 In this video, we are going to look at how we can use resonance to determine the stability of a given set of molecules. So what we have out here are three different anions and we need to figure out which amongst this would be the most stable. So how do we do that? Let's start by looking at this particular anion. Well, out here I have this oxynatom and this oxynatom has this lone pair of electrons. But this oxynatom is connected to a single bond, right? Now this doesn't follow any of our resonance patterns that we have seen earlier. For the lone pair of this oxyn to resonate, it has to be attached to either some double bond or an empty orbital, right? So the lone pair of this oxynatom are going to stay 100% over this oxynatom. They are what we call localized electrons. They are localized over the oxynatom. On the other hand, if you look at this particular anion, then out here this oxynatom is connected to a double bond, right? So this perfectly fits into our resonance picture and one of these lone pairs of the oxynatom can resonate with this pi bond. One of these lone pairs can come over here while these pi electrons can move over to this carbon atom, right? So this will lead to the formation of a new resonating structure that's going to look something like this. We'll have a new double bond out here and because this pi electrons moves over to this carbon atom, so there will be a new lone pair over this carbon which will also lead to the formation of a minus 1 formal charge over this carbon atom, right? Now because this lone pair is connected to a single bond so it can't resonate further, but because it's connected to this double bond, so it can definitely resonate back. So it can resonate with this pi bond. It can resonate back and this will give us our original molecule, right? So what's happening is that these lone pair of electrons as well as these pi electrons, they're not confined or we should say localized to this oxynatom and this carbon-carbon bond, but in fact they are delocalized over this whole system, right? So therefore in reality this molecule that's given, this anion, is actually not going to be like this. In reality it's actually going to be a mixture of both of these resonating structures. It's going to be a hybrid of both of these structures and it's going to look something like this. Now if you notice in the resonance hybrid I have drawn a partial negative charge over the oxynatom. Now this would make sense because this lone pair of the oxynatom is getting pushed into the system so it's going away from this oxynatom. It doesn't stay purely over this oxynatom all the time. So therefore this oxygen should not have a full minus 1 formal charge. The formal charge should be lower than minus 1. So this oxygen will only have some partially negative formal charge, right? Now this delocalization also leads to the formation of a negative charge over this carbon atom so our resonance hybrid should also have some partial negative over this carbon atom, right? So to summarize what's happening is that because of this delocalization, because of this redistribution of electron density, this negative charge that was present over the oxynatom spreads out and gets shared between both the oxynatom as well as the carbon atom. So if you compare this molecule and this one then out here this negative charge is purely localized over this oxynatom, right? So there's a high charge density over this oxygen. So therefore if we now throw some H plus ions, if we react these molecules with some H plus then because this oxynatom has a high charge density, so therefore it's going to react much more with this H plus ions compared to the atoms in this molecule. Out here both the oxygen as well as the carbon atom has a much lower charge density and therefore it's much less reactive to this H plus, right? So therefore to summarize because of this delocalization the negative charge over the oxynatom gets distributed throughout the system leading to a decrease in the charge density which is charge per unit volume which makes this molecule much more stable and therefore much less reactive, right? So therefore 2 is going to be definitely more stable compared to 1. So if we come to this particular molecule this oxygen is connected to this double bond. So this can definitely resonate and this will lead to the formation of a new resonating structure that's going to look like this. But this time because this lone pair on this carbon atom is connected to another double bond. So we can also have a resonance between this lone pair and this pi electrons. This pi electrons are going to come and shift to this carbon atom. So this time I'll have one more resonating structure that's going to look like this, right? So therefore in reality this anion, in reality it's actually going to be a hybrid of all of these three structures and in reality it's going to look something like this, right? Now as you can see this negative charge is being spread out even more in this particular molecule the negative formal charge is in fact carried over three atoms out here compared to only two atoms as in this molecule. So therefore the charge density for this molecule is going to be even lower making it more stable and therefore less reactive, right? So 3 is definitely going to be the most stable of these three ions. So therefore the key takeaway out here is that a greater delocalization will ultimately lead to a lower charge density thereby making the molecule more stable and therefore less reactive, right? Now a greater delocalization also means more number of resonating structures. We had three resonating structures for this particular molecule and only two for this one. So therefore we can now go ahead and make a rule. We can say that greater the number of resonating structures, greater the number of resonating structures for a given molecule, greater will be the stability of the molecule, right? Now that you have understood how stability works, which amongst these three do you think will be the most stable? Well I'm sure you must have guessed it by now, it's going to be this one, right? We can draw loads of resonating structures for this molecule. This pi bond can shift over here so we can get a resonating structure that's going to look like this and then this pi bond can move over here and so on and so forth, right? However, if we look into this particular molecule, we can only have one other resonating structure in which this pi bond is going to shift over here. So we can only have one other resonance structure for this particular molecule. So therefore because the delocalization is going to be lower out here compared to this, there has to be a pi bond in this, compared to this. So this is going to be more stable compared to this one, right? Now out here this cation, the empty orbital of this cation is connected to single bond, right? There aren't any double bonds out here so resonance can't happen out here so this is going to be the least stable of these three. So one is going to be more stable than two and two is going to be more stable than three. One final question, which of these two do you think will be more stable? The phenoxide ion or the acetate ion? The answer to this question can actually be found in the next video but before you check that out, be sure to give this a thought and be ready for some surprises.