 Okay, is it visible now? Okay, I find it. Okay, yeah. So like I said, last class, we have discussed alkali metals. Okay, we have seen few trends, properties of the molecule and all those orders, right? And I have told you to go through the preparation and properties of various compounds of sodium like NaOH, Na2CO3, NaH, CO3, et cetera, from NCRT, right? So I hope you have gone through with that. Okay, next, this one thing which is left in this chapter is diagonal relationship that I have written already. Okay, so you write down the heading first, diagonal relationship. And I think probably we have already discussed this in periodic properties if you remember, right? So what is diagonal relationship you see first of all? See, when I look at the few elements like suppose if I take lithium, right? And then potassium, sorry, sodium, here we have magnesium and all, right? So group one elements we have lithium and sodium. Correct, the elements of group one. Now, when you see the position of beryllium magnesium, so beryllium is here in group two, just a second, beryllium is here, magnesium is here, right? And then we have P block elements that is boron and carbon also, like this. So here you see like few properties if you discuss when you go from left to right, left to right, electronegativity decreases. Sorry, electronegativity increases. Left to right if you go, electronegativity increases. And how would you remember this? Because we know fluorine is present at extreme right of the square table. So I know fluorine is the most electronegative element. Correct, so electronegativity increases. Top to bottom if you go in a group, then electronegativity decreases. Why again? Because in halogen, fluorine, fluorine, bromine, iodine, again fluorine is the most electronegative element going down the group electronegativity decreases. So this is the order follows by the all the groups, elements of all the groups, correct? Now if you compare the electronegativity of magnesium and lithium, those elements which are present diagonally to each other like this or beryllium and carbon also, right? Like this, diagonally to each other, placed diagonally to each other. You see when you are going from left to right, okay? When you are going from left to right and then top to bottom, then only you'll reach to magnesium. So left to right electronegativity increases. So what we can say, the electronegativity of beryllium is more than to that of lithium, okay? And again, we can say the electronegativity of, sorry, the electronegativity of is less than to that of electronegativity of beryllium is again more than to that of magnesium. Okay, what we can say from lithium to beryllium electronegativity increases and then beryllium to magnesium electronegativity decreases, correct? So from this thing, I'm just taking an example of electronegativity to make you understand, okay? If you consider this order of electronegativity, we can say that the electronegativity of, electronegativity of lithium and magnesium is comparable, okay? They have very close value of electronegativity. From this to this electronegativity increases and then decreases. So we can say lithium and magnesium will have almost same electronegativity, right? And this is true for beryllium and carbon also, right? Boron and, how did I do this? After beryllium, we have magnesium. Beryllium, boron, aluminium, carbon we have here and then SI and like that, okay? So the, see boron, the electronic configuration is 1S2, 2S2, 2P1, right? Beryllium is four, so it is 1S2, 2S2. So this is the thing, right? So like you see the elements which are present diagonal to each other have similar properties, we can say, right? So these elements which are placed diagonal to each other are, are assumed to show diagonal relationship, right? So lithium and magnesium, if I give you the value of electronegativity of lithium is it is 1.00 which you don't have to memorize and that of magnesium, it is 1.20. So they have very close value of electronegativity. Similarly, if you compare atomic and ionic radius also, right? Atomic radius is ionic radius. I'll just write down this order, ionic radius. You see atomic radius is 152 picometer and for magnesium, the atomic radius is 160 picometer. And 76 picometer and for mg, it is 72 picometer. So you see both values are very close. So what we can say the elements which exist and diagonally to each other shows similar properties and this we call it as diagonal relationship. The first point I have given you for electronegativity. Second point I have given you for size. Third point also I can give you here which is both have high value of polarizing path. Both have high value of polarizing path. Okay, so these values are not important. What is the radius we have? These values are not important but because of these three properties and various other properties are also same, we can say the elements exist diagonally to each other shows diagonal relationship. Like for example, I can show you the few similar properties of lithium and magnesium, okay? So just first point you write down the diagonal relationship, how they have similar properties. The first point to write down both lithium and magnesium, both lithium and magnesium decomposes slowly, decomposes water slowly to liberate hydrogen, to evolve hydrogen. For example, reaction you see, LI plus H2O gives LIOH plus on reaction with H2O, it forms hydroxide. Similarly, magnesium you see magnesium reacts with H2O, forms MGOH whole twice and H2O. This is again the reaction we have. Both elements also combines with nitrogen and forms nitride, right? So next one to write down both elements, all these points you have to memorize, elements, forms, nitride. Reaction you see, we have lithium plus N2, it gives two LI3N. Similarly, we have magnesium plus nitrogen and we are heating this mixture, it gives MG3. Now one more important for these nitrides, memorize, write down this, both nitrides react with water to evolve ammonia. For example, LI3N plus H2O gives LIOH plus NH3. Similarly, the reaction of magnesium you see, magnesium nitride, MG3N2 plus 6H2O gives 3MGOH whole twice plus 2NH3. Like this we have few more similar properties, like the third one you see, carbonates of both elements, carbonates of both elements decomposes on heating. For example, you see LI2CO3, when you heat this, it gives LI2O plus CO2, MG2CO3. When you heat this, it forms MGO and CO2. These are what, you see, these are mono oxides, okay? These are mono oxides, so one more point we can say that both elements form mono oxides, okay? So these are the few properties, similar properties I should tell you. And the last one, and this one is important, is the nitrates of both metal, both metal decomposes on heating and gives nitrogen dioxide, nitrogen dioxide, and oxygen. So I'll write down the reaction now, LINO3, this, it forms monoxide plus nitrogen dioxide O2. Magnesium nitrate, MGNO3 whole twice. When you heat this, it forms MGO plus NO2 and plus O2. So these are the few similar properties of lithium and magnesium we have, and that's why we call it as that these element shows diagonal relationship, okay? So you see, there are only theories in these chapters, okay? So I am not covering all those points, which is given, but the only thing we are discussing here is the important points, right? So it doesn't mean you don't have to go through all the things which is given in NCRT, at least. You must go through the process of preparation of all the elements. Like for example, I'll write down few names here, okay? That you must go through, okay? Well, if you have any doubt, you can ask me on this, but since there is nothing to understand here, you have to memorize those things. You can go through from NCRT or any other book if you have. That's why I'm not wasting our time over here. For example, you see, I'll just write down quickly and we'll move on to the next chapter. That is, suppose I'll give you the compounds of sodium. Compounds of sodium. You see here, a few compounds, and then name you have to memorize. For example, I'll write down, the first one I'll write down NaNO3. And we also call it as chili salt pitter. Next is NaNO3, ALF6. We call it as cryolite. These names you have to memorize. Next one is NaNO2SO4.10H2O. We call it as globbers salt. Next is NaNO3O8. We call it as soda pelspa, F-E-L-S-P-A-R. And the important one is NaNO2B4O7.10H2O. And we call it as borax. We also call it as tinkel. We also call it as suhaga. All these are names for this compound. But not that much. Not that much, it releases energy, but it is not that much explosive. But we use this, actually, we can mix Q compounds into this, and then it may use an explosive, or the formation of explosives. So it has few properties. And that's why we use this in explosives. These names you have to remember. Now you see for NaO8, the few preparation method you have to go through, NaO8, we have causticization. The preparation method is causticization. We also call it as Gauss' process. These are not important, okay? Another preparation method is Kastner-Kellner cell. Kastner-Kellner cell. These are the preparation method. And whenever you see these properties, in properties you have to go through mainly chemical properties, reactions. Chemical reactions you have to memorize in properties, mainly, okay? Those are the only thing which is important, okay? Similarly for Na2CO3 washing soda also, we have lee-blank process, okay? Solve ammonia soda process, okay? These are the process, lee-blank and ammonia soda process. Solve a process or we also call it as ammonia soda process, which is important, okay? All these things are given in NCRT, which you can go through, okay? So we are not wasting our time over there, okay? So again, we'll start with the second group that is group two, algaline, earth metal. These are the elements of group two, right? So group two. Okay, tell me one thing. How many of you have gone through the previous lecture? How many gone through the previous lecture? We have discussed the previous in periodic. Tell me honestly first, not hydrogen, alkali metals I have taken last class, okay? How many of you have gone through? Dedaash, Shant, Bharat, Nikhil, Liranjan, Ashutosh, Neha, Sris, Sriramya, Ashish, Shreya, Preeti, Andrew, Sridev, Ved, Advaith, Abhyay, Gaurav, Advaith. Tell me, how many of you have gone through the last session of alkali metals? Okay, fine, you must go through with that session because we have discussed all the trends in the periodic table, okay? And here also in this group, we have similar kind of trend, okay? So I am not going to explain you the logic again, okay? In that detail, right? So if you want to understand the logic, you have to go through that session, okay? Okay, so I have already, okay? I have shared the, no, NCRT, fine, but you just go through the session that I have taken last class. So the trends we'll see now, but logic will not go into that depth, okay? Because it is more or less same only, okay? Like you see, first of all, in group two, we have what elements we have? Is big, magnesium, calcium, these are the elements. If you see the atomic number of all these elements, beryllium has four, magnesium has 12, calcium has 20, then we'll add 18, so 38. And again, 18, so 56, and then 32, so 88. So difference you see here, here the difference in atomic number is eight. Difference in atomic number here it is again eight. Again it is 18, 18, and then 32. So these numbers, we call it as magic number for group two, magic number for group two. So if you know that generally we have, we know the atomic number of beryllium, magnesium, calcium, if somewhere however it is not required that much, but somewhere if it is required for a strontium, you can add these 18 to this 20, you'll get the number if you know this magic number. That is the one, no, uses of this. Am I audible properly or my voice is breaking? Tell me, am I audible properly or my voice is breaking? So if you draw the electronic configuration of all these elements, the electronic configuration of beryllium will be, it is 1S2, 2S2, right? 1S2 is the electronic configuration of helium. So we can also write it as helium initiative of 1S2 and then 2S2. For magnesium after helium we have neon, 3S2, right? So one thing you must keep in mind that the general electronic configuration I'll write down first, the general electronic configuration of the elements of group two is nS2, where n belongs to the period of that element. Like for example, you see beryllium belongs to second period, right? So here we have 2S2. Magnesium belongs to third period, 3S2. Calcium belongs to fourth period. So helium, neon, and then we have argon belongs to fourth period, so 4S2. After argon, we have krypton, right? And strontium belongs to fifth period, so 5S2. Similarly, xenon, 6S2, and then redon, 7S2, right? This is the general trend, okay? Now in this one, if you go, if I talk about the atomic radius, right? As we go down the group, as we go down the group, what happens to atomic radius? As we go down the group, so what happens to atomic radius? Tell me, as we go down the group, atomic radius or ionic radius. Atomic radius or ionic radius increases as we go down the group, right? Atomic radius, ionic radius increases as we go down the group. Now, when you talk about ionization energy, so as we go down the group, the ionization energy, what happens? In ionization energy, as we go down the group, what happens to ionization energy? Tell me, ionization energy decreases. Why ionization energy decreases? Because the size increases. Any one of you have doubt, you can ask me, okay? As we go down the group, the ionization energy decreases. We'll come back again to this, okay, we'll discuss first this only. So like I said, ionization energy, we have two ionization energy we'll discuss. That is first ionization energy, first ionization energy, and we write it as IE1. And then we have second ionization energy. And that we write it as IE2. So if I compare for one element, suppose for beryllium, if I compare IE1 and IE2, which one is more, IE1 and IE2? Which one is more for alkaline, earth metal? First ionization energy or second ionization energy? All of you know this, what is ionization energy? If you know this ionization energy, first what is ionization energy? All of you know this, tell me, yes or no? So you see, first of all, I'll come back to this, you know, slide again. If you take the example of beryllium, beryllium electronic configuration has four electron, electronic configuration is one S2 to S2. And when it goes to, when you provide IE1 to this element, it forms B plus, which has three electron, so one S2 to S1, right? Now again, in this, if you provide IE2, right? So it forms B2 plus with two electron, configuration is one S2, right? So here you see, we are trying to withdraw electron when you provide IE1, we are trying to withdraw electron from the fully filled S orbital, fully filled S orbital, right? Which is quite stable, which is quite stable, right? And here, when you're providing IE2, then we are trying to remove electron from half filled S orbital, half filled S orbital, which is comparatively not that much stable as compared to this two S2. So what we can say, since this form is more stable, hence IE1, we have to provide more energy to remove that electron, right? Hence we can say the first ionization energy of alkaline earth metal is more than to that of two. Understood? Now, if I compare IE1, IE1, alkali metal and IE2, IE1 for alkaline earth metal, which one is more, IE1 for alkali and IE1 for alkaline, which one is more? Tell me, the answer for this will be alkaline, okay? So the correct answer is alkaline, yes. And why is it so, why is it so? Because you see the alkaline earth metal, the general electronic configuration is NS2, right? So fully filled S orbital, but here the general electronic configuration is NS1, half filled. Again, this is more stable. So we can, it is difficult to draw electron with draw electron from this alkaline earth metal, okay? Now, similarly if we compare the IE2 of alkali metal and IE2 of alkaline, which one is more? IE2 for alkali is greater, because when you provide IE2 means what? You have to withdraw two electrons, right? So this configuration becomes NS1 then and this configuration becomes NS0, means we'll have NS2 configuration here, N minus, sorry. We should write down here, N minus 1, S2 configuration. And that is why the energy required to remove an electron from alkali metals, IE2 for alkali will be more. Understood? Tell me, is it clear? Now I'm going back to the last slide. Now you see here, ionization energy as we go down the group decreases. Now when ionization energy decreases as we go down the group, tendency to remove an electron will increase, right? And which is nothing but the electropositive character, electropositive nature, right? What is electropositive nature? It is a tendency to form a positive ion by releasing one electron. Tendency to form a positive ion by releasing one electron. So as ionization energy decreases as we go down the group, tendency to lose electron will be more and hence electropositive nature increases. Now when electropositive nature increases, so we can also say that metallic nature also increases. I have discussed all these things in detail in the last session. So if you have doubt in this, you can refer that, okay? And tendency to lose electron is nothing but a reducing characteristics. So reducing nature also increases. All these three things are associated with ionization energy. Can we move on? Okay, so these are the few trends we have discussed. Now you'll see the chemical properties. Chemical properties. The first thing we'll discuss here is reactivity towards water. Okay, I want you to write down all these things properly in your notebook, okay? Because these are the things which you will revise later on. So don't miss any information given over here. Okay, write down everything. Do you see in this what happens, reactivity towards water? Few elements of this, for example, we have calcium, strontium, barium. This reacts with water and forms hydroxide. Hydroxide of what type? M-O-H, whole-twice type. Hydrogen, gas evolves into this. H2 evolves, right? So if here we have the metal M, metal M, and this M can be anything among these three, right? This M can be anything. We'll get these type of hydroxide. M-O-H, whole-twice type, okay? Now in this reaction, one note you write down, and the note is beryllium and magnesium. Beryllium and magnesium forms, beryllium and magnesium forms layer of oxides. Hence, it does not react with water. Hence, it does not react with water. This is important. Next line you write down, the oxides of beryllium, the oxides of beryllium survives at high temperature. The oxides of beryllium survives at high temperature, survives even at high temperature, right, Tom? The oxides of beryllium survives even at high temperature, and hence, it does not react with hot water also. And hence, it does not react with hot water also. Next line you write down, the oxide of magnesium, the oxides of magnesium can be removed by, the oxides of magnesium can be removed by the use of amalgam. Oxides of magnesium can, just a second, oxides of magnesium can be removed by, by the use of amalgam, amalgam. At high temperature, at high temperature, and hence, it may react with hot water. And hence, it may react with hot water. So basically, magnesium, oxides of magnesium can be removed by the use of amalgam. Amalgam is a alloy of Hg, amalgam is an alloy of mercury Hg. So that can be removed by use of amalgam at high temperature. So magnesium can react with water at high temperature, but beryllium does not react. This is an important point. Next one you write down, reactivity towards acid, reactivity towards acid. Write down, like alkali metals, like alkali metals, these elements also react with acid and evolve hydrogen. Like alkali metals, these elements also react with acid and evolve hydrogen. For example, you see the metal M, if you are taking, M plus H2SO4, it forms MSO4 plus H2. Even if you take HCl, it forms MCL2 plus H2. Again, you see balance reaction is not required here. Just you need to know, the reason I gave you today, it forms a layer of oxide. See, you have the magnesium or beryllium surface like this. So when you put this into water, it forms a layer of oxide on its surface. It forms oxide first. This is the layer of oxide. It forms oxide first. And this layer of oxide does not let the water to further react with the magnesium metal or beryllium metal. That's why hydroxide does not form. So this is the reaction we have. Again, in this also, one note you write down, write down into this beryllium, that is BE, is amphoteric in nature. Amphoteric in nature. Amphoteric in nature. It reacts with both. It reacts with both. Amphoteric means what? It reacts with both acid or base. You see the reaction of beryllium with NaOH. The acid reaction I have already written here, it forms Na2BeO2 plus H2. This we call it as sodium beryllate. B-R-Y-L-L-A-T, sodium beryllate. Okay, this is one point you must remember. Now, the next one I write down, reactivity towards hydrogen. These elements react with hydrogen directly and forms hydride of MH2 type. What kind of hydride is this? What is the nature of these hydrides? What is the nature of these hydrides? Tell me. I need. Okay. So write down this point. These hydrides are hydrides. I need. I need. Okay. So write down this point. These hydrides are hydrides are ionic in nature, ionic in nature, except, except what? BEH2 and MGH2. These hydrides are covalent. Covalent hydride. We have already discussed this in the hydrogen type chapter. Selen is both MH and MH2 type. Both. Except these two, BEH2 and MGH2. S-block elements forms ionic hydrides. Okay. S-block elements forms ionic hydride. Okay. This is the property of these hydrides. Next write down. Next write down. These hydrides are good reducing agent. Good reducing agent. See, let me tell you one thing. Let me ask you this question that hydrides of group 2 is reducing or not. They won't ask you this. But when you see the question, these informations might help you getting the answer of those questions. Okay. That's why these properties are important to remember. Okay. You have to keep these properties in mind. Yes. BEH2 is purely covalent. MGH2 is partially covalent, partially ionic. Okay. But both we consider into covalent hydride. Right. These hydrides are good reducing agent that technically you see this CAH2 calcium hydride is technically known as hydrolith, technically known as hydrolith and used for the large scale preparation of hydrogen. You see these elements I have written the preparation method here directly combines to form this. Okay. In this one last point to write down the note here about beryllium hydride, BEH2. Write down BEH2 does not form, BEH2 does not form by direct combination of beryllium and hydrogen. BEH2 does not form by direct combination of beryllium and hydrogen. It is formed by, it is formed by reacting. It is formed by reacting beryllium chloride reacting beryllium chloride with lithium aluminium hydride with beryllium chloride and lithium aluminium hydride. The reaction you see this reaction here we have BECl2 reacts with LiAlH4. It is lithium aluminium hydride and it produce BEH2 plus AlCl3 plus LiCl. This is the reaction we have BEH2, AlCl3, LiCl. You see again I am telling you this reaction is not balanced. Okay. If you want to balance this reaction you can write down to BECl2 and to here. Again I am telling you this in organic chemistry and in organic chemistry balance reaction is not required. You should know under what condition what reaction gives you what product. This is important to memorise. Right. So must remember beryllium does not form hydride by direct combination of hydrogen. Yes, LiAlH4 is a very good reducing agent. Very strong reducing agent it is. LiAlH4, NABH4. Next should I write down reaction with halogen. Write down alkaline earth metal or write down elements of this group. Elements of this group, this group combines directly with halogen and combines directly with halogen and forms MX2 type and forms halide of MX2 type. Okay. Again in this you see BECl2, if I write down about beryllium chloride, BECl2 is covalent in nature, is covalent in nature, halides of elements are ionic in nature. See I have written here BECl2 actually it should be beryllium halide, BEX2. Means halides of beryllium are covalent in nature. That is what the point I was trying to make. It's not about only chloride, all halides of beryllium are covalent. See in organic chemistry there is no mechanism. Like when you look at the reactions if you go through the portions with time you can understand what product could be formed in this particular reaction. So you will understand this or gain this thing of with experience you can say easily at what product we are going to get. There are few reactions which are a bit confusing. For those reactions you have to memorize. Okay. But with experience when you look at look into the reaction or when you study more about inorganic chemistry then you can predict the product. Okay. Except few reactions because it's a chemistry so we always have exceptions in this. Okay. But no there is no any kind of mechanism. However somehow we can explain but it's not given anywhere. Right. So with basic understanding you can do that. Okay. Can you tell me why beryllium halide are covalent and other are I need? Anyone can explain? Yeah. Right. That is what I was trying to I wanted to listen. Okay. The according to Fezzan's rule like we have explained this for hydride. Same logic is true for halide also. Okay. So if they ask you this question you have to use that term that is polarizing power. Okay. Small size high polarizing power leads to covalent character. Okay. So like in hydride here we can write down again specifically if you want to write down the ECl2 is covalent. MgCl2 is partially covalent, partially ionic and ionic. And all of the halides are completely ionic in nature. Because the large size. Right. So this is the reason. Now here you see the next thing that right down here the properties of this that the solubility of these metal halides decreases down the group. Next point you write down solubility of these metal halides decreases down the group. Okay. So solubility order if you write down it is maximum for BX2. Then we have MgX2, CaX2, SRX2, etc. And why is it so? Because we know as the size increases the solubility energy decreases. Right. So as the size of cation increases I have discussed all these logic in the last session. Okay. As size increases, solvation energy decreases. Size increases means size of cation. Right. Less solvation energy, less will be the solubility. What is solvation energy? Can you tell me guys what is solvation energy? What is the difference between solvation and hydration energy? Solvation energy is the amount of energy released when one mole of any substance okay dissolved in any solvent. Right. Hydration is yeah correct Ashish correct. See like again I let me tell you one thing energy whenever you try to define it is an extensive property. Okay. Try to be specific over here. You know I always emphasize in these small things. Okay. You have to be very specific. When Bharat you said we say the energy released. Okay. So what is the amount of substance you are putting in? That's the main thing that you have to you know you have to mention. Right. The point is one thing you must keep in mind whenever you are defining an extensive property you always have to define the amount of substance. Right. Since energy is an extensive property so you say what when one mole of any substance is all into any solvent then the energy release at a given temperature is known as solvation energy. Right. But that solvent if that solvent is water then in that case the energy we get is nothing but the hydration energy. Right. So you have only the difference of solvent. If solvent is water hydration energy if solvent is other than water then it is solvation energy. That's the only thing. Right. I hope you understand this. Right. So as the size of cation increases solvation energy decreases in some book they have written here hydration energy also here. Right. So that is also you can write hydration energy but that signifies that the solvent we are taking is water that is the only thing. Right. Okay. Now one last point I'll just give you here which you have to again memorize at high temperature at high temperature around you know 1200 Kelvin around 1200 Kelvin the ECL2 that is beryllium chloride exist in dimer form dimer form for example you see if I now draw the structure here now we have a coordinate bond like this and like this beryllium definitely have vacant orbital and chlorine has lone pair of electron so that lone pair electron donates the vacant orbital of beryllium and we get a dimer form of BECL2 at high temperature. See in all these you know the theory that we have discussed so far if you observe one thing the compounds of beryllium which is the first element of group 2 has different behavior altogether. Right. All other elements perhaps in somewhat in a similar manner but the behavior of beryllium is different from the other elements molecules of the elements. Right. So one thing you just keep in mind whether it is alkali metals or alkaline methods earth metal group 1 or group 2 always you keep one thing in mind that the first element of both group will have the different nature. Right. That's why if you have gone through NCRT they have given in any other book all the books they have given this day that abnormal behavior of lithium right abnormal behavior of beryllium have you seen this kind of heading in the book abnormal behavior of lithium abnormal behavior of beryllium. So you must remember one thing the elements which is present at the first condition of any group will have the abnormal properties and why they have abnormal properties because of their small size comparatively high electronegativity and the difference in the number of electrons. Correct. So actually you see any elements the properties of chemical properties you see are some what to this physical property also any elements if you have they have different chemical properties because of different number of electrons present in the inner shell. Okay. If the number of electrons is different in the inner shell their properties towards their chemical properties will be different. Correct. So you see in all these elements when you go from top to bottom in a group obviously the outermost electronic configuration is same but the number of electrons present in the penultimate shell or inner shell is different right because of this difference in electrons they show different chemical properties. Okay. Second point they have the smallest size the first element of any group has the smallest size right they have high electronegativity so with all these reasons because of all these reasons the first elements of any group shows abnormal properties. Correct. So suppose why I have discussed this thing suppose they have given you this question that which of these molecule does not exist in monomer pump right or suppose the question is like this which of these molecule does not exist and the options are BCl2, MgCl2, CaCl2, understand this. You see if the question is which of these molecule does not exist let me write down I'll just finish my point here which of these molecule does not exist whether you know this thing or not suppose the first option we have is BCl2, MgCl2 then CaCl2 and the D option is SRCl2. Now since we have discussed this just now so you can directly say sir the answer will be option A because we know this exists in dimer form correct yes the answer will be option A guys you there am I audible what is the answer the answer will be option A right yeah so answer will be option A and since you know this since we just now we have discussed that BCl2 exist in dimer form correct but suppose one for one minute you just think that you don't have any information regarding this all these elements right then what you will do in this kind of question okay so one thing if you know that the elements the first elements of any group shows abnormal behavior right so there are very high chances that if MgCl2 exist then CaCl2 and SRCl2 both will exist correct but since beryllium has the abnormal behavior so there are high chances that BCl2 won't exist since we know that there is only one option correct right so if you have to choose one out of these four I will definitely go with A since I know that beryllium is the first element of group 2 and it has abnormal behavior did you understand my point yeah like it's better to go with the first option if you are going with some hunch right so better you go with the first element the compounds of first element most of the time you will get the correct answer if you go with the first element right but again if you know this sign otherwise if you want to attempt this question then I will go with A since I know B has the abnormal property that is the point I was trying to make anyways coming back to the next thing here right next you write down oxides and hydroxides we will talk about some compounds of these elements okay so write down the first here again oxides and hydroxides write down except beryllium and radium except beryllium and radium you see I am just giving you this idea okay these kind of thing you can think in the exam okay but it is not like always you will get the correct thing but what I am trying what point I am trying to make is what but in this question if you want to attempt this question you should go with B, C, and 2 right but many a cases I have seen that you will get the right answer right but in few cases since it is chemistry in organic chemistry there are exceptions okay so in few cases you may get the wrong answer also right so it is not a you know it is not a thumb rule okay we are just trying to predict some answer into this question see first of all I if you get this question okay if you get this question I will give you the assignment you will understand okay if you get this question the first thing if you know the answer fine second thing if you do not know the answer and you do not want to attempt this then also it is fine but most of the students are if they do not know the answer then also they want to attempt because the question is what which of these does not exist so it looks very simple right okay we have one of these is the answer with this we have and no there is no calculation only theory okay just look at the question and mark the answer you will get 4 marks okay so these kind of question is very difficult to you know if you do not know it is then also it is very difficult to bypass this kind of question usually students attempt like they will get something and they will attempt this kind of question but if you do not know or if you don't want to attempt fine if you do not know the answer and you want to attempt then you should go with this BCL 2 that is what the point I am trying to make okay anyway so we will not waste our time over here it is not that big thing also anyway write down oxides and hydroxides the point is beryllium and radium accept beryllium and radium all elements forms oxides of oxides of accept these two oxides of M O type oxides of M O type okay so the general reaction if I write down metal reacts with O2 forms M O right on beryllium and radium are highly electropositive in nature they are more stable in peroxide form and they are more stable in peroxide form hence form peroxides of hence forms peroxide of M O2 type M O2 type peroxide their forms right so BAO2 is a peroxide it is not an oxide it is a peroxide write down next these oxides normally we are talking about this oxide okay this you let it be here only write on these oxides are white crystalline solids these oxides are stable and white crystalline solid not that much important point but just you write down except BAO except BAO which is amphoteric in nature except BAO all other oxides all other oxides are basic they combines with water basic in nature superoxide we get in case of alkali metals okay just go through my last lecture last session right basic in nature and then you write down combines with water with H2O and forms hydroxide with H2O gives M OH whole twice and this is an exothermic reaction so energy heat energy releases in this if I ask you the solubility order of these hydroxide what is the answer tell me the solubility order of this BE OH whole twice MG OH whole twice CA OH whole twice SR OH whole twice what is the solubility order I will not use blue further okay what is the order of solubility of these hydroxides increase in atomic number okay solubility increases you see here what happens the correct order is this solubility and why is it so the reason behind this is as we go down the group correct so as we go down the group the size of cation increases right and when the size of cation increases I'll just write down here as we go down the group down the group size of cation increases when the size of cation increases right so bond length also increases and hence the lattice energy decreases lattice energy decreases when lattice energy decreases its thermal stability also decreases and hence the solubility increases all these things you can relate solubility increases will be easily soluble if you have two atoms bonded with a strong bond high lattice energy then it is difficult to break this bond and solubility will be less in that case right but when the lattice energy is low right so it's thermal stability is also low and it will be easily soluble correct Devdash hydration enthalpy if you see we are talking about solubility now right hydration enthalpy of this will be more see for dissolution there are two three things involved we have lattice energy also we have hydration enthalpy also right so the point is which factor is dominating at which point okay that's why in organic chemistry what happens you have to be very clear that what fact you have to apply or what logic you have to apply when if you talk about the hydration enthalpy or solvation energy obviously smaller ion gives you higher hydration enthalpy or solvation energy correct yes yes yes the point here is what you have actually two thing here one is lattice energy and the other one is hydration energy or solvation energy so with respect to lattice energy you will get this order with respect to hydration energy you'll get reverse order right the point is when this kind of contradictory facts are there then mostly in chemistry the things become experimental right and the final result depends on the factor which is dominating on the other one right so here in this case the lattice energy factor is more dominating over the hydration energy factor and that is why in the book if you see they will not mention the two contradicting factor they'll just give the dominating factor and according to that they'll give you the result did you understand this why they give only dominating factor because according to that only we are getting the final result that's why they are discussing only those did you understand this where are where are HSR guys I can see only 3 days where is Shreya yeah right but that's what you cannot say anything into that see in chemistry like I many times have said this yes many times I have said this in chemistry first we have the fact right and we try to explain those facts with the best possible explanation correct so here it's not like physics where we have logic and then we get some final result we have first we have final result and those results we're trying to explain the best possible way no no no is the only way is to memorize this that here that's what I said here you have to think about the lattice energy factor when you talk about that you know hydration energy of halide and all there you have to talk about the hydration of that you know hydration energy factor or solvation energy factor okay so that's what the thing so that is why it is when you are studying this you'll feel like if I understood this you can do this easily but when you get the question then you start thinking about all possibilities that's why I always says that you in organic chemistry the difficult part is that you should keep this in mind that at what point you have to apply what concept that is what the difficult part suppose if I give you this question and if I ask you the order of the solubility order and I tell you assume the lattice energy factor you will easily give me this answer but if I not give you this you will start thinking about hydration energy and some of you will think about lattice energy then you'll have the confusion if you know this factor this that this lattice energy is dominating over here then easily charge and size if you're comparing always will go with charge factor if we have any contradiction between charge and size will go with charge factor okay so next you write down carbonates and bicarbonates carbonates and bicarbonates just right down the elements of group 2 forms carbonate of forms carbonate of MCO3 type which is insoluble in water insoluble in water but in presence of CO2 but in presence of CO2 it dissolves dissolves and forms bicarbonate and form bicarbonate reaction you write down like MCO3 plus CO2 plus H2O it gives M at CO3 always just a second yes sir yes sir well guys we'll take a break now five minutes okay so we'll be back in five minutes I have some you know work to do take a break for five minutes okay fine we'll take a break now we'll start in then ten minutes okay okay we'll start in ten minutes take a break oh guys can we start the class now can we resume it can we resume the class tell me guys you there so you see bicarbonate is formed when in presence of CO2 right you see first of all this carbonates are insoluble in water that's all I've written already you see what is the stability order of these carbonates if I write down BECO3 MGCO3, CACO3, SRCO3 what is the stability order of these carbonates say in this what happens as we go down the group the percentage is any character increases right as we go down the group the percentage any character increases down the group and hence the solubility also increases sorry stability also increases down the group percentage any character increases and hence stability increases okay the reason is same B2 plus is the smallest size right so high polarizing action presence rule high polarizing action and hence more covalent nature more covalent nature now if I ask you the solubility order in this the solubility order will be reverse of this like you see the solubility is maximum for BE2CO3 this is the solubility order reverse of this now why this order we have for solubility that see you see we have actually cation and an ion okay CO32 minus I'll write down here C double bond O minus O minus this is the carbonate ion okay and we have a cation like we have BE2 plus MG2 plus and all right this anion is a large anion the size of this is large large anion right so large anion is stabilized with a stabilized with large cation okay more will be the charge dispersion when both cation and anion are large more will be the charge dispersion and hence more will be the stability so BE2 plus being the smallest cation right that will not stabilize the carbonate ion right since the the reason of this is what large anion CO32 minus a large anion so BE2 plus ion which is the smallest cation we have that will not stabilize the carbonate ion and hence and hence what happens the CO32 minus or BECO3 its strength is less okay right so what happens the reason simply you write down large cation stabilize large anion stabilize large anion and hence BECO3 is least stable and that is why it is easily soluble and solubility will be maximum for BECO3 did you understand this solubility of BECO3 is maximum now the next thing and the last type of compound we have here in this chapter is sulfate the formation of sulfate these elements forms sulfate of MS4 type sulfate of MS4 type by the action of by the action of oxides and these are normal oxides M4 by the action of oxides hydroxides we can take oxides hydroxides and carbonates and carbonates on sulfuric acid sulfuric acid you see the reaction when oxides reacts with sulfuric acid H2SO4 it forms MSO4 plus H2O when hydroxides which is MOH whole twice reacts with H2SO4 it forms again MSO4 plus H2O when carbonates MCO3 reacts with H2SO4 it forms MSO4 plus H2O plus CO2 this is the reaction ok now few trends will see first of all stability stability of BSO4 and then we have MGSO4 CASO4 and last SRSO4 right so the stability increases down the group if you go down the group stability increases right this is the order of stability stability increases down the group ok and why increases stability because large cation because of the size of cation right down size of cation the same reason large cation can stabilize the large NIN SO4 2- and hence the stability will be more ok now when the stability is more its solubility will be less or minimum the solubility increases reverse solubility down the group decreases so bottom to top if you go solubility increases since stability decreases now this solubility is because of the large energy of solvation of solvation of smaller cation smaller cation right the solubility reason is this right now one last thing in this write down on heating on heating sulphate dissociates dissociates into into corresponding oxides for example you see MSO4 when you heat this it forms MO plus SO2 plus O2 this is the reaction now you see again in this trend in this chapter we have discussed ok after this like the theory that you have you can go through in the book what you have to study in NCRT I'll just write down that ok however NCRT all things you have to study line by line everything you have to study and try to make notes of that whether you know scribble in the book itself or you write down the notes that is your choice but you have to do those kind of things ok so that you can revise at the time of exam ok so the thing you have to study from NCRT I'll just tell you now is the abnormal behavior of abnormal behavior of beryllium this you have to study abnormal behavior of beryllium I am not discussing this these are complete theory nothing to understand or you know to explain here you have to go for diagonal relationship also diagonal relationship this is actually you just need to know the examples ok because they ask this question which of these elements pair of elements shows diagonal relationship like that they have asked the question so diagonal relationship reason I have already discussed the same reason we have beryllium shows diagonal relationship with aluminium so that is given in NCRT you can go through ok now you see you have to study next for calcium calcium its compound right if you have any doubt in any of these you can come up to me for this for with your doubt ok but since I am not discussing this because there is nothing to understand here right now few compounds of calcium you have to again study that is quick lime CAO you have to study preparation and properties properties when I say it means important thing is chemical reactions chemical properties ok then we have slag lime CAO H whole twice slag lime is the solid when it is in aqueous form we call it as lime water ok we have to study calcium carbonate CAO 3 hello are you there can you audible now can you get me audio is there are you able to listen to me are you listening me are you listening me guys yeah ok we will do one thing see we had this chapter we have almost what you have to study from NCRT is abnormal behaviour of beryllium diagonal relationship calcium and its compounds like quick lime that is CAO CAO H whole twice calcium carbonate gypsum and plaster of ferris right again I will repeat this CAO that is quick lime CAO H whole twice slag lime calcium carbonate CAO 3 calcium sulphate that is gypsum and plaster of ferris these things are given in NCRT or any other book that you are referring you must go through this chemical reactions and properties and preparation you have to do right next class we will start with boron family correct let me know if you are fine with it just give me a thumbs up