 So, yeah, so we have discussed this question already that f is more, fluorine is more electronegative than O, then the highest magnees fluoride is MNF4, whereas the highest oxide is MN2O7. The reason is what? The less size, the small size of fluorine and then also oxygen has tendency to form multiple bonds, right, but the same tendency we don't have with fluorine, that is also one reason we have, okay, so that is what the logic behind this question, okay. Next question you see, question number 30 we have discussed, okay, this question you see, a solution of glucose in water is labelled as 10% by weight, what would be the molality of the solution? Solve this one, what is the answer, done, guys are you there, okay, I'll do this, I think this is the same question they have also asked once in the previous year question and we have solved this question already, do you see, first of all, a solution of glucose and molecular mass of glucose, okay, so solution suppose if we assume, it's a very basic question, 100 gram, okay, so in this what we have, solute and solvent, okay, do this then, solve by weighting them, is it 0.62, yeah I think it's close, what are we done, okay, if solution is 100 gram, so in this the glucose which is the solute we have here, that will be 10 gram and solvent whatever it is, is 90 gram, okay, so molality is equals to number of mols of solute which is 10 divided by 180, molecular mass of glucose, number of mols divided by mass of solvent, this is in gram into 1000, this is what we have done, right, so you will get 0.62 mol per kg, unit is always important, okay, unit you don't forget, okay, next question you see, all the you know, complete theory, I want some keywords into this, ideal solution what are the keywords are there, in short just you write it down, ideal solution that obeys Rawls law, right, so Rawls law again is a keyword here, one more line whatever you can write here, see actually the solution definition is what, if you have A types of molecule where we have A interaction and we have BB interaction, like suppose this question is for 5 marks, okay, 5 or 6 marks, each one of these will have 2 marks, like 1 to 2 marks here, 2 plus 2 plus 2, like this it will be, okay, so you have, if you write down only one probably you will not get the whole marks, right, so it's right that ideal solution are those solutions which obeys Rawls law, but what is Rawls law that also you have to mention over there, okay, and second thing is what, how a solution forms, we have one solute and one solvent, where we have suppose A is a solute, B is a solute, so here we have A interaction in pure phase, BB interaction in pure phase, but when you mix these two, we will have what, AB interaction, right, so Rawls law is what, the interaction here which is AA interaction and BB interaction and AB interaction, if all these 3 types of interaction are same in all the terms, right, in all the terms means what, the energy exchange is exactly equal, the volume change is exactly equal, right, the entropy change is exactly equal, when all these interaction are exactly same, AA, BB, AB, then the solution forms, we call it as ideal solution, right, that is how we define ideal solution in terms of the interaction thing, and AB interaction is more than A and BB interaction, then it is positive deviation, less than negative deviation, right, all these concepts we have discussed in the class, but like they ask ideal solution, fine they follow Rawls law, but try to write down one or two more line here, like this interaction thing you can include and you can build your answer, I hope you understand this, okay, your answer was correct by the way, what is aziotrop then, yes, aziotrop is a constant boiling mixture, okay, the mixture that boils at a constant temperature, okay, and what is the second part, can we separate the aziotropic mixture, can we separate the aziotropic mixture, by simple distillation we cannot separate aziotropic mixture, okay, any example of aziotropic mixture, any example, any example of aziotropic mixture, Benzene tolu, yes, the point is when you write down the definition of aziotrop, okay, you must write down one example, or whether they ask it, ask it or not, you must write down the examples, whenever any kind of definition they ask, you support your theory with definition, sorry with examples, okay, must write examples, don't forget that, okay, now the third one, what is osmotic pressure, right, osmotic pressure is what, it is the external pressure applied on the, on the what, more concentrated side, the two sides we have, one side we have low concentration, another side we have high concentration, so osmotic pressure is always applied on, external pressure applied on more concentration side, so that the flow of solvent molecule is restricted, okay, yes, the excess pressure on the side of solution that prevent the net flow of solvent molecules, okay, again in the exam when you write down the definition, okay, the exact definition you have to write, whether you have to include a semi-permeable membrane also, right, and then you can define this osmotic pressure, okay, I suggest you to go through all those, you know, things that we have discussed in the class, or even in NCRT, you have to write down the definitions same as is given in NCRT, the definition you have to write down from there, anyways, these are the three questions which is, which was based on definitions, okay, next one you see, what is Rawls law, the excess pressure on the side of solution that prevent the net flow, okay, the, which is the pressure that is actually motioned into the product of mole fraction and vapor pressure, what is the solution is equal to the product of its mole fraction and its vaporation, what is Henry law, what is Henry's law, how can you relate, how does Rawls law become a special case of Henry's law, okay, see, suppose we have a solution of two components A and B, okay, and here what happens, the partial pressure of A is directly proportional to its mole fraction, right, and what we can write, P A is equals to some constant K X A, now when X A becomes one, right, when X A becomes one, it means what, it means what, all A molecule is present in the solution, there's no vapor of it, right, so in this case, or what we can say, mole fraction of A is one, it means there is no B present over here, right, if we write this, X A is equals to one, it means there is no B present here, it means the pressure exerted by A is the pure, is the, you know, is the vapor pressure of A because vapor pressure is what, is the pressure exerted by pure A, or pure component, okay, there's no other component mixed into it, so when X A is equals to one, the partial pressure of A becomes the vapor pressure of A and then we can write K is equals to P naught A, right, if you substitute this here, the partial pressure of A, it becomes, it becomes P naught A X A, similarly for B we can write P B is equals to P naught B X B and the total pressure if you calculate P T is equals to the partial pressure of A plus the partial pressure of B, which is nothing but P T is equals to P naught A X A plus P naught B X B, this is what the Rawls law we have, but you see, till here if you see, it is nothing but the Henry's law for solution in liquid, right, this is Henry's law in liquid, but when we take a special case of volatile components, volatile components means what, one of the component will be in the vapor phase, it is not present in the liquid, right, so X A becomes one in that case, if B is volatile, X B will become one if A is volatile and with this, we are getting this which is nothing but the Rawls law, right, so from this we can say from Henry's law, we have defined Rawls law in case of volatile components, that's what the question is understood, this is what the question is, in case of volatile solvent will apply the condition and we get a Rawls law, so this is only true when the solvent is, when the solution containing volatile components, which is derived from the Henry's law, okay, anyways, what is the answer in the second question, one gram of a non-electrolyte solute dissolving this, tell me the answer, I think it's a formula best question, 256, what is the unit, yes, tell me, gram per mole, yeah, it's correct, all of you have got 256, okay, I won't solve this, I'll just write down the formula, the formula we use here is what, del tf is equals to kf i kf into m, you cannot, m is the modality, you can directly write here also, del tf is equals to kf into m, i value is what, is one, why, because the solution is non-electrolyte, this is mentioned here, okay, you substitute modality formula, you'll get the molar mass, just you'll solve that, okay, 256 gram per mole is the answer, all of you have got the right answer, next question you see, this one, which acid of each pair, Sean here, would you expect to be stronger, the acidity of these two, yeah, because chlorine has more strong minus IE effect, okay, correct, first one is right, this molecule is the stronger acid, what about these, second one, why acetic acid per week, per week, tell me, why acetic acid, what happened, see here, when you compare the acidity of phenol and acetic acid, okay, so the conjugate base of this acid will be phenoxide ion, O minus, and for this the conjugate base will be acetate ion, O minus, okay, now here what happens, this negative charge is equally, you know, actually, you know, this, we are actually comparing an acid and alcohol, acid and alcohol, and when you're comparing the acetic behavior of an carboxylic acid and alcohol, obviously the carboxylic acid is more acetic than the alcohol, okay, if you talk about the resonance effect here, right, then we can say that phenoxide, this phenol is more acetic than this, okay, but since it is an acid and this carbonyl group that we have, it has very strong minus IE nature, this carbonyl group has very strong minus IE nature, and this phenyl group also has minus IE nature, okay, so because of that, all these negative charge or conjugate bases getting stabilized, right, so here what happens, usually we say what carbonyl group has more minus electron with drawing nature than the phenol, than this phenyl group, hence this is more stable and hence the carboxylic acid, this acetic acid is more acetic, acetic than the phenol, this also you can explain like with this theory, you must feel that it is a bit confusing, right, but it is also since we cannot compare directly the alcohol and acid, okay, so what we do, we can check the K value or PK value of acetic acid and phenol also, you see I'll just write it down here for acetic acid, for acetic acid the PK value is 4.76, PK I am talking about, for phenol, acetic acid or the phenol its value is 10.02, PK value, right, so when PK value of acetic acid is more, is less, so its K value will be more, right, so we can write here for acetic acid has more K value than phenol and hence the molecule which has more K value is more acetic, is more acetic and hence the acetic acid is more acetic than phenol, so this we can define easily with the help of PK value, actually you see this phenol and this acetic acid which is nothing but ethanoic acid also are not really compatible, the acetic nature are not really compatible, why because one is acid and other one is alcohol, right, but in both cases what happens the conjugate base of the acid is stabilized, right and hence and here also in phenol the conjugate base having the electron density there having delocalization tendency, so mainly for this acid and alcohol mainly we compare with the help of K and PK value, right, so this is what you have to keep in mind, one more thing I will just give you here, what is the, I'll just give you one question, what is the acetic, you know, acetic acidity order of this these compounds HCOH if I write formic acid and benzoic acid and one more which is CS3COOH in this three, what is the acidity order, tell me three to one, okay so I'll just number it here, this is one, this is two and this is three, this is what the benzoic formic and acetic, okay, all the answers are wrong, all of you, this is four because yes, see two things here it is important and this you have to memorize, okay it is important for JEE also, okay and and this question they can also ask in board exam also, first thing is what, I'll just write down here, all aliphatic carboxylic acid, all aliphatic carboxylic acid, less acidic, aromatic, aromatic or we can say cyclic, cyclic would be a better word I think, I think, cyclic carboxylic acid, no aromatic only you write, carboxylic acid, except one which is HCOH, okay, formic acid is the member of aliphatic carboxylic acid, we have a straight chain compound, okay and this we have aromatic carboxylic acid, benzoic acid, okay so all aliphatic carboxylic acid except this formic acid are less acidic than aromatic carboxylic acid but what happens in HCOH, here we have equal resonance, right so in case of equal resonance the conjugate base is highly stable and that's why it can lose this H plus ion easily, okay so what you have to keep in mind simply if you keep this thing in mind that aliphatic carboxylic acid is less acidic than the aromatic carboxylic acid except formic acid, so when you have this question obviously we know this formic acid is more acidic than the benzoic acid which is more than any aliphatic carboxylic acid, here in this case we have acetic acid, order will be 1, 2 and 3, this is one kind of exception you can keep in mind, is it clear, okay so here the point I'm trying to make is what this particular you know concept you have to memorize, second thing is what we can compare this acid, acid and acid, carboxylic acid we can compare with various electronic effects, eye effect, resonance whatever, right but when we have two different compounds like we have alcohol and acid then it will be difficult to compare on the basis of the electronic effects, okay so in this case it is mainly explained by pKa and ka value, mainly in these cases question becomes slightly what we can say, experimental or we can say database question it is, when you know this pKa value you can do this, right so all these questions we cannot compare through electronic effects, okay so that's the thing understood, okay we'll see the next question write down the product ZNHg, this is concentrated HCl, ZNHg concentrated HCl, one thing is I think missing here, one correction I think we should write, this benzoic acid we have and with this benzoic acid this reagent is not there, by mistake it is written it is a printing mistake here, it is bromination actually, Br2 in presence of FeBr3, this is the reagent we have, what is the first reaction, the first reaction is clementine reduction, yes so C double bond O converts into CH2 propane, second one is what, acetyl chloride, rosin mon reaction reduction, right rosin mon reaction so it gives you aldehyde, so CS3, CHO, what is the last one, beta bromobenzuic acid, correct why meta, why meta, not ortho or para, para sorry, okay, oh sorry, COH reduced density at ortho benzuil, correct, since COH has electron withdrawing nature and we know in resonance and COOH withdraws electron through resonance, right and we know resonance has major impact or effect on ortho and para position, right so if this group will withdraws electron so it withdraw more from ortho or para position, so what we can say the meta position becomes comparatively more electron rich here, right and that's why the bromine will attack at the meta position, so we'll get meta bromobenzuic acid here in this reaction, this is simply rosin mon reaction, the product here will be CS3, CHO, here it is propane, CH3, CH2, CH3, clementine reduction and this is meta bromobroduct which is nothing but the bromine here and COOH, okay, so correct, now conversion, do this conversion, propanone to propane to all, done, LIALH4, okay, so we need use reduction, propanone is this CH3, C double bond O, CH3, when we reduce this, it converts into, since it is in ketone, it converts into 2 degree alcohol, right, so we'll have the product here, CH3, CHOH, CH3, propane to all, okay and how this conversion is possible, we can use any reducing agent, H2 nickel we can use, we can use NABH4, we can also use what LIALH4, any of one of these reagents we can use and we'll get the desired product, ethanol 2, 2 hydroxy propanoic acid, ethanol, CH3, C double bond OH and from this we have to convert into, you see we have to convert into carboxylic acid, right, so if I use HCN first, then we'll get cyanohydrin, see if you want to convert this aldehyde into acid, this is one of the best way, CHOH, cyanohydrin and we know this cyanide group under acidic hydrolysis, H plus H2O converts into COOH, converts into COOH, right, but there is one more thing here, you see HCN we have used and then we use this, so this convert into CH3, CHCOOH, so we have also here 2 hydroxy propanoic acid, 2 hydroxy propanoic acid, understood toluene to benzoic acid, it's a direct reaction, toluene is this and we use what oxidation with KMNO4, H plus, this converts into COOH, toluene to benzoic acid, yes with CS3-MGBR also you can do, that way also possible, which one, the last one is the direct reaction of toluene to conversion into acid, okay, so it is actually a direct reaction mechanism you can see or you know there is one more thing you can see, I just written it in the one single step, actually what happens this toluene which is the same reaction actually, when you heat this with KMNO4 in presence of COOH, KMNO4 in presence of COOH, this converts into salt of potassium, COOK will get here, insertion actually, insertion of oxygen takes place, H2O goes out, H plus goes here and then when you do the acidic hydrolysis H plus H2O then this K comes out and will get benzoic acid, so this is the complete reaction we have, which you can simply write here all the reagent and then in one single step you'll get COOH, okay, this actually follows electrophilic substitution reaction type it is, okay, this KMNO4 provides oxygen to this molecule which joins here on this carbon atom and hydrogen comes out and finally it converts into acid, this one you tell me what is the test for these two we have, if you have doubt in this I will share one big image for this mechanism, I'll give you, send you on WhatsApp, okay, this reaction if you have doubt I'll send you on WhatsApp, okay, once we'll finish this class, okay, after the class I'll send this one step by step reaction given over there, you can see that, okay, I'll send you the image, yes, yes, I'll send you the image or you do this I'll just send it to you now only, how do we convert this pentane to on to pentane 3O, what is the test we have here? Idoform, right, you see this Idoform test they have asked many times in the exam, right, both Idoform, correct, both we can distinguish with Idoform test, you must, you know, this Idoform test is very important, okay, must keep this in mind, next question you see, oxygen is a gas but sulfur is a solid, what is the answer, why oxygen is a gas but sulfur is a solid, yes, correct, sulfur exists in polymeric form, okay, oxygen is a diatomic molecule, oxygen is diatomic exist as O2, right, O double bond O because and why it forms a double bond because of its small size but sulfur exists in polymeric form with all single bonds, polymeric structure and hence it is stable, single bond is more stable and hence it is solid in nature, okay, in polymeric structure without single bond, why O3 acts as a powerful oxidizing agent, why O3 acts as a powerful oxidizing agent, because Ojohn O3 produces, what, yes, O3 produces nascent oxygen, nascent oxygen, that's why it is a powerful oxidizing and nascent oxygen we, you know, write like this in a bracket, yeah, and that's why it is a powerful oxidizing agent, correct, next question you see, what is the answer, first one, what class of drug is aranatidine, yeah, second one is detergent, yeah, aranatidine is an anti-acid, anti-acid, water contains dissolved Ca2 plus iron out of soaps and scintillator which will you use for cleaning clothes, it's detergent, right, second one is detergent and this is again the, you know, experimental thing which is the following is an antiseptic, 0.2% phenol is an antiseptic, this is important also, I have seen this question in a neat exam also, okay, 0.2% phenol is an antiseptic, okay, so this is anti-acid, you have to memorize this, this is detergent and this is antiseptic 0.2% phenol, this one, give reasons, ethyl iodide undergoes SN2 reaction faster than ethyl bromide, simple one, living group than bromine, correct, iodine is a better living group than bromine, yeah, less electron repulsion and it is a better living group also, that you must write, yes, next question, 2-butanol is an optically inactive, why? It forms resting mixture, yeah, it's correct, you see, if the question is, so first of all 2-butanol is this, CH3, CH2, CCH3, OH and H, this is 2-butanol, obviously this molecule has one carol carbon and hence it is optically active, right, so when it is optically active, so it can produce two isomers, one is D, other one is L or we can also say plus or minus and when we take the mixture of these two, equimolar mixture of these two, it becomes optically inactive, however this is not mentioned over here, it simply says plus minus 2-butanol is optically inactive, it is very understood here that we are taking equimolar mixture of these two, number of moles of D and L or plus or minus is same, then it forms a resmic mixture and we know resmic mixture is what, it is optically inactive, so that's the question, understood, third one, done this one, this one, this one you do, partial double bond forms due to resonance and that kind of resonance is not possible here in alkyl halide, okay, so here what happens in halogen, resonance is possible, so because of resonance, partial double bond character is there and hence the bond length here it is lesser than the bond length in alkyl halide, okay, correct, your answer is correct, what is meant by crystal field splitting energy on the basis of crystal field theory, right the electronic configuration of D4 in terms of T2G and EG, tell me what is crystal field splitting energy, crystal field splitting energy is the energy difference CFSE, crystal field splitting energy is the energy difference between, yeah, between the two sets of D orbital or we can say between T2G or EG orbital, no this is not, yeah we can say T2G and EG orbital or the best way to say is what, energy difference between the lower energy D orbital and higher energy D orbital, right, lower energy D orbital and higher energy D orbital that also we can say, now D4 configuration means D orbital has four electron, right, so in case when this delta O, it's not not, it's delta O, O stands for what, octahedral complex, if you remember in case of tetrahedral we'll write del T over there, tetrahedral, so it is not del naught, it is del O octahedral complex, is greater than P, what is P here, what is P here, what is this P, tell me, average pairing energy, okay, so when this del O is greater than the average pairing energy then the distribution of electron does not follow Hund's rule, right, we say the energy difference is large and it does not follow Hund's rule, Hund's rule whatever you say, Hund's rule, okay and hence the distribution of electron will be what, T2G and EG, we know this T2G will have then four electron and EG will have zero, this is the distribution of electron in case of this and in the second case if when this delta O is less than P then it follows Hund's rule, right and in this case T2G and EG orbital we have, T2G will have three electrons, one, one, one and then one, EG will have zero, stress, EG will have one in the second case, stress, right, next question, IOPSC name of this ion, tetra, clodido, nickelate ion, oxidation state is two, correct, but don't write this two, I understand you have written the numerical two here, but you have to mention the oxidation state like this, right, understood, negative ligand where ends with 8, 8, we always change this into 8O, 8O, 8O like that, I have mentioned this in the last class also, that clodido we don't write, we write clodido, okay, clodide, from clodide it is clodido, clodate, clodato, sulfate, sulfato like that, if it is the negative ligand, okay, so tetra, clodido, nickelate, two ion, the name of this will be nickelate, two ion, since it is an ion so we have to write down here the ion, okay, hybridization is what, yes it is sp3, how do you find out hybridization here, you see, nickel is 28, so we have argon 4s2 3d8, ni2 plus will be what, argon 3d8, so 3d8 you see the, this is 3d orbital, 4s4p, 3d has 8 electrons, 1, 2, 3, 4, 5, 6, 7, 8, okay, so there is no pairing against the Hunt's rule because the ligand is weak, okay, ligand is weak, so since we have 4 chlorine, so these 3 orbital will go into hybridization, 4 orbital, and hence it forms sp3 hybridized orbital, okay, why I have done this, because they can also ask you how many unpaired electrons are there, answer will be 2, whether the molecule is magnetic and sorry diamagnetic or paramagnetic, answer will be paramagnetic, okay, because it has unpaired electron, correct, so sp3 hybridization shape of the complex will be tetrahedral, understood, so like this they can ask you any question, one coordination compound they'll give you, they can ask you name, they can ask you hybridization, they can ask you shape, they can ask you magnetic property, they can ask you the number of unpaired electron, right, anything, they can ask you the property of ligand, whether it is weak or strong, accordingly, okay, so this is what you have to do if the question comes like this, okay, and the we have done, in the class we have done many examples for tetrahedral, octahedral complex, okay, so if you go through all those questions, you will get one from there only, okay, that is more than enough, next question you see, rate of a reaction becomes 4 times the temperature changes, this calculate the energy of activation, solve this one, yes, correct, 52.863 joule, 52.863 kilojoule, okay, I'll write down the formula only here, formula will be log of K2 by K1 is equals to minus EA divided by 2.303R into 1 by T2 minus 1 by T1, this is the formula we have, okay, and here in this question it is given the rate of a reaction, okay, and if this expression is given somewhere and here if del H is written, right, if you write here minus del H divided by this and all expressions are right, so when you write instead of this EA, when you write del H here, then this K1, K2 becomes the equilibrium constant, but if you write EA, it is the rate constant, right, so all these value you have to substitute and you will find the value of EA which is 52.863 kilojoule is the unit, next question you see, write the product here, yes, correct, the first one is benzene with S3PO2 and H2O reduction takes place and from this disonium salt will get benzene reduction here, okay, and this gives 2 comma 4 comma 6 tribromo enylene CS3 CO NH2 gives you 1 degree amine which is nothing but CS3 NH2, in this nitrene forms and the reaction proceeds, the intermediate is nitrene here, yeah, right, subtract one, tell me the answer, then 1.93, no I don't think so, 0.61 is correct, you see, you must have mistaken over here this two term, when you add these two half equation, the overall equation becomes what? Cu2 plus aqueous plus 2 Ag solid gives 2 Ag plus aqueous plus Cu solid and when you write down the NERS equation for this, E cell is equals to E0 cell minus 0.0591 by 2 number of electron log of concentration of Ag plus square divided by concentration of Cu2 plus, right, all these value you substitute, E0 of the cell is given 0.46 minus 0.0591 by 2 log of concentration of Ag plus is what? 10 to the power minus 3 square, 10 to the power minus 6 divided by 10 to the power minus 1, so this becomes 10 to the power minus 5, so the expression here it will be 0.46 plus 0.0591 by 2 into 5, this is what the answer, you'll get 0.6077 approximately is the answer, 0.6077 volt, is it the expression resist, understood? Usually in this kind of question, students makes mistake over here, here, this is square term, you must check whether the, you know, this charge is same or not and it is the best way to write down the half reaction, both half reaction you write down, balance charge in both the reaction, right, and then you add both, with this you'll get N value also, right, and the exact coefficient also, so it's 0.0591 only, okay, next question you see, examples of each, what is associated collides, okay, write down the definition of associated collides, write down the certain substance, certain substance, I'll write down the first one, certain substance forms true solution forms true solution at low concentration, at low concentration, but as the concentration increases, as the concentration increases and becomes greater than and becomes, oh, I'm sorry, just a second, okay, so I'm sorry, got some call, but as the concentration increases and becomes, becomes great and become greater than, greater than critical C-R-I-T-I-C-A-L, critical missile concentration, at this time, when the concentration becomes more than this critical missile concentration, then the particles get associated with each other, then the particle gets associated and forms collides, forms colloidal particle, I'll write down, and that is how the colloidal solution obtains, right, example we can write, the example is soap, and we know the soap is the salt of sodium and potassium, salt of fatty acid, sodium or potassium salt of fatty acids, okay, so this is the associated collides we have. Now, lyophilic, I have done this in the glass, lyophilic collides are what, when the particles from dispersed phase are solvated by dispersion medium, right, particles from dispersed phase are solvated by dispersion medium, example we can write, we have a starch solution, solution in H2O, adsorption you already know, any example of adsorption, any example of adsorption can you tell me, it is a surface phenomena and all those you can write down, any example of adsorption, yeah, correct, yes, yes, right, first one is right, all of you have done the answer, then give me an example of adsorption, water in silica gel, you can say that, and we can also say the adsorption of gases at activated charcoal, right, gas at activated charcoal, you must have, if you recall the hydrogenation process, there we use catalyst as platinum and palladium, remember, H2O with PTPD, we usually write, yeah, that also correct stress, see what happens here, when you use this platinum and palladium, this platinum and palladium has high affinity towards hydrogen gas, okay, and the hydrogen gas because whenever you are using gases, okay, for reaction we have to provide some surface here, so that on that surface the reaction takes place, so what happens is this hydrogen get adsorbed at the surface of platinum and palladium and there on that surface where the hydrogen gets adsorbed, there only the reaction takes place, that is the purpose we use this platinum and palladium for the reaction, okay, so these catalyst has high affinity towards hydrogen, hydrogen gas get adsorbed at the surface and the reaction takes place on that, that's the purpose, okay, next one you see, zone refining and vapor phase refining, what is that, see, zone refining actually we use just two points I'll write down, okay, you also write down these two points, zone refining we use when very high degree of purity is required for high purity I'll write down, okay, high degree of purity in short I'll write down, high purity, right, also what happens, the impurities have lower melting point than pure metal, this is also useful when impurities have lower melting point than pure metal, then we use zone refining, okay, what is this vapor phase refining, vapor phase refining in this what happens the solid molecules or the solid particles converted into vapor and this takes place by chemical reactions, converted into vapor by chemical reaction and when you heat this vapor at high temperature, this vapor at high temperature gives the metal, so basically in vapor phase refining process chemical reaction takes place, we convert solid into vapor and then the processor will use the further processor to find out the, to get the metal into this process, next one you see, standard electrode potential of daniel cell is given, del G we have to find out, tell me the answer, tell me fast, viscist and purvik, you have chemistry exam on 2nd of March, when you have chemistry exam, viscist purvik and stress you also tell me or the coming Friday, this this Friday, right, okay, this Friday, what about stress, 12th of March, is it stress, you got minus 2.12, what is the formula we use here, del G is equals to, del G naught is equals to minus nf E naught cell, E naught cell, n value is 2, minus 2, f is given 96500 and 1.1, so simply if you multiply this, you will get minus 212.3 kilojoule, all of you are getting this, minus 212.3 kilojoule, you are getting 2.12, 212 it is and don't round it off, write the exact value, okay, minus 212.3 kilojoule or minus 212.300joule, correct, okay, next question you see, what is the order of the reaction? Order means the sum of the exponent of the concentration term in rate law expression, okay, the rate law is given here, order is equals to 1 by 2 plus 2, which is nothing but 2.5, now suppose one thing, if they don't give you this expression and they ask you what is the order of the reaction, what is the answer then, if suppose this rate expression is not given and they ask you what is the order of the reaction, this reaction is given, what is the answer then, see if the rate expression is not given, then we cannot identify the order of the reaction, then the answer will be cannot be determined because we know order is a what, order is a experimental quantity, okay, if rate expression is not given, then we cannot say the order of the reaction is this, yes, unless if the rated money step is given fine or rate law expression is given, right, even one thing, if they tell you that this is what, this is one step reaction, then also we cannot say because we do not know in this reaction whether the rate is independent of A or B or depends on both A and B, right, so one thing, if the rate determining step is not given or if rate expression is not given, then the order cannot be determined, okay, must take care of this thing in mind, okay, what is the answer of the second one, t half for first order is equals to what, 0.693 divided by, okay, that's what, right, just you have to solve this, whatever answer you are getting, I think it's correct, tell me this one, what is the relation we have, we'll just write down the phase diagonal which is equals to root 2A, right, and this will be equals to in terms of radius, we can write for R, A is equals to what we write, the relation for FCC to root 2R, this is the relation we have, and A you have to find out, R is given, your answer will be A is equals to 353.5 PM decometer, next question, structure of the molecules, I will draw, okay, because I can't see, we have for XEF6, we have one lone pair, if you draw the steric number, if you find out, we'll have F here, and one lone pair will have here, this is the structure, XEF6 we have, no, okay, sorry, so, okay, my bad, this lone pair I'll draw here, one fluorine we have here, and one lone pair on this, XE here, right, 8 for xenon, 7 into 6 divided by 8, so it is 50 by 8, so we have 6 bond pair and one lone pair, so that's what we have here, okay, S2, S2O7, so we'll have S double bond O, one oxygen we'll have here, S double bond O, 3, 4 we are left with, so double bond O, OH, then double bond O, OH, this is the structure of S2, S2O7, okay, symmetrical structure we'll try to draw this, yep, next one, for interhalogen compound forms, what a general composition can be assigned to them, interhalogen compounds, tell me next, see this interhalogen compounds are formed by two halogen atoms, formed by the combination of actually, formed by the combination of two halogen atoms of different electronegativity, different electronegativity, okay, I2, we cannot consider this as interhalogen compound, okay, two different halogen atoms we must have, there are four types of interhalogen compound forms, we may have Mx type, Mx3 type, Mx5 type, Mx7 type, these four types of interhalogen compounds may form, okay, where this M is what, M is the compound having, or I'll write down here, M is the less electronegative halogen, and X is the more electronegative halogen, okay, these four types of compound possible, okay, ICL3, ICL5, IF7, all these are the compounds we have, okay, next question we see, explain the mechanism of the following reaction, the first step is what, write down the name of the first step, what is the first step, three steps we have here, step one, step three, the first step is protonation, and the last step is deprotonation, second step is formation of carbocation, so the first step is CH3CH2OH with lone pair on the oxygen atom, so this lone pair will attack onto this H plus, and it forms CH3CH2OH2 with positive charge on this, okay, formation of carbocation means what, removal of H2OOH2 plus, and in this it takes the bond pair of electron and forms CH3CH2 positive charge on it, plus H2O, and finally CH2CH2 positive charge, from here one of the hydrogen atom donates its bond pair of electron and comes out as H plus, so this is deprotonation, next one you see, thermoplastic and thermosetting polymers, what is the example of thermoplastic polymer, yeah, any example of thermoplastic polymer, cross link polymer, yeah, so cross link polymer is beccalite, thermoplastic is PVC, polyvinyl chloride, yeah it's correct, so all these you know biomolecules, the definition with one example you must remember, okay, thermoplastic is PVC, thermosetting is beccalite, what is biodegradable polymer, any example of biodegradable polymer, biodegradable polymer, it can be starch, nucleic acid, which can be decomposed by microorganism, right, starch, nucleic acid, proteins, cellulose proteins, yeah, pH, PV also a biodegradable polymer, correct, yes, yes, correct, correct, so all these you see this kind of example they ask with some, you know, definition they ask with some example, okay, next one you see, write the equation involved in the following reaction, last class only we have done this, okay, two types of reaction with chloroform and with carbon tetra chloride if you remember, with chloroform will get cellulose dehydrate and with carbon tetra chloride will give cellulose acid, remember, yeah, so we have done this already, you see like we have, if you try to recall the all the questions that we have solved, they have asked very, you know, common type of questions only, name, reaction, primatimer, aldol condensation they have asked, this kind of reaction if you see there are five, six reactions which they usually ask every time, okay, so you must revise all these, okay, Williamson synthesis means ether formation we have, this is also we have done, these two reactions we have already done, you must go through this, okay, yeah, it's correct, correct, why does presence of excess of lithium makes LICL crystal pink, yes, correct, formation of F-centers, yeah, it's right, formation of F-centers and just you write a bit of F-centers also, okay, that is it, not required much, because these questions are just one one mark question, one mark in this A part and one mark in this B part, okay, so with F-centers it's fine, but if you have time just write down what is F-centers, that is it, formula is PQ, that is also correct, very basic question it is, formula is PQ, one last question for today, one more we'll see after this, condensation polymer, any example, condensation polymer, one example tell me, yes, correct, nylon 66, nylon 6 also a condensation polymer, right, one last question, I have seen all this question, you can do this, so I'm not discussing all those, one last question for today, this is the IUPSE name, you can do it easily, the last one, 2 bromo, 4 clodo, you see Purvik, you have written after clodo also you have given dash, which is not true, which is not right actually, 2 dash bromo, dash 4, dash clodo pentane, that is it, after clodo you don't have to write down dash over there, right, okay, so okay, so we are done for today, okay, so also Purvik and Visist, all the best for your exam, okay, what I feel that you have done your preparation very well, okay, the only thing which I am concerned with, don't make any silly mistake like you have done just now Purvik, okay, read out the question carefully, don't round it off, don't round off the question, write down the answer till the still second decimal point, okay, don't assume any questions, do the calculation properly, okay, keep that unit thing in mind that whatever terms you are writing down, unit you must mention, suppose some questions you are not able to solve, just you write down the given data and then the related formula just write it down, okay, these few things that we kept on, you know, discussing these, you know, two, three weeks, if you keep in mind these small, small things, you will definitely get good marks, it's not like that you do not know, you know almost everything, right, the only thing is what, try to use that keywords in your answer script, okay, so that is how, build your answer properly, that is what the concern I am, the concern I have, okay, it's not a thing that you do not know, anyways, so you'll do well, I know, okay, keep that small, small things in mind, all the best for your exam, see you soon, take care, bye-bye.