 So, there are mainly 6 types of conjugated systems type 1 type 1 is what type 1 we have phi sigma and vacant p orbital, we will see 1 or 2 examples parallely with this. So, we will have pi electron then sigma and then vacant p orbital. For example, you see if I take this one CH2 double bond CH single bond CH2 positive. So, this positive charge means what the 1 electron of carbon which is present in the p orbital that has been donated to the other molecule right that is why it has 1 positive charge in this. So, if you see we have a pi we have a sigma bond pi sigma and since 1 electron is donated. So, we have vacant p orbital over there. So, how do we draw the resonating structure of this to make you understand I will just draw this structure as this carbon carbon sigma bond and then carbon with 2 hydrogen here and this carbon has 1 vacant p orbital 2 hydrogen we have here 1 hydrogen we have here and this carbon also has this p orbital. So, in this we have this pi electron overlapping of this p orbital and since this is a vacant orbital we have. So, there are possibilities that this pi electron may jump to this carbon atom right and since the all the both bond pair of electrons will come over here. So, this carbon will have then vacant p orbital. So, the another structure if I draw here ok. So, this is the resonating structure here and remember one thing I forgot to tell you that we always separate the resonating structure with this double headed arrow ok. 2 resonating structures are always separated by this double headed arrow right which means these 2 are inter convertible we can convert this into this also and this into this also ok. So, all the resonating structure are converted by this. Now, to you know right because this is for you for your understanding ok this orbital thing that I have drawn here is for your understanding ok you do not have to do this in the exam. The thing is what if you see this molecule pi sigma and positive charge. So, this one is a conjugated system right. So, when it is a conjugated system. So, we can draw the resonating structure of this ok and what type of conjugated system it is it is pi sigma lone sorry vacant p orbital type of conjugated system type 1 that we have discussed ok. So, what happens here that this pi electron will jump over here right and this carbon will have the positive sign. So, the structure is C H 2 C H double bond C H 2 and here we have the positive charge. So, these 2 are the resonating structure ok and both this structure will contribute to the real structure ok. Now, if I draw the resonance hybrid or the real hybrid of this structure that will be sigma electron will write as it is right. So, in the first structure we have double bond between these 2 carbon atom right first 2 carbon atom. So, we will have a dotted line because here we have the single bond here we have single here we have double. So, here we will have the which represent the partial bond it is neither single bond nor double bond ok it is partial bond we have here we have delta positive here we have delta positive. So, this is real structure of this molecule ok. So, actually it exists like this, but to show reactions we represent we take resonating structure ok R S ok. Similarly, one more is a more one more example if you see here for type 1 suppose we have a double bond here and a positive charge ok. Again the same type pi sigma positive charge means vacant P orbital. So, what happens the resonating structure of this will be this pi electron comes over here and we can draw the resonating structure as this here we have the positive charge ok. So, this is another structure. Now, one thing you must keep in mind whenever you are drawing the resonating structure all resonating structures will have equal charge I will write on this here all R S must have equal charge means the point is if you have one positive charge here and you are drawing the resonating structure of this molecule in all the resonating structure we must have only one positive charge. If this is not there it means you have done something wrong in the in drawing the structure of resonating structure of that particular ion ok. Here also you see one positive charge we have here. So, this is the positive charge right. So, the point is for the vacant P orbital we must have positive charge present here pi sigma positive charge if it is there it is a conjugated system and we can draw the resonating structure. So, if it is a positive charge present over there then you can do this easily you can understand this easily that it is a resonating structure sorry it is a conjugated system and you can draw the resonating structure. But I will give you few examples also where we do not have the positive charge, but is still it is a type one conjugating system. So, that one I will give you one example here you see third example B H 2 ok. B H 2 if you see the configuration of this boron here we have one vacant P orbital present. However, we do not have charge here whenever the boron is associated with three bonds suppose B H 3 right. So, if you understand this boron has five electron configuration is what 1 S 2 2 S 2 2 P 1 this is S orbital and this is P orbital. S orbital has two electron P has one and this is there in ground state. So, you excited state what happens one of this pair of electron jumps over here and the structure will be the 2 S orbital will have one electron then and out of this 3 P orbital 2 P orbital is having unpaired electron. So, with all these unpaired electron hydrogen will give it is one electron and forms the bond right and we get B H 3 molecule ok 3 hydrogen 3 electrons gets paired B H 3 molecule. The point here it is what whenever boron is bonded with three bonds it has one vacant P orbital ok. So, boron you see here 2 hydrogen and 1 bond here 3 bonds. So, this is the vacant P orbital will have here. So, this type of system also pi sigma vacant P orbital type 1 conjugated system we have. So, here also we can draw the same we can draw the resonating structure ok. So, what will be the resonating structure I will show you here you see this pi electron jumps over here right and it forms a resonating structure which is C H 2 single bond C H double bond B H 2 right. So, now in this state boron has how many electrons 2 2 4 and 4 8. So, it has 8 electrons which is excess of electron it have. So, boron will have then negative charge right and since this carbon has generated its electron here. So, this carbon will have positive charge ok. So, now again you see these resonating structure here it is what it is neutral because there is no charge present here. Here also you see positive negative will cancel each other no charge present. So, these two are the resonating ok. So, all these are type 1 conjugated system ok. So, you must take care of that for type 1 we do not require positive charge always whenever you have this kind of molecule it also has one vacant P orbital and resonance is possible here ok. Now, type 2 conjugated system what is type 2 conjugated system? It has pi bond in sigma and then lone pair present in the P orbital pi sigma and lone pair ok. For example, you see C H 2 and lone pair on it ok. So, now you see in this state I will draw 1 or 2 more example here. Now, in all this structure you see this carbon has how many electrons 2, 4, 6, 8 electrons ok. For carbon itself if you see 1 electron we have here, 2 electron we have here with 2 hydrogen and 2 electron we have here. So, this carbon has side electron. So, it is excess electron that is why we have one negative charge here right. Similarly, here also we have the negative charge and here also we have the negative charge ok. Electrons of cartons only we are counting here. One carbon electron here one with one each with the 2 hydrogen atoms 2, 1, 3 plus 2 pi electrons with this carbon. So, on negative charge on it. So, if you draw the resonating structure here that will be what separated by double heated arrow right. This lone pair comes over here and this pi electron goes onto this carbon atom ok. So, it is CH2 negative charge CH double bond CH2. So, these are the resonating structure. This one again this pi electron comes over here this pi electron will come onto this carbon atom. For this one this lone pair comes over here this lone pair comes over here. We can draw one more resonating structure in which this lone pair comes over here and this pi electron comes over here. So, these are the resonating structure we have ok. If you try to draw the resonance hybrid of this that will be like this. And pi electron is delocalized from this carbon to this carbon. So, we will have this delocalization of pi electron like this. Partial negative charge will have here partial negative charge will have here ok. For this one if I draw the resonating structure it will be CH2 CHCH2 and we have a pi electron here delta negative delta negative ok. So, like this we can draw the resonating structure type 2 conjugated system is this ok. We can draw one more example like this we can draw another structure is what can we draw the another resonating structure is it possible not possible because we do not have conjugation here you see pi sigma sigma lone pair. So, you do not have conjugation over here ok. So, another resonating structure is not possible ok. However, we can shift this pi electron here also and this pi electron here that also gives you the same kind of structure ok. So, if you draw the structure next one from here it will be you shift this electron here we will have a double bond here the negative charge here that is what the structure is. So, like this we can draw the resonating structure again. So, this is the type 2 conjugated system ok. The next one is third type which is nothing but pi sigma and unpaired electron only one electron for example CH2 double bond CH single bond this one ok. Now, in this if you draw the resonating structure here homolysis actually takes place because one unpaired electron we have to retain. So, here we have two electrons one electron will jump on this carbon atom and will denote this by half arrow and another electron will jump over here half arrow ok. And the RS you get here will be CH2 with one unpaired electron CH double bond another example you see is not a double bond. So, if you draw the resonating structure of this one from this one electron comes over here half arrow. So, the another structure will be this this is the resonating structure we can draw ok. One more example we will see what happens here this will come here half arrow half arrow and this gives all these are resonating structures or contributors fourth one type 4 in this we have vacant orbital sigma and lone pair ok. So, this is what sigma and then lone pair for example CH2 positive single bond CH ok. So, this CL has lone pair of electron on it CH2 CL. So, in this you see the resonating structure this lone pair comes over here and we will get CH2 double bond CL with positive charge on it two lone pair will be as it is. For example, if you see CH2 positive single bond OH this oxygen has lone pair it comes over here and it will be CH2 double bond OH with positive charge on the oxygen atom. You see the charge separation is same all the resonating structure we must have we must have equal charges ok type 4 type 4 we have done type 5 type 5 what we can have we can have lone pair and this reorbital. So, we have lone pair present sigma bond and then deorbital vacant deorbital right lone pair sigma bond vacant deorbital example you see here we have negative charge for this if I draw the structure this will come over here CH2 SH. So, since it has vacant deorbital the sulfur can expand it is octet ok. So, expanded octet it can show ok. So, all these are different types of resonating structure and you know resonance we have here. One more important thing that I forgot to tell you here this case is nothing but the case of back bonding you must have done this in chemical bonding orthodontic table back bonding because of back bonding only we also have defined the acidity of HCl HBr and Hi the Lewis acid behavior of these molecules this I have explained this on the basis of the Lewis acid behavior of these acids on the basis of back bonding also we have done ok. This kind of different you know resonating structures possible or an conjugated systems are there which you have to keep in mind and then you can apply the concepts of resonance in the given questions ok. We have one last type of conjugated system here which is pi sigma pi conjugation this is a very common one pi sigma pi conjugation examples you see 1 3 butadiene right pi sigma pi conjugation the resonating structures of this will be this will come here and H2 here we have negative sign here we have positive sign overall the molecule is neutral and so here also it is neutral first example second example these are very common type of you know conjugated system we have. So what are the different types of conjugated system we have pi sigma vacant p orbital or we can say pi sigma positive charge right but why I am not saying positive charge because we have seen in case of BH2 also there we have vacant p orbital though we do not have charge present over there the resonance possible right. So in case of pi sigma vacant orbital vacant p orbital pi sigma lone pair pi sigma free radical right and then we have lone pair sigma vacant p orbital lone pair sigma vacant d orbital and pi sigma pi conjugated. So all these 6 types of conjugated system possible all these 6 types you have to keep in mind and then only you can understand that in the given molecule or ion the resonance is possible or not ok and one more thing the molecule must be planed because you know the shifting of electrons from one point to another point requires parallel orbital ok if one molecule goes out of plane right then the orbital is not parallel then the shifting is not possible right. So that is why we say that in planed molecule all the p orbitals are parallel and hence this kind of shifting is there for example let me draw this structure of benzene this is benzene right. So here we have p orbital and in this p orbital the overlapping takes place like this ok and that is why we have pi pi and pi orbital ok why because all these p orbitals are parallel if one molecule goes out of the plane then this parallel p orbital will not be there parallel orbital will not be there and hence the shifting of electron or delocalization is not possible ok. So that is why planed molecule is required and hence and then system must be conjugated two things we have to keep in mind. Now another thing is what here also you can understand that I have taken that these two orbital is overlapping why not these two why not these two why not these two ok the point is these two orbital can also overlap because all are parallel only any two you know consecutive p orbitals may overlap here and if these two overlap will have pi bond here these two will overlap will have pi bond here these two overlaps will have pi bond here and that is why we have a delocalized pi electron present here ok. So that is the thing planarity and the system must be conjugated for resonance these two things you have to keep in mind sigma bond has nothing to do with resonance position of atoms will not change it will be as it is pi electrons and lone pairs are inter convertible ok all resonating structures will have same charges if it is neutral then all the resonating structure of that particular molecule will be neutral if it is positive charge then all r s will have net positive charge ok right. So this is it for resonance ok on the base of this we will define another effects electronic effects that we will see in the conversation thank you.