 Okay, hello guys. Good afternoon. All of you please type in your name so that I can understand who are there in the sessions. How are you? How is the preparation going on for board? It's still starting for JEE or you have shifted your focus onto the board exams. Okay, so when you are planning to start for board, I think you are left with 20 days, no? 19 or 20 days. I think 3-4 more days you can go for JEE, but then you have to focus more on your board exam. It's from 2nd of March, right? So for NACLS boards are going on. Cool, you see, today in this session we are not going to discuss about, this is not actually a problem solving session. We are going to discuss some strategy or the question that has been asked in board exam, what kind of questions they ask, what are the sections are there. So basically you have all this idea, right? You must have gone through the past year questions, right? So you must know the pattern of questions, what kind of questions they ask, right? So those things, but let me go quickly with all of these things. Okay, so for your 12th board exam, I can say there are 3-4 sections we have in this board exam, right? So I'll just go quickly with this, okay? Section 1 contains question number 1-2-5. Okay, today we'll see some questions of, which has been asked in this board exam, previous year question 2 has 17 section questions we'll see, but we'll go through with this first also, right? So question number 1-2-5, the question they ask here in this section that also we'll discuss and, you know, what kind of questions you have to look for this particular section that also we'll discuss today. Okay, so question number 1-2-5 carries one mark each, then question number 6-10 Question number 6-10 carries 2 marks Right, then question number 11-22 3 marks, question number 23, it's abstract question, right? Carries 4 marks, and then question number 24-26, I think 26 questions are there, carries 5 marks each. So this is the, you know, distribution of marks we have, okay? So first of all here you see the first section here, question number 1-2-5 it is actually, it carries 1 marks, right, in this section, right? In this section we have basically a reaction-based question, okay? They'll give you a few reactions, like one reaction, it carries only 1 marks. So in this section we have a reaction-based question, they'll give you some reaction and they'll ask you what is the product we have here, it's not like organic reaction. Suppose like one question we'll see here, when we discuss that, when we start discussing the questions. Suppose if I ask you the simple thing here, like when N2, or if I write down when KMNO4 reacts with, reacts in with some sulfate, with some salt in acidic medium, plus some salt if I write down and write down the whole equation in acidic medium. Then what is the product of magnies we get here, this kind of question, okay? One question I'll show you, what kind of question they have asked here. So this is here, in this section we have inorganic questions or the reactions related to inorganic. It is not like organic questions that you have to find out the major product, it's not like that. One question what I have seen here, they mainly ask from IO-PAC nomenclature, okay? They'll give you the compound, you have to write down the IO-PAC name of that, okay? Effect of catalyst, you can say effect of catalyst, effect of catalyst. Or I'm just giving you one example, okay, so that you can understand the type of question that they ask here. They can also ask you what is the effect of pressure in liquefaction of gas, oh sorry, that's in 11th class. What is the effect of pressure in solubility of a gas? How the temperature effects the solubility of a gas, right? All these kinds of questions, like the effect of catalyst in activation energy they can ask, or Gibbs free energy, right? This kind of question they can ask here, okay? And then the last one, they also ask a little bit of question of colloids here, okay? So like this they ask questions from all those chapters, very straightforward and basic question here. That is one mark only, okay? And while you write down the answer, you don't have to write down much here, just you write down the exact answer that is it. You don't have to, no, explain the things here. Now you see question number 6 to 10, that is two marks. And here they can ask you the comparison of acidic basic strength, acidic basic strength. They can also ask you like, for example, they can ask you the basicity of amines, okay? They'll give you the compound and you have to compare the basicity of amines. Similarly, acidity of any carbosolic acid they can ask you, right? A few organic reactions also, here they generally give organic reactions. Here they'll give you organic reactions generally. And you have to write down the product, okay? Sometimes they also ask you the mechanics, right? So they won't give you some complex mechanism. They won't ask you here the name reaction and all, right? Name reaction when I say it means canizaro, aldol, all this parkin reaction, all this, right? They may ask you these reactions, name reactions here in five marks question, okay? So here they'll ask you some simple reaction. Like, suppose, nitration of benzene, you know, halogenation of benzene and all those kind of questions then they ask you, okay? Sometimes they also ask you what is the electrophile here, right? So this, when you have two sections in one question, like what is the product here? Give mechanism. So the product of the reaction contains one mark and mechanism contains one mark again, right? So like this, the distribution is there. Organic reaction they can ask you. They can also ask you structural formula of organic acid, like what is the structure of S3PO4? They can ask you what is the structure of, you know, S3PO3 like that, okay? You have to write down the structure of it that, you know, and sometimes they also ask you what is the structure of formula of S3PO4 and what its basicity, those kind of questions, okay? They also ask you here, what are the different types of cells? Electrochemistry, okay? Cells and the reaction involved in those cells, like mercury cell, voltaic cell and all, right? Sometimes they also ask you, give the reaction of, you know, voltaic cell or mercury cell like that. The reaction is also important, okay? If you ask you only reactions, okay? Then if you can draw the diagram of the cell, that would be better, okay? Draw the diagram, need diagram and mention everything in the diagram, okay? Here I said that they ask IOPAC, you have to do the nomenclature of the compound, but in this section when you have two marks question, they'll give you the name of the compound and ask you, you know, write down the structure of the compound, molecular form, not molecular formula, structural formula of the compound, okay? So the opposite thing they'll ask you. All these things I'm giving you, you know, with reference of the question that has been asked in the previous exams, right? Question based on intermolecular force also they ask. What kind of force involved in this solid and that solid like that? It's a factual question, but you should keep that in mind. We'll see the examples on it, okay? Now question number 11 to 12 carries three marks, okay? So in this section, they may ask you give reason of the following, while this particular compound is more acidic than this one, okay? So they can ask you reason over here, okay? They can ask you, you know, numericals, basic numericals they can ask you here. Actually, let me tell you here, when they ask you the numericals, however, you can solve the numericals of both since we have done for JE, you can solve these numericals easily. But if you are not able to solve the numericals, so if you write down the data itself only in the answer sheet, you'll get one marks for that, okay? And then you have calculation and in the last, you have answer, okay? So for answer, you'll get one marks. For all the calculation, you'll get again one marks. The kind of distribution is there, okay? But the point is if you do not know the numericals, at least you write down all the data given, and then what all formulas you can, you know, apply regarding those data given in the question, okay? So basic numericals also they can ask, they can also ask you here coordination compound, okay? The coordination compound about coordination number, few definitions they can ask you, right? What is coordination number, what is ligands, okay? Types of ligand they may ask you. And then in the coordination compound, they may also ask you the isomerism part, okay? What kind of isomerism these two compound shows, okay? This kind of question, they may ask you, okay? And one more thing, isomerism, there are many questions possible. Magnetic behavior also they ask, magnetic behavior. You may get two, three, you know, different section in one question itself, okay? So that they can distribute the marks in those sections, okay? They may also ask you here about SN1, SN2 reaction, reactivity thing you have to explain here, SN1, SN2, the difference between physios option and chemis option, right? This kind of question they may ask, okay? Catalyst, about catalyst also they may ask you, okay? And in, like I said, in the transition element, okay? Or extraction, they may also ask you, explain these terms, right? Froth, flotation, zone refining, chromatography, this kind of question they may ask you, right? So this is section third. This is an abstract question. You will have a paragraph on the basis of that you have to answer those questions. 24 to 26 carries five marks, which is the most important, right? And in this, they can ask you many things. They can ask you numericals also here. They can ask you numericals here. They may ask you name reaction also with mechanism, okay? Name reaction with mechanism. Like I said, aldol, canizaro are the most important one, okay? Name reaction with mechanism. And like that, they also ask you a reason-based question. Reason-based question, okay? So this is all about the questions of board exam. This kind of question generally they ask, okay? So like this, you can prepare. The point I'm trying to make here, it is what? That the best way to prepare for board exam is to solve past, at least past five-year question. If you solve past five-years properly, okay? Whether it is numerical or theoretical question, if you solve past five-years, you can do very well, very good in your board exam. Why? Because if you solve this question, you actually covered all the portions, okay? If you see, and what I observed, if you solve past-year questions, five-year questions, you will get, you know, 20 to 80% questions will be, you know, not exactly same, but you'll feel like you have done this kind of question, okay? Data may be different from here and there, but similar kind of question. And one more thing here, like when you are, when you practice for this, so since you are in habit of solving multiple choice questions, where you know, you scribble things in the question paper or your answer booklet, and then you mark option A, B, C, and B, okay? So you have to come out of that frame of mind that here you cannot scribble, okay? You have to explain things in a proper manner, okay? Like suppose, for one question, if you have to write down the definition, definition should always be supported by, supported by the examples, okay? Like if they ask you coordination number, right? You write down the definition of coordination number, and with one example, you can show that the definition, the coordination number of this metal is this in this particular compound, right? What is geometrical isomerism if they ask you? So write down the definition, and then show that with an example, okay? So that is always good to support your, you know, carry with some examples. That is always good. Because nobody is going to read line by line of your answer. They just wanted to, you know, focus on the keyword, right? Like if CO2 comes out, it comes out with an effervescence, right? So that you have to write down that effervescence word should be there. So those kind of keyword you have to keep in mind. That kind of, that is the preparation that I think you need to do. Because when you look at the question, you will feel like, okay, you know probably all the question. Conceptually, what is the question all about? What is the reason behind this? But here what happens? You have to explain those things in the words, in the answer sheet, okay? So you have to build your answer. You cannot go from here to there, right? Simply, you have to narrate your answer in a proper way so that anybody will, if they read, they will understand that what is the point of trying to make, okay? It should be in a proper manner. So keep that in mind, okay? Especially this five-mast question, you must when you, if you have any practice paper or a previous year paper, you try to write down the answers, okay? Properly, like suppose you're writing down the test only here, right? So this habit you have to build otherwise in the exam, and you have written exams in the schools, you must have that habit. But see some questions now. These questions are of one-mark question, right? This is one-mark question. I'm just giving you the question which has been asked previous year, right? It's a one-mark question. You see, the question is the main product of the reaction when 2-butene reacts with chloroform. Just a second. Can I have a student name? What's your name? Thank you, thank you. Link I already sent. Oh, I think this is the question. Yeah, this is the question. That was the different one. This one. Correct, correct. That's the new one. That's the new one. That's the new one. That's the new one. This one. The one-mark question. We will not discuss the whole paper. Okay, just we'll see the type of question and how to write down the answer. So that we'll see. The question is write the formula of the compound of phosphorus, which is obtained when concentrated HNO3 oxidizes P4. Okay. So it's there in nitrogen family. Okay. And if I write down the reaction here, it's even P4 reacts with HNO3. And this is concentrated acid. So the product here we get is H3PO4, NO2, and H2O. Balance reaction is not required. So the compound of phosphorus, which is obtained when concentrated HNO3, oxidizes P4, answer is H3PO4. And nitrogen dioxide. The question was, what is the oxide of nitrogen we'll get here, right? So that's the question. If you take here diluted HNO3, you'll get NO, nitrogen monoxide. So there are a few reactions of this given that dilute, constant dilute, H2SO4, concentrated H2SO4, those kind of reaction you must go through. Okay. But here it will be H3PO4. The answer will be H3PO4, right? First question you have already done. Let me check that. What is the question? Main product of the reaction would be 2-butene chloroform with NaOH. Yeah, it's correct. We'll get butanoic acid. And we have 2-butene. So methyl will be there at first. So this one is correct. Okay. But that was not the board question by mistake I've given you. So you see the next question. Sorry. Question number 2. You see this one. All these are one-mark question, right? IOPSE name of the following. Tell me, what is the IOPSE name? I think, tell me, what is the name of this one? IOPSE nomenclature. See, all of you have written, but you have to take care of the comma and the dash. So name is correct. The numbering will be 1, 2, 3 and 4. So the name of the compound will be 2-methyl-3-bromo. 2-methyl-3-bromo. Then, butene-4-all. 2-ene-4-all, right? This is one possibility. Or, if I do the numbering like this, 1, 2, 3 and 4, then the name will be what? Name will be, oh sorry, I have, this should be in the alphabetical manner. Okay, this should be like this. Now, the name will be what? Again, 2-bromo, 2-bromo, then 3-methyl-3-methyl-2-ene-1-all, right? So these are the two possibilities. And here, the name is 3-bromo-2-methyl. It's not 2-methyl. So, here you see the sum of low canes is 3 plus 2-5 only, here also 3 plus 2-5. When the sum of low canes is same, we'll follow what? We'll follow the alphabetical sequence, right? So we should do the numbering in such manner so that the halogen group, which is bromo, gets the lower position, right? So, we will do the numbering from this side, like this. So this is the correct name. So, this is obviously not correct. So, here you see another thing. Whenever you separate a letter with a number, we always use dash over there. When we are separating two numbers, then we use comma. 2 comma 3, die, something will write. So, two numbers are always separated through comma, right? And a letter and a number is separated by dash, right? Here it is O here. So that's why we have dropped this terminal E. We don't write this terminal E here, right? So these are the few small, small things you have to keep in mind. If you don't write this dash comma and all, your whole answer will be wrong. Okay, you will get zero into this, right? So keep that in mind. Don't make any mistake in these kind of questions. Okay, these are again one more question. Next question I will show you. Tell me this one. Effect of adding a catalyst on activation energy and GIFs free energy of a reaction. Okay, the answer is, this is just saying, Lord, sign here, activation energy decreases. Cushion decreases. Both decreases. What about GIFs free energy? Okay, you see, first of all, when it is a catalyst, so for JE we have discussed that there are two types of catalysts, positive and negative. Excuse me, but if it is not mentioned, we assume positive catalysts only, right? We assume positive catalysts. So, and for the board also, whenever catalyst is written, we assume positive catalysts only. So we know the positive catalyst or catalyst decreases the activation energy, right? And the reaction proceeds with a faster rate, correct? So first of all, this activation energy in presence of catalyst will decrease and all of you have done this and Del G, that is change in GIFs free energy will remain unchanged. Del G depends upon, it depends upon equilibrium constant. Equilibrium constant. And hence we know the equilibrium constant does not change with catalyst, okay? So it only depends on, equilibrium constant only depends on what? Equilibrium constant depends only upon, equilibrium constant depends only upon temperature, correct? So if temperature is not changing, the equilibrium constant will not change, okay? So obviously activation energy decreases and Del G remains unchanged, right? Answer is this. How many of you have gone through previous year board questions? Honestly, tell me, how many years have you seen? Do solve it again. At least five years. See, one more question. Which one is allylic halide? At least 15, 2015, 16, 17, 18. Halide is this. Okay, this is again one more question. Okay, so these kind of questions, they ask very straightforward questions they ask in this question, number 125. And you don't have to write so much. Just you write down the answer. So next question I will show you. See this two marks question I will show you. This one you see. This is for two marks question. Are in the following compounds in increasing order of the acetic strength? Nitro phenol crystal. Nitro phenol. Is it decreasing or increasing order? Okay, they ask increasing. Okay, increasing order. Correct. The question is in increasing order. Okay, so generally this is also correct, but generally we'll write down in left to right in increasing order. Yeah, so these are a few things that you must keep in mind. Increasing order decreasing order. Read the question properly before answering. Okay, now I'll draw the mechanism of this. Okay, and you have to tell me the product when this reacts with these two. Okay, we have CH2 C double bond C acts with what? H3O plus OH positive charge on it. Mechanism of this will be what? This pi electron, which is a weak electron, it attacks onto this hydrogen and then oxygen will take this bond pair of electron. Okay, so what happens? Positive charge plus H2O. This is what the product we have, which is nothing but CH3 CH2 plus. This is the mechanism. Write the structure of the products when butane to all one, two, three, four, butane to all reacts with CRO3 and then this butane to all reacts with SOCl2. Tell me the product. What is CRO3? What is CRO3? Chromic oxide. What does it do? What is that two chloroform butane? Oxidizes. So this will oxidize and what is this alcohol? It is 2-degree alcohol, right? Secondary. So we know the secondary alcohol oxidizes into ketone. Primary goes into aldehyde and this produce the chlorination. So the product will be, all of you have got the right answer. So you see the nature of the question. When you see what they have asked into mass question, the acidic strength. So here in this section they can also ask you the basic strength of any questions or whatever we have done in GOC. Any of those, not very vast, but those kind of questions they can put over here in this section, two-mark section. So basically you have to use the concept of various electronic effect which is plus R, minus R, plus I, minus I. Right, aromaticity. These concepts you have to use to solve this kind of question. Mechanism, it's not required, it's not there for CR-O3. But it is an oxidizing agent. Okay, that is what we have to keep in mind. Like KMNO4, those are oxidizing agents, right? K2CR2O7. So this is also an oxidizing agent, right? And we know that alcohol, primary alcohol is 1-degree aldehyde without any change in the number of carbon atoms. Any change in the carbon chain. If you have secondary alcohol, this converts into ketone, right? Partial oxidation. Otherwise, it finally converts into acid. That's the thing you have to memorize. So chromium oxide gives you partial oxidation. This one you solve. See, very basic numerical for two months. This kind of question, I think you can solve. But you know, you have to have this idea that what kind of question they're asking. Accordingly, you have to prepare. What is the answer? 4.5, 10, 20. How did you do this? Can you explain a little bit? Yeah, correct. So like I said, you can easily solve this kind of question. Like if you do not make any calculation mistake, you will obviously get the right answer, right? They're very basic question that they ask, okay? No, there's no any concept of void. Very straightforward question. If you have basic understanding of mole, you can do this kind of question. So answer is correct. 4.5 into 10 to the power 22. Number of unit cell. These many unit cell we have, right? So we have basically number of moles, right? So number of moles into Na gives you number of atoms. And one unit cell, there are four atoms. Accordingly, you can find out by unitary method. Okay, so answer is correct. 4.5 into 10 to the power 22. What is this? Tell me. Write the name of the cell, which is generally used in hearing aids. The reaction technique. Please let me know in the chat what in this cell. Tell me. Just a second. What is this cell? Okay, mercury cell. It's correct mercury cell. And what is the reaction we have here? Or can you tell me the product, what we get the product at anode and cathode? Mercury cell is right. Mercury cell, again, we are talking about mercury cell now, but you have to go through all those voltage cell and all, which is important. And then reactions also. What reaction takes place at anode? What reaction takes place at cathode? Right? So, yeah, zinc or mercury will get... No, we get mercury at cathode. HGO will get reduced at cathode. Right? And we get HG liquid at cathode. ZN will get oxidized at anode. So these reactions of anode and cathode, you must go through. So it's not about only mercury cell, all those cells which are given, when you go through the previous question, you will understand which are important. Okay? So for all those cells, anode and cathode reaction, you must remember. Yeah, correct. So that's what you have. It's there. The reaction is given in NCRT. The reaction is given in NCRT. This is again one type of question that we have, which they ask in the exam. Another one I will show you. Okay, can you write down the, you know, molecular formula of all these? Tell me the molecular formula of this. Can you just mention my name on that? Yeah, sure. When I speak, I will, I will let you know the combination will come. Not that. Sodium disynido, all right. What is this, all right? Okay, you see the compound is this. Compound is sodium, all right. And that is one. So what does this one represent? This represents the oxidation state of metal. So whenever you write down the formula, you just cross check what is the oxidation state of metal you are getting in that formula. Dysynido. Why have you written this? Ido over here. Can you tell me why this Ido is there? Why this Ido is there? Can you tell me why it is not Ide? Dysynide, all right. Remember, we have done this and we, whenever negative charged ligand, if you have that ends with suffix Ido, right? Ito, sulfato, like that. Ato, like that, okay? So negative charged ligand always ends with this, like this Ido, Ido, Ido. That's right. Anyway, Cn, right. Dysynido means Cn is whole twice. Or it is nothing but it is gold, AU. That hit close, right? And here this sodium metal we have, so Na, we'll write down here because it is the positive charged metal, right? Na plus, we'll write down left-hand side. Now, if you cross check the oxidation state of this, it is plus one. So here we have minus one. Here we have minus one. And this cyanide is again minus one. So minus two will go that side. So it will be plus one. So that's correct. Okay, so whenever you write down the molecular formula, just cross check the oxidation state given. Similarly, you see here tetraamine. So we'll write here what NH3? Amine is NH3, so it is NH3, 4, right? Tetraamine, chloride. So chloride is Cl, nitrito. What is nitrito? Nitrito is NO2. Platinum sulfate, platinum 4 sulfate. Platinum is the metal. So we'll write down Pt here. And sulfate is SO4. Okay, if you cross check the oxidation state of platinum, it is minus two. So here we have plus two. NO2 is minus one. This is also minus one. This is neutral, minus N minus. This platinum is in plus four oxidation states. What does this N platinum means? What does this N platinum means? What does this N platinum means? Yeah, it means the ligand NO2 in which the N atom is bonded with the metal, platinum. Okay, so it is involved in bonding. Now the point you have to understand here that if suppose, if it is written O platinum like this, O platinum, then instead of writing down NO2, it's always better to write ONO. So this will be wrong in that case. It means nitrogen is involved in the bonding. So when O platinum or any other metal is written, then better NO2 you write like this ONO. It means oxygen is the donor atom here. It is involved in bonding. This one, based on the nature of intermolecular forces, classify the following solids, silicon carbide and argon. Silicon carbide covalent, silicon carbide covalent argon molecule. Argon is molecular solid fine. Silicon carbide is covalent solid, that is also fine. But what is the difference between this silicon carbide and molecular solid? Yeah, correct. That also you can say covalent and weight dispersant forces respectively. What is the difference between a covalent solid and molecular solid? Okay, you see, silicon carbide, your answer is correct. Silicon carbide is covalent solid. Covalent, we also call it as network solid. Covalent or network solid. Argon, sorry, AR. Argon is the molecular solid. Molecular solid, right? And Sandhya here rightly said, because this molecular solid, it actually contains the intermolecular force here we have in this kind of solid is London dispersant force. London dispersant force, right? Which is very weak intermolecular force, right? Very weak. But in network solid or covalent solid, we have covalent bond. We have covalent bond. You know, what is London dispersant force? What is London dispersant force? What is London dispersant force, guys? Second one is correct. It gets oxidized. Zinc gets oxidized. Forms Zn plus 2, which is, you know, correct. But what is London dispersant force? Yeah, correct. First of all, this London dispersant force is actually, you know, defined for inert gases only, right? Suppose if I take the example of helium. So helium molecule and in this we have two electrons and both electrons are revolving around the nucleus. So two electrons we have these two electrons are continuously moving around the nucleus. So it is possible that at some instant two electrons will be on this side. So on this half of the molecule, this will be slightly negative and this will be slightly positive. Similarly, if the two electrons here in this atom for an instant, it will be here. Sorry, if it is here. So what happens? This side will be slightly negative. And then there is a charge. There is an electrostatic force developed between these two and which is nothing but the London dispersant force. So you see this is because of the relative movement of electrons. Okay. So for a fraction of second also this kind of, and you know, this electron you cannot even imagine or you know, you cannot find out the position of an electron. But whenever for a fraction of second also this kind of alignment is there, then there will have a very weak interaction, electrostatic interaction, which holds for, you know, even fraction of fraction of a second. But this kind of force we call it as London dispersant force. Which is actually defined for inert gases only. So that's the thing. Covalent bond already. Okay. This is the thing. Now this one we have done. ZNO turns yellow that you correctly said when you heat this ZNO it converts into ZN2 plus plus half of O2 plus 2 electron. It gets oxidized into this which gives yellow color over here. Right. Now what is meant by groups 12, 16 compounds I'll give an example. C, what is this C? Answer? Yeah, it is semiconductor like ZNS, ZNS, CDS, all these elements. This gets semiconductor. This is because of oxidation. Understood. Okay. Full you can see the question is what is meant by the groups 12 and 16 compounds. See actually you take elements of group 12. Right. And elements of group 16. And the compound that you get here by this use that compound. So group 12 we can take ZN, CD, we know it is the after 12 we have boron family. Right. It is a transition element ZN, CD and all. Right. And group 16 elements we have oxygen families of sulphur. ZNS, CDS behaves as a semiconductor. Understood. See the next one. One thing is just, okay, before solving this question, just one thing I want to discuss here. You see, this is 11th one. So this three marks question. Right. So you have actually three sections, three different question here, one, one marks each. Right. So when you have this kind of question, so you cannot write down two, three, four, five lines for answering one particular question. Only one line you have to write, but you have to be very specific. One to two line you can use, but you have to be very specific over here. Right. You cannot write down about like here. If you don't mention this random dispersion force and covalent bond, you won't get marks. Okay. So don't write randomly anything. Try to be specific. Since you know the three marks question, so each ABC contains one, one marks each. So accordingly you have to build your answer. So this one. Okay. So this is a numerical. I think you can solve this. Try this. That's 45.54. This is to the exact calculation. Don't approximate all this. You will get, you won't get marks here. What is the electrons? How many electrons flow? Answer is correct. 45.54 is correct. But what is the electron for flows? Option B. Sorry. Question D. Only two electrons. Only two electrons. Simon, what happened? Check. Okay. This one you are getting the wrong answer. I'll do this. See, first of all, how do we solve this question? So first of all, like I said, you have to write down the data. So what I'll write, I'll just write down the thing, the way you should write in the exam. Given cell reaction and that is 2FE3+, which is an aqueous plus 2I- gives 2FE2+, aqueous plus 2I2, solid. This is the reaction given. And what are the data, next data? E0 is given. E0 of the cell is 0. Temperature is around 8 Kelvin. And it is given, 1 Faraday is equals to 965 double 0 Coulomb per mole. These are the data given. Now if I write down the half reaction, the two half reaction, all these things you have to write down. Two half reactions. So FE3+, is going into FE2+, so this is what the reduction we have. FE3+, right, is going into 2FE2+, 2FE3+, we have here. So this will take two electrons here. So 6-2 is equals to 4. And this is also aqueous. And this is also aqueous. This is also aqueous, correct? So you see the number of electrons involves here is 2. Now then another reaction is what? 2I- and it converts into I2, correct? So this has to lose to electron. So, oh sorry. So when the number of electrons involve here is 2, n value we get, right? So, and all these, you know, state also you have to write down. I- is in, it's not given in the question. It is solid, so I'll write down solid here. All these things you have to write down. This is how we write down the two half reaction. Now we know del G is equals to standard gift energy. Del G0 is equals to minus nFE0 cell. All these values substitute here, you'll get the answers. Whatever you get, that is right. Okay, so all these data you have to write down. Now how many electrons flow through a metallic wire, if a current of this again for question B, again I'll write down the data. So I is equals to 0.5 ampere. T is given is equals to 2Rs, right? So one Faraday again this is given. So we know what we can write Q is equals to IT, Q is equals to IT, so that will be equals to 0.5 into 2, okay? So this is the charge we have and if you, it is in R, right? So better we convert this into second, into 3600. So you will get the charge here. So what we can write that if any electrons okay 96500 coulomb of charge, okay? Two things you have to understand here. We take time always in second here right? And this is the coulomb we have. 9650 coulomb of charge when passes through when passes through the you know the number of electrons in this case will be Na electrons 6.02 into 10 to the power 23 correct? But what charge we have here? 36 into this. So these many charges are nothing but 3600 coulomb itself. So 3600 coulomb charge gives you Na 96500 into 36 So this is 3600 coulomb okay? So when you solve this you'll get the answer and here the answer you will get is around 2.246 into 10 to the power 22 electrons. This is the answer we get. Is it clear? So you always remember the definition of coulomb okay? 96500 coulomb of charge is the charge when any electron passes through or we can say the charge on any electron is nothing but 96500 coulomb. Don't forget to write down these reactions okay? The data given you'll get one marks here right? Then if you got the right answer you'll get another one mark and one marks here. Next question you tell me this one. Okay. What type of isomerism is shown by the complex this linkage correct? Yeah. Next one B. So one marks in this. I have done this question in the class. Please I'll have some water and come. I understand actually I am not you need actually that's it. Thank you. Okay so this was again linkage is done. But we have to draw the box diagram and is there an electron in this pairing? Why is that pairing and there's no pairing? CN is a strong ligand correct? CN being a strong ligand and CL minus is a weak so CN initiates pairing of electrons opposing the Hans rule. So yeah it's correct which is right. I have done the same question in the class you can go through. You should have the information of strong and weak ligands which ligands are strong which ligands are weak. Generally when oxygen and nitrogen are the donor atom ligands are strong ligands generally we have this right? What is the answer for the third one C? Why are low spin tetrahedral complex? See in tetrahedral complex what happens the orbital energy or technically if you want to say the orbital spinning energy is not sufficient enough and hence the pairing does not takes place. Okay generally low spin complex we observe in octahedral complex okay why because in tetrahedral complex the orbital spinning energy is not sufficient and hence the pairing of electrons against the Hans rule does not take place. Yeah that's the reason here we have low splitting energy splitting energy and hence the chances of pairing against the Hans rule is very less or rare that's why we don't observe low spin tetrahedral complex. Correct? See here you have to write down the difference between these. Again you see this 14 number is for 3 marks right? This is for 3 marks so you don't have to write much into this because ABC we have 3 options 3 questions 3 separate questions each question contains one mark okay so on one point you have to write down here. In multi-molecular solid what happens I'll just give you the information here in multi-molecular solid in multi-molecular solid the aggregation of large number of molecules or atoms or atoms okay but in associated collide what happens we'll have large number of ions not atoms or molecules this is the difference we have right so in associated collide the ions you can memorize like this water is an associated liquid why is it an associated liquid because we know there isn't hydrogen bonding possible right so it is delta plus delta plus delta negative right when it is associated with another oxygen molecule like this so here also we have delta negative delta positive and delta positive so you see there are charts present here so this we can consider as an ion which is not exactly true but to memorize this you can understand associated collide means it is related with ions multi-molecular molecule is what molecule is molecular and it is not a charged species right it is neutral so it can be an atom or molecules right coagulation you know already it is settling down colloidal particles for this peptization the conversion of precipitate into colloidal soil by adding a small amount of electrolyte okay peptization when we add a small amount of electrolyte into it coagulation is the settling down of colloidal particles homogenous catalyst the reactant and catalyst are in the same phase right homogenous catalyst is what the reactant catalyst in the same phase that's why it is homogenous heterogeneous means reactant and catalyst are in the different phase okay so these things dispersed phase dispersion medium of milk what is the answer of this dispersed phase dispersion medium of milk the dispersed phase is liquid here and dispersion medium is also liquid both are liquid similarity between species option and chemis option I have done this in the class you can go through right the chemical method of which this soil is prepared from FeCl3 what is the similarity between these two you can write on these are the surface phenomena okay FeCl3 it is prepared by the hydrolysis of FeCl3 right this is a liophobic soil this you must remember this is a liophobic soil and this is prepared by the hydrolysis of FeCl3 plus H2O which gives you FeOH whole price plus HCl 3 moles of HCl you see you are getting one question from this surface chemistry right so all these things you have to memorize next one you see again this is a numerical question so you see this is the second numerical question we have here 3 marks okay so we roughly we will get 2-3 numerical questions in this section 3 marks question solve this with the proper method okay 96 minutes you are getting right the answer for this question which is given here is 96.96 minutes so if you write 96 you won't get one mark over here okay for 96 always do the calculation up to at least 2 digits you see the question all of you have got the answer what you have to do initially if I assume 100 right so when 25% decomposition is there so 80 is what 75 100 minus 25 and then we will find out t is also given 20 minutes so we will find out k right we will find out k with the formula which is 2.303 divided by t log of a0 by 80 a0 by 80 this k will find out from here and then again we will use here in the second set of data when 75% of the reaction is completed okay like before starting this you have to write down again properly that let the initial concentration is 100 and concentration after time t according to the given question is 100 minus 25% of that will be 75 time given is 20 minutes don't forget to write down the unit also here like any terms without unit you don't write always keep this in mind unit you have to mention okay so from here you will get k and then this k will use for the another set of no data which is nothing but t is equals to 2.303 divided by k log of a0 is 100 and 80 is when 75% complete it is 25 right so here it will be 2.303 divided by t log of a0 is 100 and 80 is the concentration left so k will substitute in the form of this which is nothing but 2.303 log 4 divided by k is a0 will substitute this will not get k you calculate from here substitute here you will get the answer you will get 96.96 minute the point is the questions are easy you can solve this but so let's start you should have this habit to write down the expression first then the formula use you should write down since the reaction is given it is the first order kinetics so the expression of k and t is this and then you have to put the data and solve the question okay so all of you who got 96 minute you won't get one mark for this because they won't consider this as the correct answer don't round off here okay this is not the objective question don't round off do this one remember this is CH3 COOH this compound is actually CH3 COOH don't forget that CH3 COOH done what happened tell me the answer what happened guys see first of all the reaction is acid when reacts with NH3 right the reaction of ammonia it is so what happens in this as I have already told you that in ammonia all this NH bond are polar in nature correct so those can react with any acid as H plus and forms H2O correct so this CSL in the acid the lone pair of oxygen will take H plus from NH3 and forms H2O and convert into amide that is CS3 CO NH2 acetamide will get first and then will get ammonia okay so I am here got acetamide, ammonia and ammonium chloride see the first one the product A would be H2O goes out and it would be CS3 COOH and H2O correct plus H2O will get the another one the second product when this is allowed to react with BR2 COOH so this is the preparation method of 1 degree amine correct 1 degree amine and it converts into CH3 NH2 right we have done the mechanism also in the class if you remember carbene forms and this will rearrange itself and onto this nitrogen atom and it forms CS3 NS2 and this CO groups goes out so this is B and when this is allowed to react with CHCl3 with alkoholic COOH chloroform with alkoholic COOH this converts into isocyanide this is also a direct reaction CS3N C will get here all these reactions I have done in the class right you check your notes you will get the mechanism also Hoffman Hoffman bromination these are very straight forward reaction you don't have to write down the mechanism correct 3 mask question you have 1 so 1.5 and 1.5 will get here right so no mechanism required just you go through the various reaction given in the book or we have done that in the class you can go through with the notes also right so all these questions which contains 3 months you don't have to write down the mechanism simply write down the product here now here you see it is a dijonium salt correct and when this reacts with NaNO2 then clearly it is easily understandable from here that any BF4 forms and then this NO2 N2 also goes out you see the reaction of dijonium salt if you remember in most of the reaction nitrogen gas forms right Na combines with this BF4- will get any BF4 also and this NO2 comes over here this is nitrobenzene NO2 will get first nitrobenzene and then when this is allowed to react with FEHCl reduction takes place here and this nitrobenzene converts into Benjaminine NO2 converts into NH2 another last thing here this when it is react with CWOCl again the same thing in pyridine solvent from here forms HCl and CS3CO attached with this so here the product will be NH CWOCl this is the last product we get right guys all these are very direct reactions you must revise you must revise all these ok we cannot discuss the mechanism now but you should know that under which condition what reaction will get what kind of product ok so you must revise this don't miss this kind of question it actually helps you in JE also in April next question this one give reasons red phosphorus is less reactive than white phosphorus is polymeric and so if it is polymeric then what happens its intermolecular force it forms dimer see red phosphorus is polymeric in nature fine and the second thing white phosphorus if you see it has a tetrahedral geometry correct white phosphorus is this so because of this tetrahedral structure we have more angular strain here we have more angular strain ok and since red phosphorus has polymeric structure so we have long chain compound and hence it is what it is less reactive with white phosphorus what happens it has this structure in which we have more angular strain so it is not at all stable that's why it is comparatively more you know reactive than red phosphorus yeah your reason was correct reason was correct but one thing also for white phosphorus you also mentioned it has more angular strain ok that also you can mention it is white phosphorus what is the answer for B tell me the reason guys electron gain enthalpies of halogen are largely negative what is the reason yes they are electronegative they can easily accept electrons to gain noble gas configuration yeah so that's the reason electron gain enthalpies of halogen are largely negative since they have very high affinity of electron to gain noble gas configuration right tell me C N2O5 is more acidic than N2O3 what is the reason N2O5 has two lone pairs which can donate them and it is acidic correct oxidation is correct oxidation is general oxidation is state reason you can easily write or you can also write the percentile oxygen character of N2O5 is more than N2O3 right so see when you have oxides of similar atom like in this case N2O5 and N2O3 then we can also explain the acidity of those oxides depending on its percentile oxygen character if the percentile oxygen character is more its acidity is more N2O5 nitrogen is in plus 5 oxidation is state which is the maximum right and N2O3 nitrogen is in plus 3 oxidation is state so if the central atom has more positive value of oxidation is state so it can attract more electron towards it right and that's why it is more acidic so any one of these you know you can write the reason behind this okay next question you see three marks question acidity of enylene reduces its activation effect it forms resonating compound so electron and I don't know correct the lone pair of nitrogen is involved in the reaction right okay I will write down the reaction your logic is also correct you see enylene right so acidity of enylene and you know acetylation always takes place in presence of peroxide anhydride sorry right so enylene is this we have NH2 anhydride is CS3 C double bond O O C double bond O CH3 okay so what happens here is this lone pair of nitrogen is involved in the reaction and this lone pair will attack on to this carbon atom and this pi electron goes here so here the product here it is NH2 CH3 then we have O C double bond O CH3 and this NH2 will have the positive charge so this since this lone pair is involved hence it get deactivated the reason is what due to resonance you see it is there in resonance now what happens next this bond breaks so we will get here NH2 CH3 then C double bond O positive charge and this molecule will come out as CH3 C O O minus now finally what happens I will just write down here only so that you will understand finally what happens this H plus will also come out and forms with this so you will get NH positive charge here because of this only it is its activity decreases or we can say the activation effect decreases next question CS3 and 2 NH2 is more basic than this the reason behind this is what here the lone pair is available but here the lone pair is involved in resonance so it is not available that is why this is more basic NH2 although ortho-paradirecting group yet N-line on nitration gives a significant amount of metanitro N-line see what happens in this NH2 group of N-line this is ortho-paradirecting fine but on nitration what happens it gives metaproduct this is because the aromatic amine that you have which is nothing but N-line is highly reactive in nature and it reacts with acidic hydrogen of nitrating agent forms N-lineum ion which gives metaproduct so that is what I will write down here if you have N-line which is nothing but this NH2 lone pair on it on nitration so we know nitration we always do in acidic medium so H plus so it forms NH3 plus here so this becomes NH3 positive charge because of this ion only when NO2 plus reacts it really gives metadirecting product which is nothing but NO2 here and NH2 H plus will come out eventually so we will get a metadirecting product so the answer for this question is what if you want to write down this is because of aromatic amine which is nothing but N-line is highly reactive and it reacts with acidic hydrogen which is H plus here of nitrating agent and forms N-lineum ion first so intermediate is N-lineum ion because of that N-lineum ion we get metaproduct over there next question now this is here we have 3 marks question now you do this one a 10% solution by mass of sucrose in water as a solution so first question it is a 5 marks question take care of 1 mark question here marks we have with one option also remember write down the data first always okay data you have to write down and since it is a solution question so I always suggest you to write down the data and try to understand the question that what is the unknown what we have to find out and with that you try to apply the formula that you have already started in this chapter then it is KF not KB guys this KF is for solvent okay it is given no goes in water see a poor wig first of all it is KF not KB and whatever it is KF or KB it is defined for solvent not for solute you understand and solvent is water okay see I will do this first of all you see 10% solution by a mass of sucrose in water has a freezing point this the freezing point of 10% glucose in water the freezing point of pure water is this you see first of all the freezing point of pure water is given and when you add sucrose into this water its freezing point decreases to it is denying it here but you have to write down the data like it is data given write down all the data I am solving this question directly that decrease in freezing point delta TF is equals to it is the delta T of solvent which is water here minus the delta T of solution and what is solution here 269.15 when you add sucrose into this so that difference will be if you would reference these two that will be 4 delta TF is 4 and we know delta TF is equals to I went of factor into KF into M where M is the molality M is the molality I since these are non-electrolyte whether it is sucrose or glucose I is always one for non-electrolyte okay if you use the expression here the delta TF is 4 KF you let it be molality is given 10% solution by mass of sucrose okay if you take 100 gram of sample right then the mass of sucrose will be what 10 gram and that of solvent is 90 gram this is sucrose and this is solvent so that will be molality is what the number of moles of solute so 10 gram divided by the molar mass of sucrose is 342 given it is the number of moles divided by mass of solvent is 90 gram so that must be multiplied by 1000 this is the molality this expression is the molality right so when you solve this you will get KF and that KF is 12.30 unit whatever it is you just write it down now the thing is calculate the freezing point of 10% is freezing point of glucose we have to find out since KF value is not there so we will first find out the KF value that we have done already now the freezing point of glucose will be what the delta Tf again we will write this will be equals to KF into M right KF we have calculated 12.30 into molality will be 10 gram of glucose we are assuming divided by the molecular mass of glucose is 180 this gives you number of moles this whole divided by the mass of solvent which is nothing but 90 into 1000 right so when you solve this you will get the value as 7.7 ok then the temperature of solution from this we can write down that will be delta T of solvent which is nothing but 273.15 minus 7.7 and when you do this calculation the answer will be 265.45 Kelvin this is the answer we can yeah close correct 265.6 ok now this definition you know molality, abnormal molecular mass you know already so these two I am leaving what is the answer of this one 30 gram of urea if we use the expression the formula I will write down here p0 of solvent minus the pressure of solution this divided by p0 of solvent is equals to xb by xa or mole fraction will write so that will be equal to mole fraction ok so I will write down xb the mole fraction of the urea so this is the formula you have to use here all other values are given 23.55 since you have got this I am not solving this 23.55 this difference you know ideal and non-ideal solution ok next question we will see this is actually the 5 marks question ok you have to write down all those you know formula data given and always calculate the 2 decimal point ok don't round it off next question you see one more question we will discuss yes this one this is the last one we are discussing for today this is also 5 marks question 1sino cyclohexanol yeah it's correct and OH will be there on the ring what does this diisobutyl aluminium hydride do that converts CN into CHO if you remember we have done the mechanism also DIBLH diisobutyl aluminium hydride this is a larger base 2 in 1L yeah right so the product here you see we have OH and CN CN minus we will attack on to this carbon this pi electron goes here which takes this H plus we will get this product you second one I will discuss few things about it this is actually DCOONA plus NaOSCO soda lime it is see actually this reaction you can write the same product when you have acid here instead of Na if you have H here right this reaction we call it as D-carboxylation reaction D-carboxylation reaction and in this reaction we heat this acid or salt of this acid with soda lime so what happens here you see if you have see the product here it will be this only plus Na2CO3 forms you will get benzene but I am talking something else if you have an acid like this Z double bond OOH when this is heated with soda lime where we have NaOH right so what happens this lone pair of this oxygen will attack on to this hydrogen and takes out as H plus right so we get here what we get here carboxylate iron and when you heat this carboxylate iron it forms so what happens this lone pair comes over here and this is over here so we will get a negative charge here negative charge plus CO2 now this step is the RDS a rate determining a step and this reaction yeah that is what and this reaction is actually it forms by the formation of a carbon ion negative charge okay rate determining a step forms a carbon ion here right now this what does this CAO does here do here CAO this CAO combines with this CO2 and forms CACO3 this is what the purpose of using CAO we have here this reaction actually what happens as soon as this CO2 forms CAO combines with this and CAO3 so this reaction always goes into forward direction right you are removing this CO2 completely continuously with the help of the CAO hence this reaction goes into forward direction right now here what happens the H plus from the H2O that you get here will get here H2O also from that only H plus comes out and attacks onto the this negative charge so point I am trying to make here it is what first of all with this reaction will get H2O now from this H2O only H plus attacks over here right and we get OH minus back correct so the concentration of OH minus initial and final is same in this reaction right same in this reaction this act as the catalyst because we recover the same amount of OH minus at last in this reaction or what we can say the rate of change of concentration of hydroxide iron is equals to 0 ok so this we also call it as heating effect heating effect of acid ok there are 2-3 more cases we have here this is the one case we are discussing here is it clear no one is simple this converts into CH4 simply with diisobutyl aluminium hydrate understood right but all these things actually not required for bodies just you have to know this whenever you heat whether the salt of acid or acid itself in presence of soda line right it converts into alkane with lesser number of carbon atom because 1 carbon from here it goes with CO2 the number of carbon atom in the chain decreases here yeah correct now the first one is tolerance reagent we can use benzoic acid and phenol what we will do for that say for benzoic acid and phenol we use FeCl3 write down with FeCl3 phenol gives violet color but benzoic acid does not show any color actually right we use neutral FeCl3 or simply ferric chloride FeCl3 we use to distinguish benzoic acid and phenol phenol gives violet color with FeCl3 benzoic acid does not give any color does not give this reaction yeah that also you can do okay so this is how they ask the question you see the few things you have to build in your habit in these days okay like I am always saying that data you write down formula you write down it's not like what you know you have to write down everything related to the question asked in your answer sheet then only you will get the answer it is not like this where it is so obvious we don't need to write so depending on the number of marks the marks allotted to that question you have to build your answer right and never forget to write down the unit okay and those things right always calculate till the second point of the decimal right don't approximate don't round off the values okay and one last thing like I will tell you here okay that you see the type of question that they have asked in one mark two mark three mark and then five marks question one more thing monomers in you know polymer chapter monomers of all those monomers given over there that you must memorize okay for given a polymer what is the monomer okay that also they ask in one or two marks question okay so analyze the type of question that they ask and then you try to understand that in this particular question what is the one mark question we can have two mark question we can have and then try to especially when you have theoretical questions try to write down don't think simply try to write down and build your answer like I have to write down these two line first and the third one is this like this you have to go right always support your answers with of examples that is the most important part okay so the best way for the preparation I have given you I think fair bit of idea as far as chemistry is concerned in physics also the same thing units and all you have to take care of and those numericals you have to do right so the best way is to solve the previous year at least five year questions just go through and don't read it out theory question you write you just go through and then you write on in your own words that gives you of you know a fair bit of idea how to write down the answers okay so last five year question you must solve you can do much much better into that okay do you have any module paper for board exam it's there in the market I guess did you buy yes yes yes we can use any at co3 also do you have any of you know practice workbook for board exam or module for board exam it's it's there in the market I think yeah also that is fine that is fine we just need to solve that completely don't look at here and there and you follow and need to have to solve that keep those things in mind what examples you can put in like those things you can add from your site yeah sample paper okay so you have enough time I think 15 days is enough you can do it easily okay so I'll wind up the class here only okay let's see next class next class we can have next class what do you want us to take the next class or you will prepare for your board exam no I think we don't take J level now you want J level class correct then we'll take then we'll take the next class okay we'll discuss some reactions of polymers and I'm correct we'll take that okay so next class we'll discuss those we'll discuss board level questions only of polymers and biomolecules okay okay see you soon take care bye bye and thank you all yeah I'll do that we'll do that we'll see some more questions reasoning based questions we'll see that