 get your favorite dishes delivered. Okay, hello everyone, good morning. Please type in your name. Okay, so last class we were discussing about GOC and that to resonance we have discussed how to compare the stability of resonating structure. Okay, few applications of this resonance we'll see. Okay, and there we have the calculation of bond order and then aromaticity. Actually there are three, four application of resonance we have. So just you write down your application of the resonance. So first application we have the comparison in comparison and calculation of bond order, calculation of bond order. Second one we have aromaticity, little bit aromaticity I have discussed in the last class and I told you we'll discuss in today's next class. Okay, today we'll see some examples of aromatic compounds and anti-aromatic compounds. Those examples you have to keep in mind. Okay, then we have, we already know this plus R plus or minus R or we also call it as plus M and minus M group. We have mesomeric effect that comes under resonance only. Stability of intermediates we have discussed stability of carbocation with respect to I effect. Okay, stability of intermediates. So here also it is the same thing we have. Like plus I is electron releasing group, stabilize carbocation and de-stabilize carbonyl. So here also we have plus M group, electron releasing group, stabilize carbocation de-stabilize carbonyl. Okay, like that we have here also. Okay, and then last one we have heat of combustion and heat of hydrogenation that is, let it be that is not required here. So first thing that we are going to discuss today is the bond order. Okay, so how to calculate bond order in the resonating structures, right? So first point to write down, we'll just see this bond order thing and this aromatic, aromaticity, these two we'll discuss, okay? And then some of you have asked me to discuss that bond, what we say, molecular orbital theory, VBT and all, some concepts of chemical bonding. Okay, that also we'll discuss today itself. So these two then concepts of, little bit of concepts of chemical bonding we'll discuss. Okay, and then if we get time, we'll discuss tautomers also. Tautomers and percentess in all content. Okay, we'll just do the revision of that because that is also an inform, that is nothing but information. If you know all those cases, you can do the question. Okay, if the question comes directly from there. But if you do not know, you cannot do that question. Okay, it is, there's, okay. So that's the point. We'll discuss percentess in all content. Tautomers will discuss little bit, what all cases we have, condition and different cases. And then these two plus, some concepts of chemical bonding. That's why this class is for four hours. Okay, so we'll try to finish all this. Okay, so first of all, you write down the first thing we have bond order. See, bond order, we have discussed this in chemical bonding, you must have done this. And we calculate bond order of various molecule or ions. Right, the most common we have in this one. O2, 2 minus O2 minus O2 and O2 plus. This question has asked many times in the exam. Okay, the bond order of this. Anyway, so here we are not going to discuss this. The point is what? We calculate this bond order according to what? MOT, that is molecular orbital theory. Right, and the formula of bond order, we have discussed here is what? Number of electrons in bonding molecular orbital minus number of electrons in anti-bonding molecular orbital divided by two, right? This is the formula we have. Correct, so this formula we'll discuss when we do the MOT, right? Today only we'll discuss chemical bonding cause we'll do this here. But here since we are talking about application of resonance, so the molecule in which resonance is applicable, resonance is there. How do we calculate the bond order of that? Like, suppose if I give you formic acid or formate ion and if I ask you to calculate the bond order of CO bond in a formate ion, then what is the answer? That is what we are going to discuss today. So first of all, you see the bond order is what? It is the effective number of bond between the two atoms present in the molecule, okay? The definition of this, you see, it is the effective number of bond present between the atoms molecule, okay? Effective number of bonds present between the two atoms in a molecule. Why we are calling it as effective? Since we're talking about resonance, suppose if I draw here benzene, right? Benzene is this and the resonating structure of benzene in the other structure we can draw is this one, right? If it is one and two, this is one and two. So what will be the bond order of this, the C1 and C2 carbon in benzene molecule? You cannot say one or two because why this thing is there? Because resonance is possible into this. That's why we are calling it as effective number of bonds, okay, since resonance is possible here, right? And usually when the molecules in any molecules, when resonance is not possible, we usually say it is a number of bonds present between the two atoms in a molecule, right? We also have this information that bond order is directly proportional to what? Bond energy. Bond energy or bond strength also we call it as, right? More will be the bond order, more will be the bond strength and inversely proportional to bond length, right? So in resonating molecules, if you know the bond order, you can also compare the bond energy of that particular bond, right? Bond strength of that particular bond. Now, how many of you or you must have this information that since we're talking about effective number of bonds or bonds between the two atoms, but bond order of any molecule can be fractional also. This information you have already, right? That the value of bond order can be fractional also. Integer, fine, but it may have some fractional value also. And this we have discussed again in molecular orbital theory, we have seen this, that bond order can be any fractional value also, right? There are two values, different types of value of this bond order possible. The first one is integer, right? And this is very easily, like we know this, you can understand it easily because the number of bonds, so it can be an integer, right? But it can be fractional also, right? And that we have seen in MOB. See, this fractional value of bond order is possible when resonance is possible, in case of resonance, right? So in case of resonance only, the fractional value of bond order will possible, otherwise it won't be possible, right? Integer value will be possible when there is no resonance. So we have discussed about this bond order and we know the fractional value is also possible, but there we haven't discussed about what, why this fractional value is possible. So the fractional value is possible because of resonance, because of delocalization of electrons, pi electrons, right? Now, you see how to calculate this bond order. Now the main thing is the calculation of bond order, okay? So, see the calculation of bond order is possible when we have equal contributors, okay? For unequal contributors, since the stability will be different, okay? So bond order of the molecule of the two atoms, two bonds between the, or two atoms present in the molecule, the bond order calculation is difficult over there when we have unequal contributors, right? So that we'll discuss later on. The first case we are taking here that calculation of bond order when we have equal contributing structures, equal contributing structures. Or we can also write it as equal contributors, okay? So here you see the first example. I'll try to make you understand with one example here. Suppose the example I am taking, that is formate ion. Formate ion is H C double bond O, O minus. And the other structure of this will be H C single bond O minus double bond O, right? So now, here you see if I ask you, suppose this is first atom and this is second atom. And we have to calculate the bond order, bond order of we can say carbon oxygen bond. See in this molecule, again, like I took the example of benzene ring, similarly you cannot say the bond order of CO bond is one or two here. Understood? Why you cannot say? Because you see in one, two, this carbon oxygen bond if you are considering, here we have double bond, here we have single bond. So you cannot say one or two, right? Why this thing is possible because of resonance, right? So in this case, the formula of bond order, the calculation of bond order, we do by this formula and this formula you write down. Bond order is equals to the sum of summation of total bond, sum of the sum of summation of total bond, total bonds in all resonating structures divided by summation of number of RS, means total resonating structures, simply you can write down, total number of resonating structure. Have you seen this formula? Yes or no? Yes, have you seen this formula? You haven't done this bond order calculation in resonating structures, okay anyways. So you see basically if you know the bond order, you know the stability also, that's why we are doing this. So you see here, the formula is the total number of bonds in all RS and what is this bond? Like suppose we are calculating the bond order of CO bond, right? So we'll take the total number of bonds between carbon and oxygen in all RS. So you can take one, two and three, right? Or you can also take, suppose you're considering these two carbon and oxygen, so two bond here and one bond here, right? So what we can write, two here present and one here we have. So two plus one, total number of bonds between the carbon and oxygen, one and two, is two in this structure, one in this structure divided by the total number of RS. Total number of RS is what, one and one, two. So three by two, that is 1.5. The bond order is 1.5, order is 1.5. So this is the bond order of carbon and oxygen bond, okay? Another example you see, second one. Just I'm leaving some space here. Okay, we'll use this space later on. I'll just give you one more trick here. So just you see the second example. Suppose I'll give you this carbonate ion, okay? That is C double bond O, O minus, O minus. And you can easily draw the three different structures of this. These are the three different structures we have. Now, suppose we have to find out the bond order of this carbon and oxygen bond, one and two, one and two, one and two. So bond order, according to this formula will be what? See, Rithvik, we have to calculate the bond order of carbon and oxygen, right? So any of these bond you can take, either this carbon oxygen or this carbon oxygen. Any one of these we can take. So I'm taking this one, one and two, one and two. So number of carbon oxygen bond present in this structure is two. And number of carbon oxygen bond present in this structure is one. So two plus one divided by the total number of resonating structure, two. So two plus one by two, 1.5, right? Similarly, here you see bond order will be what? Two, three and four, four divided by three. So four by three is the bond order, okay? Another one you see? Double bond, positive. You draw the resonating structure of this, okay? And tell me the bond order of this bond, one and two. Any one, carbon-carbon bond. Bond order of carbon-carbon bond. And this structure possible here. So bond order will be what? Two, three and four. So again, it is two plus one plus one divided by three resonating structure, three. So four by three, right? How it is zero, Vaishnavi. See, this is the conjugated system. Sigma pi and a vacant P orbital. Positive charge means what? We have vacant P orbital. When this bond comes over here, right? Here we'll have the positive charge, okay? So four by three is the, what we say, bond order here. This example, fourth one, S double bond O, O minus double bond O. Tell me the bond order of sulfur oxygen bond. What is the answer? Five by three, five by three is the answer. We'll get five, we'll get total bond really what? Two plus two plus one. So bond order will be two plus two plus one divided by three resonating structure possible. And this minus charge will be on this oxygen, this oxygen, this oxygen. Three different resonating structure. So that will be five by three. Now for this one, what is the answer? Bond order, you have to calculate for carbon and carbon. So this one, this one you don't do because again, you can draw the resonating structure, calculate the number of resonating structure and the number of bond between the carbon atoms, any two carbon atoms you can take. Why any two? Because we are considering all example you see that we have discussed so far and this all we are discussing for equal contributors. Okay, this formula, this formula is applicable only for equal contributors. Okay, if the equal contributors are not there then we do not calculate the, then we cannot calculate the exact bond order. Okay, we'll take then some approximation there, but this formula is applicable only for equal contributors. You see all these examples, the molecule has equal stability and equal contributors. These all are equal contributing structure. Okay, so here you see the, in this molecule, again you have to draw the resonating structure, calculate the number of RS, take any two carbon atom and count the number of bonds present between the carbon atom that you can do already. But I won't ask you to do this because this is again time taking. I'll give you one more formula here, one more trick that we use to solve this kind of question. And the trick is what, you see bond order we calculate in case of equal contributing structures. Okay, the another formula of bond order and it is just a trick we have, it is not the formula, it is the number of sigma plus pi bonds present in the molecule divided by the number of sigma bonds. Sigma plus pi by sigma you have to calculate. Right, and one important thing that you have to keep in mind here that the sigma and pi bonds, we only calculate for the atoms which is involved in resonance, right? Like you see this example, this carbon hydrogen bond is not involved in resonance. This sigma bond we don't count, right? We always count the number of sigma and pi bonds for those atoms which are involved in resonance. Is it clear? So you write down this condition properly here, number of sigma and pi bonds is, we calculate only for those atoms which are involved in resonance. Okay, wait a second, I'll explain, just give me a minute. How will you know if any compound in an equal contributor? Okay. Okay, so some of you have doubt in equal contributors. Actually you see, I'll just come back onto this formula. Okay, just I'll discuss this equal contributing thing once again. Okay, see, suppose if I take the example of this, okay, I'll take this example, see. Here we have negative charge, double bond and double bond. Negative charge alone there, I can say. Now, when you draw the resonating structure of this, there are five different resonating structure possible. You see first of all, okay. Now in this, you see the resonating structure you'll draw like this, this lone pair comes over here and this pi electron goes here, right? So here we have the double bond, here we have negative charge and here we have the double bond. Another one is what this negative charge comes over here and this pi electron goes here, right? So here we have double bond as it is, double bond will have here then and here we have the negative charge. Again, this negative charge goes here, this pi electron goes here. So here we have negative charge, double bond, double bond. This comes over here and this goes here. So double bond, double bond and negative charge. Is any other structure possible in this? What have any other structure possible? If this bond comes over, this pi electron comes over here, this will come here and then we'll get the first structure only. So we're talking about this example, right? Now in all these structure you see, one by one, the negative charge is present on all other, all carbon atoms. Suppose this is the first carbon, suppose this is one, two, three, four and five. So here the negative charge on first carbon, here the negative charge on third carbon, negative charge on fifth carbon, negative charge on second carbon and then the negative charge on fourth carbon. And we know all these resonating structure will contributes into the hybrid structure, real hybrid, right? They contributes their property to real hybrid, right? Now the point is when this negative charge is present on all the atoms present in the ring, it means that this negative charge is distributed uniformly or equally over the five carbon atoms, right? So when equal distribution is there, then the resonating structure is said to be what, equal contributors. So you can conclude two, three things into this. How do we understand equal contributors are there? The charge, whatever it is, positive or negative, this charge must be delocalized uniformly over the entire molecule, right? Or equally over the each atom present into the molecule, right? Then only it is equal contributors. Now when you draw the resonating hybrid of this, so since the negative charge is present over the, on all the carbon atoms, so it is equally delocalized. So we'll write down this ring here. Means the pi bond is delocalized equally. And if you write down the charge on each carbon atom, so one negative charge is distributed among five carbon atoms. So each carbon atom will have charge minus one by five, minus one by five, minus one by five, minus one by five, and minus one by five, right? So this is what the real hybrid of this structure, actual structure is this. All these are resonating structure. So when the charge distribution is equally over the each atom present in the ring, that the molecule is said to be what? Then the resonating is said to be equal contributors. Is it clear? So whenever you have to find out whether the equal contributors are not, just you draw the resonating structure and see whether all these charges are present on each of the carbon atom or not. Now, if you see all those examples that we have done, there the charge is present on each of the molecule, right? Here also you see this is present in oxygen, oxygen and oxygen. This is the alkyl group, we don't consider this, right? Or hydrogen, if it is there, we don't consider this. This example we have discussed just now. Another thing what you can conclude here, since the distribution of charge is uniform, right? And it is equally distributed over the carbon atom. So these kind of molecule is said to be aromatic compounds, aromatic compounds. When the distribution of charge is equal, then the compound is aromatic, right? Now, like I told you in the last class, for aromatic compounds, Huckel's rule will follow, right? Huckel's rule. And in Huckel's rule, we'll find out what? The number of pi electrons, if it is equal to 4n plus 2pi, then it is said to be aromatic. That is what we have discussed last class. So you see here, any of this molecule you see, a number of pi electrons you calculate, 2 plus 2, 4, and we calculate this lone pair also into this, right? 2 plus 2, 4 plus 2, 6 pi electron we have here. So for n is equals to 1, Huckel's rule is followed, and that's why the compound is aromatic. So actually this Huckel's rule is given, see actual thing for aromaticity is this only. The planar molecule and uniform distribution of charge. When the charge is uniformly distributed, that is only possible by resonance, right? By resonance because pi electron we are considering. And resonance is only possible when the molecule is planar. That's why you see, when we discuss today the aromaticity, there we have condition. The condition is what Huckel's rule must be followed and the molecule must be planar. This kind of condition we have. So those thing is necessary for resonance to take place. So the actual thing or the basic thing for aromaticity is what? Uniform distribution of charge or equal distribution of charge in planar molecule, right? So from this only we get aromaticity, right? And when we observe, we observe this what? Whenever we have equal distribution of charge, the molecule is highly stable, right? And with this only observation we have given this number of pi electrons formula Huckel's rule. But since it is just an observation, so it is not true for all kind of compounds. Like I've given you one example in the last class, there we have 4n plus 2 pi electron, but the molecule is not aromatic. Since it is what? It is non-planar. If you remember I have given you this formula, this compound. This one I have given you last class and you see the number of pi electrons follows Huckel's rule, right? But this is not an aromatic compound because the molecule is not planar. Now you see if the molecule is not planar, then obviously this resonance is not possible. The charge or the pi electron will not be distributed equally over the entire molecule and that's why it is not aromatic. Is it clear? The compounds, if you have to find out, you can also do this, try to draw the resonating structure, try to analyze the structure and whether this pi electron is equally distributed or not, if it is equally distributed, then the possibility of aromaticity is there. Since, and then you have to check the molecule is planar or not, anyway. So all these molecules you see, coming back to the topic we are discussing now. All these molecules you see here, in all these molecules you see oxygen, this negative charge is distributed equally over the oxygen atom, minus half, minus half here. Here it is minus 2 by 3, minus 2 by 3, minus 2 by 3. Here you see this positive charge on all the carbon atom. Here, then here, and then here. All these are equal contributors. I hope it is clear now. What is equal contributors? Now you see, like I said, that these kind of structure, this one you see. Now in this, if I have to find out the border of these two, what will be the answer? See a bond order of suppose, okay, I'll take this one. Let me write this 4 and pi only, 4 and pi, the bond order you have to find out. So what we'll do according to this formula? Two plus two, four, five, six, and seven. So it will write two plus two, plus one, plus one, plus one divided by, the number of resonating structure is five, you got five, five. So answer will be seven by five, right? So when you draw this resonating structure, you can calculate the bond order. And whatever the bond order of four and five carbon we have, that will be the bond order of all of the carbon atom if you check, why? Because the equal resonances there, equal contributors are there, right? One and five also you will calculate, you'll see the same values. You see one, two, three, five and seven, seven by five. For all of the carbon atoms, the bond order will be same. And that's why we are using this formula for only equal contributors. The other point is, it is very difficult in the exam to draw all this resonating structure and find out, not much difficult, but it is time taking, right? So what we do, we do not use this way to find out the bond order. But what we'll do, you see the last one, I have given you this formula, you have to find out simply this one, sigma plus pi by sigma, okay? Only one thing you can do take care here that sigma and pi is the number of, excuse me, number of sigma and pi bonds present between the atoms which takes part in resonance, okay? Like I said, this sigma bond you don't consider because it is not taking part in resonance. Only we consider what? Sigma bond present here and pi bond present here. So according to this formula, if you calculate the bond order of this particular molecule, the bond order of this molecule, you see how many sigma bond? One here and one here, two sigma bond and one pi bond divided by sigma two, three by two. Same value we are getting, right? For this one, you see, if you have to calculate bond order from this formula, one sigma bond here, two, three, right? So three sigma and one pi divided by number of sigma three so four by three. Here you see one, two, three, again three sigma and one pi, so four by three. Three sigma and one pi by three, so it is four by three. Yes, so with this formula, you can calculate directly. You don't have to draw the resonating structure, okay? Here you see, number of sigma and pi. All these carbon atom is involved in resonance. Number of pi bond, obviously two, number of sigma bond, one, two, three, four, five, five sigma divided by five. That's why it is seven by five. This one you calculate bond order, number of sigma bond is what? One, two and three. This we do not consider because it is not taking part in resonance. So one, two and three, three sigma, two pi divided by three, so it is five by three. Is it clear? Just a second, okay? So I'll give you some question on the same concept. You have to find out the bond order quickly. Okay, let's write down the question here. C L, double bond O, double bond O, double bond O and O minus. Phosphorus, double bond O, O minus, O minus and O minus. Benzene ring, bond order of benzene ring. CH3, double bond O, O minus. NH2, CN, double bond NH2 plus. The bond order of carbon and nitrogen here, carbon nitrogen. Yes, first one is seven by four correct. Second one, five by four correct. Three by two, right? Three by two, right? Fifth one, oh sorry, made a mistake here. This is H, this one is H, not nitrogen. This is CH, the last one is those three by two, okay? So like this, you can calculate the bond order, okay, of all these molecules. This is just, let me write it down. Bond order will be four plus three by four, seven by four, bond order is again four plus one by four that is five by four. For this, we have six sigma, three pi, divided by six, nine by six, which is three by two. For this, we have two sigma, one pi by two, three by two. And for this, we have one plus one, two sigma, one pi by two, three by two, okay? So like this, we can calculate the bond order and then we can compare the stability of the two compounds also. If more will be the bond order, more will be the bond strength or bond energy and hence the stability, right? If you observe all these, you know, bond order, this is nothing but the average value we are calculating, right? Total electrons or total bonds divided by number of structural possible, total bonds, number of structural possible. So this is nothing but the average distribution by electron, right? And why we are taking average here, kind of average only you can understand because the distribution of electron is equal, equal distribution is there. Why? Because equal resonance is there, equal contributors is there, right? That's why it is average thing. But when the equal contributors are not there, the distribution is not equal, then we cannot take average also, right? So in case of unequal contributing structure, you see the next type, in case of unequal contributing structure, in case of unequal contributing structures, we cannot use that formula, okay? Because that is nothing but the average, okay? And in this case, we cannot calculate the exact bond order, cannot calculate the exact bond order, okay? We can, you know, speculate here that what should be the bond order and which, like, we can say that the bond order should be in this range, okay? And closer to this value, okay? We cannot say the exact value here, right? Why? Because the contributors are not equal, so we cannot take average of those electrons, right? Some of the resonating structure will be what? Some of the resonating structure will be, like, behaves as major contributors and some of the minor contributors. And we know the more stable resonating structures are major contributors and less stable is minor contributors. But in the previous case, all resonating structures are equally stable, right? So they contributes equally. That's why we are calculating bond order over there. Now you see this example if I take, the first example in this case, it is CH2 minus C double bond O and H. If I draw the resonating structure of this one, the structure will be CH2 double bond CO minus and H, right? Now if I ask you here, what is the bond order of carbon oxygen bond in this molecule, in this molecule? It will be two, C double bond O2. In this molecule, the bond order of carbon oxygen bond is what? It is one. So what we can say, the bond order of carbon and oxygen bond in this structure, like when you draw the hybrid structure of this, the real structure, will lie in this range, greater than two, one, and less than two. Is it clear? Greater than one, but less than two. So we can say this, that the bond order will lie in this range, okay? Exact value we cannot say. Yes? Now, if I ask you, the bond order will be closer to one or two, what will be your answer? The bond order will be closer to one or two. What is your answer? Closer to one, why it is one? Because again, the structure is more stable, yes. The point is, in this case, the bond order will, you know, find out the approximate value of bond order according to the relative stability of resonating structure. And that we have already done. Which rule we are applying here to find out the stability? See, number of pi bonds are same, charge separation is also same, right? Then we use what? Negative charge on more electronegative element is more stable, right? That rule we'll apply on more electronegative, right? So according to that, this structure, which is B and A, the stability of B is more than to that of A. This means what? This is the major contributor, major contributor. And this one is the minor contributor. And this will contribute more into the real structure, resonance hybrid. That's why the bond order of the real hybrid will be closer to one. So for real hybrid RH, bond order will be closer to one. So the value of bond order will be somewhere between one to 1.5, that we can say. Understood this? So like this also, you can compare two different molecules, the bond order of two different molecules. They'll give you a question like this only so that you are able to compare if you know this. Okay? And then according to the bond order, we can find out the stability of the molecule. Second one you see, one last example, if I have these molecule, chlorine, and then this is alpha and this is beta. The bond, what is the bond order of alpha? What is the relation of bond order of alpha and bond order of beta? Which one is more, which one is less? Alpha is less, why? Because in this resonance is possible, right? So when resonance is possible, and here we do not have resonance, alpha is less. Tell me what happened? Bond order between two carbons, order between this carbon and chlorine bond, CCL, this alpha, this alpha, this alpha, this beta here. Carbon and chlorine, what is the answer? Yeah, perfect. When you draw the resonating structure, right? Here we have partial double bond characteristics, no? This we have when this lone pair comes over here and this pi electron goes here. So this bond will have partial double bond characteristics. So this bond order will be between one and two. And since here there is no resonance possible, the bond order is one only here. So bond order of alpha is more than this. That's why it is stable, more bond order, more stability. Yes, right, it's on the there. So like this you see, we can compare the bond order of two different molecules, and then according to that stability also, correct? If we have equal contributors, exact value of bond order we can calculate. For unequal, we'll just do the speculation or approximation to find out the bond order, right? If you have to compare the stability of two resonating structure, then we have eight rules that we have discussed last class. According to that, we'll find out the stability of resonating structures, okay? So this is it for in this one. Now the next thing we'll have we are starting is aromaticity, aromaticity. Now what are the conditions for aromatic compounds we have? So write down conditions for aromaticity. The first condition we have, the molecule must be, the cyclic molecule must be there, and it has planet structure, cyclic molecule and planet structure. Or we can also say must have, must have means conjugated system we can say indirectly, must have available P orbital, P orbital for delocalization, for delocalization. The delocalization of electron is important for aromaticity. And for that we must have vacant orbital, P orbital. This is the one condition. Second one is Huckel's rule that the molecule must has four N plus two electrons, where N is equals to zero, one, two and so on, any number, any integral number. Third one must have conjugated system, which is nothing but for resonance it is possible. Must have conjugated system, okay? These are the conditions we have for aromaticity. And like I said, these are the basic rule, but there are some compounds in which they does not follow these rules, but they are aromatics, aromatic, right? So those compounds you have to memorize, okay? You have to keep in mind. I'll give you enough example into this, okay? First of all, if I take the example, some aromatic compounds you see. This one, we have already discussed this just now and negative charge here. Negative charge and lone pair will count in pi electrons while calculating this pi electron will count lone pair also into this, okay? That's what you must take care of. So how many pi electrons we have here? Four and six. So six pi electrons. And six pi electrons follows Huckel's rule when N is equals to one. So that's why this compound is aromatic, simple one. Aromatic. If you consider benzene ring, benzene ring is also an aromatic compound because it has six pi electrons. Six pi electrons, aromatic. Pi electrons we have into this, positive charge. How many pi electrons? Tell me the number of pi electrons. Positive charge will not count. Only one pi bond, so we have two pi electrons. And two pi electrons, again, you see for N is equals to zero, it is possible, right? So this is also an aromatic compound. If I take the example of naphthalene, naphthalene is this. Last class also we have discussed. Electrons we have here, 10 pi electrons, right? Pi bonds, 10 pi electrons. Again, for N is equals to two, it is possible. So aromatic, right? Aromatic. One more example, all these examples are important, okay? Double bond here, double bond here and we have double bond here. Positive charge here. How many pi electrons we have here? Six pi electrons. So six pi electrons, so it is also aromatic. All these are aromatic compounds. See this compound, the name of this compound is propelium ion, okay? The name is also important, you must remember. One, two, three, four, five, six, seven. Seven carbon atoms are present in propelium ion. All these are, you see, all these are aromatic compounds. If there are simple compounds where there is no exceptions or planner molecule, is there, just you calculate number of pi bonds, right? And bonds, if it follows Huckel's rule, then it is aromatic compounds, okay? And all these atoms, you see, none of the carbon atom is sp3 hybridized, okay? None of the carbon atom is sp3 hybridized. Atoms is sp3 hybridized. So that is also one necessary, because you see, if the carbon atom is, suppose if I write down this example, you see, see how you can relate this, suppose if I give you this molecule, since pi sigma pi conjugation is there, so you can draw the resonating structure, and that structure will be double bond here, positive charge here, and negative charge here, right? And the resonance hybrid of this molecule is this, okay? We have from this carbon to this carbon. You see, in this structure, the electron or pi electron is not distributed over the entire molecule. There's no uniform distribution. Why uniform distribution, excuse me, why uniform distribution is not there, because these two carbon atoms are what, sp3 hybridized. And we know sp3 hybridized carbon atom does not take part in resonance, right? So that's why this molecule, if you have any sp3 hybridized carbon atom present in the molecule, directly you can say the molecule is non-aromatic, okay? Because the resonance or complete delocalization is not possible, okay? That's the thing. So this is not an aromatic compound, okay? And you don't have to count number of pi bonds and all in this, okay? You don't have to even consider whether the molecule is planar or not. Just you see carbon atoms, sp3 hybridized, non-aromatic. Here you see all the carbon atoms, none of the carbon atom is sp3 hybridized, okay? That is one point you keep in mind. Now, one note you write down here, write down. In polycyclic system, in polycyclic system, still not the example that we have done, that is very straightforward, based on those concepts, because we know you can do that. Now some exceptions and some different kinds of compounds we'll see. In polycyclic system, polycyclic system means what? When the system contains more than two ring, when more than two ring is present, Huckel rule, now there are slight modification into this. Huckel's rule is applicable only for peripheral pi electron. And this is just an observation, okay? There's no logic behind this, okay? Like I said, still studies going on, research is going on in aromaticity. So things are not explained now, not properly explained now, okay? So this is just an observation that when we have polycyclic system, we'll take this or that, like this here, okay? So here, if I take one example of, so this you must keep in mind, this is when we have more than two ring. Now you see this molecule, double bond, double bond, double bond, we have here alternate double bond, then it is here, this is the ring, right? Now in this compound, if you calculate the number of pi electrons, one, two, three, four, five, six, seven, eight, right? So for eight, so we have what? If you calculate all these pi electrons, then the number of pi electrons will be 16, right? But this molecule, this molecule according to 16, you see it follows 4n pi rule, 4n pi electrons, which is for anti-aromatic, a, a. If the molecule contains 4n pi electron, it is anti-aromatic, okay? So according to this, you see it is anti-aromatic in nature, okay? But the thing here, it is what? That we can draw the resonating structure of this. When this pi electron goes here, this pi electron comes here, and this pi electron will come here. So the resonating structure, when you draw, you will get this. We have double bond here, double bond here, and here. And all these bonds will be as it is. We have double bond here, okay? Now the rule is what? In polycyclic system, when we have more than two ring, right? In that case, we apply this Huckel's rule only for peripheral, what? Peripheral pi electron, right? So this electron here also you see, this is not the peripheral pi electrons, because it is present between two molecule. Or peripheral means what? You have to start like this. You have to go like this. All these edge, you have to follow. And along this edge, whatever pi electrons you are getting, that you have to count, right? This one, from here to here. So two, four, five, six, seven, right? Seven into two, 14, which is true for this. In this molecule, if you see resonating structure, this pi electron, which is in between, this pi electron is non-peripheral, is non-peripheral. So out of eight pi electrons, eight pi bond, one pi bond you don't have to consider, we consider only one, two, three, four, five, six, seven, right? So in this case, the number of pi electrons, this non-peripheral, we will not consider for Huckel's rule. So number of pi electrons for Huckel's rule into this will be seven into two, 14 pi electrons. And then for n is equals to three, it follows Huckel's rule. And hence the compound is aromatic. Is it clear? Okay, so now like one rule that I have given you, that is Fry's rule, number of benzenoid form, right? So if you calculate the number of benzenoid form here, it will be one, two, and three. We have three benzenoid form. And here we have only one and two benzenoid form. So according to that Fry's rule, this should be more stable. But like I said in the last class, that when aromatic seed is there, that will overrule the all other concepts of stability that we have, right? So since this compound is aromatic, that's why this is more stable than this one. Is it clear? Next we'll write down, there is a compound, general name called enylene, A double N, E, N-E, and enylene. These are actually cyclic compounds which contains alternate pi bonds. Okay, like you see this example, this is, we have alternate pi bond present, just a second, something, one, two, three, double bond here, and then we have double bond here. So this is enylene, general name is enylene. But we write, this is what, the name of this compound is 14 enylene. This 14, this number represents the number of carbon atom present. This 14 represents the number of carbon atom present. So one, two, three, four, five, six, seven, seven pi bonds, 14 pi electrons compound is aromatic. How many pi electrons, what is the name of this compound? What should we write the name of this compound? What is the name of these two compounds? Number of carbon atom is three, three, six, eight, 10, 12. So it is 12 enylene. Similarly, this one is three, six, eight, 10, 13, 16. This is 16 enylene, general name. Electrons here, one, two, three, four, five, six. So it is, it has 12 electrons, right? So this one is anti-aromatic, four and pi. This one has one, two, three, four, five, six, seven, eight. Two, three, four, five, six, seven, eight. So we have again 16 pi electrons, anti-aromatic. No, this one is anti-aromatic. This compound you see, this one I have discussed. Last class, this is 10 enylene. This is non-planet structure. Like I said, the repulsion between here and here. These two point, this structure is non-planet. And hence it is not aromatic. Hence it is not aromatic. So these are a few examples of enelins we have that you must keep in mind. Now we'll see another examples of where the molecule is. I'll just write down the structures here. You have to tell me the number of pi electrons and then the behavior, aromaticity, and aromatic or anti-aromatic or non-aromatic, anyone. Non-aromatic if it is, any SQ3 hybridized carbon will be there, okay? You see a few examples. This is boron. You must take care that resonance also, you have to consider resonance is possible or not. This is nitrogen. Tell me all these first. Anti-aromatic, just you write down AA. AA for anti-aromatic. Aromatic is A and non-aromatic is NA. That's it. Anti-aromatic, aromatic and non-aromatic. First one has four pi electron, correct. Because boron has electron deficient. It has vacant P orbital, but no lone pair on it. So four pi electron, it is anti-aromatic. What about this one? Six pi electron, aromatic. Because it has four pi electron. So it is anti-aromatic. What about this fourth one? Fourth one, Kushal is saying aromatic. Vaishnavi, non-aromatic, non-aromatic. Purvik, aromatic. You see this carbon atom is SP3 hybridized, Kushal and Purvik. So it is non-aromatic, SP3 hybridized, non-aromatic. What about this one? How many pi electrons in fifth one? See, actually, these two pi bonds, we have four electron, and this lone pair is involved in resonance. But this lone pair is not involved in resonance. Not in resonance. Okay, so this also you should know. So two, four, six. It has six pi electrons, so it is aromatic. Electrons you can consider, but since this pi electron is perpendicular to this plane, so this is not involved in resonance. This is involved in resonance, this pi electron. Right, so this is aromatic compound. Understood? Sixth one is aromatic. How many pi electrons? Sixth one, how many pi electrons also you tell me? Number of pi electrons also you tell me? Only six pi. You see, one, two, three, four, this lone pair is also in resonance. One, two, three, four, five, five pi electrons. So we have 10 pi electrons present here. And the compound is aromatic. The lone pair is involved that you must know. Right, this lone pair is involved in resonance. What about this? How many pi electrons? Seventh one, seventh one has two pi electrons. So it is aromatic for n is equals to zero. Right, what about this one? How many pi electrons? Six pi electrons, both negative charge will consider here. So it is also aromatic. This one, ninth one, ninth one, this one. See, oxygen has two lone pair. Out of two lone pair, one lone pair each on each oxygen is involved in resonance. Two, four, six, it has six pi electron. The compound is aromatic. So this lone pair will not be there in the question. Okay, you should know this. Oxygen bonded with two bond, so two lone pair. Nitrogen bonded with three bond, so one lone pair. This information you should have, right? So it is also aromatic. What about this one? The electrons are there. But you see this carbon atom is what? This carbon atom is sp3 hybridized. So this will not involve in resonance. So it is non-aromatic, okay? This one, we have done this. It is aromatic, two pi electrons. This one, four pi electrons, it is anti-aromatic. Yes, two pi electrons in 11th. Ninth one, this one, right? See, Vaishnavi in this ninth one, oxygen has two lone pair of electrons, each oxygen, right? But both this lone pair is not involved in resonance. One lone pair is in the plane of this ring and other lone pair is in the perpendicular plane. So out of the two lone pair present on each oxygen, only one lone pair is involved in resonance. So one lone pair here, one pipe on here, so two electron and one lone pair here. So two, four, six pi electrons it has, right? So it is not like all lone pairs you have to consider. You should know this that out of the given number of lone pair or the number of lone pair which is present, out of that, how many lone pair is involved in resonance? Only that electron you have to consider while you are calculating the number of pi electrons, okay? Or if you are trying to figure it out that whether Huckel's rule is applicable or not, okay? So this you should know, like some examples of nitrogen, you see this lone pair also, we have a lone pair here, right? If I put positive charge here now, suppose in this molecule if I put positive charge here, like this one you see, positive charge, double bond, double bond and one lone pair. What about this? Tell me, it has four pi electrons, correct? Two plus two, four pi electrons. Because this lone pair is in the perpendicular plane. This is not involving resonance, this lone pair, right? Two plus two, four pi electron, now it is anti-aromatic. But when I put negative charge here, then two plus two plus two, six electron we have then it becomes aromatic. Clear? If you put negative charge here then six pi electrons, it becomes aromatic. Understood? Okay, now some more examples you see. Tell me these four. A has six pi electron. This A has six pi electrons. So it is aromatic. What about B? B also has two plus two plus two, six pi electrons. It is also aromatic. What about this? Yeah, correct. Third one you tell me. Is it anti-aromatic? Yes or no? So this one is non-aromatic. That is what, like this is how you are making a mistake. You make a mistake in the exam. This nitrogen is sp3 hybridized. So sp3 hybridized won't take part in resonance. Non-aromatic. What? Nitrogen is sp3, no. What about this? Pyridine, eight pi, how many pi electrons? Six pi. See, pyridine has six pi electron. I have discussed this many times in the class that this lone pair is not involved in resonance. And that is why this is pyrrole and this is pyridine. The best city of pyridine is more than to that of pyrrole because this lone pair is involved in resonance but this lone pair is not involved. It is available on the nitrogen atom. Correct. This six pi electrons, aromatic. Okay, so must remember this. Best city characteristics also and aromaticity also. Oxygen here with two lone pair, double bond, double bond and here we have double bond O with two lone pair, unknown pair on nitrogen. How it is, see again you are making the same mistake. What I said, the number of lone pair present onto the atom, out of that first of all you have to judge how many lone pair are involved in resonance. You are considering all lone pair is involved in resonance? See whenever nitrogen is bonded with single bond whenever nitrogen is bonded with single bond then the lone pair present on nitrogen is involved in resonance. In general it is true but there are some different cases also. In this one you see the number of pi electrons here pair of this oxygen and here we have two and two. So we have six pi electrons here. In this one you see one lone pair, two plus two, six pi electrons. One lone pair, two plus two, six pi electrons. Again one lone pair, two plus two, six pi electrons here. This lone pair can go here, this will go here and then again when this comes here resonance is possible. See this, you just take care of one thing here. This bond, this pi bond is here with this bond. It is not with the nitrogen. So this is a single bond. So whenever the nitrogen is bonded with all single bond then the lone pair of nitrogen is involved in resonance. Right here also you see six, six. Oxygen only one lone pair is involved in resonance. That's all these electron molecules has six pi electrons all are aromatic. Double bond O. See that this double bond O we don't count. We only see the compound that aromaticity, the condition is what? Ring must, there must have, we must have cyclic ring. So when you only consider cyclic ring because of this double bond what happens? This carbon is not SP3 now, it is SP2. So that is the benefit of this double bond O we have. If this double bond O is not present, you write down OH here, single bond OH then all these compounds are non-aromatic then. You understood? So the double bond outside the ring that we don't consider, okay? Because of this double bond, this carbon becomes SP2 and then it is involved in resonance. That is the only benefit we have. So you see this kind of questions like they have given one question like seven, eight questions they have given structure. They have drawn in J.E. And they have asked how many of these structures are aromatic, okay? It is there in the previous year paper you can go through, I don't have that question now otherwise I would have given you. But we have discussed so many structures so far, okay? On aromaticity. So if you keep all this structure in mind, if you revise all this, you can do those kinds of questions easily, okay? All those factors that we have discussed in this. So this is one type of question whether you have to decide whether the compound is aromatic, non-aromatic or anti-aromatic, okay? Sometimes they ask question like this also. This question you see. Concept is aromaticity only but different types of question. Now, suppose if I draw this structure, double bond, double bond here, you have to find out which nitrogen atom, which nitrogen atom donated first. Means if H plus is there to attach with this nitrogen atom, where this H plus will attach on this nitrogen or the another nitrogen? Suppose this is first, this is second. On which nitrogen atom the H plus will attach first. Reason, why it is second? Yes, because the first is involved in resonance unavailable for bonding, this lone pair. This lone pair is not involved in resonance. Another way what you can say the same thing, another way what you can say if H plus attached over here, right? Then the aromaticity has been lost, right? So the compound will become comparatively unstable then, okay? If H plus will attach here, aromaticity will be maintained and hence the compound will be still stable, right? And the second thing which is the same thing we have, this nitrogen, this lone pair is involved in resonance. So electron density is less here. Tendency to take H plus will be lesser than this nitrogen. So the second nitrogen will be protonated first. This one, this kind of question is required in reactions actually, like which bond will break and how the bonds will break. There are two possibilities, right? Since chlorine is more electronegative, so this will take this bond pair of electron, one possibility is this, and it forms this product. Positive charge double bond, double bond plus Cl minus. Another possibility is what if this bond is taken by, taken up by this carbon atom, then the possible structure will be this, negative charge here, this one and this one plus will get Cl positive charge on it. Which one of these is possible? This is the first one and this is the second one. Which one of these is possible? Why first one? Purvik, Prithvik, second one is correct, Sundariya. Why first one? Yeah, you see, when this bond breaks, this compound is highly stable because of aromaticity. This is aromatic compound. This is not aromatic, right? This is anti-aromatic, four pi electron. This is anti-aromatic. So however, like you see, chlorine being more electronegative element, since aromaticity is here, then the chlorine will not take electron, but the carbon atom here will take electron and gains stability. So this is how the bond breaks, okay? So you see the aromaticity will always be preferred, right? Whatever the condition we have. This is, yeah, it is unstable, but since it is aromatic compound, it is highly stable, so this one will be favorable. Even here also, chlorine will be a little bit more stable than here, but this is highly unstable, anti-aromatic compound. So we'll prefer this one. This one is right and this one is not possible. For example, you see, we have chlorine. How this bond break, either this carbon will have positive charge or negative charge. This carbon will have positive or negative charge. Yeah, if it is positive, if it is positive, this is not right, then the ion is tropelium ion with this positive charge on this, the ion, we call it as tropelium ion, which is aromatic, right? So here, this Cl will take this bond pair of electron and goes out as Cl minus and here we'll have the positive charge because we are getting aromatic compound here, okay? So that's another type of, see this kind of question you will require in reaction. Suppose you have to form some product and how the bond breaks. So aromaticity, you have to also keep in mind. Okay, one more question you see, double bond here. This compound has high dipole moment, dipole moment. So what is the reason of this dipole moment? See the point here is what? Okay, you think on it and tell me first. Why Purvik? Why so? Yes, that's the reason. See, the reason of dipole moment is what charge separation, okay? Actually, when you have some positive negative charge center, then because of this difference, we'll have the dipole moment, right? So in this molecule also, since the dipole moment is there, so we must have some movement of electrons, right? Because of that, the charge center will create and because of that, we'll have dipole moment. So the point is this double bond here, right? If this double bond comes over here, then here we have the negative charge, then negative and pi electron, four pi electron, the compound is become then anti-aromatic, right? If this pi electron comes over here, so we have negative charge here and positive charge here. Because of this separation, negative and positive, this ring is also aromatic and this ring is also aromatic, right? And this the movement of electrons or pi electrons takes place like this and because of that, the molecule becomes aromatic and that is the reason of dipole moment, the separation of charge. Double bond here will have the positive charge, here we have negative charge and this double bond will be as it is. Now you see, this ring has six pi electrons aromatic and this ring has two pi electrons, again aromatic. The shifting of electrons takes place like this, leads towards aromaticity, highly stable and this because of this charge separation, the dipole moment is there, is it clear? Okay, so this is it for aromaticity. Okay, we have done so many examples, those examples you must remember and then some reactions of aromatic compounds, we have benzene and all that is there in alkyl halide also, there in nitrenes also, amines also. So those reactions you can revise, right? So those reactions, the mechanism based on aromaticity only, you must have remembered that ortho-substituted product we get when the double bond breaks the ring. So when the double bond breaks, so that will leads towards what? Less stability because the aromaticity is being lost over there. You gain aromaticity again, the H plus comes out and again we'll get the double bond. So that all those reactions based on aromaticity only, all the molecules which is aromatic will not tend to lose its aromaticity because aromatic compounds are the most stable form we have, okay? Yeah, we'll take a break at 12.30, okay? We'll take a break of half an hour because I will also take lunch, okay? So 12.30 we'll take a break, then we'll start at one o'clock. Will it be fine or you want to take it to the break now only? Okay, so we'll take a break at 12.30. Now we'll start this chemical bonding concept can we discuss now? And after the break, we will discuss that in all tautomerism thing. What we should start in chemical bonding? Tell me what is the doubt you have any specific topic because there are many things I cannot start from the beginning, okay? Any specific topic you can mention you'll start from there. VSEPR you know all, right? Okay, hybridization and MOT. Okay, so we'll start with valence bond theory, okay? That will contain hybridization also and then we'll move on to MOT because if you have to discuss MOT first we have to understand hybridization and then magnetic behavior in all, right? So we'll start with VBT valence bond theory. VSEPR you know all, right? Okay, from VSEPR onwards. So we'll start then VSEPR then VBT and then MOT. So we'll go a bit faster then, okay? See valence shell electron pair repulsion theory that is VSEPR it actually deals with the shape of the molecule. Shape, geometry and bond angle mainly with shape and bond angle, okay? So write down the heading first. VSEPR valence shell electron pair repulsion. As the name suggests in the valence shell whatever the electron pair we have there we have some repulsion into that, okay? It deals with the shape of the molecule, okay? So there are some theory of this particular concept we have. So three, four points we have you just write down quickly all those points. First point the shape of the molecule, shape of the molecule, shape of the molecule is determined by, is determined by the repulsion in valence shell. Repulsion is this first of all we have lone pair, lone pair repulsion, right? Then we have lone pair, bond pair and then we have bond pair, bond pair. Lone pair, lone pair repulsion will be maximum then lone pair, bond pair, bond pair, bond pair. Region of this is lone pair of electron occupy more space around the nucleus then we have more repulsion. In this also we have one more thing if you have multiple bond, multiple bond means double bond or triple bond has more number of electron so obviously multiple bond will produce more repulsion, will produce more repulsion single bond. These are some points we have of VSEPR, okay? Now how to calculate the geometry and shape? To find out the geometry and shape why I'm giving you this theory because on the base of this theory only we'll do. So if you do not get this you will not understand that, okay? So just two, three points write down quickly. We'll calculate the total number of electron pairs number of electron pairs on central atom, on central atom. Second one, multiple bond here, multiple bond will be treated as single bond. Place the electron pair around the central atom, around the central atom in such a way they experience minimum repulsion, minimum repulsion, okay? Now you see in this, number of electron pair you have to find out first, okay? And that number of valence electron. Like you see, suppose if I take one example with the help of example only we'll discuss. You know, when the number of electron pairs is two then the shape will be linear. All these things, you know. Number of electron pair is two, shape will be linear, electron pair is three, shape will be trigonal planar. Okay, just to write down this first. Let me write down this. Suppose the first case we have number of electron pair is two. Now in this electron pair we can have bond pair also and lone pair also. Electron pair whenever we say we mean both, bond pair and lone pair, right? So the possibility is what since electron pair is two so for any molecule we should have at least two bond pair. One bond pair is not possible for a molecule, right? So number of bond pair is two then lone pair will be zero then only two plus zero will be two, right? So in this case the general form of the molecule will be AB2 type, right? And the geometry for this type of molecule will be linear and this will be AB like this. Bond angle will be 180 degree. Example we have a BEH2, CO2, et cetera. We'll write down this few four, five examples. If you have electron pair three then in this one possibility is what? The first possibility is we have three bond pair and zero lone pair. In this case the type of the molecule will be AB3 type because we have three bond pair and the geometry will be what? Our shape will be trigonal planet. Another possibility is what? We should have, we have two bond pair and one lone pair. The type of the molecule will be AB2 type because the lone pair, the number of bond pair is two only, so AB2 type molecule, right? In this case, the shape of the molecule will be bent shape or V shape or angular shape. Anything you can see. So AB3 type molecule if you see it is like this, A, B, B and B, AB3 type. If you have one lone pair then one B you just remove this one B and you place a lone pair over here, right? This is what? So while we are talking about the shape of the molecule, shape of the molecule, this lone pair you don't have to consider. You just remove this lone pair and then you see it looks like V shape, right? Angular shape, bent shape or V shape. That's how the shape is V shape, angular shape or bent shape, right? So the point is, however, the geometry of this molecule is trigonal planet only. Trigonal planet. Geometry is trigonal planet. Here the geometry and shape both are trigonal planet. So one thing you must keep in mind, here if you talk about the bond angle will be 120 degree and since we have one lone pairs, here we have bond pair, bond pair repulsion but here we have lone pair, bond pair, lone pair, bond pair, bond pair, bond pair repulsion. So because of this, the lone pair will suppress these two bond pairs because the repulsion is high. So bond angle here is less than 120 degree. Is it clear? Right? So like this, there are many other possibilities. Electron pair when you have four, right? Then we have bond pair four, lone pair zero, bond pair three, lone pair one, bond pair two, lone pair two, like this, okay? So like that we can form different, different compound but the bottom line of all these stories is what? Just one thing you have to keep in mind. Whenever lone pair is zero, right? Whenever lone pair is zero, then the shape and the geometry of the molecule will be same, right? Whenever lone pair is zero, then shape and geometry will be same. Whenever lone pair is present, then the geometry will be same. You see for both the molecules, right? If I take one example here for bent shape, it is SO2, right? And for this AB3 type, suppose I'm taking SO3. So for both the molecules, the geometry is trigonal planar only, right? But the shape for this, since it contains one lone pair, it is bent and for this trigonal planar, why? Because there is no lone pair understood. So the point here is what? If electron pair is same for all the molecules, for the given molecules, if the molecules has same number of electron pair, then their geometry will be same, right? Shape will be different according to the number of lone pair present, right? Why it is different? Because the lone pair bond pair repulsion, right? One more example I'll write down and then we'll see some, one more electron pair four I'll write down, then we'll see some examples directly. Suppose we have electron pair four. So one possibility is what? That we can take bond pair four and lone pair zero. Another possibility is what? We can take bond pair three and lone pair one. Another possibility is what? We can take bond pair two and lone pair two. Possibly? Possibly, right? Three lone pair and one bond pair is not possible, okay? So in this case, you see, whenever you have four bond pair and zero lone pair, so this, the geometry and shape will be tetrahedral. I'm not writing down geometry and shape here. It is tetrahedral, geometry and shape. Both will be same since there is no lone pair. Now, when you're talking about the geometry of this, geometry will be tetrahedral only. Geometry will be tetrahedral only because the number of electron pair is same. For this also number of electron pair is same, so geometry will be tetrahedral. But the shape of these three molecules will be different. Okay, if three and one we have, then you see, if I write down the example here or four and zero, it is CH four. CH four is tetrahedral geometry. And how do we write down this structure? Carbon, hydrogen, right? Hydrogen we have here. Three-dimensional structure I'm drawing. Another hydrogen is here, and one more hydrogen is here. Bond angle is 109 degree, 28-minute tetrahedral. Now what happens if you want to draw this bond pair four and lone pair three? Suppose I'm taking the example for this is NH three. For this one, it is NH three. For this one, it is H two O. We know in H two O, oxygen has two lone pair, NH three nitrogen has one lone pair. So structure of NH three will be what? Out of this one bond, in place of one bond, we'll have one lone pair, and then we have three hydrogen. I'm drawing it simply only. And then if you draw H two O, so H two O will be like this, and then here we have one lone pair, and here we have another lone pair. So if you compare the bond angle here, if it is theta one, so theta is less than theta one, and here this theta two, it is less than theta even. If it is theta here, and if it is theta two. Understood? Because we have lone pair, two lone pair we have a repulsion over there. The point here is what? For all these molecule, whether it is CH four, NH three, and H two O, number of electron pair is four. So geometry will be tetrahedral only. But according to this lone pair, when this lone pair is present, now you see, if you try to understand this, see whenever we find out, we try to find out the shape of the molecule. Lone pair we don't consider in shape. So if you see the structure here, I'll just draw it in the next part, NH three. What is the shape of NH three molecule? What is the shape of NH three molecule? NH two O? NH three and H two O? NH three is pyramidal, and how it is pyramidal you see, if I draw the three dimensional structure here. So this is the dotted line going away from you. Hydrogen, hydrogen, and hydrogen. So this bond, it is coming towards you actually, and this is the bond going away from you. So this plane, this is coming out of your screen, and this is going inside your screen. And here we have nitrogen. Here we have nitrogen. And this nitrogen is connected with all these hydrogen. So it is like, it is coming towards you, going into the plane, and from center, this night on right side of your, and one lone pair is this present here, just above the nitrogen atom. So when you see this molecule, we have a triangle, and nitrogen is above triangle, right? See nitrogen in this triangle is not in the same plane. So when you imagine this, we have a triangle, and in the upper plane of this triangle, the nitrogen is present here, right? So this looks like a pyramidal, right? This looks like a pyramid, right? This is a pyramid with triangular base. That's why we are calling it as trigonal pyramidal or simply pyramidal only, trigonal pyramid, or pyramidal, we also call it as simply. Why trigonal? Because the base is a triangle, correct? X2O when you draw. So X2O, the structure is this with two lone pair. And this, well, like I told you, that when we try to find out the shape of the molecule, we don't consider lone pair. So you imagine this lone pair is not at all present here. The shape is what? Bent shape. That's where the shape is bent. Now the point here is what? Well, while we are judging the shape of the molecule, lone pair we don't consider. We only consider the bond pair, right? But geometry, if number of electron pair is same, geometry is same. Is it clear now? Now suppose we'll see, like this we have electron pair, five electron pairs, six and seven. So this, you must have the notes of that. Okay, you can go through. Now suppose if I give you the structure that is H2SO4 or you just try to find out for SO4 2 minus. Sulfate ion. What is the geometry and shape of it? SO4 2 minus geometry and shape. See, there are two methods of it. One thing, if you know the structure, you can draw the structure simply. Now the structure of this will be what? S double bond O, double bond O and O minus O minus. Do we have any lone pair present on the sulfur? Present on the sulfur, right? Louis structure, if you draw, you will understand that. Now, the theory that I have given you in the beginning, how to determine the geometry and shape. The second point I gave you, you can go through that. A multiple bond is treated as a single bond. So if I ask you, how many bond pair present in the molecule? What is your answer? How many bond pair? How many bond pair in this? How many bond pair present? Yes, the number of bond pair is four. Why it is four? Because this, any multiple bond will consider this as a single bond only. So one, two, three, four, four bond pair, right? So there's no lone pair, only four bond pair, right? So its geometry will be what? Tetrahedral and shape will be tetrahedral. Geometry and shape will be tetrahedral since there is no lone pair present. So like this, if you know the structure, you can easily find out the number of bond pair. Only thing you have to keep in mind, we don't count this as two bond pair, it is one bond pair of electron. Any multiple bond we consider as one bond pair. The another way, because structure is difficult to memorize, the another and easier way to find out the geometry and shape of any molecule is to find out the number of valence electron. Suppose it is N, right? Now what we do, this N should be divided by eight. You will get some quotient here, right? And then you will get some remainder. This quotient gives you the number of bond pair. This remainder gives you the number of electrons in the lone pair. Take care of this thing, R is the number of electrons in the lone pair. So number of lone pair, if I have to find out, number of lone pair will be R by two. Is it clear? So electron pair is what? Electron pair is Q plus R by two. This is the number of electron pair. Now for this one, if I apply this rule here, for SO4 two minus, sulfur has six valence electron, oxygen has six valence electron, six into four plus two negative charge, so two for that. So it is 32. Now this 32 will divide by eight. So it gives four, 32 and zero. So we have four bond pair and zero lone pair. So number of electron pair is four. And when the number of electron pair is four, geometry and shape is tetrahedral. Understood? So this by eight method we'll find, or in hybridization also we'll do the same thing. Try to find out the shape and geometry of this molecule. IO3, F2 minus, what is the shape and geometry of this? What is the shape and geometry of XEO2, XEF4, ICL3, H2O, CH4, NH3? What is the answer? Tell me, what is the number of valence electron here? It is seven, so seven into three because two floating, we have seven into three plus six into three plus one negative charge, so one for that, it is 40. So 40 divided by eight gives you five bond pair and zero lone pair. So this is what trigonal bipyramidal is, geometry and shape, both. Because the lone pair is zero, geometry and shape will be same, trigonal bipyramidal. So XEO2, number of valence electron, eight plus six into two, that is 24. So 24 divided by eight gives you three bond pair and zero lone pair. So this will be, oh, it is 20, no, six into two is 12, 12, oh, it's 20. So 20 divided by eight. So it is two bond pair and two lone pair, right? Two bond pair and two lone pair, electron pair is four. So geometry will be what? Tetrahedral, since four electron pair we have. But shape will be what? Bent shape, or V shape. So hybridization also, wait. Little hybridization, let me finish this first. XEF4, eight plus seven into four, 36. So 36 by eight gives you four bond pair and, right? So four bond pair and two lone pair, the geometry will be what? Octahedral and shape will be square planar. Like this you can do this, okay? X2O, if hydrogen is the outer atom, what is the valence electron we have here? X6 plus two, eight. When hydrogen is the outer atom, the last three cases will divide this eight by two, not by eight, right? So here I'm taking eight here in all these compounds. When hydrogen is the outer atom, so we'll take two, not eight here, because hydrogen can have maximum two electrons. So it has four bond pair and zero lone pair, geometry and shape will be tetrahedral. Sorry, it is H2O, no? Right, so we have only one mistake. See, we are getting four here, right? But you see there are only two hydrogen atom present. So we can say out of four, here we have fours, out of four we have two bond pair and two lone pair. So geometry will be tetrahedral and shape will be bent. For CH4 you see, carbon it is four, hydrogen four. So eight by two, so we have four bond pair and zero lone pair, geometry and shape will be tetrahedral. Similarly, for this one it is, geometry will be tetrahedral and shape will be pyramidal or trigonal pyramidal. Okay, this one we have already discussed. Okay, now this bond pair, okay, now this bond pair and lone pair, the sum of this bond pair and lone pair, we also call it as steric number, which is nothing but the electron pair, is nothing but the steric number. Now, when steric number is two, right, hybridization will be SP, steric number and hybridization. If it is two, hydrogen is SP, three, SP2, four, SP3, right? If it is five, then SP3D, six, then SP3D2, like this, okay? If it is four, then DSP2 is also possible, which is there in coordination compound. This hybridization will get in coordination compound when inner D orbital is involved, okay? So when you have this number, that is electron pair or steric number will be same. That will be the sum of number of bond pair and lone pair. So if it is five, then its hybridization will be SP3D, in case of five, like here you see it is four, so hybridization for this is SP3. If it is six, then hybridization is SP3D2, sorry, if seven, then it is D3. If here it is four again, so hybridization is SP3 for all these molecule, is it clear? So we'll just, the molecule, given molecule, we'll divide it by eight. We'll find out the number of bond pair and lone pair. The sum of that will be the steric number. If the steric number is two, hybridization SP3, SP2, four, SP3 or DSP2, five, SP3D, six, SP3D2, seven, SP3D3, that is it. Like this we find out the hybridization. Understood this, what? One more point and this is the last one, write down hybrid orbital, hybrid orbital never forms. See there we can understand this also, there are explanation of all this that we can also discuss, but we don't have that much time and in the last time, January you are going to write down JMM exam, so there's no point going into those details. Just these points you keep in mind, hybrid orbital never forms by bond. So if you have any organic compounds, suppose this given, all these compounds, like this suppose there are compounds given. So in this, if you have to find out the hybridization of each carbon atom, this, all these carbon atom, this nitrogen, these two carbon atom, okay? So in this kind of question, since we cannot find out the number of valence electron and then all those things. So when the structure is given like this in organic chemistry, then you just count the number of sigma bond, number of sigma bond. If number of sigma bond is three, SP2, number of sigma bond is two, SP, number of sigma bond is four, then SP3. These are the only three general things you'll get in this kind of question. So for this carbon, what is the hybridization here? Number of sigma bond is C double bond OOH. So one sigma with double bond, one here, one here. Three sigma bond, this carbon is SP2 hybridized. We only count the number of sigma bond. This carbon atom, we have four sigma bond, SP3. This carbon atom, we have four sigma bond, SP3. All these carbon atoms in the ring are SP2 hybridized because we have one sigma bond, SP2, SP2. This is also SP2. This is also SP2, SP2, and SP2. This carbon atom, again, SP2, one pi bond, this carbon atom, SP2, right? What about this nitrogen? This nitrogen has one, two, three sigma bond and one lone pair, so it is SP3. Understood this? This kind of question you'll get in hybridization. Okay, and then you can solve this. Any doubt you tell me, otherwise we'll take a break now. Take your lunch then. Yeah, in hybridization, we count lone pair also. That's why you see NH3. You see NH3 for NH3, what is the steric number? We have three bond pair and one lone pair, steric number is four, SP3. So one lone pair we are counting here, then only it is four, okay? Okay, so we'll take a break, we'll start at 120.