Can you draw the energy diagram of this? Just you show you can can you draw the t2g and e g how it goes. You see this the axis is same the axis is same here y axis is the energy and x axis is the distance decreases between the metal and ligand right. So we have five orbitals like this 1 2 3 4 and 5 when the ligands approaches the metal the average energy increases like this and then it splits into two parts where the t2g orbital goes to the higher energy state and e g orbital comes to the lower energy state. This is the average energy right and this distance we call it as 0.4 delta t and this one is 0.6 minus 0.6 delta t delta t this t stands for tetrahedral draw this got it exactly same it is the diagram and all again you see here the distribution of electron will be first goes goes into e g orbital and then it goes into t2g orbital. So everything will be exactly reverse of octahedral complex. One experimental relation we have this is also they have asked in the exam delta t is equals to 4 by 9 of delta o octahedral complex right. So generally splitting in octahedral complex is more than the splitting in tetrahedral complex this you must remember right. So this this relation you must remember delta t and delta o relation. What is dsp? You know it is not dsp2 hybridization it is tetrahedral we are assuming. So it is sp3 square planner will discuss after this okay understood dsp2 is square planner so we will discuss that later okay not now got it one question we will see because the next we have to discuss square planner it will take some time okay so that we will not discuss today okay and in this again the distribution of electron I am not telling you because I have done it for this octahedral complex similar way we can do it they won't ask you to calculate this cfsc and all correct. So one question you write down this one Fe NO Fe NO H2O 5 2 plus last one for today H2O 5 2 plus you tell me the hybridization of it okay geometry and unpaired electron then yeah formula of cfsc is same for all the complex. See one thing you see this NO if you remember I have given you this NO as a neutral ligand if you remember NO I have given you as a neutral ligand okay but this NO may behave as NO plus also or NO minus also in some cases okay so in this molecule this NO is behaving as NO plus this is this is very important okay this NO ligand is very important and this particular example you must keep in mind okay so NO is a positive ligand over here right so one point in this you write down NO is a positive ligand first of all you can do this it's hybridization will be sp3d2 I'll write down the answer here hybridization will be sp3d2 okay only one electron get paired into it okay you can try this at home I have I don't want to do this again here I'll just give you the answer here sp3d2 3 unpaired electron we have here and it is paramagnetic paramagnetic what is important here that you need to write down for this particular complex you write down this note here it is sp3d2 only one electron get paired all the electrons won't get paired here okay write down the NO into this the complex the given complex is form in the given complex is form in brown ring test form in form in brown ring test okay next line for the confirmation of nitrate ion for the confirmation of okay forms in brown ring test for the confirmation of nitrate ion so this is what you need to keep in mind this complex forms and it confirms the presence of nitrate ion okay that is why this is important and one more important thing if you do not know whether it is positive ion positive positive ligands here or negative ligands you cannot find out the oxidation state and then whole thing will be wrong okay so this is one thing that you need to you know memorize this NO in this complex is behaving as a positive ligand okay you try this at home sp3d2 3 paramagnetic behavior okay only one electron get paired and then we'll discuss this in the next class okay if you could not do this yes tell me so we'll wind up the class here okay next we'll start the square planner and then isomerism we'll discuss into this so next class we'll finish this chapter okay next class mostly will be physical offline class tetrahedral energy diagrams here you see the it's exactly opposite of octahedral complex why because the ligands are approaching the metal approaching the metal in between the axis they are not you know along the axis so obviously the orbital which is non axial orbital will have more repulsion you see here ligands are in between x y y z and z x axis okay so obviously the non you know axial orbital will have more repulsion because the ligands are more closer to those orbitals right more close to those orbitals so that's why this thing we have that all these t2g orbital which is non axial okay all these t2g orbital which are non axial orbital will definitely have more repulsion than easy orbital it's just we need to see the approach of ligand to the metal whether it is close to which orbit the orbit which is sorry the orbital which is closer to the ligand will definitely have the more repulsion so what we can compare here ligands are in between the axis so non axial orbital will have more repulsion that's that's what it is let's share understood Aditya understood the orbital is this t2g is what t2g is dx y dyz and dzx orbital eg is what eg is dx square y square and dz square orbital this is what the thing we have what is sp3 hybrid i'm not getting you yeah so why would electrons even more see when the ligands are you know see the orbital will have some energy the d orbital orbital will have some energy right so when the ligands are approaching the orbital so out of the five orbital there are few two or three orbitals which will have more interaction with the ligand so obviously they will disturb that interaction will disturb the degeneracy of orbital that's why the splitting takes place understood yeah the need of the d orbital is what because the approach of ligands towards the orbital is different the ligands are approaching the orbital in different manner right they are they are not approaching the ligands the all the five orbitals is the same manner dx y dyz dzx orbitals are you know more closer to ligands so that is why they experience more repulsion and that is why it's happening got it anyways anything if you have any doubt we can discuss this in the next class also okay understood see in the next class okay if any if you have any doubt we'll discuss that and then we'll move on into this okay another topic okay guys thank you bye bye okay thank you