 So, we have seen that anything that has a lone pair of electron can push this lone pair into a pi system and in this way it can increase the electron density of this kind of pi systems and these are what we call the placer groups. So therefore anything that has a lone pair of electron, if we have something like an O- or even something like OCOCH3 because all of these have lone pairs which can be pushed into a pi system, so all of these are what we call our placer groups, right? Now the ability of each of these groups to donate electrons is not the same. Some of these groups are better at donating electrons compared to others. So how do you figure that out? Let's find out in this video. So what I have out here are two placer groups O- and NH- and as you can see they have this lone pair which they can donate to a pi system thereby making them placer, right? So the question is which of these two will be able to donate these electrons much more easily? Now as you know, oxygen is much more electronegative compared to the nitrogen atom, right? An oxygen atom loves electrons much more compared to nitrogen. So therefore these lone pair of electrons will be held much more tightly by the oxygen atom, by the oxygen nucleus compared to nitrogen, right? So therefore it's going to be much easier to donate these electrons from nitrogen compared to oxygen. So using this rationale we can say that because nitrogen is less electronegative, it can donate electrons more readily compared to the more electronegative oxygen atom. So the placer of NH- will be greater than that of O-. Let's take another example. Between OH and NH2, which one do you think will be the stronger placer? Well using the same rationale we can say that because oxygen is more electronegative than nitrogen, so these electrons will be held much more tightly by the oxygen atom compared to nitrogen. So therefore electron donation from this nitrogen atom will be much better compared to oxygen, right? So NH2 will be a stronger placer group compared to OH. Now what about these four groups? Which of these two set of functional groups will be stronger placer? The negatively charged O- and NH- or the neutral OH and NH2? Well I'm sure you must have guessed it. Because these functional groups are negatively charged, so they are relatively unstable because as you know in nature, neutral compounds are more stable compared to the charged species. So both of these groups, they don't want this negative charge over them. So therefore they are much more likely to push electrons away from them compared to these neutral molecules, right? So therefore these groups are going to be stronger compared to this. So I can go ahead and put these groups out here, right? So these are going to be stronger compared to this. Let us now take a look at these placer groups. Now even out here we have an oxygen and a nitrogen atom with this lone pair of electrons. But unlike in these cases where this lone pair was localized over the oxygen and the nitrogen atom. In this case these lone pairs can resonate with this C double bond O, right? Even out here this lone pair can show resonance. So these lone pairs are not localized but they are delocalized. They are not exclusively over the oxygen atom. So this is going to decrease the electron density over the oxygen atom or we can say that these lone pairs will be less available for donation. So these groups are definitely going to be weaker than these ones, right? Now between OCOCH3 and NHCOCH3 in both these cases these lone pairs are delocalized. So if we have to compare between these two then we again need to take the help of electronegativity. Because oxygen again is more electronegative than nitrogen, so these electrons will be more tightly held to the oxygen nucleus compared to the nitrogen. So this whole group will be a stronger placer compared to this, right? So overall these are going to be weaker than these groups and between these two NHC double bond O CH3 will be stronger compared to OC double bond O CH3, right? Okay, let's take one final example. So what I have out here are halogens. Now halogens are highly electronegative, right? Fluorine in fact is the most electronegative element in the periodic table. So these electrons are very tightly bound to the fluorine atom. So the placer effect of fluorine is relatively poor. Now if we had to compare the placer effects of these halogens, where would your answer be? Well, I can go ahead and say that as we go down the period as we go from fluorine to iodine, we know that the electronegativity decreases, fluorine is the most electronegative element followed by chlorine, bromine and iodine. So therefore the electron donation from iodine should be the easiest compared to fluorine, right? So therefore the placer effect of fluorine should be less than that of chlorine, should be less than that of bromine, should be less than that of iodine, right? However, it turns out that this order is actually incorrect. The correct order turns out to be exactly opposite of this. So fluorine in fact turns out to be a stronger placer group compared to chlorine, compared to bromine, compared to iodine. Now to understand the reason behind this anomaly, we need to look deeper into the carbon-carbon pi system. So whenever we are talking about electron donation or electron withdrawal, in general we are talking about electron donation or electron withdrawal to a carbon-carbon pi system. Now as you can see this carbon-carbon pi system, in fact let me bring this down a little bit, these carbon-carbon pi system are made up of two p orbitals, right? Now oxygen, nitrogen and even fluorine, all of these belong to the same period as carbon, so their lone pairs, let me in fact go ahead and draw the lone pair of fluorine. So the lone pair on this fluorine atom also lies in a 2p orbital, right? So the lone pair on fluorine as well as on oxygen and nitrogen belongs to a 2p orbital. So the orbital which is going to overlap with this pi system also belongs to an energetically similar 2p orbital. However, as we go down the group, as we go from fluorine to chlorine, chlorine is of a different period compared to fluorine, right? Chlorine belongs to the third period, so this lone pair in case of chlorine will belong to a 3p orbital. Now as you can see there's going to be a greater mismatch between these two orbitals, there's going to be greater energy difference between these two, so there isn't effective overlap between these 3p orbital of chlorine and the 2p orbitals of this carbon-carbon pi system. Now this energy mismatch keeps on increasing as I go from chlorine to bromine and in case of iodine it's going to be the highest. If we have iodine, the lone pair of iodine will belong to a much bigger 5p orbital, there's going to be a much greater energy mismatch and therefore the extent of overlap is going to be even poorer. So therefore as we go from fluorine to iodine, the placer effect to a carbon-carbon pi system keeps on decreasing. Now a quick note that I'd like to add out here is that because these halogens are really poor placer groups, so resonance effects are actually not that important when it comes to halogens and inductive effects becomes more dominating compared to even resonance. So keep this in mind, inductive effects may become more important than resonance when it comes to halogens.