 Hello, good afternoon. All of you, please type in your name in the comment box. Hello, Andrew, Trideb. OK. Shiram Ganeha. Just a second. OK, today in this session, we are going to discuss some basic level problem of a gaseous state, that is states of matter. And then we will also discuss S block elements. In organic chemistry, we'll discuss all those important topics of S block elements. And then you can go through your NCRT book. OK? OK. Lakshya Brindha is also there. Correct. OK, so the first question you can see on your screen, this question belongs to states of matter, gaseous state. OK? We'll quickly solve 8 to 10 problems onto this, and then we'll move on to S block elements. OK, S block elements, mainly we have theory into that. OK, so those who are not watching this video right now, and since I know that some of you are watching, they are in first class also is going on for Rajinagar students. So I have told them they can watch it later also. It's theory only, so they can copy it down whenever they want. OK. So anyways, we'll start with this S block element, sorry, gaseous states of matter first. Resolve few questions, then we'll move on to S block elements. So OK, hello there Das. Is it your name? Gajodhar Singh, cool and all. OK, solve this question, guys. First question, above a lot of gas released at the bottom of a lake, increased to eight times its original volume when it reaches the surface. Assume that atmospheric pressure is equivalent to the pressure exerted by a column of water of 10 meter high. What is the depth of the lake? Solve this question. OK, what is the answer? Tell me first one. OK, Tridev is getting C. Now you see this question. Suppose the lake is this. And according to the question, a bubble of gas released at the bottom, OK? So at the bottom, a bubble of gas has been released here. And when it comes up at this surface, its volume becomes eight times its original volume, OK? So this height, we have to find out the depth of this lake, OK? So here you see the pressure here is what? Suppose this pressure is P from this atmospheric air, right? Here the pressure is P. And since as we go down here, the pressure increases, OK? So here the pressure is suppose P1 at the bottom of this lake. So P1 is equals to what? We can write atmospheric pressure P plus pressure due to, pressure due to, height of water. So this pressure is due to the height of water column, OK? Or liquid here, the water that you have, this height, OK? Height of water, correct? You see here, we can apply a boil slow easily, correct? So here the pressure here it is P into the volume here if it is V. So here it should be 8V. So P into 8V, PV is equals to P1, V1 is equals to P2, V2. That will be P1 into V. So here we get the relation as P1 is equals to, means the pressure at the bottom of the lake is 8 into the atmospheric pressure. And here if I substitute 8P is equals to this value I substituted here. It goes to P plus pressure due to the height of the water. That suppose I assume is P dash. So P dash is equals to what? 7P when you solve this, right? Now you see the question. Assuming that the atmospheric pressure is equivalent to the pressure exerted by a column of 10 meter of height, OK? Atmospheric pressure is what? This P. And this is equals to the 10 meter of height of the water, means what? This line means what? If you take 10 meter of height of water, the pressure exerted by this 10 meter of height of water is equals to the atmospheric pressure, OK? So P is equals to 8 meter of height of this we can take, OK? So in terms of height, if I write down, P is 10 meters. So P dash will be equals to what? 10 into 7, 70 meter. Hence, the answer will be option C. Is it clear? Any doubt? If you have, you can ask me, otherwise I'll move on. Next question you see. Second one, what is the answer? Prithee is getting A. Andrew is getting A. Thirudev, what is the answer? Where is Neha, Lakshya, Brinda? Tell me the answer. The first one, P versus P. Yeah, it is. We have discussed this. Thirudev, you see, first of all, you just need to write down the expression of PV is equals to NRP. And then one by one, we'll try to draw the structure. The first one is PV versus V. So obviously, here we have to take this temperature constant, right? So if temperature is constant, then PV is equals to constant. So when you draw the graph of this PV versus V graph, this won't change with volume at constant temperature. So obviously, this will be a straight line, because it is constant. PV is equals to constant. Even if you write down pressure also here, the graph will be same, volume or pressure. Now in the next one, you see P versus V graph at constant temperature. When you draw P versus V graph, then what happens is, again, constant. So this P versus V graph gives you what, a rectangular hyperbola, right? And that will be like this. V at constant temperature, the graph will be like this. Next one is P versus 1 by V. According to this only, we'll have some constant K into 1 by V. So this graph will be what? It will be a straight line passing through origin, right? OK? So A will be 2, D will be 3, and C will be 1. Option A is correct. Next question, you see. And D is getting C. Devdas, D. Preeti is getting C. Thridheb is getting D. What we have to find out into this? What is the meaning of mole fraction, the fraction of total pressure? What is the meaning of fraction of total pressure? The fraction of total pressure means we have to find out the partial pressure. Fraction of total pressure is nothing but partial pressure. Total pressure, suppose we have Pt is equals to partial pressure. So we know what partial pressure P of any component is equals to the mole fraction of that component into total pressure Pt. Suppose the component is A. So partial pressure of A is equals to the mole fraction of A into total pressure. And mole fraction, we can find out since we have equal weights of ethane and hydrogen. So we have ethane and hydrogen, equal weights. So I assume ethane is x-gram, and hence, hydrogen will also be x-gram. So when I convert this into number of moles, so it is x-by, molecular mass of ethane is what? 30. And this is x-by 2. So we know the number of moles. So mole fraction of hydrogen, that is, suppose if it is A, it is suppose B, and it is A. So mole fraction of hydrogen, that is xA, is equals to the number of moles of hydrogen divided by the total number of moles, x-by 2 plus x-by 30. So when you solve this, you'll get 15-by-16 mole fraction of hydrogen. So the partial pressure is equals to the mole fraction, 15-by-16, into total pressure Pt. Hence, the answer will be option B. Is it clear? What is 2-by-th, what is 2 Pt? What is that 2? See, there is one more method. You can solve this question. I'll just tell you that short method we have. See, in this type of question, usually, we have ethane and hydrogen here, ethane and hydrogen. So this is the second method, easier method we have. OK. And we have ethane and hydrogen, so usually in this kind of question, sorry, this was again, usually in this type of question, what happens? The mixture that they give in this type of question, the molecular mass of those gases are multiple of each other. For example, you see the molecular mass of ethane, C2S6 is 30, and molecular mass of hydrogen is 2. Usually, they give like this only, multiple of each other. If they are not, then also you can do, but that will be a bit calculative, OK? This is the method by which you can solve. Assume the mass and then calculate number of moles, mole fraction, and then you can find out with this formula. But now you see, since they have given equal mass, what I assume here, what I assume, whatever the molecular mass you get here, you take the higher value as the mass of each of these. Suppose 30 is the higher value in these two. So what I assume, we have 30 gram of ethane, and since the condition is what, they have equal weights. So if we have 30 gram of ethane, so hydrogen must be have 30 gram only, OK? So when you have 30 gram of ethane, so if you convert this into number of moles, you'll get one mole here, and here you'll get 15 mole. So mole fraction of ethane, we have to find out. So that will be the number of moles of ethane divided by total number of moles, which is 15 plus 160. This is what the mole fraction of hydrogen here, and hence it is 15 plus 60. Is it clear? So it is actually a bit easier way to do this kind of question. You don't have to write down, you don't have to use your pen also to solve this question. OK, when you do like this. For this, you have to do this kind of calculation. Yeah, write them, OK? So both way you can solve, I will prefer this one, right? Because it takes lesser time. Next question, what is the answer? Done? Tridev and Devdas is getting D. OK, D is the correct answer. So you see this question, the mixture of N2O4, and what we assume, this has 100 ml of mixture, volume I have assumed total, 100. So in this, if I find out the volume of N2O4, N2O4, that will be what? 2 divided by 3 into 100, because 2 is N2O4 and 1 is N02. So total is 2 plus 1, 3. So volume fraction of N2O4 is 2 by 3 into 100 total volume. Similarly, the volume of N02 is equals to what? 1 by 3 into 100. And if the density of the mixture is, suppose D of mixture we have to find out. So what we can write, the density of mixture into the volume of mixture, which gives, this gives what? The mass of the mixture, mass of what? The mixture is equals to what? The density of N2O4 into volume of N2O4 plus density of N02 into volume of N02. This is what the equation we have. Total mass of N2O4 plus mass of N02 is nothing but the mass of mixture. So density of N2O4, we know the molecular mass of N2O4 is what? Molecular mass of N2O4 is 92. So its density will be 92 divided by 2 because we know 2 into vapor density is equals to molecular mass. The formula we are using is what? 2 into VD, vapor density is equals to the molecular mass. This is what we are using here, the formula. So density of N2O4 we have got. Similarly, the density of N02 will be what? Its molecular mass, that is, M of N02 divided by 2, which is nothing but 46 by 2. So density of N02 is equals to 23. All these value you have to substitute here. So you will get here what? Density of mixture into volume of mixture is 100. Is equals to density of N2O4 is 46. Into volume of N2O4 is 200 by 3 plus 23 into 100 by 3. So when you solve this, the density of mixture we'll get here is 38 point. Whatever the unit we have, that we can write down. The answer will be option D. Understood? Next question. Solve this one. What is the answer? A, Prety is getting C. OK, C. The question is through two ends of a glass tube, length of 200 centimeter, SCI and NS3 are allowed to enter. At what distance the whiting of an ingrowite will first appear? So suppose we have a glass tube of length 200 centimeter. This is the length we have. From one end, HCl is allowed to pass. And from the other end, NH3. So when these two reaction takes place, NH3 plus HCl, what is this reaction? What kind of reaction is this? We'll get NH4Cl. What kind of reaction is this? Can you tell me? Action is this? Tell me. Is it a combination? See, NH3 is a base. HCl is an acid. So it is an acid-base reaction. Yes, it is an acid-base reaction. So here, from this example, you can understand, all acid-base reaction does not give you salt plus water. It is not necessary that acid-base reaction gives you salt plus water only. Acid and base will give salt plus water only when base contains OH ion, OH molecule, and acid contains H plus ion. Correct? So salt, if somebody asks you, acid-base reaction gives you salt plus water. True, but not always. That is only possible when acid and base both contains H plus and OH minus ion respectively, then only we'll get salt plus water. It is not always true that acid-base reaction gives salt plus water. That is another thing which is not required in this question. So here you see, NH3, HCl gives NH4Cl. This is the white fume, white fumes or white ring. The point is, from this side, we are introducing HCl. And this side, we are introducing NH3. L is given, L is 200 centimeter. Now, HCl and NH3, if you compare, which will diffuse more, HCl or NH3? Obviously, the rate of diffusion is inversely proportional to a square root of molar mass. It means the gas which has higher molar mass, that will diffuse with a lesser rate. And hence, the distance traveled by that gas is, distance traveled by the gas which is higher molecular mass is lesser than the distance traveled by the gas which has less molecular mass. So if you see the molecular mass of HCl is 36.5, and that of NH3 is 17. So obviously, when these two will meet, NH3 will travel more distance because rate of diffusion of NH3 is more than to that of HCl. Hence, NH3 will travel a larger distance than HCl. Suppose what happens at this point P, which is not the middle point of this tube, at this point P, a two gas mates and white ring or white fumes appears over here. And this distance from this end is suppose A. From this end, it will be 200 minus A. Obviously, this distance will be lesser than the half of this value. It means this distance should be lesser than 100 centimeter. So from HCl, the answer would be less than 100 centimeter. So which is only option A. That is how we can neglect option here. We'll solve also this one, how to get this value. Option B and C are same. That's why those are not the answer. Like this also, you can eliminate B and C. Since we know the options are, the question has only one correct twice. So here what we can write, the distance traveled divided by time. So what I will write, the rate of NH3 divided by the rate of HCl, this will be equals to under root of molar mass of HCl divided by molar mass of NH3. And this will be equals to the distance level by NH3. That will be 200 minus A divided by time. This is divided by the distance level by HCl, which is A by T. So when I take these two expressions, the molar mass of HCl is 36.5 divided by 17, so root over of it is equals to 200 minus A divided by A root over of it. This you have to solve. You will get A. And hence the answer will be 81.1. When you solve this, you'll get this one. Under root. Oh, sorry. This root will not be there. This is not there. Correct? Option A is right. 10th one. What is the answer in this one? Got it. What is the answer? Done. What happened? Should I do this or are you still trying? 25. It's a random guess or you have done some calculation or you are getting a close value today? OK, see this question. See, the question is what in a mixture or we have a mixture of we have a mixture of NH3 into H4. In a sealed container, temperature is given. Total pressure is given here. For this mixture, the total pressure is given, which is 0.5 atm. Temperature is also given. 300 Kelvin. Volume we do not know. So we'll assume the volume as V. Now in this mixture, it is not given that how many moles of NH3 and N2H4 we have. So what I assume, we have N1 moles of NH3 and N2 moles of NH4. This I assume, OK? So once we have this information, we can apply ideals of gas law, which is PV is equals to NRT. Where pressure is what? 0.5. Volume we assume as V only. So it is V. Number of moles is what? Since we have a mixture of these two. So number of moles will be N1 plus N2 into R. And temperature is 300 Kelvin. This is equation one. This is the first part of this question. OK? Now what happens? This mixture we are heating at 1200 Kelvin. This mixture we are heating at 1200 Kelvin. So temperature becomes now 1200 Kelvin. Pressure also changes in this. What is the pressure now? Pressure becomes 4.5 atmospheric. Now number of moles is what? Because volume will be same. It is there in the same container. So volume is nothing but the initial volume we have. That is V only. OK? But what will be the number of moles? Now you see this equation. Two moles of NH3 gives four moles of product here. One out of four moles, we have one mole of N2 and three mole of H2. So what we can say, two gives four moles. How many moles of NH3 we have assumed? N1. So N1 moles gives what? Two N1 moles of product, twice of this? Right? Similarly, one mole of N2H4 gives two plus one three moles of product. So N2 moles gives you what? One gives three. So N2 gives three and two. Is it clear? This step? So here if I write down the total number of moles, N is equals to two N1 plus three N2. Is it clear? Till here, tell me. Is it clear now? Yes, guys, tell me. OK. Now you see the question, what is the percentage of N2H4 in the original mixture? OK? So percentage of N2H4, N2H4 number of moles we have assumed, right? So it is N2. What we have to find out? N2 divided by N1 plus N2. This is a fraction. N200 gives you percentages. This value we have to find out in this question, right? We have to find out this. So consider this expression now. So now coming back to this here, we can again apply PV is equals to NRT for the second case when we have heated this. So pressure is 4.5. Volume is V. Number of moles is 2N1 plus 3N2. R will be as it is. And temperature is 1200 Kelvin. This is equation 2. OK? Now, we'll take the ratio of these two so that everything will get cancelled. OK? And I'll take what second divided by 1. So what happens, you see? Left hand side, second by 1, gives you 9. And right hand side, we have 2N1 plus 3N2 divided by N1 plus N2. And here, N1 plus N2, right? N2, 4, right? So now just we have to solve this and get the expression. One thing is what? You can solve this for N1 and N2, and you'll get the answer. But I'll solve this in a different manner. OK? Like you see, we have 9 by 4 left hand side. We have 9 by 4 is equals to 2N1 plus N2. Can I write this expression as 2 into N1 plus N2 plus N2 divided by N1 plus N2? So this becomes 9 by 4 as it is. And this N1 by plus N2 will divide in both this expression. So we'll get 2 plus N2 divided by N1 plus N2, which is nothing but N2 divided by N1 plus N2 is equals to when you solve this, you'll get 1 by 4. So when you multiply this, this will be equals to what? 1 by 4 into 100. And that gives you 25%. So answer will be option T. Any doubt? Let me know. Sri Ramya, Preethi, understood? Devdas, are you there? Next question we'll see. Solve this one. What is the answer? What is the answer? Tell me. I'll do this. You see, at STP, we are having 0.48 gram of O2 diffused through a post partition in 1,200 second. What volume of CO2 will diffuse? In the same time under the same condition. So 0.48 gram. First of all, you tell me, can we convert this mass into volume? If I ask you, what is the volume of O2 has been diffused? Can you tell me? The volume of O2 diffuses what? Since STP, we know 32 gram of oxygen at STP is equals to 22,400 ml. Since options are given in ml, I have taken in ml. So if the mass is 0.48 gram, then the volume of O2 diffuses what? 22,400 divided by 32 into 0.48. When you solve this, the volume diffused here, you'll get is 336 ml. This is the volume of O2 diffused. So since time is also given, so we can find out the rate of diffusion of O2. That will be the volume diffused divided by the time 1,200. In the same time, suppose the xml or if I take CO2, the volume diffused, we have to find out that is v divided by 1200. So we can write rate of O2 divided by rate of CO2 is equals to molar mass of CO2 divided by the molar mass of O2 root under, which is nothing but 44 by 32 root under. Now when you simplify this expression, you will get rate of O2 is 336. This 1200 and 1200 will get cancelled divided by v is equals to 44 divided by 32 root under. And when you solve this, approximately, you will get 286.5 ml. This calculation you have to do. Hence, the answer will be option A. Clear? Next question, this one. After this, we will start as block. Tell me the answer. Question number 8, getting B. The derivative is getting B. Assuming the same pressure of each case, what is the mass of hydrogen required to inflate a balloon to a certain volume v at 100 degrees Celsius if 3.5 gram of helium is required to inflate the balloon to half the volume v? So simply we can use PV is equals to nRT. So suppose if you have hydrogen first. So the mass of hydrogen we have to find out. So what we'll write for hydrogen, if we use PV is equals to nRT. The pressure is what? We did not know. So let it be as it is. Volume is v, p into v. Number of moles, I can write this in terms of mass. So it is a mass of hydrogen required divided by molar mass of hydrogen is 2. R, you let it be. And temperature is 2700 degrees Celsius. So it is 373 Kelvin. This is for hydrogen. Similarly, for helium, it is p into volume is v by 2 because half of this we have. Number of moles is mass of helium. It is 3.5 gram divided by the molar mass of helium is 4 into R into T is again 25 degrees Celsius plus 273, so it is 298. Now, again, this is 1 and this is 2. We'll simply divide 1 by 2. So when you divide 1 by 2, left hand side will get cancelled, will get 2 over here is equals to m into 373 divided by into 3.5298 into, this is what you have to solve. And once you solve this equation, you'll get mass is equals to 2.8 gram. Hence, the answer will be option p. Got it? Fine. So we'll move on now to S block elements. OK. In HSR, I have discussed a few things into this in HSR batch. So all those HSR students, you have to wait for 10 minutes. Just you can revise in this 10 minutes. You don't have to write anything. I'll quickly finish those 10s few topics that I have already done and then we can continue. So first of all, you see, we are going to start alkali metal. And this alkali metal, we have two different groups. That is group 1 and group 2. Group 1 is alkali metal. This is not alkali metal. This should be S block. OK, so we are going to start S block. S block contains two groups. Group 1 is alkali and group 2 is alkaline earth metal. We'll quickly discuss about alkali metal first, because like I said, here you have to memorize things. OK, there are a few trends where you can put some logic and understand. OK, but when it comes to preparation and properties thing, those you have to memorize. OK. So alkali metals, you see, the elements are present in alkali metals. Those elements are, these are the elements present. OK, hydrogen, lithium, sodium, potassium, rubidium, cesium, and potassium. OK, so here I'll write down the atomic number. The atomic number of hydrogen is 1. That of lithium is 3, 11, 19, 37, 55, and then 87. These are the atomic number or number of electrons. OK, in all these metal, OK, generally what happens, we define alkali metals are those metals which reacts with water and form hydroxide, right? So alkali metals, these are the definition of alkali metals are what? These are the metals reacts with water and form hydroxides. Hydrogen, however, present in group 1, but hydrogen we don't consider as an alkali metal, not an alkali metal. Why it is not an alkali metal? Since it does not react with water, H2. That's why hydrogen is not an alkali metal. Now here you see the difference in atomic number. If you see the difference in atomic number here, it is from here to here, it is 2. Then it is 8, again it is 8, 18, 18, and then 32, OK? 2, 8, 8, 18, 18, 32. These numbers we call it as magic numbers, magic numbers, OK? So if you remember the atomic number of any elements present the top of the group, by adding these numbers, you can find out the atomic number of the elements present in the lower part of the group, right? That's why it is important. OK, so now the next point here it is what? If you see the electronic configuration, next one will see the electronic configuration. For hydrogen it is 1S1, for lithium it is 1S2 and 2S1. So 1S2 is the configuration of helium. So we also write it as helium and then 2S1. Similarly, for sodium you see it is 1S2, 2S2, 2P6, 3S1. This part is the configuration of neon. So we can write neon into neon and then 3S1. Similarly when you write down the electronic configuration of potassium, helium, neon, then we'll get argon, 4S1, krypton, 5S1, xenon, 6S1. And in the last we have radon, 7S1, OK? So when you look at the outermost electronic configuration of these elements, OK, so the outermost electronic configuration is nothing but nS1, where this n represents the number of period. Like you see hydrogen present, lithium present in second period, so 2S1, sodium present in third period, 3S1, potassium, fourth period, 4S1, 5S1, 6S1, 7S1, OK? So general electronic configuration is nS1. For all these S-block elements. One more point you write down here. These elements has very low ionization energy. These elements has very low ionization energy. Hence, very reactive, very reactive, and never found free in nature. These are also metals, OK? In periodic table, if you go from left to right, the metallic property decreases, OK? So these are very good metals, all these elements, OK? So these are few general things that you should keep in mind. Now the important point here, which you have to understand where you can put some logic and understand, that is the general trend of alkali metals. So write down the heading next, general trends. So the first point here is atomic, OK? So in this you write down, as we go down the group, we go down the group, atomic radius, atomic radius, as well as ionic radius, atomic radius, as well as ionic radius increases. So if you write down the order, lithium, sodium, potassium, rubidium, cesium. In fact, for ions also, Li plus, Na plus, K plus, RB plus, CS plus, OK? This is the order of radius, atomic radius and ionic radius. Why atomic radius increases? Because you see, due to addition of an extra shell, you see here, this is the first shell, second shell, right? Then third shell, fourth shell, fifth shell, sixth shell. It means when you are going down the group, we are adding an extra shell into it, since the number of electrons are increasing. So all these electrons fills into the orbital and it is going away from the nucleus of these metals, OK? That's why the radius is increasing. So as we go down the group, it is true for all the groups actually interactable. Whenever you go down the group, the atomic radius and ionic radius increases. Yes, there are a few exceptions, but general trend is this only, down the group, atomic radius and ionic radius increases, OK? What happens if you go left to right, right? If you go left to right in a period, OK? Left to right. What happens left to right? Due to, sorry, due to effective nuclear charge, due to effective nuclear charge. What is effective nuclear charge? I have discussed this in periodic properties, OK? So you must go through and revise. Due to effective nuclear charge, atomic radius decreases, right? So in a period, in a period, the alkali metals, alkali metals will have largest size. That's the very first point you should understand, OK? Alkali metals will have the largest size. Now on this size only, there are a few more properties which are trends which are related to the size factor only. That's why I have discussed first the size thing. Now you see the second thing here is ionization energy. Ionization energy. All these terms I have discussed in periodic properties. You must revise that, OK? So write down the definition for this ionization energy. It is the amount of energy required. Write down quickly. It is the amount of energy required. It is an amount of energy required to remove an electron from the outermost shell, the outermost shell of an isolated gaseous atom. It is the amount of energy required to remove an electron from the outermost shell, from the outermost shell of an isolated gaseous atom. Now you see the meaning of this is what? Suppose we have an atom, but it is not circular. You can assume this as suppose we have an atom. This is the nucleus of the atom. And the outermost shell has an electron. Nucleus will have the positive charge, right? So we must have an electrostatic force of attraction here. Positive and negative charge. We must have electrostatic force of attraction. This is an isolated gaseous atom. Now if you want to remove this electron, right? So what you have to do, first of all, you have to overcome this attraction pull. This you have to break, right? You have to overcome this attraction pull. Then only you can remove the electron, right? So for this you have to supply some energy into it, right? You have to supply some energy into it. And with this energy, first of all, you will break this attraction pull and then the electron comes out from this shell, right? So this energy which is required to remove an electron from the outermost shell is the ionization energy, right? So we can have another electron present in the inert shell. For that also, you can provide some energy and the electron you can remove, OK? The point is the energy required, energy required to remove first electron from the outermost shell of an isolated gaseous atom, all those are the same. But the energy required to remove first electron, we call it as first ionization energy, IE1. And the energy required to remove second electron, second electron, we call it as IE2, right? So now we will compare first and second ionization energy of these elements, OK? For example, if I take sodium, OK? Na. Now this Na has 11 electron. So that will be 1S2, 2S2, 2P6, 3S1. Once you provide ionization energy into it, which is IE1, first ionization energy, Na becomes Na plus and Na plus will have 10 electrons and its configuration will be what? 1S2, 2S2, 2P6. Now this is the configuration of neon, which is an inert gas, right? And this configuration is highly stable, right? Because of this stable configuration, it is very difficult to remove further electrons, right? Or the electrons from Na plus. So what we can say from this comparison that for alkali metals, the energy for second or second ionization energy, IE2, is greater than the first ionization energy, IE1. Is it clear? Why it is greater than? Because when it loses one electron, it gains inert gas configuration, which is highly unstable, and it is very difficult to remove an electron from this configuration, right? And if it is, you have to remove an electron, you have to supply a very high amount of energy for this. That's why IE2 is greater than IE1. Is it clear? Anastu, tell me. Yes. Now, the second point here, it is what? You see, the attraction between these two is what? The force we can write, F is equals to K, Q1, Q2, by R square. This is the electrostatic force of attraction between the two charge. R is the distance between the two charge, right? So when this distance increases, right? When this distance increases as we can say, R increases force of attraction decreases, right? In this case, when you have an electron present here, the distance between this charge and this charge is nothing but the radius of that atom, right? So as we go down the group, the radius increases, right? Means R increases, R increases, so force of attraction decreases, and when force of attraction decreases, it is easy to remove an electron, right? Means ionization energy decreases. When this force will become weak, with less energy only, you can remove electrons, right? That's what the next point you should write. So you write down the next point. As we go down the group, as we go down the group, then what happens? Ionization energy, sorry, as we go down the group, first you write down size or atomic radius, size increases, which means ionization energy decreases. And when I say ionization energy, it means we are talking about first ionization energy, right? OK, so ionization energy decreases. With this only, we can also understand one more term here that is electropositive character. Electropositive character is nothing but the tendency to lose electron and becomes positively charged. Tendency to lose electron, right? As ionization energy decreases, it is easy to remove an electron, right? So as we go down the group, we go down the group, electropositive character, electropositive character increases, since it is easy to remove electron. And this is nothing but the metallic character also. Tendency to lose electron is nothing but electropositive and metallic character. So metallic character also increases as we go down the group. Is it clear till now? These are the trends you have to keep in mind. Let me know if you have any doubt. Clear understood? Tell me. Can we move on? Next point you write down, that is oxidation state. Oxidation state. Write down into this. These metals have tendency to form. These metals have tendency to form. Univalent positive ion, univalent positive ion have tendency to form. Univalent positive ion and forms and shows, sorry, and shows plus one oxidation state always and shows plus one oxidation state. For example, you see all the compounds of alkali metals like Na2SO4, NaCl, K2SO4, KBr, Ki, et cetera. In all these compounds, the metal always have plus one oxidation state. OK? So next thing you see, that is hydration energy. Write down, the definition of this is what? It is the amount of energy released. Amount of energy released when one mole, a substance, is dissolved in water. See the term hydration, it is related to the solvent. Right? If the solvent is water, then if the solvent is water, then the energy released, we call it as hydration energy. Hydration energy. But if the solvent is not equals to water, means except water, if we have any other solvent, then the energy released in this process, then the energy released in this process, we call it as solvation energy. Hydration energy and solvation energy are nothing but the same thing. Hydration is associated with water as a solvent and solvation is we have any solvent other than water. OK? So degree of hydration or like coming back to this point as the size of cation increases, then degree of hydration decreases. Degree of hydration decreases. It is inversely proportional. So basically we can write size of cation is inversely proportional to degree of hydration. Larger ion is difficult to dissolve. Right? Simple thing you can see. If you put some large molecule into water, it is very difficult to dissolve in compared to the smaller molecule. Right? So that's the thing. Size of cation increases, degree of hydration decreases. OK? So hydration energy will follow the same pattern as we have this. As size of cation increases, hydration energy also decreases. So order of hydration energy will be Li plus, Na plus, K plus, Rb plus, and then Cs plus. OK? So we have maximum hydration energy we have for Li plus. Hydration energy order. So till now I have discussed in HSR 11 class I have discussed till here. So we'll go further now. Yes, correct. So we'll start next trend here. And that is the next one we have is reducing nature you write down. Reducing nature. OK? Now reducing nature you write down. What is reducing nature? It is to reduce other and oxidize itself. So reducing nature is what? To get oxidized, oxidize, to get oxidized and reduce others. See, oxidation process, right, oxidation you try to understand. We have already discussed redox reaction. So oxidation is what? It is a tendency to lose electron, tendency to lose electron. And then only the oxidation number increases, which has the symbol of oxidation. Oxidation number should increase. Oxidation is a tendency to lose electron and this will be easier when we have low ionization energy. All these are associated. If the ionization energy is low, tendency to lose electron will be more, right? What we can say here, that as ionization energy decreases, ionization energy decreases, reducing nature increases. Reducing nature increases. And ionization energy decreases when we go down the group, right? So we can say, as we go down the group, reducing nature increases, right? But this order is there in free state, means gaseous state. This order, order of reducing nature, if I write down, the order will be this. It is minimum for Li. Then we have Na, K, rubidium, cesium. This is the reducing nature we have, but this order is there in free state or gaseous state. Because if you remember this term, ionization energy, we define only when the atom is isolated gaseous atom we have, right? So that's why this order is true in case of free state or gaseous state, right? But in case of solution, first of all, one point to write down, reducing nature is explained, right all here. Reducing nature, reducing nature is explained, explained on the tendency of, tendency of losing electron in gaseous state, right? In gaseous state, we define this by ionization energy. Next, write down, write down, however, the tendency to lose electron, tendency to lose electron in solution is explained by its oxidation potential, oxidation potential. And this oxidation potential, I'll write down like this, this oxidation potential depends upon, depends upon the hydration energy, hydration energy, okay? More hydration energy, more will be the oxidation potential. What we can write, more hydration energy, more will be the oxidation potential. See, oxidation potential is the, you know, energy required to oxidize itself, okay? Means suppose one molecule, one thing just to keep in mind, if you have two metal, metal one and metal two, if the oxidation potential of metal one is more than to the oxidation potential of metal two, then the metal wall will go under oxidation, right? If these two combination we have, if these two combination we have, then this will go under reduction. Means for whatever the metal we have, if oxidation potential is more than the other metal present in the solution, then the metal which has more oxidation potential will go under oxidation. Whatever is more, that process will take place for that particular metal. The point here it is what? If two metal we have in a solution and one metal having more reduction potential, then that will go under reduction and other will go under oxidation. If two metal we have in a solution, having one metal will have higher oxidation potential, then the metal which has higher oxidation potential, that will go under oxidation, right? So only one thing you have to keep in mind that if oxidation potential is higher for a metal, then on that metal oxidation takes place. If reduction potential is higher, then on that metal reduction takes place, correct? So you see here more hydration, more will be the oxidation potential, okay? More oxidation potential means what? The oxidation of that metal will take place, right? And when oxidation takes place of that metal, it means it will reduce others and that is nothing but the reducing nature, right? So basically in solution, the reducing nature of metal is defined by its hydration energy. That is a point you have to keep in mind. In gases state, the reducing nature is defined by what? Ionization energy, but in solution, reducing nature is defined by its hydration energy. So that is the next point you write down, that in solution, hence in solution, the reducing nature is defined by or explained by hydration energy, hydration energy. And we know hydration energy is maximum for Li plus, okay? Hence, the order of reducing nature in solution will be reverse. And that will be maximum for Li plus, then Na plus, K plus, RB plus, CS plus, et cetera. Is it clear? Understood? Okay. Next is write down, reactivity towards air, reactivity towards air. Write down, alkyl metals are very reactive in nature. Alkyl metals are very reactive in nature and on exposure to, on exposure to moist air, they forms oxides, peroxides, super oxides, oxides, peroxides and super oxides. So, you see the metal on exposure to moist air, they form oxides of this type. Further, they forms peroxide of M2O2 type and further, they form super oxides of MO2 type. So, this is oxide, this is peroxide and this is super oxide. All these three types of oxides forms. Now, you see, when we talk about oxides, oxides which is the normal oxide or simply if I write oxides, general formula is oxides are M2O type. Okay, this is the general formula of oxides. M2O type. For example, you see, Li2O, Na2O, K2O, RB2O, CS2O. In all these oxides, lithium mainly forms this kind of oxides. Okay, so mostly lithium forms these oxides, mostly form. A little bit of sodium also form these kind of oxides, but generally, these oxides are very rare to form. Okay, mostly if they ask you which elements forms oxides will go with Li2O. To some extent, sodium also forms this very rarely forms. Okay, and the next point here you have to remember that these two are colorless. Like I said, these are facts. Okay, so you have to memorize this. This is, the color of this is yellow or we also specifically if you write it is pale yellow. RB2O is bright yellow and CS2O is orange. These colors you have to again memorize. And the most important thing you have to memorize here is all compounds or all are ionic in nature. All these oxides are ionic in nature. Okay, next one you see peroxide. Peroxide that is M2O2 type. Here in this oxide the charge on oxygen is O2 minus. Okay, in peroxide we have O2 2 minus. Right? O2 2 minus. This peroxide is mainly formed by sodium. Mainly formed by that is Na2 Na2O2. Okay? And this is yellow in color. Yellow in color. These are also diamagnetic in nature. Magnetic in nature. Important point it is. Okay? Next is superoxide. Superoxide. Superoxide is this O2 minus type. General formula is MO2. Generally formed by potassium, KO2, rubidium and cesium. This contains an odd electron bond. An odd electron bond. Means between oxygen atom we have three electrons bond. One, two, three. Three electrons are present into this. With all these oxygen atom has two lone pair and we have a three electron bond. That's why we call it as odd electron. Because of this odd electron bond, odd electron bond, these are paramagnetic in nature. These are paramagnetic in nature. See all these superoxides are strong oxidizing agent. Few property you just write down. This is also a strong oxidizing agent. And these are also strong oxidizing agent. Next one point to write down for peroxides and superoxides. Write down for peroxides and the stability increases. As we go down the group. In other way what we say is tendency to form peroxide and superoxide increases as we go down the group. Since they are getting stable. So stability of peroxide and superoxides increases as we go down the group. So this is an important point we have understood. One more reaction we will see and then we will take a break. Next write down reactivity towards water. Activity towards water. Write down alkali metal reacts with water forms hydroxide. Alkali metal reacts with water forms hydroxide and evolve hydrogen. Reaction you write down. We have 2m OH plus H2 2H2O. Hydrogen gas evolves. Alkali metals reacts with water forms hydroxide and evolve hydrogen. Write down the reactivity of alkali metals with water increases down the group. This you must remember. Write down the reason behind this. The reason you write down. Write down the reason can be explained by the reason can be explained by the kinetics of the reaction. The reason can be explained by the kinetics of the reaction. And when I say kinetics of the reaction and when I say kinetics it means we are talking about the rate of the reaction. Kinetics means rate. Thermodynamics if I say it means energy. Energy involved. So this particular reaction and the reactivity order is explained by the kinetics of the reaction. So what happens next write down next line as we go down the group the melting point decreases melting point of the metal. As we go down the group the melting point of the metal decreases and it is spread over spreads over the surface of metal sorry melting point decreases and the metal sorry and the metal spreads over the surface of water easily and the metal spreads over the surface of water easily which increases the contact area and hence the rate of the reaction which increases the contact area and hence the rate of the reaction. The rate of the reaction you must understand. If the contact area is more contact area means what contact between the two reactant. If that area is more then the rate of reaction will be more and as we go down the group the melting point of metal decreases and that metal spreads over the surface of water during the reaction. So because of this decrease in melting point and hence the rate of reaction increases that is why as we go down the group the reactivity of alkali metals with water increases because of large contact area that's the point we have. Understood? Is it clear? Guys let me know quickly. So can we take a break now? Okay so we are taking a break now we will start at 7.30 we will resume the class at 7.30 fine Cool take a break now we will resume the class at 7.30 Okay guys can we start the class? Are you there? Are you there? Can we start? Let me know. Reactivity towards water we have discussed. The next thing we have reactivity towards halogen reactivity towards halogen. Write down alkali metal reacts with hydrogen sorry reactivity towards halogen write down alkali metals directly reacts with halogen and forms halide directly react with halogen and forms halide okay the reaction you can write m plus x2 gives mx see this metal and halogen bond write mx since this is an this halogen atom is more electronegative more electronegative and hence it has tendency to drag the bond pair the electron of metal towards its side right so as we go down the group ionization energy decreases right ionization energy decreases hence the bond formation is easier okay so reactivity of alkali metals towards halogen we can write as we go down the group as we go down the group ionization energy decreases and hence reactivity increases reactivity means reactivity of a metal so order of reactivity if you write down that will be l-i-n-a-r-v-c-s order of reactivity towards a particular halogen this is the order of reactivity towards a particular halogen atom order of halogen towards a particular alkali metals is this it is maximum for fluorine then chlorine then bromine and then iodine okay and this is because the decrease in electronegativity as electronegativity decreases reactivity also decreases stability of what stability of stability of mx or what tell me stability of halide you are asking for a metal halide you see for like suppose l-i-c-l-n-a-c-l k-c-l r-b-c-l etc for all these metal halides as we go down the group the size increases and hence stability decreases stability decreases right size of cation is increasing and hence the stability is decreasing bond length will be more that's why the stability is less next you see reactivity towards hydrogen towards hydrogen right on alkali metals reacts with hydrogen and forms hydride of hydride of m-h type hydride of m-h type the nature of these hydrides are what these are ionic hydrides discussed in hydrogen chapter these are ionic hydrides okay if you see the stability order of these hydrides stability order that will be maximum for l-i-h the same factor size increases stability decreases l-i-h n-a-h k-h r-b-h etc right and this is because of the size of cation size of cation stability order or lattice energy also if you say the order is same smaller size higher will be the higher the lattice energy okay next one more important order we have that is ionic character the order of ionic character is this l-i-h minimum then n-a-h k-h r-b-h why this order we have because with smaller cation you see a smaller cation charge density high positive charge density with high positive charge density high will be the polarization polarization in electron cloud of the anion again I have discussed this in periodic properties what is polarization and polarization leads to what covalent character phase answer it is right higher polarization higher will be the covalent character so for a smaller cation it is more covalent than ionic that's why the ionic character the order will be this and reverse of this will be the covalent nature yeah correct the same thing it is the phase answer only smaller size of cation high will be the polarization high polarization means lower ionic character and higher covalent character same thing phase answer next slide down solution in liquid ammonia solution in liquid ammonia write down this write down alkali metals when dissolved in liquid ammonia alkali metals when dissolved in liquid ammonia not in liquid ammonia gives gives a deep blue solution deep blue solution which is conductive in nature this is important okay this one is important which is conductive in nature and has a strong reducing properties has a strong reducing properties this solution this solution a magnetic in nature all these points are important okay in Jay they have asked question on this now you see the reaction here metal M and combines with ammonia NH3 we will get here M NH3 X plus plus electron NH3 Y minus in this what happens metals are these okay metals loses electron and that combines with that combines with this one combines with ammonia and hence it is called as hence it is called as ammoniated electron this is only responsible for the color for the the blue color of the solution that we get okay this electron is ammoniated electron and we also call it as solvated electron this is important what color of solution we get answer is blue color solution is diamagnetic or paramagnetic paramagnetic okay what is ammoniated electron ammoniated electron are those electrons which are you know which comes out from metal and combines with ammonia in the liquid ammonia solution and these electrons are responsible for the blue color of the solution right can you move on next next you see hydroxide hydroxides right down normal oxides oxides which is M2O type normal oxides react with water reacts with water to form hydroxide hydroxides all these all these hydroxides are white crystalline solid white crystalline solid you see the reaction suppose I have Li2O plus H2O and it gives 2 Li OH this kind of hydroxide we get now this dissolve in water dissolve in water with evolution of heat exothermic information now the thermal stability the order of thermal stability of these oxides thermal stability you know what is thermal stability thermal stands for heat okay this thermal stability is the stability towards heat stability towards heat okay so the order as we go down the group as we go down the group size increases and hence stability decreases so order of stability is this Li OH then NaOH KOH RBOH etc all these solutions are base okay NaOH is a base NaOH is a base okay so the basic order if I ask you what is the basic order here so basic order or basic nature or basic order of this follows this order CS OH maximum then we have RB OH KOH and we have Li OH the reason for this is reason you see due to large size right down due to large size of cation due to the large size of cation less lesser will be the polarizing action lesser will be the polarizing action lesser will be the polarizing action and hence less will be the distortion of hydroxide ion of hydroxide ion Just a second just a second you wait. Okay, so what did you write? Due to large size the lesser will be the polarizing action lesser polarizing action means less distortion of electron cloud of hydroxide ion and hence Will have strong cation and iron bond will have strong Cation and iron bond and hence and hence the basicity Will be more and hence the basicity will be more Okay, so the basicity order you write down It increases down the group. So that's what I've written here basicity order increases down the group next you write down carbonates and bicarbonates the last one here is is Carbonates and bicarbonates see this alkali metals Forms carbonates of carbonates of M2 co3 type Okay, and Bicarbonates of MH co3 type down next as Electro-positive character increases You can say that also pretty there not an issue. Okay, that also you can say Okay, write down as electropositive character increases the stability of carbonate increases right and the order of Stability is this li2 oc2 co3 Then na2 co3 because we know electropositive character increases down the group So k2 co3 RB2 co3 and then we have CS co3 You see one important point you have to keep in mind here That this li2 co3 all these carbonates right down all these carbonates are stable towards heat and Easily soluble in water all these Carbonates are stable towards heat except you write down this one first right on all these carbonates are Stable towards heat except Li2 co3 this is the exception Not stable towards heat All these carbonates are stable towards heat except li2 co3 and soluble in water and soluble in water see Write down the reason of this li2 co3 li2 co3 is not stable since not a stable because of a small size of cation because of a small size of cation comma it produces Very high polarizing action this polarizing action this strong polarizing action Then this polarizing action weakens the carbon oxygen bond of carbonate ion weakens the carbon oxygen bond of co3 2 minus ion and hence and Hence the bond dissociate the bond dissociates and co3 and You write on next line then this reaction and this li2 co3 on heating it converts into li2 o and Co2 this is that dissociation takes place right this is important for it is cs2 only I have missed this two over here. It is cs2 co3 My mistake. I have written that okay. It should be cs2 co3 Correct, so for bicarbonates you write down bicarbonates next time write down lithium due to its less electropositive nature Lithium due to its less electropositive nature does not form does not form bicarbonates in Solid state all these bicarbonates are found in solid state and And soluble in water Just this reaction you write down for bicarbonates all these bicarbonates are found in solid state and Soluble in water when you heat this bicarbonate you will get carbonates H2o and CO2 now see we have done for the Today's class What is your homework is I'll tell you Okay, because all those trends we have discussed now the point is in NCRT you have to study These points what I have written over here abnormal Behavior of lithium lithium why is it so because a small size of lithium okay abnormal behavior of Lithium you have to study Okay, then there are a few compounds of sodium it is given in NCRT Compounds of sodium and there's nothing to understand in this compound of sodium for example NaOH It's preparation and properties. It's given Okay, preparation method and properties you have to do that Then baking soda that is NAHCO3 Baking soda then sodium carbonate and a2 CO3 Right, so these three compounds you have to go through it is given in NCRT probably if it is not there you can refer any other book But you have to memorize all these things the method and all okay the compounds of sodium their preparation method There are various process of preparation Preparation and properties these things I expect you to finish By the next class, okay If you have any doubt you can ask me we'll discuss that and then we'll move on to the next chapter Is it clear understood? To the important point in these kind of chapter is the trends trends. They generally ask in exam Okay, some preparation method also they ask but that you have to memorize completely You cannot do anything in it. Trends you can put some logic and then you can understand. Okay? Okay, so we'll see you in the next class. Thanks for joining all of you. Take care. Bye