 Now, this is just something you know these electron how the electrons are distributed outside the nucleus or you know there is a probability of finding electron as you you know like to call them can be you know more clearly be taught in physical chemistry course. Here I will give you just a brief idea may not be exactly a exactly clear idea, but you know it it is something which will which will give you an idea to go forward. So, how are the orbital steps of course, we have learned that steps are different, but then if you want to see that electron distribution of course, my physical chemistry colleague will will perhaps kill me if I try to define like that, but let me do it anyway you are going to learn better in physical chemistry course. So, if this is the 1 s electron we are talking about then 1 s electron is distributed with respect to the nucleus if the nucleus is at the center we would like to think like that the probability of finding the 1 s electron is going to be like this. So, the so, the probability of finding 1 s electron towards nucleus is quite high. If you look at 2 s electron probability at finding 2 s electron at the nucleus is not that much of course, there is a node at at this distance after that there is a probability of finding this 2 s electron is quite high. If you look at 3 s electron it is you know one more node added and you can see this if you compare 2 s versus 2 p you can further see of course, 2 s has some component very close to the nucleus over here 2 p does not have really that much component very close to the nucleus ok. So, the penetration is actually meaning that how much these orbitals are contributing towards or you know how much close they are towards the nucleus. So, as you can see 1 s is very close to the nucleus and thereby the penetration of 1 s is going to be the maximum and it can neutralize the positive charge most efficiently right as you can see for the other for example, 3 d 3 d is distributed or the maximum probability of finding 3 d electrons are really far from the nucleus. Therefore, the neutralization ability of 3 d electrons are going to be kind of least out of all these orbitals shown in here. Now if you compare directly 2 s versus 2 p you over you kind of you know overlap these 2 s and 2 p electron and then this is what the distribution curve you see as you can see penetration of 2 s electron is greater than 2 p electron ok, because you have a very good component of 2 s very close towards towards this towards this nucleus. So, the penetration that means, the you know the positive component is getting neutralized by these electrons at the orbitals the penetration of the 2 s electron through the inner cell or inner core is greater than that of 2 p electron because the latter vanishes at the nucleus. So, the 2 p electron vanishes at the nucleus whereas, 2 s electron are still there. Therefore, the 2 s electrons are less shielded than 2 p electrons ok. 2 s electrons are basically penetrating much more compared to 2 p electrons. Another thing you really need to understand very simply is you know the d electrons repeal each other much more compared to p or s electron. How is that? So, you know it is like involvement right. How much involved you are in a particular thing? Well let me take give you an example of your hostel rooms right. In your hostel if you have let us say you know 3 people in a room it is possible or 2 people in a room or 3, 4 people 4 people in a room ok. Now if out of the 4 people 1 or 2 of them are involved in something it could be music it could be studying it could be sports it could be anything outside you know outside the normal thing you would like to do. So, if they are too much involved they it is less likely that they will have much more time to interact with the third or the fourth you know roommates right. So, the if the first and second person in your hostel room are involved in something deeply they are less likely that they will disturb you right, but on the other hand third student or third mate and the fourth one are not being involved in anything that way. So, it is more likely that these 2 guys third and fourth guy will be having you know more number of fights or more number of disagreement right. So, this is what we are trying to see in D electrons. What is happening is D electrons are not really involved with the with the with the nucleus D electrons are kind of not capable of neutralizing or engaging with the nucleus. And therefore, the D electrons will be repelling each other very strongly or will have more conflict. On the other hand if you look at comparatively let us say S electron they are much more penetrating they are much more involved with the nucleus. And therefore, the S orbital having 2 electrons will repeal less compared to D orbital having 2 electron. So, what I am trying to say is 2 electrons if it is in D orbitals those 2 electrons will repeal each other more strongly compared to 2 electrons in the S orbital. Because S orbital is more engaged with the more engaged with the with the nucleus right. So, 2 electrons present in the same D orbital repeal each other more strongly than do 2 electrons in the same S orbitals. So, that is what I am trying to say it is hostile roommate. The electrons present in F are obviously much less influenced by the nucleus as compared to D. So, electrons in the S orbital is much more involved with the nucleus in the P orbital it is little bit less involved with the nucleus. The orbital electrons are less further less involved F electrons are even I kind of least involved with the nucleus and therefore, the repulsion among the F electrons will be much more among each other will be much more compared to any 2 electrons in any other orbitals right. So, when you are trying to calculate or trying to understand any effect or any effect by these different electrons you have to you have to essentially think of a bigger picture you have to you have to think how much penetrating there you have to think how much repulsion they will be having with each other and so on we will come to that. So, now I will discuss order of filling of orbital you have studied quite long back I guess that order of filling should be 1 S should be fill out first and then 2 S then 2 P 3 S 3 P not 3 D then then after 3 P should be 4 S then 3 D and this is where I think we need to understand little bit better why after 3 S 3 P gets filled, but after 3 P not really 3 D is getting filled, but 4 S is getting filled technically speaking 4 S is having more energy compared to 3 D because the principal cell is of higher number right 3 S since it is the electron in 3 S should be more or more stabilized means less or energy compared to 3 P. So, 3 S should be filled out first then 3 P after 3 P then it should be 3 D then 4 S, but in reality what we see that 4 S is filled out before 3 D of course, first thing we need to understand that 4 S and 3 D are having very similar energy. If anything we would like to think that 3 D is having lower energy compared to 4 S, but for that we must we must think of something which is which we try to discuss just briefly like 5 minutes back we were discussing. So, the D electrons are less penetrating right D compared to let us say S electron. So, compared to 3 D electron 4 S electron is going to be more penetrating and therefore, repulsion of electrons in 4 S orbital will be less compared to repulsion in the 3 D electron right. So, 4 S and 3 D first of all they are having very little difference in energy. Since the repulsion in 4 S is going to be much less compared to 3 D what will happen that 4 S is going to get preferably filled out compared to 3 D. So, this is an exception and this is how it can be perhaps explained. Therefore, occupation of orbitals of higher energy can result in a reduction in the repulsion. So, in this case if you occupy the electrons in the 4 S orbital it is going to ripple less consequently what we see is 4 S electron is getting 4 S orbital is getting filled out preferentially over the 3 D. Once again I think this is kind of becoming a disclaimer you have to really consider all other contribution not only you know numerically we have to really we should not be seeing that 1 S 2 S 2 P 3 S 3 P 3 D instead we should be looking at their relative energy and the consequences that that is going to going to have if you fill out these orbitals let us say 4 S versus 3 D. In this case once again the repulsion among the electrons determines which one is going to be preferentially filled out ok. Now, let me show you some other data which I think I like it very much where it is clearly showing how kind of electrons are getting filled out. 1 S you have one electron filled out for hydrogen let us say 1 S 1, 1 S 2 2 electrons filled out compared to this one you are seeing the dark curve ball of the same size 1 S electron getting filled out. Now over here for the lithium you have 1 S 2 and 2 S 1 so the 1 S orbital is this one 2 S orbital is this one. So, it is you know overall you see the distribution of electrons kind of you know not so accurate picture once again, but this is something you can visualize to basically understand for 1 S 2 2 S 2 you will see that 1 S of course, is very dark 2 S orbital also becoming very dark. If you see the boron that is 1 S 2 2 S 2 and 2 P 1 now one of the P orbital is getting occupied by electrons electron and then from 2 P 1 if you see 2 P 2 2 P electrons are getting occupied then 2 P 3 2 P 4 you are seeing the P electron clouds are getting more and more deeper and therefore, the overall if you see neon that is 1 S 2 2 S 2 2 P 6 this should be something like this. This is again this is an approximate picture this is just to give you a feeling how perhaps the electrons are distributed with respect to the nucleus in reality this picture perhaps you know in reality this does not exist ok. I think I have discussed this. Now of course, what we find that experimental data show that B block elements are of the form 3 D N 4 S 2 that means, 4 S electrons are getting filled out preferentially compared to 3 D electrons with 4 S orbital fully occupied. For example, scandium atomic number which is 21 by electronic configuration for scandium would be this is the argon configuration and then 3 D 1 4 S 2. So, it should be 4 S filled out first 2 of them and then 3 D 1 if atomic numbers scandium titanium let us say if you have then it would be 3 D 2 and 4 S 2 and so on. This order is followed in almost all cases that means, 4 S 2 electrons will be distributed first or given first and then the remaining electron if any should be going to your 3 D electrons. 2 atomic configuration actually kind of violates this general principle what are those cases you might will know that is chromium and copper. What is happening in these cases? In this cases it should be you know according to what we were discussing it should be 3 D 4 4 S 2 electronic configuration. Of course, now this is going to be a half filled full filled configuration or the zone where you can have 3 D 5 electronic configuration to gain the further stability if the electronic configuration will become now 3 D 5 4 S 1 not 3 D 4 4 S 2. In case of copper it is a similar situation where according to what we were discussing earlier the you know electron repulsion in the D orbital it should be 3 D 9 and 4 S 2, but it is not it should it is 3 D 10 and 4 S 1 because the D orbital if gets one more electron it becomes 3 D 10 which is a full filled electron configuration. So, the cases where you have half filled or full filled electronic configuration attainment then you are going to see the deviation from the standard 4 S filling first and then 3 D electron. As atomic number increases energy of 3 D orbitals decreases related relative to both your 4 S and 4 P. So, we are trying to say that 3 D should be having comparable energy with 4 S, 4 S should be having lower energy compared to 4 P that is fine. Since the D electrons can repeal each other strongly compared to 4 S electron this repulsion energy or the destabilization energy preferentially allows 4 S electron to be filled out first, but once D electrons getting filled out more and more what happens is D orbitals is is no longer having a higher energy compared to 4 S the energy of D orbital 3 D orbital decreases compared to 4 S. So, in that case the filling order should be as expected or normal 3 D should be filled out first and then 4 S and then 4 P. Now, another thing to understand see overall what we try to say is 4 S should be filling out first then 3 D, but removal what will be the removal order. So, this is what we now need to understand for transition metal series your filling order should be 4 S and 3 D, removal order should be again 4 S and 3 D. This is what I think it is little bit contradictory I think you would expect since 3 D is filling out later on. So, the removal time 3 D electrons should go first, but that is not the case 4 S fills out first 4 S gives out first. How is that possible or what is the explanation? Let us say take the examples of titanium. So, it is 3 D 2 4 S 2 fine 4 S filled out first and then 3 D filled out when titanium 2 plus we are thinking then it should be according to the filling order 3 D orbitals electron should go out that is not happening what we are seeing is in reality 4 S electron is going out not that 3 D electron. How can you explain this? This is one of the small topic we need to discuss. So, now once again I think you have to see the clear picture before and after electron removal. During filling out what is the consequences we have understood that D electron repeal each other strongly therefore, D electron should not be filled out first that is the filling time. So, the 4 S electron fills out first 4 S electron fills out first because 4 S electron is kind of more stabilized compared to the 3 D electron right. Now once you are trying to remove let us say you have removed 2 electron titanium 2 now you have to see this overall stability. 4 S electron is having lower energy or S electron in general is having lower energy mainly due to the fact that it is more penetrating. It is more penetrating therefore, it is getting more stabilized. Now when you remove 2 electron either from 4 S irrespective of 4 S or 3 D overall what is happening is your attraction of the nucleus towards the electron is going to be much more right electrons are less right. So, the atomic number remains same the attraction of nucleus towards the outer sphere electron or towards all the electrons is going to be much more compared to the normal or neutral situation. Since the z star that means effective nuclear charge is going to increase dramatically this will give you know give the stability ok. This will reduce the shielding of the electron. So, there will be some sort of contraction right. One has a small effect in 4 S because 4 S is already very close to the nucleus comparatively the stabilization of this the contraction will cause the stabilization to 3 D are much more compared to your 4 S. So, what is happening here is after removing 2 electrons the effect of contraction is going to be much more felt at 3 D compared to your 4 S because 4 S 4 S is already lower in energy and the lower energy is the reason why they are going to be they were filling out first. Now during removal since 4 S is already kind of neutralized or 4 S is already having lower energy the removal effect is going to be felt much more at that d orbital. So, d orbital energy will be much more stabilized and removal of the electrons will be much more easier from the 4 S compared to your 3 D. 3 D energy is going to be stabilized 3 D electrons are going to be stabilized compared much more compared to 4 S. So, the stabilization will lead to the removal of electrons from the 4 S not from 3 D ok. So, that is what we try to summarize in here contraction in d orbital causes a considerable decrease in energy. This decrease is evidently enough to lower the energy of 3 D well below 4 S in the ion that results from this. The removal of electron from 4 S will be easier compared to 3 D electron.