 Once again, we have different cations and we need to compare the stability, the relative stability of these cations. So how do you do that? Well, if you look at this particular cation, it's a benzyl cation. There's an empty orbital on this carbon atom that's directly attached to a double bond, right? So we can have resonance out here. These pi electrons can move over here and we are going to have a new resonating structures and then these pi electrons can again shift and so on, right? So this cation is stabilized via resonance. Let's look at this particular cation. Now, this is a really famous cation and maybe you are familiar with it. It's called the tropolium ion. The tropolium tropolium ion. And if you look out here, you can see that these pi bonds can also undergo resonance, right? However, what's even more important out here is that the tropolium ion is cyclic and it's planar and it's fully conjugated. All these pi electrons can move in the whole plane. It's fully conjugated and in such systems, we need to look out for aromaticity, right? Now, if you count the total number of pi electrons that are getting delocalized out here, so we have 2 plus 2 plus 2, so we have 6 pi electrons out here that are getting delocalized and this falls under the Huckel's 4n plus 2 system, right? So if you recollect, if the total number of pi electrons in a cyclic planar and fully conjugated system equals to 4n plus 2, where n can be 0, 1, 2 and so on, so that is if the total number of pi electrons is equal to 2, when we put n equal to 0, we are going to get a total of 2 pi electrons or if the total number of pi electrons is equal to 6, if I put n equal to 1, the total number of pi electrons will be 6 or 10. If you put n equal to 2, the total number of pi electrons will be 10 and so on. So if the total number of pi electrons that are getting delocalized in a cyclic planar and fully conjugated system, if the total number of pi electrons that are getting delocalized is either 2, 6, 10 and so on, then the compound actually turns out to be aromatic, right? And aromatic systems are extra stable. So a tropolium ion is aromatic. Now, if you look at this particular cation, even this is cyclic planar and fully conjugated, even this can have resonance out here. But if we count the total number of pi electrons here, then it's going to be 2 plus 2. So this only has 4 pi electrons that are getting delocalized, right? Now, we know that if the total number of delocalized pi electrons is a multiple of 4, so if the total number of pi electrons equals to 4n, where n is equal to 1, 2, 3 and so on, so if the total number of pi electrons are 4 or 8 or 12 and so on, then such cyclic planar and fully conjugated systems actually turn out to be extra unstable and they are what we call the anti-aromatic compounds, right? So this particular cation is anti-aromatic. Now, if you look at this final cation, there's a positive charge on this particular carbon. There's an empty orbital out here, but there aren't any pi bonds around it, right? So this can't be stabilized via resonance, but it does have these hydrogen atoms that are attached out here at the alpha position. So this can get stabilized via hyperconjugation, right? Now, between A, B, C and D, A, B and D, they are relatively stable. They are stabilized due to resonance or hyperconjugation or because they are aromatic. So these are going to be more stable compared to C, which is extra unstable as it is anti-aromatic, right? Now, between A, B and D, because B is aromatic and aromaticity leads to extra stability. So therefore, B is going to be more stable compared to A and D, followed by C, right? Now, between A and D, because A is stabilized by resonance, it can delocalize this positive charge much better compared to hyperconjugation, right? Hyperconjugation involves breaking this carbon-hydrogen sigma bond, which are relatively much more difficult to break compared to these pi bonds. So delocalization is much better in this particular molecule compared to this, which will ultimately make A more stable than D, right? So the correct order would be B greater than A, greater than D, greater than C. Now, out here, instead of a cation, I have these different anions and you need to compare the stability of these anions. So how will you do that? Why don't you pause the video and try and find out the correct order? Well, out here, I have a carbon anion, I have this lone pair on carbon atom that can get stabilized via resonance, right? Now we have a CN group, CN is C triple bond N and this is an electron withdrawing group, right? So this can withdraw electrons via resonance and if you draw the resonating structures, you'll realize that CN can bring about a positive charge at these very specific positions. Right? In fact, in one of the resonating structures, we can have an empty orbital out here and this lone pair can overlap with that empty orbital. So therefore, this CN group can delocalize this lone pair over it, right? Presence of a CN group at this pair position will help in delocalizing this negative charge from carbon even more. So CN is stabilizing the carbon anion via the minus R effect. Now if you look out here, we have an NO2 group and NO2, let me draw the proper structure of NO2. So NO2 is also an electron withdrawing group. It's a strong minus R group and it can withdraw electrons from the benzene ring, right? However, this time if you draw the resonating structures, you'll see that this NO2 brings about positive charges only at these positions, right? So there won't be any positive charge or an empty orbital right under this carbon atom. So this lone pair of carbon cannot overlap. So the presence of an NO2 at the meta will not aid in delocalizing this negative charge of the carbon, right? So therefore, even though NO2 is a strong minus R group, it's stronger than even CN, but this effect doesn't help in delocalizing this negative charge. So the minus R effect of NO2 is not that important out here, right? In fact, NO2 is also an electron withdrawing group via induction. So the more important effect will be the minus I of NO2, right? It can also withdraw electrons across sigma bonds via induction. So therefore, between A and B, because the cyan O group is helping in delocalizing this negative charge even further via resonance, while NO2 can at best only pull some electron density away via induction. So therefore, A is going to be much more stable compared to B, right? So A is going to be more stable compared to B. Now let's look at C. Now C is a cycle. It's cyclic and it's planar. And as you can see, this lone pair can also get involved in resonance, right? So this is a cyclic planar and fully conjugated system. And in such systems, we should be careful about aromaticity and anti-aromaticity, right? Now if we count the total number of pi electrons that are getting delocalized out here, so we have two electrons out here and this lone pair will also get converted into pi bonds. So we have a total four pi electrons that are getting delocalized. We have four pi electrons that are getting delocalized. And because this is a multiple of four, so this system will be anti-aromatic, right? So we have an anti-aromatic compound in C. Now if you look at D, even this is cyclic planar and fully conjugated, this lone pair in the pi electrons can get delocalized over the whole cycle. And this time if we count the total number of pi electrons, we are going to have one, two, three, four, five, six. We are going to have six pi electrons that are getting delocalized over my cycle, right? Now this follows Huckel's 4n plus 2 rule. So what we have out here is an aromatic compound, right? So D is aromatic. Now because aromaticity leads to extra stability, while anti-aromaticity leads to extra instability, so therefore D is going to be the most stable amongst all of these and C will be the least stable, right?