 In this video we are going to talk about the inert periphery effect. Now we know that in certain groups like groups 13, 14 and 15, the heavier members of the groups exhibited a different oxidation state, that is a lower oxidation state compared to the group oxidation state. For example in group 13 the group oxidation state is plus 3 that corresponds to the total number of valence electrons, however the heaviest member thallium was more stable in plus 1. Like that in group 14 and 15 the heavier members like lead and bismuth were found to be more stable in the lower plus 2 and plus 3 oxidation states instead of plus 4 and plus 5. Now you can see that these two oxidation states differ exactly by two units, right? So my question is where does this difference of two units come from? I mean what does this two units refer to? Well it looks like the two refers to the two ns electrons that are somehow not participating in bonding. Yes, in all of these groups we can see that the lower oxidation state corresponds to the number of electrons in the p orbital. But what is happening to our ns electrons in these elements? I mean what makes them inert in these heavier elements? Well this reluctance of the ns electrons to participate in bonding is something that we can attribute to inert per effect. And that's what we will study in detail in this video. When we go down the periodic table a couple of things happen. Firstly, the nuclear charge increases because we have more protons in the nucleus as its size increases and at the same time newer shells get added with successive periods and with that the inner electrons also get filled like this, right? Now these inner electrons do two things. First they get attracted towards the nucleus obviously because they are oppositely charged but at the same time they also push the electrons in the outer shells away from themselves due to electron-electron repulsions. So because of this the outer electrons or more specifically the valence electrons get pushed by these inner electrons and they get farther away from the nucleus such that overall they experience less effective nuclear charge. In other words, the inner electrons shield the outer electrons from the influence or the pull of the nucleus and this is what we call shielding. Shielding refers to the core electrons repelling the outer electrons and thereby decreasing their effective nuclear attraction. But here's a question, do you think all the inner electrons would shield the outer electrons in the same effective manner? What I mean is, suppose you have the valence electrons in the fourth shell and you have completely filled inner shell. So let's say n is equal to 3 and you have 3 electrons in the fourth shell. Now how many electrons can be accommodated in this shell? That would be 2n square number of electrons which is nothing but 18. Now these 18 electrons are spread across different sub-shells like 3s, 3p and 3d sub-shells. So the question is, do you think that the electrons that are present in the orbitals of these sub-shells would all have the same shielding impact? Pause the video and think about it for a moment. Well the answer is no. This is because these orbitals in these sub-shells have different degrees of shielding depending on their shapes and how close their electron density is to the nucleus. Ok, let me explain. We know that the atomic orbitals have different shapes, right? The s orbital is spherical in shape whereas the p orbital has a dumbbell shape. Now from the shape of the orbitals we can see that s orbital has maximum electron density towards or near the nucleus. And when you compare an s orbital with a p orbital you can see that the electron density in p actually increases as you move away from the nucleus, right? And in the case of d and f orbitals it increases as you move further away from the nucleus. For example, if you compare an s orbital with a p orbital, which among the two do you think can shield the outer electrons better? So let's assume that this is a nucleus and here is your s orbital, right? Spherical in shape. And how would a p orbital look like? In a p orbital the electron density increases as you move away from the nucleus. So like a dumbbell shape, something like this. Let's say that you have a valence electron somewhere far away. Let's assume that this is the distance and you have valence electrons here, ok? So let's assume that there are three valence electrons. This is a very rough representation folks. So remember that this is not to scale and don't blame me if it's not really nice. Now would the s or the p orbital be able to shield these outer electrons or valence electrons better? Now the s orbital has more electron density near the nucleus and that means it will be able to shield these outer electrons much better from the nucleus, correct? But what about the p orbital? So if you take a separate image here, clearly you can see that the p orbital will not be able to effectively shield these valence electrons from the nucleus as much as the s orbital. Now what do you think would happen to the d and f orbitals? So if you look at their shapes they are even more diffused. That means they do an even more poorer job of shielding the outer electrons from the nucleus. And this is why the correct order of shielding effect is s greater than p greater than d greater than f. Obviously 2s orbitals will shield better than 3s orbitals which will shield the outer electrons better than 4s orbitals. Ok now back to a question. If we have 3 valence electrons in the fourth shell and if you look at how the inner electrons shield these valence electrons that would be the electrons in the innermost shell or the first shell would shield them the most because they are closest to the nucleus followed by the electrons in the second shell and lastly by the third shell. And even if you look at the third shell electrons you can see that the electrons present in 3s orbitals would shield the most followed by 3p and least would be by 3d electrons. Ok now what is the consequence of the poor shielding effect with the d and f orbitals? I mean what happens to the valence electrons? So because of the poor shielding effect as I said before these valence electrons experience greater nuclear charge. So they are pulled closer or strongly towards the nucleus and it becomes difficult for them to participate in chemical bonding. Especially the ns electrons because once again they are closer to the nucleus right? And this reluctance of the ns electrons to participate in bonding is what we refer to as the inert pair effect. Thus in essence only the np electrons become available for bonding and this gets reflected in the stability of the lower oxidation states. That is plus 1 oxidation state in case of group 13 plus 2 oxidation state in case of group 14 and plus 3 oxidation state in case of group 15. Now remember folks this stability of the lower oxidation state is just one of the consequences of the poor shielding effect of the d and f orbitals. In the next video we will see what other properties get affected by this particular phenomenon.