 Hello friends, good morning and welcome back to Centrum Academy again. Friends, last class we have discussed resonance and application of resonance we were discussing. And the first application we have already seen that the calculation and comparison of bond order. So, in that we have two cases, first is when we have equal contributing structure, that is we have already discussed and we have seen that we are taking the average value in equal contributing structure. So, here today we are going to discuss about the comparison of bond order in unequal contributors. Well, the resonance or we can say the pi electron or lone pair is not equally delocalized, that is nothing but the unequal contributors. So, today topic we are going to discuss here is the comparison of bond order in unequal contributors, in unequal contributors. Friends, if you remember last class we have written here calculation of bond order in case of equal contributors. Here I am writing down comparison of bond order. So, the point here why I have written this comparison because in case of equal contributors, in case of equal contributors we cannot find out the exact value of bond order, unequal contributors we cannot find out exact value of bond order, exact value we cannot find out, but we can compare we can compare the bond order, why is it so? And here also we cannot find out since we cannot take the average value, the reason behind this is what? The lone pair or pi electrons, lone pair or pi electrons are not completely delocalized. Remember in case of equal contributors what we were doing? We were just calculating the number of bonds right pi bond and sigma bond divided by the number of it as the rating structure which is nothing but the average value. But here we cannot take average value since the electron pairs are not completely delocalized. So, now here you see if you take one example with this example I will try to explain that is CH2 negative C double bond OH ok. If I draw the resonating structure of this, resonating structure will be this negative charge or lone pair comes over here and this pi electron will go on to this oxygen atom. So, the bond here it is CH2 double bond CO minus H ok. It is a conjugated system ok, type of conjugated system we have also discussed negative charge or lone pair sigma bond pi bond ok. So, it is a conjugated system ok. Now, the point here it is if I calculate or if I ask you what is the bond order of carbon and oxygen bond here ok and the bond order of carbon and oxygen bond here ok. Here since we have double bond, bond order is 2. Here we have single bond, so bond order is 1 ok. So, obviously since the electrons are not completely delocalized we cannot take the average value of this 2 ok. So, what we can say exact value we cannot find out ok we cannot get the exact value, but we can one thing we can obviously say that the bond order of carbon and oxygen bond will be in this range right. It is greater than 1, but less than 2 this is the range of the bond order ok. Now, if I ask you whether this bond order is closer to this value or this value ok, that is one question they may ask you. So, how do we assign these and how do we decide this? First of all you see whether the bond order is close to 1 or 2 it depends on whether this resonating structure is more stable or this resonating structure is more stable ok. Who like whatever resonating structure I have drawn here depending on the stability of this and this we can say the bond order is either close to 1 or close to 2. Close to 1 means what? Less than 1.5, close to 2 means what? More than 1.5 ok. So, again what happens we can compare the stability of these two easily ok. And if you see the number of pi bonds are equal ok. And we know there is one rule which is what? When the negative charge present on more electronegative atom it is more stable and positive charge present on less electronegative atom is more stable ok. Here we have negative charge on carbon atom, negative charge on oxygen atom. So, since oxygen is more electronegative element. So, this particular ion here it is more stable ok. So, this form is obviously more stable right, this one is less stable ok. And hence we can say since the stability is more for this one where the bond order of carbon oxygen bond is 1. So, we can say the bond order of carbon oxygen bond is close to, this is what how? We can compare the bond order of the unequal contributors or resonating structure in which the pi electrons are not completely delocalized. So, always remember one thing in this case we cannot find out the exact value but we can compare the bond order of the resonating structure ok. The structure which is more stable the final bond order will be closer to that value right. So, this is the thing we have to sum up all this discussion exact bond order we can always find out for equal contributors for unequal contributors we will compare the bond order of the resonating structure ok. Now, one more thing one more example I will write down here this is the second example here also we have known there we have to compare the bond order of this bond which is alpha and this is beta suppose ok. So, we have to compare the bond order of alpha and bond order of beta ok. So, you see here we have resonance possible right, but here what we can say resonance is not possible but here we have resonance possible right. So, because of resonance what happens here this lone pair comes over here and we will get a partial double bond character ok. Like you see if you draw the resonating structure of this one this is how we will draw the resonating structure ok and then furthermore we can draw further resonating structure but we have we are concerned with the bond order of carbon chlorine bond ok this bond alpha bond. So, I am just drawing one resonating structure here right. So, you see here in this particular molecule since we have resonance possible. So, bond order will be what bond order will be according to the same logic we can say it will be in this range depending on the stability of the molecule right. It will be either close to 1 or close to 2 that is another thing. But here the bond order is always 1 right in this case the bond order is always 1 for carbon and chlorine I am talking about. There is no way so that the bond order is more than 1 because there is no resonance here right and we can say since the bond order in this range. So, what we can say here the bond order of alpha is more than the bond order of beta right. Bond order of alpha is more than the bond order of beta ok. So, this is how we can compare the bond order. One more thing we can compare if bond order is more bond strength will be more more bond order means what more will be the number of bond between the two atom. So, more will be the bond strength. So, the next thing we can write here that the bond order is directly proportional to bond strength and inversely proportional to bond length. Sometimes they also ask you to compare the bond length ok. So, obviously the bond length order if I write down here. So, bond length of alpha is less than bond length of beta. Bond strength of alpha is more than to bond strength of ok. So, this is the you know the whole thing that we have discussed either they can ask you the bond order they can give you option like close to 1 close to 2 like that ok maybe 1 and none of these they can you know give you options like this ok. But depending on the stability we can say the bond order is either close to 1 or 2 or whatever it is. So, here also if we write down the comparison of bond strength. So, bond strength of which one is more since we have only one molecule and in these two if we compare the bond strength ok. Since this is more stable right suppose I write down this is molecule B and this is molecule A right. So, this is more stable. So, where we have single bond right. So, bond strength of A is more than to the bond strength of B and bond length will be of A is less than to the bond length of right. So, like this we can compare the bond order bond length and bond strength ok. So, these are two examples we have discussed I will give you some more examples in which bond order we have to compare. So, we will discuss this one by one you see here on this oxygen we have lone pair present and hence the resonance is possible ok. And because of resonance what happens this double bond will convert into single bond ok. So, obviously this carbon-carbon bond order if you calculate it will be it will be in between 1 and 2 right, but here we have lone pair on this oxygen right, but resonance is not possible. So, the bond order of carbon-carbon is 1 only here right. So, here the bond order is 1 and here the bond order will be less than 2, but greater than 1. So, obviously the bond order of first one is more than to that of second one ok. Now, here you see again this nitrogen has lone pair sorry one lone pair this nitrogen also has one lone pair here we do not have resonance. So, bond order is what 1, but here the bond order is again less than 2 greater than 1. So, bond order of first one is more than to that of second ok, here we do not have resonance. So, bond order is 2 here we have resonance I have already discussed. So, bond order is less than 2 greater than 1 this question only ok. So, this is the bond order of this one is more ok. Now, this bond carbon-carbon bond we are comparing here we do not have resonance bond order is 2 here we have resonance possible right. So, bond order between 1 to 2 ok. So, bond order of first one is more here we do not have resonance bond order is 2 here it will be between 1 and 2 here it is exactly 1 ok. So, order of bond order will be this right we have resonance possible here because oxygen has lone pair. So, bond order of this will be between 1 and 2 and here the bond order is 2. So, bond order is this ok. Again you see here we have resonance here also we have resonance here we do not have resonance. So, bond order of this will be maximum and since we have more conjugation here pi sigma pi sigma pi. So, more resonating structure we can draw right and what happens in this resonating structure you see I am drawing taking this one this is the resonating structure ok. And how it forms this pi electron comes over here and this pi electron goes here in a second structure this lone pair comes over here this pi electron comes here right. So, there are 3 resonating structure possible right and in all these resonating structure the number of carbon oxygen single bond is more 2 resonating structure will have carbon oxygen single bond and one will have carbon oxygen double bond. Even in this molecule there is only one structure possible here to here right. So, if you compare this two since the carbon number of resonating structure containing carbon oxygen single bond is more hence the bond order of this will be lesser than this one order will be this here it is completely 2 here it is lesser than this because it has more number of resonating structure containing carbon oxygen double bond here we have only one resonating structure containing carbon oxygen single bond ok. So, all these examples you can understand it is not always necessary that due to resonance the bond order increases ok. It depends on the type of molecule we have we may have because of resonance bond order may decrease or increase depending upon the molecules ok. So, what we can say here one more thing that what we can say that the bond order of carbon oxygen bond is lesser than this molecule since there are 2 resonating structure containing carbon oxygen single bond ok. This is how we compare one more example we will see which is important here. Suppose this is alpha now if you compare the bond order of this one. So, the answer here it will be bond order of beta is greater than the bond order of why is it? So, I will tell you here you see this molecule and this molecule will have steric repulsion because of this molecule and this molecule will have steric repulsion ok. So, because of steric repulsion what happens the molecules change in its plane to minimize the steric repulsion ok. One of the molecule this side will go in a different plane ok. So, this molecule becomes in that case this molecule becomes non-planet. So, in non-planet molecule resonance is not possible ok. So, here this lone pair since it is non-planet. So, lone pair will be is localized means what? It is there at the nitrogen atom only the localization means what? If it is spread over the entire molecule ok. So, lone pair is localized resonance is not possible. So, bond order here it will be 1, but here it will be between 1 and 2 right. So, obviously the bond order of beta is more than that of. A few examples you have to keep in mind especially this one where the steric repulsion is there ok according to that we have to check the bond order. So, this is it for the calculation of the comparison of bond order of unequal bond reducers ok. So, I hope you understand all these ok. So, in the next session we will see another application of resonance which is nothing but the mesomeric effect. Hello friends so far we have discussed the application of resonance and that is nothing but the calculation and comparison of bond order ok. So, second application we have of resonance is mesomeric effect right. We also call it as resonance effect, mesomeric effect or resonance effect ok. So, what is mesomeric effect? Mesomeric effect is the electron donation, electron withdrawal, withdrawal through resonance, electron donation or electron withdrawal through resonance ok. This is also a permanent effect right and in this effect there is no involve like the sigma electrons are not involved like we have also seen in inductive effect only sigma electrons are involved ok, but here we do not have any sigma electrons, but only pi electrons lone pair electrons are involved in this. It has nothing to do with sigma electrons ok. So, that is the difference between resonance and mesomeric effect and inductive effect we have there we have sigma electrons involved here we have pi electrons and lone pairs of electrons are involved ok. You see like i effect where we have plus i and minus i plus i is electron donating effect and minus i is electron withdrawal effect ok. So, similarly this resonance effect is also there are two types it can be plus m effect or minus m effect plus m or minus m plus m we also call it as plus r minus m we also call it as minus r ok plus m effect or minus m effect ok. These are the few you know points or you know key points I have already I have discussed here. Now, like I said that there are two types of mesomeric effect it can be plus m also or minus m also depending on electron withdrawal tendency or electron donation tendency ok. So, first effect we are going to discuss here which is the plus m effect and i effect you see plus i effect is electron releasing group right electron releasing effect it is. So, plus m also it is electron releasing tendency it releases electron ok and when it is possible when it is possible you see if the group is this x y attached to any other group. So, if you have pi bond present between first and second atom then it has electron withdrawing tendency where the condition is what the electronegativity of y is more than that of x this is the condition we have if there is multiple bond present between the first and second atom. First atom means what the atom which is attached to the molecule multiple bond like I said then have x triple bond y also again the same condition atom with which the group is attached has some vacant orbital ok this is the vacant orbital in all these cases electron withdrawing tendency is there and the group will withdraw electron from the molecule. Like I said first example if I write down here and O2 ok you see between the first and second atom we have lone pair present and the electronegativity of oxygen is more than that of nitrogen. So, it has electron withdrawing tendency and only pi electrons are involved this pi electron will come over here and this lone pair this pi electron will come on to this electron on to this oxygen ok. So, the resonating structure of this will be this is the resonating structure another structure we can draw where this pi electron comes over here and we get this positive sign and everything will be as it is next and the last resonating structure will be where this pi electron comes over here and this comes here. So, these are the resonating structure we have ok and all these structure you see it has the sigma electrons are not all involved only pi electrons are involved here. So, if they ask you how many resonating structure possible we have 1 2 3 4 and this is the one total 5 resonating structure possible here ok. So, now you see if you see one thing if you observe will have positive sign present at ortho position right then at para position and then at again ortho position ok. So, in this case what happens whenever what we can you know conclude here whenever an electron withdrawing group is attached to the benzene ring right and resonance if it is possible then we will get carbocation at ortho and para position ok. So, if electron withdrawing group EWG through resonance electron withdrawing group if an electron withdrawing group through resonance is attached a benzene ring the benzene ring forms at ortho position. So, if I tell you the ortho and para position in a benzene ring if any group is attached. So, there are two ortho position here this one is ortho this one is also ortho this one is para position this one is meta this one is also meta. So, we have two ortho two meta and one para position in a benzene ring and whenever an electron withdrawing group through resonance is attached the benzene ring will get positive charge at ortho and para position and hence any nucleophile if you have they have more tendency to attack on ortho and para position in comparison to the meta position ok. So, that is the thing we will see some more examples in this one mistake I made here just make the correction see I have taken electron withdrawing group by mistake I have written here plus M plus M R electron releasing this is minus M. Minus M effect are electron withdrawing effect if I have written here electron releasing just change that it is electron withdrawing that is the thing I have taken this example this is electron withdrawing example by mistake I have written plus M here. So, you just change this minus M effect minus M group you should know who are the groups are there which shows minus M effect and plus M effect ok. So, for example you see if you have this C H 2 positive charge on it shows minus M electron withdrawing tendency if you have B H 2 it has vacant orbital minus M effect you can have sin i group C triple bond N minus M effect you can have aldehyde also minus M effect you can have acid also minus M effect you can have C triple bond O O R also minus M effect you can have acetyl chloride also minus M effect you can have M H 2 minus M effect you can also have C double bond O O P H minus M effect all these groups you see they can show minus M effect one more example I can write down here C double bond O N H P H minus M effect ok. So, you should know all these groups shows minus M effect ok, but it is you do not have to memorize this because the logic here it is what the we should have multiple bond between first and second atom right and the second atom must have more second atom should be more electronegated in the first atom right this is one case another case is what we should have vacant P orbital vacant orbital present under the first atom positive charge means what vacant orbital we have here B H 2 also here vacant orbital here right. So, these two are the example of the first atom contains vacant orbital this is again multiple bond second atom has more is more electronegated than the first second atom more electronegated first is the same thing we have here right. So, all these groups shows minus M effect ok you should memorize this and by common sense also you can say that whether it is plus M or minus M.