 Now, like I said the resonating structures or contributors we have two types of contributors ok. So, write down next contributors or resonating structures we have two types one is equal contributor other one is unequal contributor. The equal contributor example we have see both molecules look like identical right this one equal contributor even if you draw the benzene ring this is also equal contributor benzene ring ok. The stability of R S in case of equal contributor is generally more than the stability of unequal contributor ok. Stability will see that comparison, but unequal contributor the example is C H 2 negative C double bond H it is a conjugated system. So, in benzene ring can't you apply pi bit ma pi n. You can. Will it end up to be the same thing as now? Yes it should you can draw ok and when you shift this here this becomes negative this becomes positive double bond here we have positive here you are negative this negative comes over here this pi electron goes here double bond positive. What about the one in the middle here like positive and negative won't that be a separate resonant. Which one you are talking about this one? Double bond positive and negative this one. No it is also not. See this look like different structures right, but the positive negative charge here is highly unstable ok and when this comes over here see you can take it this way also. We don't draw like this ok we just continuously shift this electron there like this this comes over here this comes over here this comes over here and we will get this. So, we will we don't write this actually ok so it converts into this and these two are the resonating structures. So, when you draw the hybrid of it you will draw the entire molecule. So, if they ask you know if they ask you how many r is possible for benzene ring. So, we will contribute we will count all these structures, but this two structure we are talking about I am talking about this two structure. This to the contribution of this and this is same in the real structure. So, this two are equal contributing structures ok. This one the stability is lesser than this one even any charge species is lesser stable than these two structures right. So, there is stability is over, but if you talk about the stability of these two it is same and its contribution is also same. So, which type is that one this one here it will be pi sigma lone pair or pi sigma any type we will consider. So, we can start at one time conjugation then end up with different conjugation. In between you can have any kind of conjugation if it is conjugation shifting of it like this understood this. Now, this molecule you see this comes over here C H 2 double bond C O minus C S 3. It is unequal contribution ok the contribution is not same even if you take this example double bond O this part is not involved in resonance. This part is not involved in resonance. Why? I have conjugation. Why? Why? Why? No this side the right side of the molecule is not involved in this that is why. Why is it here? First one. This part is not involved in resonance. We have conjugation only here not this part. So, it is unequal contribution. When contribution when the resonance is there in the entire molecule H 1 actually associated with it is not the variance here variance chain is this only variance chain is this only. Now, one thing you see like in equal contributed structures the stability is same exactly same and hence the contribution is also same right. But here since the stability is different so contribution is also different and that is why it is unequal contributors. So, in this case we can talk about a major contributors and major and minor contributors. Major contributors is that structure which is the most stable structure among the resonating structure that you have drawn. And it contributes maximum into the real structure. Minor contributors the stability is less. And here major and minor is not possible because all R s has the same stable. It is all R equal contributors. No, no I am talking this side if you draw this it is fine but these two are equal contributors. We will have that we will have 6, 7 different rules to find out the stability that we will do now. So, next write down see like I said here the one of the structure is major contributors and other one is minor contributors. So, which one is major or minor that depends upon the stability of the R s. So, to find out the stability of R s we have 6 or 7 different rules 8 different rules we have. So, according to that rule we will find out the stability of R s and this is the stability of R s comparison of the same molecules of ions. We cannot compare the R s of this and this same molecule to R s we are comparing right. So, write down the write down next the factors affecting factors affecting relative stability of R s resonating structure factors affecting the relative stability of R s. So, we will follow we will you know compare the relative stability of R s in the same manner like first we have to apply rule 1 then rule 2 and rule 3 like this behind. So, rule 1 in this you write down rule 1 you write down the structure which has more number of pi bond is more stable. So, first example in this suppose the molecule is this. Can you draw the R s in this? Can you draw R s in this one? Suppose this is A and this is B which one is more stable A or B? A because it has more number of pi bond. So, stability of A is more than the amount of B. What is the R s for this one? This comes over here. Second one has more number of pi bond. So, B is more stable. Just you have to follow the rules in this. First rule if fails then we use the second rule third rule and so on. Now, the property of this is what suppose you have to draw the R s of this two molecule. So, R s will have a structure like this single bond O and this pi bond is delocalized from this carbon to this oxygen to this carbon. So, we will have a pi bond characteristic here. Now, which one is more stable here? A is more stable. If the question is the bond order of this C 1 C 2 carbon if you want to compare the bond order of C 1 C 2 carbon. So, we can definitely say that the bond order of C 1 C 2 carbon is closer to 2 since this is more stable. This contributes more into the real structure. So, it is closer to 2 and we can say the bond order of C 1 C 2 carbon lies in this range and if they ask the question for equal contributors remember we cannot find out the exact value of bond order. We can give this range and option also will be in range only whether it is closer to 1 or closer to 2. Got it? First of all the bond order here is what? 1 here it is 2. So, it lies in this range 1 to 2 and since this one is more stable which has bond order 2. So, we can say that bond order is closer to 2. So, it must be more than 1.5 but less than 2 because if both are equal then bond order is what? 1.5 average of the 2. Since it is not equal. So, this is more stable. So, closer to 2. Yes, we can see. Understood this? Rule number 2 you write down. And remember always from these rules we will compare the relative stability of RS of a given molecule. We cannot compare the RS of the two different molecules. See is this C 1 C 2 bond order C 1 C 2 bond order C 1 C 2 bond order C 1 C 2 bond order C 1 C 2 bond order. We will do this bond order part also later on. I am just giving you an idea. Rule number 2 you write down. When the number of pi bonds are equal. When the number of pi bonds are equal then the stability decreases with charge separation and stability decreases with charge separation. Charge separation means when you have one molecule is charged another one is neutral. So, neutral one is more stable in the other way we can say. So, for this the example is you see this one, benzene pair will have O H like this again and its RS is what? Yes. We have done this already this comes over here and this comes over here. So, it is resonating structure is O H positive. Another example if you see N H 2 nitrogen has one lone pair it is also electron releasing. This electron pair comes over here this pi electron shift on to the star one. Now, in this two example if you compare how many pi bonds we have here? Three. How many pi bonds we have here? Three. So, first rule we cannot compare then we will have a charge separation here in this plus minus. Here it is no charge right. So, this one is more stable A is more stable than B. Got it? Similarly, in second one again A is more stable than B. There is no charge separation. So, by charge separation you mean like presence of charge. Yes, charge present. Because that in the next. Oh, no it is neutral. Yeah. So, it is like the next resonating structure of that the minus moves one more down. Yes. So, both of those we cannot compare using the second rule. With this rule we cannot compare. Okay. For that we have another you know. Okay. Next time rule number three. Which is the actual physical distance between the charges does not matter. That is also a factor we have we will see that another rules we have. But that is not this rule. No. Like if you want to compare from rule one and rule two another R is if you draw. These two structure we cannot compare with rule one and two. For that we have another rule. Okay.