 great hello one I am both it and instructor of code CH 105 welcome to the class my contact address is department of chemistry then I wish you can reach see by email that is the day IITB dot AC dot IN, okay. Once again, it is DMIT, DMAITI at the rate, IITB dot AC dot IN. My office phone number is 022-2576-7155 and my cell phone number is 098-7155. Please feel free to contact me whenever you need. So, this course is mainly on inorganic chemistry, right. I assume that all of you have had some sort of introduction in inorganic chemistry. Basic introduction I will not take you through. I will mainly focus on the contact or content of these course. Well, let us start with some simple understanding. It is optimist sees the glass half full. The pessimist sees the glass half empty, right. But the chemist which we think you know hopefully better of all of them, chemist sees the glass completely full, of course, half in liquid state, half in vapor, right. With this note, I will briefly go into the periodic table. Periodic table is something all of you have come across. It is the gathering of all the elements that we have seen so far. It is an ever evolving process. For example, still after 105 elements which we are thinking for quite long, that is the maximum number, there could be a lot of other elements which are recently developed or recently discovered. Not only that, the list does not really stop over here. It is as I said ever increasing and thereby their accommodation in the periodic table has to be also be there, right. Based on their atomic number and their properties, as you know, these atoms are organized in a particular fashion, ok. I will come back to that later. First, let us look at that, try to look at the nomenclature of the unknown entities, unknown atoms, right. So, lot of other, lot of new elements are getting discovered. If something like 114 atomic number containing element is discovered, what will be the nomenclature? The nomenclature for the earlier elements are already done. So, we do not have to worry about their noble creatures like hydrogen, helium, lithium, beryllium, these, you know, carbon, nitrogen, so on. But if a new element which is having atomic number, let us say more than 100, 105, 100, you know, so on is discovered, what would be the way to nomenclature them in terms of IUPSE, ok. The simple rule here is given. As you can see, 0 should be pronounced or should be stated as nil, 1 should be un, 2 should be by and so on, up to 8 should be oct, 9 should be n. So, for example, if some element is having atomic number 114, then it should be named as un, un, quadium, ok. So, it would be u, u, q. So, this m should be at the end of it. Anyway, un is 1. As I said, for 1 you should use un. Second one will be another un, un, un, quad for 4 we have quad and it ends with IUM. So, the name for 114, element number 114, atomic number 114 should be un, un, quadium. If it is 118, I am sure you can name it by now. So, it would be un, un, octium, right. 11 that is un, un, 8 that is octium, un, un, octium, right. Well, this is a small homework for you. The money which has recently been discovered, although money has been discovered long back, let us assume that money has been recently discovered to be an not yet identified super heavy element. Let us say money is an element, heavy element, what should be its, you know, IPSC nomenclature. Simply, the IPSC nomenclature you can say it is un, octium, right. Money, the proposed name for this should be un, octium. Let us move on. The factors that affects the atomic orbital energies, that is something we need to quite understand before we try to understand the periodic table. Of course, principle quantum number wise, if you look at 1s should be having lower energy than 2s, then 3s and then 4s and so on. 2p should be having lower energy than 3p, 4p, 5p. But still there are few factors we need to really understand quite detailed, right. So, atomic orbitals energies are affected mainly by few factors. One is nuclear charge, what is the charge at the nucleus? That is very simple. You can determine that is the atomic number, let us say and shielding by other electrode. This is what we need to simply understand how it can affect the atomic orbital energy, ok. So, in an ideal world what we see is number of protons and number of electrons are equal. It is equal, right. In any atoms if you see number of protons and number of electrons are equal. But still you see that different electrons, let us say outer sphere electrons for different atom will be feeling or will have different feeling, why is that? That is mainly due to the fact that the electrons that the inner electrons are affecting the outer electrons indirectly, right. Let me try to give you an example. So, if you have atomic number, let us say 3 that is lithium, right. Lithium is having 3 electrons and 3 protons of course. Now for the third electron that means that the outer sphere electron the most outside electron will have 2 electrons that is inside. Now these 2 electrons which are inner orbital electrons will try to neutralize the nucleus charge because you know it is a positive charge and negative charge should be neutralizing. In an ideal world 3 proton and 3 electron should be kind of canceling each other, but in reality that does not happen because you know that is because of the fact which we are going to discuss now. Let us say another one having you know atomic number 5. So, 5 proton and 5 electron once again they should neutralize the outer one 1 s 2 2 s 2 and 2 p 1 that is the p 1 electron will be having total 4 inner electrons, right. Yeah 1 s 2 and 2 s 2 these 4 inner electrons will be trying to neutralize the positive charge at the nucleus, right. So, thereby the fifth electron for this atomic number of elements having atomic number 5 should have 4 electrons trying to neutralize the 5 positive charge, right. So, the way atomic number will be 3 containing element or atomic number 5 containing element will face the nuclear positive charge is going to be different, ok. Let us look at we will come back to that again. So, higher nuclear charge increases nucleus electron interaction and lowers the sub level energy that is quite understandable. If you have a higher nuclear charge at the nucleus that will try to attract the electron close to its close to it nucleus electron interaction will be higher and thereby their sub level energy will be minimized, right. Now, since these atomic orbitals so called S, P, D and so on can have different orientation can neutralize the nucleus charge to different extent we are going to see the extent to which the nucleus is attracting the outer sphere electron that is also going to be varied. For example, if some electrons are penetrating too much that means are able to neutralize the nucleus charge more effectively then the neutralization will be much more felt and therefore the attraction between the nucleus and the outer sphere electron will be less. So, S electron being more penetrating that means S electron can be neutralized and penetrating the positive charge at the nucleus more effectively and thereby the penetration power gives more neutralization of the positive charge and in term what happens I mean you will have very little attraction between the nucleus and the outer sphere electron. So, the shielding these neutralization of this positive charge by the inner sphere electrons are called shielding it is just like a battlefield if you are you know the kings are protected by the different layers of soldiers. Now, the soldiers which are at the outmost zone will be facing the hit they are the one who are fighting right they are the outer sphere electrons comparable to outer sphere electron. Now, if this inner sphere electron or the soldier in the inner sphere zone or soldier which are close to the king can protect can neutralize or can protect the nucleus very effectively. Therefore, the king will not have much attraction towards the outer sphere those soldier or electrons ok. So, this is the shielding is basically you know you try to protect something from you try to protect the outer sphere electrons from the nucleus right that is what is shielding you have nucleus which is attracting the outer sphere electron. Now, this attraction can be minimized by this inner sphere electron based on their penetration power based on their neutralization power ok. So, this is something you know something equivalent to what I would like to call is true love what is what is that it is a it is something like you know when everything settles down how much attraction still left for between these outer sphere electron and the and the nucleus. So, there are that is a lot of factors one of those factor is inner sphere electron inner sphere electron try to neutralize the positive charge after those neutralization how much attraction still left at at the outer sphere electron right. So, this is what is called you know z effective or the effective nuclear charge z star that is something very important and I would like to call very simply just finally, it is a it is a true love right. So, what we tried to discuss so far is simply it it is really matters the the inner electron really matters a lot if you want to talk about the outer sphere electron outer sphere electron although they are not directly attached with the you know with the inner sphere electron, but inner sphere electrons are the one who are going to take care of the outer sphere electron. If inner sphere electrons are very much penetrating then the attraction between your nucleus and the outer sphere electron going to be very little right ok. So, we will come back to that. So, what is effective nuclear charge then z star effective nuclear charge z star is your total atomic number whatever it is let us say 5, 5 atomic number 5 that is hydrogen helium lithium beryllium then boron right 5 5 minus the shielding total those 4 electrons let us say if we are talking about the 5th electron. So, the 4 electron how much shielding that that is having that is going to be the effective nuclear charge or z star for the 5th electron ok. So, the inner electrons that is the one which matters most. Now, how to calculate the z star we should have a clear idea it is more than basic understanding how z star is getting affected by different orbitals. We should have some sort of clear understanding how to calculate the z star of course, it is not possible to calculate exactly, but some sort of theory perhaps can be you know can be summarized or can be brought to estimate the z star ok. So, there are no uniform ways available to determine z star. We can have z star calculation let us say you can follow the procedure I am trying to discuss over here. Actually if you look at different books they might will be discussing different ways to calculate the z star, but let us follow the procedure what we are trying to discuss over here. It is the procedure I think it is also in this text book which is going to be done you know very standard text book for you is Sriver Atkins this inorganic chemistry book. If the electrons that means the outside electron or the whatever electrons you are interested in if that electrons electron is S or P orbital then you can calculate the z star as follows ok. All electrons in higher principle cell let us say you have 1 S 2 2 S 2 2 P 1 electronic configuration. You want to calculate the z star for 2 S electron you also of course, you can calculate the z star for 2 P electron just for example, you want to calculate the z star for 2 S electron right 2 S second electron right. So, the fourth electron you would like to calculate the z star for right. So, therefore, the 2 P electron ok. So, any electron let us say in this case you would like to calculate the z star for 1 S electron not 2 S electron ok. 1 S electron means. So, you have total electronic configuration 1 S 2 2 S 2 2 P 1 you want to calculate the z star for 1 S electron. That means, the outer sphere electron that is 2 S and 2 P you do not have to worry about you just have to worry about the electrons in the same orbital or that of the lower atomic number ok. So, lower principle number N minus 1 principle cell. Now, all electrons in higher principle cell contribute 0 to sigma. So, electrons that is outside you do not have to worry about that. Each electron in the same principle cell should contribute 0.35 I will come with an example very soon. Electrons in the N minus 1 cell should contribute 0.85 electrons in deeper cell that means, N minus 2 N minus 3 and N minus 4 and so on should contribute 1 to sigma ok. Let me show you one example then it should be clearer ok. So, for fluorine for example, you want to calculate the z star for the fifth electron ok. So, it is 1 S 2 2 S 2 2 P 5 this P 5 fifth electron you want to calculate the z star 4 right. So, the formula you have to follow is simply z star equals z minus shielding the sigma right. So, the screening constant for one of the outer electron 2 P screenings constant for one of the outer electron 2 P that means, this P 5th electron is. So, the same quantum number is 2 S and 2 P. So, the fifth electron you are considering. So, you should not be calculating the fifth electron. So, you should take 2 P 4 electronic configuration because you are trying to calculate the z star for the fifth electron. So, overall in the principle cell 2 you have 2 S 2 and 2 P 4 total 4 plus 2 6 electron 2 S 2 2 P 4 that means, 2 2 S electron and 4 2 P electron should contribute 0.35 each this is the same principle cell as that of the one we are interested in. So, 6 time 0.35 that is 2.1 the one below below the principle cell n minus 1 should contribute 0.85. So, it should be 0.85 times 2 because 1 S orbital has 2 electrons right. So, 0.85 times 2 it should be 1.7. So, the sigma or the shielding constant should be the combination of all these effects. So, 0.1.7 for 2 electrons plus 2.1 for the 2 S and 2 P electron overall sigma is going to be 3.8 as you can see and the z star should be equal to 9 that is the atomic number of the fluorine 9 minus the sigma which is 3.8 that is 5.2 right. Now, once again that rule is very simple just to remind you anything outer sphere outside the zone we are interested in should contribute to 0. If for example, in if you are interested in 1 S electron then 2 S and 2 P electrons they should not contribute anything since you are interested in fifth electron. So, all the electrons below this should be considered anything above it if there is anything above it should not be of any interest for calculating the shielding constant. After that you should see the same principle cell quantum number principle cells that is 2 S and 2 P and each of these electrons should contribute 0.35 anything less 1 less that is n minus 1 should contribute 0.85. If there was more electrons below this in this case it is not possible then that should have contributed to 1 that means effectively it will be neutralizing all of the positive charge that is over there ok. What essentially we are talking about we are trying to tell you that for each electron we add technically we are also increasing 1 proton at the nucleus this proton and electron should be neutralizing each other, but in reality they do not neutralize each other. To what extent this electron that is negative charge is neutralizing the positive charge that is what we are trying to calculate. If it is if it is the electron which is little bit outside or the electrons we are interested in in the same cell those electrons are there then they should not be neutralizing the nucleus charge completely. The one which is deeply buried or you know inner sphere electron will be neutralizing the positive charge of the nucleus effectively. If it is n minus 1 cell electron there will be neutralizing the positive charge of the nucleus little more effectively compared to the outer sphere. If it is n minus 2 or n minus 3 essentially they should be neutralizing one electron should be neutralizing one positive charge right. So, that is what they are contributing 1 to the sigma. So, this calculation or this mode of calculation whatever we have discussed here right now is valid for the electrons if we are calculating for p and s orbitals right. Now if we are interested in discussing the effective nuclear charge z star for d orbitals or f orbitals why they are different of course their penetration you know their shapes are different and their ability to neutralize that nuclear charge is completely different and actually very less. So, therefore, the calculation varies little bit over here. For example, first thing remains same anything any electron outside the electron which we are interested in should contribute 0 they should not be affecting in the calculation each electron in the same principle cell as before for the s and p of cell should contribute 0.35 to sigma all inner cell n minus 1 n minus 2 and so on should contribute 1.0 previously what you have seen n minus 1 cell is contributing 0.85. But in these cases it should be contributing 1 not 0.85. So, all the inner cell electrons n minus 1 n minus 2 n minus 3 n minus 4 whatever possible should contribute 1.0 ok. So, it is actually calculation is very simple you can take a look at any standard textbook as we mentioned Sriver Atkins perhaps would be a good idea. So, here is some effective nuclear charge that we are looking at for the outer most electron ok. For example, hydrogen hydrogen's effective nuclear charge should be 1 ok. If you go down the periodic table hydrogen lithium sodium potassium rubidium you see there is very little increase if any or very little value for the z star ok. So, what essentially happening here is every time you are going from top to bottom ok you are increasing a new cell as we are saying the since the cell number is increasing inner cell can effectively neutralize the positive charge inner cell electrons can effectively neutralize the positive charge and thereby overall what you are seeing the z star remain almost constant ok as specifically if you can see potassium rubidium cesium they remain constant ok. Now, if you look at periodic table carefully and try to work from left to right ok. If you are working from left to right let us say for series for lithium, beryllium, boron, carbon, nitrogen, oxygen, fluorine and neon what is happening here your atomic number increases 3, 4, 5, 6, 7, 8, 9, 10 right of course. Now, quite interestingly your effective nuclear charge is also increasing dramatically for that does mean simply that does mean that the electrons each case electrons are increasing and each case your atomic number is also increasing the electrons cannot neutralize the nuclear charge effectively. It is the same principle cell ok where electrons are entering since the same principle cell electrons cannot neutralize the nucleus charge effectively what essentially we are seeing is the contribution instead of one neutralizing one electron by one proton electron is only contributing 0.35 towards the neutralization of the positive charge and therefore effective nuclear charge is gradually increasing ok. So, if you look at the attraction of the nucleus for lithium towards the outer cell or outer electron let us say x whatever it is if you go down the period if you walk from left to right you will see the nucleus will be trying to attract the outer electron very effectively. That will in turn will have an effect in your size size of the atoms technically speaking if your nuclear charge is increasing or effective nuclear charge is increasing. So, effectively nucleus will pull in the electrons towards it. So, your size will be decreasing from left to right right. So, we will come back to that again. So, what we have seen in the PA dictable I think you have already learned about it. So, if you are walking from top to bottom in a periodic table top to bottom your principal cell increases. So, 1s 2s 3s 3p 4s 4 and then 5s and so on you are effective. So, the principal cell increases since the inner electron can neutralize the positive charge effectively in these cases. So, 1s 2 2s 3s 2 4s 5s since these inner electron can neutralize the positive charge effectively thus increase in this atomics you know atomics this cell or this principal cell will result in increasing size from top to bottom size should increase and in it in the periodic table if you walk from left to right then what will happen electrons are getting into the same cell 1 by 1 exactly in the same cell let us say in this case 2p. So, the electronic configuration over here is 1s 2 2s 1 then 2s 2 then 2p 1 2p 2 2p 3 2p 4 2p 5 2p 6 right. So, it is in the same principal cell the electrons are getting incorporated and therefore, therefore, the neutralization of these electrons neutralization of the positive charge by these electrons are not going to be too much and since the Z star therefore, going to be increased they will try to pull in the outer sphere electron and the size will get smaller and smaller ok. So, the Z star will try to squeeze in kind of the size the size will decrease dramatically.