 Now, this video is in continuation to our previous video on inert peripherals. So before you go ahead and watch this video, please make sure that you have seen the first video and then come back to this one because here we are going to discuss a pretty interesting question. So this video is basically inspired by a question that a friend of mine asked about inert peripherals. And since many other students might also have a similar doubt, I thought why not make a small video and clarify it here. So from the previous video, we saw that the shielding effect of the different atomic orbitals is in the order s greater than p greater than d greater than f, that is s orbital will shield the valence electrons most effectively, whereas d and f orbitals have very poor screening or shielding effect. Now I also mentioned that this shielding effect decreases as we move away from the nucleus. So that is 1s would have greater shielding effect as compared to 2s, which would have greater effect as compared to 3s and so on. So here is where my friend had a doubt. So his doubt was that, if you look an example here, so this is gallium and it has 3 electrons in the valence shell, 4s to 4p1. So if you assume or if you compare the shielding effect of let's say a 1s orbital with a 3s orbital, so which one would shield better? Still according to what we just discussed, 1s should shield better. So his doubt was that if you assume this is the 1s electron and this is the 3s electron and shielding is nothing but the electron-electron repulsion between these electrons, you know the inner core electrons of the valence electron because of which the nuclear charge, effective nuclear charge decreases. Then according to him, the 3s electrons would repel the valence electrons better than the 1s electron which is much farther away from the valence electron. So according to this, the order should not be 1s greater than 2s greater than 3s. In fact it should be the opposite that the electron farthest from the nucleus will shield the valence electron the most and the electrons closest to the nucleus would shield it the least. But that is not the right answer. The right answer is still this one. Now pause the video for a moment and think why this is the correct order. Okay so the answer lies in the size of the atomic orbitals. You see as you go away from the nucleus the size of the atomic orbitals increases. Now if you look at this, this is the 1s orbital. If you assume this is the 1s orbital, then this would be our 2s orbital, this would be our 3s orbital and where can I draw? This would be our 4s orbital. That is as the n value increases, this is the principal quantum number, the number of shells increases, the size of the atomic orbitals also increase. But all of these orbitals can still hold only 2 electrons right? Now on comparing this you can see that the electrons are most concentrated in the 1s orbital whereas they occupy a much larger area in the 4s orbital or as the orbital size increases. That is as the size increases the charge per unit volume decreases, the electron density decreases and therefore the larger atomic size orbitals cannot effectively repel or shield the valence electrons effectively. So this is why the correct order of shielding would be 1s greater than 2s greater than 3s greater than 4s and so on. And if you compare the orbitals they would be s greater than p greater than d greater than f. Now as we discussed in the last video, because of the poor shielding effect of the d and f orbitals, these valence electrons are drawn closer towards the nucleus and within that it becomes more significant for the s electrons, that is s electrons are drawn closer than the p electron. So as a result the s electrons would want to remain paired and not participate in chemical reactions. In other words they would prefer to remain inert. Now the inert pair effect becomes increasingly significant as we go down the group. That means in group 13 inert pair effect would have maximum impact on thallium. While both the s and p electrons get drawn closer towards the nucleus, it affects the s electrons more than the p electron. So if this was the original 6s and 6p orbitals and the gap between them, after inert pair effect the orbitals are drawn closer towards the nucleus, that means their energy level decreases but the energy gap between 6s and 6p is likely to be more. That is the energy gap between them is likely to increase. And because of this we need to provide more energy to unpair the 6s electrons and therefore only the 6p electrons participate in bonding.