 Hello guys, good afternoon. Hello Bhuvik, how are you? What is your exam going on? Guys, happy to name, where is the last exam Bhuvik? Hi, we just, hello Vaishnavi. Hi, Sanjana. When is your chemistry exam? Okay, and when is chemistry? Okay, chem is on first March. What about Sanjana? Vaishnavi, when is chemistry? Okay, so it's on 12th March. Correct, okay. So like we were discussing the questions which has been asked in both exam. Okay, so today also we'll continue with the same. So we'll start the session. We'll discuss first the question which has been asked for five marks today. Okay, and then we'll go on to the two marks and first one mark question. Okay, so you see this question first, we'll start from the last question. See, like I said, for the five marks question, the last three question, they ask some reactions. Okay, a reaction-based question on some numericals they'll ask. You see the question is what? The first one, 26. Illustrate the following reactions giving suitable example in each case. Hoffman-Brahman degradation reaction. First of all, you tell me the B part. How do we distinguish N-alien and N-methyl N-alien? Give me a second. Yeah. So how do we distinguish N-alien and N-methyl N-alien? N-alien forms pink precipitate, bromine water taste, basicity, azodite taste, okay. Okay, you see. To distinguish N-alien and N-methyl N-alien, what we'll do here is you see N-alien and N-methyl N-alien can be distinguished using one test that we have already done. And the name of the test is pylamin test. Amin test, write down this. This test, you write down carbylamin test. And what happens in this test? We used to heat primary amines with chloroform in alkolic KOH, right? And that gives a foul smell and forms isocyanide. Okay, this forms isocyanide, which we call it as carbylamin also, right? So in this test, what happens? This is given by primary amines only. Primary amines. This primary amines are heated with chloroform, chloroform in presence of ethanolic KOH or alcoholic KOH we can write, right? This reaction gives you a foul smell which confirms the formation of isocyanides. Because of this formation only, foul smell appears, right? So this isocyanides, which we also call it as carbylamin, right? So this is given by only primary amines. We have done the, in this, what we say, a mechanism also. And there we get, the intermediate is carbenes in this case, right? CCL2, carbenes are the intermediate. And then that reaction proceeds, we have some rearrangement, but that is not required here, the mechanism and all, right? So the reaction here it is, this test is given by only primary amines, okay? Secondary amines does not give this foul smell, okay? So that this foul smell only confirms the presence of primary amines, right? But in case of enylene, what happens? It is a primary aromatic primary amine, okay? Enylene is what? Enylene is C6H5NH2. It is aromatic primary amines, okay? So this gives carbylamine test, right? So this is aromatic primary amines, but what is the another compound we have? N-methylenylene. So if you write down this particular formula, N-methylenylene, if you see, enylene is this with N-methyl. So we have N attached with one CS3 here, this is N-methylenylene, correct? So this one N-methylenylene is a secondary amine, secondary amine, okay? This particular reaction compound does not give foul smell within this carbylamine test, but primary amine does, okay? So this one is primary amine, which gives this foul smell. And since N-methylenylene is a secondary amine, does not give this test, that is carbylamine test. So the answer is we can distinguish enylene and N-methylenylene with this carbylamine test. Second one you tell me. It is tertiary and secondary. Yeah, it's correct. Hinsberg reagent. Hinsberg reagent we use for the, you know, to distinguish secondary and tertiary amine. Secondary amine give this reaction, tertiary amine does not give this reaction, okay? And what we get into this reaction? In Hinsberg test, what is the Hinsberg reagent? Can you tell me? What is the Hinsberg reagent? Can you tell me? The Hinsberg reagent, benzene sulfonide chloride is the Hinsberg reagent and it forms benzene sulfonamide, correct? It does not, the primary amine, tertiary amine, since it does not contain any hydrogen. So this reaction is not, you know, possible in case of tertiary amine. So Hinsberg test is the answer for the second one. Now you tell me, what is Hoffman bromide degradation reaction? In short, just to tell me, you know, reaction you can, you have to write down the reactions also into this. You have to explain this reaction first and then you have to write down with the examples. Examples you have to write down after that spending the reaction. So what is Hoffman bromide degradation reaction? What happens in this reaction? RCO-NH2 with bromine and alkali forms R-NH2, correct? Yeah, so why we are calling it as degradation reaction? Hoffman bromide, why this degradation term? All of you know the reaction? All of you know the reaction, right? If you remember, we have done the mechanism also here. In the class, we have done the mechanism also. So you have to write down this thing. What happens in Hoffman bromide reaction in just theory? In this, we'll get primary amines. One carbon is less in the product, hence it is degradation and the reagent that we use bromine with alkali solution. With that, just write down three lines and then you write down the reaction, that is it. What is di-esotization? Di-esotization, formation of disonium salt from any link, correct? And what is the reagent we use here? Convergenal primary aromatic amine to disonium salt, correct, correct, right? What is the reagent we use? HNO2 and HCN, is it? We use HNO2 and HCN, confirm this. HNO2 or NANO2, resist? Yes, the reagent we use is NANO2 with aqueous HCN. Okay, that reagent you must remember, not HNO2. Next one, what is Gabriel thalamide synthesis? We've done this reaction in the class. What is Gabriel thalamide synthesis? What is this reaction? What is the major product here? Why do we use this reaction for the formation of what? For the formation of amines, yeah, right. So I'll just write down the reaction here. The whole step you have to write down into this, okay? So I'll just write down the reaction first. Just drop this off. So you'll see in this reaction, first of all we have this reaction involves the synthesis of primary amine using thalamide. Okay, and what is thalamide? Thalamide is this, the benzene ring with this C double bond O. C double bond O and this is thalamide. This structure is thalamide. Now when this is allowed to react with KOH, then we get a salt, right? H2O forms and we get a salt here, which is nothing but this. And here we have minus and K plus. Now when this is allowed to react with an alkyl halide, RX, remember here, this method is used for the preparation of primary amine. First of all, that you have to keep in mind. Now, since primary amine you have to use and that amine you'll get with respect to this alkyl group, okay? So that will see KX forms here and the product in this reaction will be, this will have C double bond O, C double bond O and here we get KX forms and we'll get N, alkyl substituted, this product we'll get, okay? Which is allowed to react with NaOH and with NaOH product we get here is sodium salt of this, O minus Na plus. Here also we have the same thing, O minus Na plus, plus we'll get R and H2. So if we have to form methyl amine, then here we'll take CH3X. With respect to this R group only, we are getting the amines here. So this we can choose according to our requirement. Whatever we have to form accordingly, we'll take the alkyl halide here and we'll get the final product. So this is one degree amine, which is, we get in this particular name reaction that is Gabriel thalamide synthesis, okay? Now going to the next part, you see this part, the R one, what I write as structure of main products when benzene-disonium chloride reacts with the following reagent. What is the product we get? These are the direct reaction. Tell me the product, benzene-disonium chloride reacts with CuCN or KCN, then H2O and then alcohol. Yes. What is the product we get in the first one? Sino-benzene, CN will attach over there, right? Nitrogen goes out, then some chloride also goes out and then CN attached with the ring. Second, the reaction will get phenol, correct? Third reaction, CS3, CS2, OH. Last part, is it benzene? What is the last one, CS3, CS2, OH? Do we get benzene or ether? N2 plus HCl forms. What is the main product, the last one in this? The third reaction, the last one I'll write down with disonium salt, which is N2 plus NCl minus when this is allowed to react with ROH, where R is CH3, CH2. So what happens is this, N2 goes out in this reaction, HCl also forms and with this benzene, OR group will attach like this, okay? So the product in this reaction will be CS6H5O, CH2, CS3. This is the product we get in the last reaction. Is it benzene? Send me the pic. Can you send me the pic just now? Right now you send me the pic. I'll just see the reaction condition. Send me on WhatsApp, okay? And then you saw the next question. I'll see this now, just now I'll see this. Yeah, I'll just check you. You do the next question, question B. It forms CS3, CH4, HCl, correct, N2 also forms. Okay, then we'll go with NCRT only, okay? We'll go with NCRT only, I'll just change this one. The answer will be benzene itself, right? This will not be there, go with NCRT, right? Benzene will form in this reaction. What is the answer B? Okay, I said, okay, I'll just number this. This is first, second and third, right? That is what you have done. We have to find out the boiling point, increasing order of boiling point. So we have alcohol and amine, correct? So if you compare alcohol and amine, which one will have the higher boiling point? Alcohol or amine, and why so? You're saying three is maximum, then one, and then two. Two maximum, then one, and then three. Just to tell me one general question I'm asking you. Alcohol and amine, which one will have the higher boiling point? Alcohol has the higher boiling point, and why is it so? Why alcohol? Because alcohol can show a strong hydrogen bonding. Yes, because of hydrogen bonding only, oxygen is more electronegative, so the strength of hydrogen bonding in case of alcohol is more than to that of amine, right? And that's why the boiling point of alcohol is more than to that of amine, right? So if you compare all these three, if you compare these three, so alcohol will have the maximum boiling point, the second one will have maximum, and then one and third, if you compare. First one is primary amine, and the third one is tertiary amine, which does not have any hydrogen present here. So in this case, we have hydrogen bonding possible, but that kind of hydrogen bonding is not possible here, because we don't have hydrogen present over here. So the boiling point of first is more than to that of third because of the greater extent of hydrogen bonding possible in the first molecule, right? So order of boiling point would be two maximum, then one, then three, in decreasing order, I have written it. Again, let me tell you, this is the decreasing order. The question is in increasing order of their boiling point. So answer will be three, then one, and then two, this is the answer for this question. Remember, if you write this, you won't get marks. Next, you tell me aniline, para-nitro-aniline, and methyl-aniline, order of basic strength. Again, suppose aniline is the first one, para-nitro is second, and methyl one is the third compound. One, two, three, increasing order of basic strength. Third one is maximum, third one is methyl-aniline. Why third one is maximum? Okay, the answer, let me write down the answer, methyl-aniline and nitro-aniline. So this one is minimum. So minimum is two, then we have one, and then we have three. This is the increasing order, right? So all of you have got this answer. Why third is maximum here? What is the logic? What is the reason? Because of what? Methyl is an electron-donating group because of plus i, because of plus i nature of this, okay? All these things you have to write down. Methyl group shows because of plus i nature, the electron density on nitrogen increases, and hence it has more density to donate electron. Electron-donating group CS3 fine, but you also mentioned plus i nature of CS3 group. Okay, answer is just fine. Next question you see, the other five marks question. Tell me, interhalogens are more reactive than pure halogen. What is the answer? Why interhalogens are more reactive than pure halogen? Yes, correct. Yes, reason is the logic you are giving is right, okay? Because the bond forms between the interhalogen compound. How this interhalogen compound forms? You see all these things you have to write down. Interhalogen compounds forms, interhalogen compounds forms by the reaction between the halogen atom. Forms by the reaction between two halogen atom, two halogen atom, okay? Example also you can write down, IF7, right? BRF5, all these are interhalogen compounds, there are many other compounds you can write down, okay? So these are the interhalogen compounds formed by the reaction between two halogen atom, and here the bond strength of two halogen atom is weaker than the bond strength of molecule of same halogen atom, right? Like we have supposed I bond strength and IF bond strength if you are comparing. Since this bond strength is weak, it is like comparatively more reactive than the pure halogens. That's the reason you have written which is correct. But just a little bit, you write down about interhalogen compounds, write down few examples, and then you support your reasoning that whatever you are giving, okay? I hope you all know this. Okay, next question. N2 is less reactive at room temperature. What is the reason? Yes, right, Povic. N2 having triple bond, nitrogen combines with triple bond with the other nitrogen atom to form N2 molecule, which has very high bond dissociation energy. And hence it is not reactive at room temperature. Yes, that is correct. Yeah. Reducing character increases from NH3 to BIS3. Third one, tell me the reason. As we go down the group bond dissociation enthalpy between central and atom and hydrogen decreases. Okay, so bond strength decreases. Means stability decreases, right? Correct. So basically what we can say, if the stability is more, it's reducing nature is less, right? And stability of hydrides like NH3, PH3, ASH3, SVH3, and BIS3. Stability of all these hydrides decreases as we go down the group, right? Stability decreases as size increases, bond length increases, that is the reason you are saying, right? So stability decreases, and hence stability decreases. So reducing nature also, reducing it increases in this case, right? So about NH3, what we can say, it is the most stable molecule hydrides and have least reducing nature. For BH3, the fact will be reversed. Least stable and have maximum reducing behavior here. Right, that's the reason for reducing characteristics of NS3 to BIS3. Okay, can you draw the structure of H4P2O7? I know, I cannot see, but just tell me the, you know, how this structure would be pyrophosphoric acid, H4P2O7. What is the basicity of this acid? Basicity is four. How many P double bond O group, how many P double bond O bonds are present in this molecule? How many P double bond O bond two? Okay. How many POH bonds are present? Four. And the two phosphorus atom are linked with one oxygen atom, right? So the structure of H4P2O7 is this P double bond O, oxygen P double bond O, and there are two OH group present at each phosphorus atom, right? So basicity is four correct, four OH are there. So basicity is four. Structure is this pyrophosphoric acid, okay? Here's go through all the structure. They have only one compound, but there are more phosphoric acid molecular there. In case of boron also, there's this kind of compound forms that also you go through, okay? They can ask you anything, any possible structure. Okay. XEF4, can you tell me the hybridization first of XEF4, hybridization geometry, is it SVP? Is it SVP? How many bond panel known there are present? SVP3D2, what is the geometry and shape? Geometry, what is the geometry? Square planar is what? Geometry, okay, square octahedral or square bipyramidal is the geometry and square planar is the shape, correct? You see, like we have this method to find out the hybridization of this XEF4. So we'll find out the steric number first. And don't do like this in the example because these are not the appropriate method, okay? This is just a trick that we use, okay? You can directly write down, you can do this in rough work and then you can directly write down the hybridization of XEF4 is this, geometry is this, shape is this and then you can draw the structure, okay? So the number of valence electron for this will be what? Eight for xenon, seven into four for fluorine, all these divided by eight. So it is 36 by eight. So we have four bond pair and two lone pair, okay? So steric number will be six, hybridization will be SV3D2, geometry will be the shape is square planar. Square planar. Now, here you see if you draw the geometry of this, we have xenon present at the center of this square and four fluorine atom will be at the corner of the square, right, one lone pair here and other one is here in the bottom, right? In this, one of these bond is going into the plane three-dimensional structure if you try to draw and this one is coming out of the plane, the dark line, okay? Similarly, we have here also into the plane, this is the structure of this. The two lone pair will be at this position so that the distance is maximum and the repulsion will be less. What is this angle, can you tell me? What is this angle? And what is this angle? This bond I'm talking about, this bond. What is this angle? Yes, it will be 90 degrees because this bond is perpendicular to this plane, square plane that you have. So all these bond angles are 90 degrees, okay? This kind of question you must take care of. If you have this information of bond angle, you must mention that the bond angle should be 90 degrees. Okay, that also you keep in mind, okay? Next question you see this one, which poisonous gas is evolved when white phosphorus is heated with concentrated NaOH solution? Write the chemical equation involves, write the chemical equation in this. Yes, PS3 is correct. Chemical equation is what, what is the other molecule we'll get? Correct. PS3 plus NaS2 PO2, okay? This kind of reaction we write down the balance reaction. Right, phosphine is the poisonous gas we have here. Next one, which noble gas has the lowest boiling point? Boiling point order. Next one, why helium? Tell me what is the reason why helium has the, yes. So as we go down the group, as we go down the group, the intermolecular forces increases and hence their boiling point also increases. Or you can simply keep in mind as the mass increases, as the size increases, boiling point also increases. And that is why helium has the least boiling point. The reason what you'll write, you can write as we go down the group, the intermolecular forces increases, right? Now they have very weak intermolecular force, that is London dispersion force, right? But whatever force is there, it is present in all the molecules. And as we go down the group, this forces increases, right? And that is why the boiling point also increases. Helium has the least boiling point among the noble gas. Next one you tell me, fluorine is a stronger oxidizing agent with chlorine, why? What is the answer? Okay, fluorine has high ionization enthalpy. Okay, you see there are two things here in this question. Like you are saying fluorine has high ionization enthalpy. Remember, whenever you are saying ionization enthalpy, right? It means you're talking about the state which is in gaseous state. So if the atoms is in gaseous form, then we can explain with the basis of ionization enthalpy, which is correct, like you are saying, Sanjana. Okay, so this is one thing. Another thing is what in case of solution, what happens, this tendency depends on three factors actually. That is electron gain enthalpy, right? Hydration enthalpy, hydration enthalpy, right? And the last one we have here, electron gain enthalpy hydration, hydration why we are saying because we are considering water as a solvent. Otherwise we can write down solvation energy. Generally we write hydration only, right? Hydration energy and then one last thing we have that is bond dissociation energy. All these three factors comes into the picture when we have to compare the, you know, property in solution in case of gaseous state with ionization energy, we can say, right? Now what happens, chlorine has more negative value of electron enthalpy, right? That we already know. Chlorine has the highest electron gain enthalpy, more negative value of electron gain enthalpy. So in case of chlorine, the electron gain enthalpy is more than that of chlorine, more negative value, right? Negative value we are considering for both, more negative value of electron gain enthalpy for chlorine. This is one thing which goes in favor of chlorine, but second thing what happens, hydration energy, in case of chlorine it is more because the size of chlorine is lesser than to that of chlorine, right? Now when you talk about the bond dissociation energy, that is also less in case of chlorine, right? So overall with these three factors what we can say, overall the energy change favors the acceptance of electron in case of chlorine, right? And that is why it is a stronger oxidizing energy. Okay, so these three factors comes into the picture in solution, right? And when you compare all these three factors separately with of chlorine and, you know, chlorine, only the electron gain enthalpy of chlorine, it is more negative to that of chlorine, but in case of hydration energy and bond dissociation energy, chlorine dominates over chlorine and hence the acceptance of electron in case of chlorine is easier and hence it is a stronger oxidizing agent, right? So these two things you can write down easily. In case of gaseous state, we'll come and see this factor in case of solution. These three terms you have to mention otherwise you won't get marks, okay? Electron gain enthalpy, hydration energy and bond dissociation energy. Understood? With the help of reduction potential also you can explain. Next question you tell me, what happens when S3PO3 is heated? What happens when S3PO3 is heated? It is a disproportionation reaction. Yeah, product is fine, product is right, Sanjana, but it is a disproportionation reaction. You see the reaction here when this S3PO3 heated, so it converts into, sorry, it is three. It converts into S3PO4 and phosphine, PS3. So here you see the oxidation state of phosphorous. Here it is what? Minus six plus three, so this is plus three. This is minus three and this one is plus five. You see plus three to plus five, it is oxidation and from here to here it is reduction. This is oxidation and this is reduction. So this is the disproportionation reaction prior to we get S3PO4 and PS3, correct? What is this product? The last one, PBS plus O2. What is the product we get here? It's O3, not O2. Done, let's sulphide and oxygen. We'll get PBSO4 plus O2. Must write the balanced reaction. Okay, so four plus two, it is six. So we'll write down two SO3 here. Two oxygen and then this. Okay, whenever you write down the reaction, it must be balanced reaction always in board exams. Next question you see question number 24. Yes, this will be numerical. The question, what is the rate of reaction? Two factors, okay, it's theory. Yeah, this one is numerical. Okay, one numerical, one theory, correct. So this question, question number 24. Factors that affect the rate of reaction? Yes, temperature, temperature and concentration, temperature, concentration, catalyst, concentration of reactant, yes. See, Purvik, Sanjana, you have to, when you say concentration, concentration of what? That's what you have to write down. Temperature is fine and concentration of reactant. You have to mention that. So temperature is one of the factor, then we have concentration of reactant and then we have catalyst also, right? These are the factors which affects the rate of the reaction. How do you define the rate of a reaction? How do you define the rate of a reaction? Tell me the definition. How do you define the rate of a reaction? Change concentration of reactant by changing in, by changing time, changing concentration of reactant in unit time. Yes, right, you can say that and then you can write down one example and then you can write down the expression of rate with respect to that reaction, right? Does it depend upon the isometric coefficient? Does it depend upon the isometric coefficient? Okay. Tell me B, what is the answer? Okay, this is saying no, Pulvic is saying yes, Cypher and I am saying yes. Okay, tell me one thing. Suppose if I write down this reaction, I'll just give you one example. N2 plus 3H2 gives 2NH3. What is the rate expression of this would be? The rate of this reaction can be right minus of D concentration of N2 by dt. That is equals to minus 1 by 3 of D concentration of H2 by dt, which is equals to plus 1 by 2 D concentration of N is 3 by dt. This is the expression. What can you say? Rate of the reaction and suppose if I write down one more reaction here, which is this one, this is the first one, second reaction if I write, half of N2 plus 3 by 2H2 gives N is 3. So rate of this reaction will be what? Minus 2 N2 by dt. Next it would be minus 2 by 3 D concentration of H2 by dt and next it will be plus DNH3 by dt. So rate of the reaction is this. With respect to N2, with respect to H2, with respect to NH3. This is again with respect to N2, with respect to H2, with respect to NH3. So if you compare these two expression here, you see the ratio of these two are same. So whatever reaction you'll write, whether you'll write this reaction or this reaction, for a given concentration of reactant, the rate of the reaction you will get the same. So second one you are getting what? Tell me one answer. The rate constant on the first order reaction increases from this to this, what the temperature changes from this to this, calculate the energy of activation. Okay, Ea you have to find out. Is it, do we have calculation in this? The correct answer is around 535 kilojoule. 535 point something, it is around. And I think there is only calculation in this. If any one of you are getting 535, 535 point something kilojoule, good. So you see the rate constant of first order reaction increases from this to this. So I'll just write down K1 is the rate constant at temperature T1, given value is 410 to the power minus 2. Like I said in the first class, you have to write down the data first, all the data given in the question, write down the top, data given. K1 is equals to this and temperature at T1. So what is T1? It's 27 degree Celsius. So 27 degree Celsius goes to what? 273 plus 27 is equals to 300 Kelvin. This is one set of data. Now when this K2 increases to twice of this value, age in 10 to the power minus 2, and the temperature is increases by 10 more. So it is 310 Kelvin, okay? So EA we have to find out. So we have this expression of log K2 by K1 is equals to minus EA by 2.303R into 1 by T2 minus 1 by T1. Is it correct? This expression is correct? Yes, it should be 8.314, because you see this energy you have to find out. Yeah, it's correct. I have taken the minus sign out, psi prime not. That's why I have written 1 by T2 minus 1 by T1. Okay, so same thing. Okay, so all these value you have to substitute here. See, whenever you write down log here, here we must have 2.303, right? You can take this R in calorie also, you'll get the same value, unit will be changed. And whatever EA value you'll get here, write down the unit of that also, important, okay? Don't miss the unit, okay? And what will be the unit? What is the unit of EA, whatever value you get? Tell me the unit of EA, R value you take here, 8.314. That is joule per mole, Kelvin. What is the unit of EA you are getting? Only joule or joule per mole? See, it is easy. Like don't make mistake like in this kind of thing. As you see, just try to understand one thing. Do we have any unit of this term? It is unit less. The unit of this value is what? This part, the unit is one by K, Kelvin. EA you have to find out, R will go this side. So we'll have unit of R here, which is nothing but joule per mole, Kelvin. Is equals to unit of into one by Kelvin, correct? This and this will get canceled. So unit of EA will be what? Joule per mole, okay? So don't miss this, you will lose one mark if you make any mistake in your unit. Don't do all these mistakes, okay? Now you substitute all this value here. T1, T2, R, K1, K2, you'll get EA, okay? The value of EA you will get is 535 point something. Point something you'll get approximately. That will be in kilo joule per mole. Understood? Next question, for a reaction A plus B gives P, the rate is given by, the rate is given by this. How the rate of reaction affected if the concentration of B is doubled? Tell me, it is a reaction affected if the concentration of B is doubled? This is four times, see I'm correct. Two B's squared, four times of the initial one. Second one, what is the answer for the second one? Why it is first order? Zero reaction, what order, Sanjana? No, we should see what happens. When A is present in large X's, right? Then what happens, actually one of the reactant is present in X's of the amount, right? Then any change, a small change into that react, that particular reactant will not affect the overall order of the reaction. Since it is present in large amount, okay? It is already X's present over in X's amount, right? So any change in that particular reactant would not affect the overall order of the reaction, right? However, A is present in the reaction rate expression, but since A is in the large amount, then any change in A would not affect the order of the reaction. And then in this case, the rate only depends on the concentration of B, which is raised to the power two, so the order of the reaction will become here two, this kind of reaction, we call it a pseudo second order reaction. Why pseudo? Because when you look at the rate expression, it should be two plus one three. But since A is present in X's, and it is there in the rate expression, then also the rate is independent of the concentration of A. So this kind of reaction, we call it a pseudo reaction, okay? So rate depends only on B, which is raised to the power two in the rate expression. So we call it a pseudo second order reaction. See, you must not confuse that you were thinking maybe that since this A is in maximum X's, so you just negate this B. That is what you were thinking, right? A is in X's, so just remove that or negate this B is squared on here. That is what you have thought. See, you cannot do that because both are present in, it's not plus or minus sign here. Both are the, these are the product of A and B squared. So you cannot neglect the one term with respect to other, if they are in product or divisor of this thing. If plus and minus is there, then we can do this, okay? So in this expression, you must take care of this thing. If one of the reactant present in X's amount, then any change in that particular reactant will not affect the overall rate of the reaction. And we say the rate of the reaction is independent of that particular reactant. This kind of reaction, we call it a pseudo reaction, pseudo second order reaction in this case. Understood? Okay, now the next one, what is the overall order of a reaction if A is present in, oh, we have done this already, right? This one, B, the first order reactions solve this question numerically. How did you do 46.2, 154, tell me how, what did you do? It's too much, 46.05, okay, let them do. Okay, 46.2 is the right answer. You see, there are two things here you can do. See, first of all, suppose the reaction, like how we solve these kind of questions in the J main test, or when we have some competitive exam if you are writing down. So what we say, here is the reactant, right? And this is the 100, we are assuming, total 100 we have here, the concentration of the reactant, and this is the product. This is the half lifetime, which is T half. So at this time when the reaction proceeds out of 100, what is left here? 50. And what we say 50% of the reaction has been completed. Correct? So this time is given in the question, which is 23.1, right? Now the question is, you have to find out the time when 70% of the reaction completes. So to complete the 70% reaction, we need suppose this much time more, right? This T suppose 75, I'll write down. So here it is T half, and from here to here, 70% of the reaction completes. So what is left here? 25. And what is react from here to here? It is also 25. So when you see from this point to this point, from here to here, if you have to go, it also takes the another T half of the same reaction, right? So from 1 to 50, 100 to 50, the time required is T half, right? And then 52, again 25, so that from here, it becomes 50 plus 25, 75. From 50 to 25, another half time required. So this time is also what? T half. So from this point to this point, the time required to complete this reaction is what? Twice of the T half, which is nothing but 23.1 into two, which is 46.2. Yes, did you remember this? We have done this kind of question in the class to solve the questions of J main style. Did you remember this? Now the point is, if you solve like this, you won't get anything in, the marks you will not get in board exam, okay? But the point is, if we have this kind of understanding, you will know the answer of the question, which is 46.2. Now you do all the calculation and in the last, when you get the expression, you solve those expression and finally the answer, you should write this only. You cannot write down any other thing because you know the answer should be the twice of the T half. So how do we do this? The method, like the method that we have to use to solve this question in board exam is this. First order reaction, T half is given. So we know the relation of K and T half, which is K equals to 0.693 divided by T half. So you put this value of T half, you'll get the value of K and we know the relation of first order log of K2 by K1 is equals to use the expression for first order and we'll take this value, we'll put over there and we'll get the answer, okay? So the relation of first order is what? We know this K is equals to, for first order reaction, we can write K is equals to 2.303 divided by T is equal into log of initial concentration C naught by C. This is what the expression we have to use, okay? C naught we have assumed, initially it is 100. C is what? The concentration which is left. So 75% completion is there. So what is the concentration left? 25, you substitute here. T you have to find out. K you already calculated from this expression, you've substituted and you'll get the answer, the time, finally you'll write 46.2, whatever the unit we have minutes, so we'll write down minutes. Did you understand this? Did you understand this? These kind of questions are very common for board exam. From one set of given data, you find out one term and that we use in the another term, another expression, you'll get the unknown, okay? So these are the three, five marks question we have which are not that tough, okay? You have done all the questions one or two, if you know, ignore all those questions you have solved. Now you see question number 22. 23rd is an abstract one. We'll see question number 22. C's option decreases with increase in temperature. What is the reason? Yes, it is an exothermic reaction. Then what happens if it is exothermic then? Yes, and we know that in exothermic reaction, as temperature increases, according to Lee-Tatelier's principle, this term you have to mention, always I'm telling you this, okay? We know that C's option is an exothermic process. And in exothermic process, as temperature increases, according to Lee-Tatelier's principle, backward reaction is favorable, right? Backward reaction is favored in that case. As temperature increases, and that's why C's option decreases with increase in temperature, okay? So the key words in this question is what? Lee-Tatelier's principle, exothermic process, and all this, right? And what happens when in exothermic process our temperature increases? This is just one line you have to mention over there. Okay, that is what my concern is. I know you know the answer, okay? But how do you write down in the answer sheet that is important? The physics option is an exothermic process, and we know according to Lee-Tatelier's principle, as temperature increases in exothermic process, the backward reaction is favored, and hence, physics option decreases with rising temperature. Next one, addition of alum purifies the water. Yeah, correct. See, one more term I just wanted to see here, that is, you know, alum behaves as a flocculant, okay? That term you should write down, alum behaves as a flocculant. And what is the, you know, property of flocculant? Flocculant are those substance which combines with the various different impurities present in water, right? And forms heavier particles, heavier molecules, right? All those impurities combines with alum, forms heavier particle, which can easily be settled due to the gravity, gravitational force, right? So when you settle down, you can easily drain off water from the top. That's how it works. That's how the water get purifies, okay? So alum behaves as a flocculant that you must write, and then the property of flocculant is what? It combines with the various impurities, forms a heavier particle, which settles down due to gravity, and we can drain off the water from the top. That's how water gets purifies. Brownian movement provides stability to the colloidal solution, how? That's right, Brownian movement is nothing but the zigzag movement, okay? Which does not allow the particles to settle down and it stabilizes the colloidal solution, right? Time at 21 with the C now. Solve this question, calculate the EMF just a second. Okay, what is the answer? 0.11 board, I think it's correct. Is the calculator EMF of the following cell at 25 degree cells here? Concentration is given, the nurse equation is what? E cell is equals to E naught cell minus 0.0591 divided by N. Log of concentration of add-on product divided by reactant in the equation of electrochemical cell that we can write down here easily, okay? So in this question you see SN plus two, which will get oxidized here, right? You must remember that we have discussed, we write anode always on the left-hand side and anode oxidation takes place. So this is what? Whatever the left-hand side we have here on the salt bridge, that is nothing but anode. And that anode we know oxidation takes place, right-hand side will have cathode. So this hydrogen electrode behaves as a cathode here, right? So this will get oxidized. So in the product side we have SN two plus and reactant side will have H plus, okay? So if I write down the concentration of product, that will be 0.001 and that of reactant 0.01, okay? Cathode is what? Cathode is nothing but hydrogen. Since we know right-hand side we have cathodes, hydrogen, E naught cell is what? The formula of E naught cell, E naught of cathode minus E naught of anode. All these things we have to write down in your answer sheet. Cathode is zero and anode is minus 0.14, which is nothing but 0.14 volt is the E naught cell. Now we can find out the EMF of the cell. E cell is equals to 0.14 minus 0.0591 divided by N value here it is two because we have two plus here. N value is two log of concentration of product, which is SN plus two, which is 0.001 divided by concentration of reactant, which is 0.01. But one thing you must take care of here, here hydrogen has only one positive charge and it is plus two. So when you write down the half cell reaction, which is nothing but SN gives SN plus two plus two electron. And here we have H plus plus one electron gives hydrogen H two. So we'll have to multiply here with two, right? So we have two H plus on the left-hand side. So this two will go on the power here, like this two. Don't miss this, okay? Usually the students makes mistake over here. So don't do this. The charge must be balanced, okay? Plus two N two so that this gets cancelled. The best way is to write down the cell reaction. Half reaction first, you add these two, you'll get cell reaction. And then you will have the value of N, substitute this and you will get the answer. The answer will be 0.1105 volt for this question. Next question you see. All these are three marks question. Give reasons. Ortholytrophinol is more acidic than orthomethoxyphenol. Done? What is the answer? Nitro group shows minus I. Does it show minus M also? Minus M or minus R? What happened? Nitro group shows minus I effect. What about minus M or minus R effect? Does it show? See, nitrophenol that you have, nitro group has both characteristics actually. Because of the difference in, you see here, why I effect comes into the picture? Because the difference in electronegativity. Okay, so if you have benzene ring OH and ortho-nitrophenol, and double bond O and O, this is ortho-nitrophenol. So you see here, this carbon is sp2 hybridized, right? And we have electronegativity difference between this carbon and nitrogen. So because of this electronegativity difference, it can show minus I, right? That's fine. But because of the resonance also, you see pi sigma pi resonance with this ring, it can show minus R also, or we can also say minus M, both possible, right? So if you have any group attached with this ring, which shows minus I, minus R both, except halogen, keep this in mind, except halogen, always minus R or minus M effect dominates the I effect, minus I effect, right? So what happens because of this minus R effect, though this molecule shows both effect, inductive and resonance, minus I and minus R, but it's minus R is dominating here. Now, because of this minus R, what happens? It withdraws electron from this ring, right? And what happens? This bond lone pair or the electron density on oxygen decreases and then it, which can lose this H plus easily, which is nothing but the polarity of OH bond decreases and removal of H plus is easy over there, right? So that's why this is acidic. And the other compound is what? Orthomethoxy phenol. Orthomethoxy phenol is nothing but it shows plus R effect. Reason is again same. You see OCS3, Orthomethoxy phenol. This shows plus R effect because the lone pair present here, right? And because this plus R effect, the electron density on this oxygen increases a bit and hence the removal of H plus is difficult. However, this group also shows minus I and plus R, both or plus M. But again, like I said, when you have these two present here except halogen, the resonance effect always dominates the inductive effect. So the property of this will explain according to plus R, for this will explain according to minus R. Minus R group increases the acidity plus R decreases and hence the ortho-nitro phenol is more acidic than methoxy phenol. Alkyl halide having a smaller. Which question you are asking? Which one? Sight remnant, which question? Oh, the third one, third one. Okay, yeah, I'm coming to that, wait, I'm coming. See, why this butane one all has a higher boiling point than diethyl ether because of hydrogen bonding. Okay, that's what I've already discussed because of hydrogen bonding, okay? Now, the third one we'll discuss, CH3, this molecule on reaction with HI gives this and CS3 is the main product and not this one. So we have to give the reason, okay? You see this one, here what happens? The molecule is this, CH3 whole price C, OCH3, with this HI, okay? So what happens, this HI will give H plus and the lone pair on this oxygen will attack onto this H plus. So it takes this H plus from HI and it forms this one. CS3 whole price, C, O, H, CH3 and positive charge on this oxygen. Now, the reaction of I minus the carbocation, we'll get carbocation here, right? So what happens here, you see? Since this is tertiary, okay, this bond will not be there. This is, see what happens? How this positive charge will get stabilized here? So there are two possibilities. What are they, you see? Either this bond will break and this bond pair will come over here. This is one possibilities. CS3, O, H forms and we'll get a carbocation here. So in this case, the possibility is this, we'll get a tertiary carbocation, positive charge on it, and we'll get methanol, CS3, O, H. Or the another possibility is what? This bond pair comes over here and we'll get methyl carbocation, right? So which carbocation is more dominating, this one or methyl? Obviously, this tertiary carbocation is more dominating. So this reaction takes place, which takes this I minus and forms this product. So the point is this reaction proceeds with SN1 mechanism. Okay, this you have to mention, SN1 mechanism. And we know SN1 mechanism like proceeds with the formation of carbocation as an intermediate, right? So we'll try to form more stable carbocation, which is nothing but tertiary, not this one primary. That's why CS3, O, H will get and we'll get tertiary iodide here. Clear? So that's the reason. Now you see next question that will see question number. Do this, how do you convert the fallout with prop one in to fluoropropane? Yes, we'll have the peroxide effect here, peroxide effect, Swartz method. What we use in Swartz method? What reagent we use? In Swartz reaction we use AGF, correct? So this F will replace bromine from the compound, forms AGBR and fluorine will attach in the compound there. So we'll get one fluoropropane, right? First one is done. Second one you tell me. Cloropangene to two-clorotoluene. How halogenation? It's two-clorotoluene, you see. We can use freddy-craft, freddy-craft reaction? Yes, correct. So that's the second reaction, freddy-craft reaction. See, these are very common reaction, okay? Yeah, right, freddy-craft localization. That's how we get two-clorotoluene. How do we get this propane nitrile from ethanol? Yes, first SOCl2, then we'll get chloride and then we can use NaCl, KCl, okay? It forms NaCl or KCl and then Cn attached over there. We'll get propane nitrile, right? Correct. Okay, next question, write the main product when n-butyl fluoride is treated with alkali-KOH. Yes, it forms an alkene, but one in is the answer. Correct. Alkali-KOH gives you alkene. Second question, in place of chlorine, we'll get OH over there, right? It's a direct reaction. We'll get 246 trinitrophenol, yeah, correct. 246 trinitrophenol. When you do the hydrolysis at Cn forms and OH attached at the place of chlorine, chlorobenzene converts into phenol, 246 trinitrophenol, yeah, so that's the answer. Third one, methyl chloride is treated with AgCn. What is the product we get? Yeah, we'll get methyl isocyanide in this, CH3NC, correct. Okay, these are the few basic questions that we get that they ask in product thing. What is the major product we get? Next question you see, solve this one. Write down the data first. You're getting BCC, element with density, this forms a cubic unit cell with AgCn, this. What is the nature of the cubic unit cell if the atomic mass of the element is 81 gram per mole? Just you have to find out Z value, simple cubic. What is the Z value you are getting? Purvic, viscist, question of the side frame, not. What is the Z value? Okay, side frame, not, yeah. If it is two, answer will be BCC. Yeah, if it is two, so answer will be BCC. So just you need to write down the formula of density and then everything is given. Substitute all the values, find out Z, okay? In this type of question, you must take care of units, okay, meter, centimeter, angstrom, whatever it is given. According to that, you just do some changes, whatever is required, okay? But unit you must take care of in this question, density-related question of solid state, okay? You must take care of that, yeah, correct. So if it is two, so we know in BCC, Z value is two, hence the nature of this cubic unit cell is what? It is body-centered cubic, okay? I'm not solving this. Next question, solve this one. 20, I don't think it's correct, check your answer, calculation. 59.6, I think it's correct. Check your calculation, guys, you see, I'm just observing this. You're making some mistake in all these kinds of questions and that is calculation mistake only, okay? So do your calculation properly. If you've taken I value here, what is the I value you have taken? Yeah, it's three, correct. You have taken I value, right? You have taken I value. See, for CACL2, I value is three, it's correct. That's not an issue. But you see the point here, it is what? It is given in question that CACL2, assuming CACL2 is completely dissociated. When complete dissociation is there, then I value is what? I value should be one. However, it is an electrolyte. But since they have told already that you have to assume this, CACL2 completely dissociated. So it should be one, yes? In general, for CACL2, I value is three, but here it is not the case, got it? That's why I always say, the smallest one information, you have to take care of this thing, read the question properly, then you start solving this. Now you use the expression of this one. Delta EF is equals to I into KF into M morality. All these values are given, Delta EF is given, I values one, KF is given, morality expression, you write down, find out number of moles and then you'll get mass also, right? So the answer for this question will be 59.6, something you'll get here. Understood? So read the question carefully. Don't assume. Can you move on? Do this one. Starch is a polysaccharide. What about maltose? Starch is correct. What about maltose? Is it mono or disaccharides? Maltose? Yeah, maltose is disaccharides. And fructose and glucose are, fructose and glucose are monoseccharides. So starch is poly, maltose is dye, fructose, glucose are monosecarides, correct? Second one, what is the difference between native protein and denatured protein? See, native protein has a definite structure, right? And it performs its biological activity, native protein. It has a definite structure and it performs the biological activity. The structure is obviously three dimensional structure, right? It has a particular shape also. Native protein has a particular shape, three dimensional structure and it performs its biological activity and responsible for its biological activity. But denatured protein, it is actually, it does not have any particular shape and it also does not perform the required biological activity. I'll just write down here. Denatured protein, two, three points. It has unfolded three dimensional structure of, structure of native protein and it does not perform. Second point, does not perform its biological activity. Third point, does not have a particular shape, particular shape, okay? But for native protein, it does have a particular shape, perform its biological activity and it has a three dimensional structure, which is responsible for its biological activity, okay? So these are the difference between this native protein, denatured protein. What is this? Write the name of the vitamin responsible for the coagulation of blood. What is that vitamin? It's very, yeah, correct. Of the vitamin responsible for coagulation of blood. What is that vitamin? Yes, it's vitamin K. Next question you see. Products of the following reaction. Yeah, it's formed cyanohydrin, OH and CN group will attach at the same carbon. This is actually the nucleophilic addition of CN minus. H3C, C double bond O, CH3. It takes H plus from HCN and the lone pair here. It takes this H plus and it forms H3C, COH, CH3 with positive charge on this carbon. Now from this, CN minus as a nucleophile attacks onto this carbon atom and we get CH3, CCN, OH, CH3. It is a cyanohydrin. Nucleophilic addition of CN minus. Now in this alkyl benzene we have in the second one. What is the product we get here? First of all, it forms with KMNO4 and COH, it forms salt of potassium, COO minus K plus with this KMNO4, COH, which on reaction with H plus, it converts or gives you benzoic acid, COH. Third reaction, CS3, COOH, carboxylic acid on reaction with NH3, heat. H2O goes out and we get amide. CS3C double bond O and H2O. We'll get amide and H2O forms. These are the three reactions we have. Next question you see. Geometrical isomers of complex this. It won't form CH2, COOH, it's a direct reaction Cypremnath, you can check the mechanism also. Alkyl benzene, alkyl benzene means what? When you have alkyl group attached with benzene, alkyl benzene whenever reacts with KMNO4 and COH, it gives you the salt of potassium over there. So the alkyl group converts into the potassium salt. Benzene group will be as it is and it converts into the potassium salt and we'll get some other product also. One carbon will go out. So that's the direct reaction. You can check the mechanism. The mechanism we are not discussing now. We will not get CS2, COOH. One carbon will go out as some other product. CO2 and something will get out also. Okay guys, we'll see one mistake we have made. You see this question? That solution question, CACL2. I think this was the question. Yeah. You see, actually I have made a mistake over here. They have given CACL2. When this will dissociate completely, actually when it dissociates, it forms CA2 plus plus 2Cl minus. So it's alpha value is one actually. That's what I made a mistake. So this complete dissociation is there. So degree of dissociation is one, right? So the number of ions after dissociation is three. So I value will be three. It's not one. So we've mixed this alpha and I actually. I mixed this alpha and I, my mistake, right? So you make this correction. Your answer will not be this. This answer you will get when you get I is equals to one, which is not true in this case. Did you understand this? Sai Premnath and Sanjana. Did you understand this thing? Okay. So when it completely dissociate, alpha is one, not I. Correct. So you must have got confused here only. That's what I also did. Right? So I value will be three, not one. Just now, Vaishnavi and Purvik told me this, right? So that's the thing. Alpha is one, I is three, correct? Okay. Now we'll come back to the same question. Okay. So the geometrical isomers of this will be cis and trans. Can you draw the structure? Yes, yes, yes. So the answer will also be changed. Okay, it's not, it will not be 59.6. So do you do that calculation? The first one will have cis and trans isomers, correct? And the structure will be what? We have cobalt attached with this and this. Another one will be, this is a bidented ligand. So the structure we can draw like this. And then one chlorine will be here. Another chlorine will be here. Yeah, if you've drawn it, fine. One structure will be this, another one will be, both chlorine will be here and one en group will be here like this. So that's the two cis and trans isomers possible. Okay, and the next one, on the basis of the crystal field theory, write the electronic configuration of D4. What will be the electronic configuration when delta O is greater than P? What is the meaning of this, you tell me first? Delta O is greater than P. What is the meaning of this? P is what? It is the average pairing energy, right? And when the crystal field is splitting energies more than to that of this, then we have T2G orbital, lower energy, EG orbital we have here. So when in this case, the pairing takes place against the Hunt's rule, right? So the structure will be this, one, two, three and four. Fine, is it correct? So in case of this delta O greater than P, low spin complex forms, low spin complex forms, and that is only possible when inertial orbital takes part into hybridization, correct? That's what the thing is. NICL4, we have done this in the class very times. Okay, and we have seen the question, this question has been asked in 2017 also. Okay, you see this one, how would you have gotten the following? The chemistry of actinides is more complicated as compared to lanthanides. Okay, you see here lanthanides, actinoids. See what happens in lanthanides actually, that the elements of lanthanides actually exist in mainly three types of oxidation state, which is plus two, plus three and plus four. This one is the most common, plus three. Most of the elements will be in plus three oxidation state in their compounds, okay? And why this, because the energy difference between among this 4F, 5D and 6S orbital is quite large, is quite large. That's why only three, mainly three oxidation state possible. But in case of actinides, what happens, their energy difference between 5F, 6D, 7S orbital is not that large. Okay, it's very less in comparison. The energy difference is very less. And that is why they may exist in plus three, plus four, plus five, plus six oxidation state. Okay, and if you talk about this one, neptunium, this particular elements exist in plus three, plus four, plus five, plus six and plus seven also. Not this plus six is not possible, plus seven. This kind of different oxidation state is possible. That's why it is very complex. When large number of oxidation state is there, energy difference is very less. They can form very complex compounds. So that's the reason we have here. Why transition element form complex compound? Availability or de-orbital? And one more thing we can see that a small size of metal ion and high charge density, it's suitable for the forming of bond between the metal and the ligand. Yeah, correct. What is the answer for the third one? Complete the following equation. Can you balance this? I'll give you this hint that it is MnO4 minus is nothing but KmO4 and in acidic medium. This MnO4 minus, it converts into what? Mn plus two in acidic medium. So one product here will be Mn plus two, right? SO3 will convert into SO4 two minus and then you can balance this reaction according to the Redox reaction method that you have. So Mn two plus SO4 two minus. Okay, and now you see the two reaction here is this MnO4 minus is converting into Mn plus two and SO3 two minus is converting into SO4 two minus. Okay, medium is acidic. How do we balance oxygen by addition of H2O? How many? Four, because we have four oxygen here. Four into two, eight hydrogen, so eight H plus will add, right? Sulphur is balanced already, four oxygen three, so we'll add H2O here, right? Now what is the total charge here? The total charge is, sorry, it is eight H plus. Total charge is seven positive, seven plus, plus seven charge we have total here. Total charge here it is what, plus two. Plus seven to plus two you have to go, so for that you have to add five electrons. Here it is minus two and here it is minus two, okay? Now one mistake I made here, two H2O to balance this hydrogen will add two H plus here also, correct? Now two negative charge here and here it is two positive two negative zero. Total charge is zero here. To make two negative we have to add two electron here. Now you see, I don't have a space here, when you add these two, so this equation I'll multiply by two, this equation I'll multiply by five so that the electrons, electrons get canceled. So when you multiply this equation by two, you will get what? On the left hand side you'll get two MnO4 minus plus 16 H plus, 16 H plus five into this plus five SO3 minus two, plus five H2O. And this gives the product side we have two Mn plus two plus eight H2O, plus five SO4 two minus plus 10 H plus. Now we'll write down the net equation simply. What is the net equation here? This 16 H plus minus 10 H plus, we are left with six H plus here, five H2O, eight H2O, we are left with three H2O here. You see, left hand side we have two MnO4 minus, five SO3 two minus six H plus, we have this only here, this gives two Mn plus two, five SO4 minus two, plus we'll get what? Three H2O, this is the reaction. Is it clear? It is just the balancing of redox reaction. Did you understand this? Okay, so can we wind up the class here? Okay, so all of you just, you know, take care of your calculation because I can see that in numerical questions, you know, but you're not getting the answer in single attempt. You make some mistake in the first attempt and then you correct it. Okay, so don't repeat that into the exam. Okay, so we'll wind up the class here only. We'll see you in the next class, right? Thank you all for joining.