 So, by now you know that resonance involves the actual delocalization of these pi electrons. So there's actual delocalization of pi electrons in resonance, while in induction there's no such actual delocalization. In induction electrons can only be pushed or pulled via these sigma bonds. So therefore in general resonance effects are much stronger compared to inductive effects. Now equipped with this particular knowledge, let us now try and solve a few problems. So what we have out here are four different cations and we need to figure out the most table amongst these. So how do you do that? Let's start by looking at this particular cation. Now out here we have this pi bond, this pi electrons attached to a carbocation, right? So we can have resonance out here, so this can be stabilized via resonance. Now if you look at this particular cation, there aren't any pi electrons or lone pair attached to this cation, so there can't be any resonance out here. This can only get stabilized by the inductive effect, by the plasi effect of this ethyl group, right? So this can only get stabilized via induction. Now we have seen that resonance effects are stronger than inductive effect, so A is definitely going to be more stable than B, so B is definitely not going to be the correct answer, right? B is definitely not going to be the most stable cation. Let us now come to C. Well out here we have a fluorine atom, a highly electronegative element attached to a carbon chain, right? So this is going to pull electrons via induction. So this in fact going to be minus i effect that's going to happen out here, right? Now pulling electron density from an already positively charged system is going to make it even more unstable. C is in fact going to be the most unstable amongst these three, so definitely C is also not going to be the answer. Let us now come and analyze this particular cation. Well out here we have an oxygen atom and oxygen is also going to be a highly electronegative element, so even this is going to show minus i, right? However, if you look at this particular molecule closely you will see that this oxygen has this lone pair of electrons and these are right next to a carbocation, right? So what's going to happen out here is that this lone pair is going to overlap with the empty orbital of this carbocation, right? So there's going to be an overlap out here and this will lead to the formation of a new resonating structure that's going to look like this, right? So in this particular cation, oxygen is showing minus i via induction but it's also showing plus r via resonance, right? Now as we've seen resonance effects are more important compared to inductive effects. So we should say that ultimately this cation is stabilized via resonance, right? So the correct answer should be either A or D. Now if you draw the other resonating structure of A, if you shift this by electrons over here we'll get the other resonating structure that's going to look like this, right? Now if you see in both these resonating structures this carbon atom, this carbon atom if you count the total number of electrons you will see that this carbon atom only has a total of 1, 2, 3, 4, 5, 6 electrons, right? So this has an incomplete octet even out here and even out here, right? However, if you look at this particular structure you can see that this carbon atom if you count the total number of electrons you can see that this has a complete octet, right? It has 1, 2, 3, 4, 5, 6, 7, 8 electrons. In fact if you count even for oxygen you will see that even this oxygen atom has a complete octet. So all the atoms have a complete octet out here. So this particular cation is actually going to exist more in this particular form. The actual structure is going to look more like this in which all the atoms have a complete octet, right? So therefore D is going to be the most stable cation amongst all of these. Let's do another problem. What we have out here is a phenoxide ion. So this is a phenoxide ion and to this phenoxide ion we have attached these different nitro groups at different positions and we need to find out the most stable out of these anions. So how do we solve this? Well, if you look at this NO2 group, this NO2, it's a minus R group, right? It can withdraw electrons via resonance. In fact, let me bring up the resonating structures so that we can see exactly what's going on. So out here we have these pi electrons which can be withdrawn by this NO2 group. And this will lead to the formation of this particular resonating structures. And the pi electron out here can shift further and this can go on and this will give us our resonating structures, right? Now as you can see this kind of electron withdrawing groups bring about a positive charge at their ortho and para position, right? So let me go and highlight this out here. We get positive charges at this ortho and para of NO2, right? Now if you look at this particular resonating structure, we have a positive charge directly under this negative charge, right? In fact, out here we even have lone pair of electrons over this oxidatum and a positive charge on a carbon signifies the presence of an empty orbital. So this lone pair can in fact go ahead and overlap with this empty orbital, forming a pi bond out here, right? So there's going to be a pi bond out here and this is going to remove this negative charge from the oxidatum. So the point is presence of an NO2 group at this particular position is going to bring about a positive charge right under this oxidatum. It's going to create an empty orbital out here, which is going to ultimately help in removing this negative charge from the oxidatum. So therefore this N9 is going to be more stable than the phenoxide N9, right? Let us now look at this particular N9. Now even out here we have NO2 and even out here NO2 is a minus R group and if you again draw the resonating structures, you'll see that this NO2 brings about this positive charges at these positions, right? It's going to bring about this positive charges at this orthoenphera of NO2. Now out here as you can see there won't be any positive charge right under this oxidatum. There won't be any empty orbital created out here. So this NO2 group even though it's a minus R group, but it won't help in removing this negative charge from the oxidatum, right? So purely based on resonance, the stability of this particular N9 is not going to be any different compared to this one, right? Let us now take a look at this final N9. So even this NO2 is a minus R group and it's going to withdraw electrons and it's going to bring about positive charges at these positions, right? It's going to bring about positive charges at these positions. Now even out here you can see that there's going to be a positive charge right under this oxidatum. There's going to be an empty orbital created out here which is going to help in removing this negative charge from this oxidatum, right? So putting NO2 out here is also going to help in delocalizing this negative charge via resonance. So therefore we can go ahead and say that B and D are definitely going to be more stable compared to A and C, right? So what about B and D? Which of these two will be more stable? Well, as you know, resonance is distance independent. It doesn't matter if you put this group out here or out here, it's going to create the exact same positive charge at these positions. So resonance wise they are going to impart the same stability. However, NO2 is also a minus I group, right? So this is also a minus I group. Both of these are also minus I. Both of these are electron withdrawing via induction. An inductive effect, as you know, depends upon distance, right? So putting an NO2 group at the ortho position is going to pull more electrons via induction from this O minus compared to putting it at Pera, right? So therefore, because this NO2 will be better able to remove this negative charge via induction. So therefore between B and D, D is going to be more stable compared to B compared to A and C, right? Now what about between A and C? Well, even out here, even though NO2 is not able to remove this negative charge via resonance, but it is still going to show inductive effect, right? So it's still going to remove some of these negative charges via induction. So this is going to be more stable compared to the phenoxide ion. And because inductive effects are weaker compared to resonance, so therefore the ultimate order is going to be D greater than B which is greater than C which is greater than A, right? Let's solve one final problem. In fact it's pretty similar to the last one. So why don't you try your hand at solving this? Why don't you pause the video and try to come up with the most stable anion amongst these? Well, out here we have a methoxy group that has this lone pair of electrons that it can donate to this benzene ring. So these methoxy groups are plus R, right? They are electron donating. So all of these methoxy groups are plus R. And because of the presence of this highly electronegative oxon atom, they're also going to withdraw electrons via induction. So all of these are going to be plus R and minus I groups, right? So whenever we have a plus R group, it's going to donate electrons. So these lone pair of electrons are going to move over here while these pi bonds are going to come over here, right? So this will lead to the formation of a double bond out here and these pi electrons are going to move over here. So this will bring about a negative charge on this carbon atom. Now if you keep drawing these resonating structures, you'll realize that just like in case of minus R, the positions that are most affected by this methoxy group are going to be this ortho and para with respect to methoxy, right? However, unlike minus R, this time I'm going to get the development of this negative charge over these positions, right? So therefore, even in case of this methoxy, I'm going to get negative charges developed over these positions. But in case of this one, I'm going to get them over these positions, right? Now as you can see in both para and ortho, these negative charges gets developed right under the nose of this oxynotome, right? So there's going to be repulsion between these two atoms, between these charges. So therefore, B and D this time are going to be definitely less stable compared to C, right? Now between B and D, because resonance is distance independent, so both of them bring about the same charges via resonance, they're of the same stability. But if you look at induction, because OCH3 is placed closer, so therefore it's going to withdraw electrons more via induction compared to putting this OCH3 in para, right? So therefore, between B and D, D is going to be more stable than B, right? So D is going to be more stable than B and C is going to be the most stable amongst all of this. Now what about between C and A? Well, even though in C, even though it's bringing about negative charges on this benzene ring, but none of these negative charges are directly under this oxynotome, right? So this resonance effects of the methoxy group at metaposition doesn't affect the stability of this NIN in any considerable way. But because this is also a minus i group, so it can withdraw electrons via induction. And this becomes the dominant factor out here. So C in fact becomes more stable compared to A. So the correct order would be B less than D, less than A, less than C.