 Okay guys welcome back to the session and till now we have discussed about resonance and you know what resonance is in which kind of system resonance is possible, what are contributors, resonating structures and the relative stability of contributors resonating structure okay. So since we have discussed about the relative stability right now the next part we have to see here is the you know the application of resonance okay there are three four five types of different you know applications of resonance we have so what are those types that we'll see first and then we'll discuss each one of those one by one okay. So we'll write down the heading here which is application of resonance like I said five types of application we have so first one is calculation comparison then we have mesomeric effect mesomeric effect in this only we discuss plus m or minus m which we also call it as plus r or minus r plus m minus m plus r minus r are same thing third one we'll discuss is aromaticity fourth one is the stability of intermediates and the fifth one that is heat of hydrogenation heat of hydrogenation heat of combustion okay we also call it as and hydrogenation and conversion. These are the five applications of resonance we have okay so we are going to here we are going to discuss first the calculation in comparison of band of bond order in the session okay the first application is bond order bond order is what it is the number of effective bond between the two atoms in a molecule okay so now for example you see suppose if I write down this one CH3 CH2 CH double bond CH2 this is suppose one molecule we have okay and this is first second third fourth carbon okay so if I ask you what is the bond order of C1 to C2 carbon number of bond is what 2 bond order is 2 what is the bond order of C2 C3 it is 1 similarly C3 C4 it is also 1 okay basically we see the number of bond between these two carbon atoms okay but in case of resonance since the delocalization of electron takes place so this kind of direct thing direct you know answer we cannot give okay we have to take care of you know a few things over here for examples of course if I write this molecule first example suppose I am taking which is H C double bond O minus okay since here the resonance is possible and the resonating structure here it will be what H C O minus double bond O okay so if I ask you to find out the bond order between carbon and oxygen okay what is your answer because I have given suppose if I give you this molecule also and if I ask you what is the bond order of carbon and oxygen you cannot say it is 1 because we do not we know but that it is not the actual the current actual you know structure of this ion okay we know that resonating structures are imaginary right the only thing which is real is the resonance hydrate the real structure okay all these are hypothetical or imaginary structure so we cannot say the bond order of carbon and oxygen is 1 since resonance is there okay how do we calculate the bond order of carbon and oxygen so we have to consider all the R as possible for this ion and according to that only we can say the resonating bond order of carbon and oxygen here okay so what we can say here you see the number of bonds the trick that we use here to calculate the bond order of carbon and oxygen the number of bonds between carbon and oxygen in all Rs in all resonating structure divided by the number of Rs resonating structure possible okay like you see here how many structure possible 1 and 2 so divided by 2 and what is the number of bonds between the carbon and oxygen we have 2 plus 1 3 bonds right so 3 by 2 1.5 okay like this we calculate the bond order okay now one thing I forgot to mention here this bond order we are now calculating for the molecule which shows equal resonance this formula is true only in case of equal contributors only case of equal contributors the first case we are discussing here in case of equal contributors we use this formula for unequal contributors what we will do we will see that later on okay just we calculate what the number of bonds present between the given 2 atom divided by total number of Rs possible okay I will write down few more examples on this okay now you see if I draw the resonating structure of this carbonate iron that will be this I will just draw this quickly because we have already discussed it the resonating structure of this will be the resonating structure of this will be the other resonating structure if we calculate the bond order of the nano oxygen here we will calculate the number of bond between carbon and oxygen how many bonds are there 2 3 4 bonds right and 3 structures are possible 1.33 is the bond order right here the bond order will be what 2 plus 1 plus 1 that is 3 1.33 bond order here it is what 2 plus 2 plus 1 5 divided by 3 1.6 okay you always remember one thing in this type the number of bonds will count will count only those bonds which is involved in resonance like you see this bond I am not counting here I am counting only 2 plus 2 plus 1 5 because this is not involved in resonance so we will count only those bonds which are involved in resonance okay now one more you know formula we can write here in short one more thing we can do which is actually the same thing but whatever you feel you know easier that you can memorize one formula I have already given you number of bonds in all r s divided by the number of r s okay here the bond order we can also write that is number of sigma plus number of pi bonds divided by the number of sigma bond okay number of sigma bond divided by number of plus number of pi bonds divided by the number of sigma bonds now according to this formula if you calculate for this one how many sigma bonds are there 1 2 3 and 1 pi so bond order will be what 3 sigma bonds 1 pi divided by 3 which is 1.33 same right for this one you see how many sigma bonds we have 1 2 3 3 sigma and 1 pi divided by 3 1.33 how many sigma bonds we have 1 2 3 3 sigma 2 pi 1 plus 1 divided by 3 sigma 5 by 3 so this is the easier one I think just you can in the given molecule just you calculate the number of sigma bonds plus number of pi bonds divided by number of sigma bond you don't have to draw the resonating structure simply here but you have to remember one thing that this all this formula is true only when only when equal contributors are there equal resonance are there okay all these molecules you see all these has equal contributors okay all the structures are equivalent to each other in all these right um equal contributors we can use this formula we can find out the exact value of bond order when equal contributors are there in case of unequal contributors we can you know find out the range we can say that the bond order will be close to this value or this value but we cannot give you the absolute value of bond order in case of unequal resonance because the distribution is not equal okay when the distribution is equal we can find out the absolute value of bond order according to this formula I'll write on few more examples and then we'll wind up this session okay you copy down this question and find out the bond order okay you can pause the video and solve this I'm just doing it now okay equal resonance is there okay bond order for this will be what number of sigma bond is five number of pi bond is two five plus two by five seven by five okay number of sigma bond here it is one two three four pi bond is three divided by four seven by four number of sigma bond is four pi bond is one divided by four five five four number of sigma bond is six pi bond is three divided by six nine by six number of sigma bond is two pi bond is one divided by two three by two remember we are taking only this part because this is not involved in resonance okay. So number of sigma 1 here it is 2, 1 pi bond is 2 plus 1 by 2. This is how we can calculate the bond order in equal contributors okay, this is the better formula we have okay. So in the next session we will see how to find out the bond order of unequal contributors however as I earlier said that we cannot find out the absolute value of bond order in case of unequal contributors but then we can see how we can you know compare the bond order of the molecule okay. So we will see you in the next session. Thank you.