 So this is the 11th question of J-min question paper 9th January of 2019, second shift, right? So question number 11 that says that the reaction is given, its equilibrium constant is given k1. You have to find out the relation between the equilibrium constant is k1 and k2. First of all you see the reaction we have is a2 plus b2 and then it gives 2ab, right? And for this it is k1. Now the second reaction is 6ab gives 3a2 plus 3b2 is k2. We have to establish the relation between k1 and k2. So first of all you already see this ab is on the reactant side and this is on the product side. To establish the relation we will just try to write down this ab on the reactant. So when I reverse this equation it becomes 2ab gives k2 plus b2. And since if I reverse this, this becomes 1 by k1, the equilibrium constant of this reaction, ok? Now when I divide it by 2 both side, ok? Or if this reaction if I multiply it by half it becomes ab gives 1 by 2a2 plus 1 by 2b2. Since if I multiply it by half equilibrium constant will get power of half so it is 1 by k1. Again if I multiply it by 6, right? 6ab gives 3a2 plus 3b2. So this becomes 1 by k1 to the power 1 by 2, 4 to the power 6. This is reaction is nothing but k2. k2 is equals to 1 by k1 to the power 3. k2 is equals to 1 by k1 to the power 3, option b is correct, right? Next question you see, question number 12. Which compound can cause the temporary hardness of water? It is because of bicarbonate of calcium and magnesium, right? Bicarbonate of calcium and magnesium causes temporary hardness in water. And so it will be option 4. I have a question, nothing to do in this one. Question number 13 you see. Which compound have maximum CFSE? Strongly Gantt, strongly Gantt produces maximum splitting into the d-orbiter, right? Since the question is maximum splitting we have so we will find out the strongest or strongest ligand of this, okay? So your cyanide ions is strongly Gantt. That's 2 a weak, Ns3 is strong but not more than this. Cl is also weak. In all these example if you see only the cyanide ions, Cn- is the strongest ligand we have. That's why the splitting in this one will be maximum, right? The more strongly Gantt, the more will be the splitting. Question number, option 1 is correct. Next question you see 62 gram of ethylene glycol present into 50 gram of water. Delta Tf it is given plus 10 degree Celsius. And how many grams of water will get piece, right? What we can write? Ethylene glycol first of all it is a non-electrolyte, okay? So the formula we have in delta Tf is equals to I Kf into M. I value will be 1 since it is a non-electrolyte. I is 1. Kf is given for water. It is 1.86 molality we have to find out, okay? So, you see here at this temperature delta Tf is 10. I is 1. Kf is 1.86 into molality is what? Number of moles of ethylene glycol. So it is 62 divided by molecule mass of ethylene glycol is what? CS2OS, CS2OS. So we have C2H442. I think this is a molecular formula. 2 and 6, right? So 12 into 24 plus 6, 30 plus 32 into 62. This is the number of moles of ethylene glycol, right? Divided by mass of the solvent which we have to find out into 1000 because this is in gram. So when we solve this we will get 186 gram, right? So initial the mass of water is 250 gram, right? And at this temperature the molality gives the mass of water should be 186 gram. The amount of ice formed would be the initial mass of water minus now mass of water whatever we have, right? So it is 64 gram. Answer will be option 1. I see question number 50. Ellingham diagram it is given. See in this Ellingham diagram one thing you have to keep in mind. This is again. So in this Ellingham diagram the important thing is what? That the line which is present below in diagram can replace the line present above in diagram, okay? For example, the reaction of magnesium and aluminium if you see. Okay, in a given range of temperature. Since this line is magnesium and this line is aluminium if you see, right? So in the reaction what happens? This magnesium can replace aluminium, okay? So what we can write the line which is present below in this diagram to replace the line present in this diagram. So here it is magnesium in a certain range like till here in this temperature till here. So magnesium can replace aluminium, okay? The question is we can extract copper due to the carbon reduction method, right? We can extract ZN from ZNO by coke of ZN. ZN can be extracted by ZNO by using 500 uses, okay? In this diagram which is not clear over here, so you can refer the diagram in the book and then you can go through the options, okay? Only thing that you need to know here is this one, okay? And we can easily conclude that the answer in this question will be option 3. ZN can be extracted by ZNO by using aluminium, okay? Because aluminium will replace zinc, okay? That's the reaction we have. You can see aluminium can replace zinc, ZN forms and will get oxides of aluminium in this at this temperature, okay? So option C is correct here. Question number 16, which is not aromatic, okay? So for aromaticity, we know the molecule must be planar, okay? Molecule must be planar and follows Huckel's rule, right? So planar molecule with 4n plus 2 pi electron. That is what it requires, okay? So how many pi electron, where n value can be what? n value can be anything from 0, 1, 2, 3 and so on. You see this one, this molecule has 4 pi electron because this positive charge we won't count. This negative charge will count, so 2, 4, 6 pi electron we have here. Here we have again 6 pi electron, right? And this is also we have 6 pi electron. So we must take care of this thing that in this 4n plus 2 pi electron we count negative charge and lone pair provided for lone pair will have some condition, right? Here this lone pair is not involved in resonance, not involved in resonance, right? So that's why we are not counting it. This is in the p orbital which is perpendicular to the plane of this thing. So this does not contribute into the aromaticity of this thing. There are only 3 pi bonds, that's why it is 6 pi electron. 2 pi bond, 1 lone pair, 6 pi. 2 pi bond, 1 lone pair, 6 pi. So the molecule which is not aromatic is very clear from this. It is a 4 pi electron does not follow Huckel's rule. That's why option 1 is correct here. It is not aromatic compound, right? 17. You see in this case the reagent is this. You see what happens here in this oxygen being more electronegative, right? So this will drag the bond pair of electron on its side. The reagent we have here that becomes this oxygen negative charge. Here we have double bond O with positive charge of this carbon atom, right? Now what happens? Electrophilic substitution reaction takes place. Electrophilic aromatic substitution reaction. And here it becomes what? NCl3-activity. This will take this to the bond pair. Okay. Now from this, the molecule that we have here, which is OHMCS3. Electrophilic aromatic substitution reaction takes place. So this group, this positive charge will get attacked by this pi electron cloud here. So here the product we get is OH double bond O. And this will be as it is. OH here and CS3 here. Para position is attacked, right? Because ortho position is already occupied. Para position is attacked. Okay. So you see option 4 clearly is the answer we have here. It is an electrophilic aromatic substitution reaction, right? At para position since ortho is occupied. Question number 18, this question. Obviously it is very clear from this question that CS3-COH molecule will form. And the rest of the group will attach under this molecule, right? Either this H will come from this nitrogen and this oxygen. Right from here to here. This is the two possibility. If this comes, if this take part in the reaction and CS3-COH will attach over here, the product will be one. If this takes part in the reaction and CS3-COH will attach over here, the reaction will be. The product will be two. This is the two possibility we have here, right? So which oxygen, which like which hydrogen will take part in the reaction that depends on the nucleophilicity of OH and NH2, right? So when you see the nucleophilicity order, nitrogen being less electronegative. Its nucleophilicity is more than to that of oxygen, okay? NH2, the nucleophilicity of NH2 is more than to that of oxygen because it is less electronegative. Hence the hydrogen of nitrogen, this takes part in the reaction, part in the reaction. And hence the product will be what? One of the H will go with CS3-COO group and CS3-COO group will attach with this nitrogen. The product will be option two easily in this, correct? This is based on the nucleophilicity order, right? Question number 19, write down the best city order. Best city order is option is not given into this one, right? Optional write down what all options are given here. The first option we have is maximum, then one, then fourth, and then second. Second option is one, then third, then second, and then fourth, three, one. The last option we have, one. These are the options. Now you see the best city of amines, these are actually amines, right? That depends on the plus i nature, plus i nature of alkyl group, del H of hydration, hydration energy basically, right? So with this combined effect, we'll decide the best city of, what we say, best city of amines, correct? You see, fourth option we have maximum plus i, right? Here we have two, here we have one, right? But in the fourth option, we don't have any hydration energy. We don't have any hydrogen bonding possible here because H is not present in this nitrogen. But here we have hydrogen bonding also and i effect also. Here we have more hydrogen bonding, less i effect. So if you see the option third, Me2NH is most basic compound we have here because of the combined effect of plus i and hydration energy. We have hydrogen bonding also, that list implies the, no, nitrogen which loses round pair of electron and then two i, plus i effect also, right? If you see this group, THNH, TH is an electron withdrawing group, right? It withdraws electron, that will decrease the basicity most here, right? The basicity of the second compound will be least, correct? So we'll have third most and second least. So we have only one option possible, which is option A. So you see the basicity of third will have maximum because of both plus i nature of methyl group and hydrogen energy also, hydrogen bonding also. Then we have first one, fourth one, then second one. All we can say is two degree amines are most basic. Then we have one degree, then we have three degree. And then the list we have phenyl amines, right? This is, we have discussed many times in that class, right? So this is the question. We have fourth one, you may have doubt in these three, first, third and fourth. So two, one, three, you, we have discussed it already. So from this you can do, logic also we have discussed now. And this, since it is a minus i group, so this basicity should be least. Second one will be least. So you can get A and D option least or not possible. And then we can do, however, in this question, the answer will be option A. Question number 20, pH of the rainwater. See, it is a very factual question, right? You cannot do anything in this, but one thing you can keep in mind that when H2O will have, it combines with the carbon dioxide present in atmosphere and forms weak acid or the carbonic acid. Because of this only it is slightly acidic in nature. And since it is acidic in nature, so its pH value should be less than 7, which is only one option given over here, 5.6, correct? So if there are two options which is less than 7 we have, then we cannot do anything. Then we have to know this value that pH of the rainwater is 5.6. Actually, if you see this there in NCRT another book, they have given the pH of the rainwater is 5.6. So it is the correct value we have. If you have two options less than... See, first of all if you know its value, pH value is fine. If you do not know, you can put some logic whether it is acidic or acidic. According to that you can say it should be less than 7 or more than 7, right? So less than 7, if you have only one option, fine. If you have more than one option, then you have to, you cannot do anything into it. If you not know the exact value, right? And in this question option C is correct, third option. Question number 21. This one is the question based... It is there in polymers, okay? That question we have amide linkage over here. So it should be phenyl. So phenyl should be in between. We are left with 1 and 3. Well, phenyl, you are right. In this one the option A is correct. Question number 22. The name of this is what? Okay, it is metmoglobinemia. Okay, metmoglobinemia. This occurs due to high quantity of the following. Okay, so this is again, it is there in chemistry in everyday life. Okay, so it is the direct question they have asked from NCRT. Okay, and it is given that when nitrate ion is more than 50 ppm, then this disease may occur, right? So the cause of this disease is 50 ppm of NO3 minus. Let's see question number 23. pH in this. Okay, so this is an aldol reaction, okay? Beta hydroxy and dehydrochetone. Right, it forms beta hydroxyl dehydrochetone in presence of a base. Right, so alpha hydrogen will take part in the reaction. So the alpha hydrogen of this compound that is CH3C, C double bond OH and OH minus from the base. This will attack on to this and takes hydrogen. So we'll get CH2 negative C double bond OH. Okay, now this attacks with, attacks on the carbonyl group of the another compound. And this bond pair comes over. This is the mechanism of aldol reaction and hence this pH, C single bond OH, pH2 attached over here, C double bond OH. Okay, beta hydroxy, aldehyde or ketone, right? pHC OH, the answer will be what? Answer is pH. pH is also there. So this H will be here. pH OH, CH2, CO. I think option C is correct. In this one, there is no, we are not heating it, right? If you heat this, then maybe condensation takes this, and we condensate this place and we form a double bond over here. H2O will go out and we get a double bond. Since it is only NA over it, so we'll not get any double bond here in between. Condensation will not take place, right? Option C is correct. Question number, 24. Identify the compound from the given which positive test of 24 DNP and positive hydrofond test. Okay, that's not from azodipe. The compound which shows positive hydrofond test must contain CS3-CO group. This compound, because we have NH2 here, may form azodipe because it reacts with some nitro compound and this H2 may go out and we'll get double bond over here. That's why this fourth one and third one for the same reason it is not possible, right? Here it does not form azodipe, but this does not give hydrofond test also. This is also not possible. CS3-CO group, positive hydrofond test, because N-CS3-CS3 we have it will not form azocompound. That's why option A is correct, right? Why this is not true? It contains CS3-CO group, but this will form azocompound here. So suppose it reacts with NO2 or something like this compound, nitro compound and H2O may go out and this N combines with double, this nitro is a double bond. There are possibility of forming an azocompound here in these two, right? That's why these two are not possible. This azocompound will not form, but this does not give positive hydrofond test. That's why third option is also not correct. First option is correct, according to the given condition. Question number 25, you see, BR2 with KOH, okay. You see in this compound, bromine KOH, it is Hoffman bromide reaction, okay? And with Hoffman bromide reaction, we know it's a direct reaction actually. We know amide group, amide gives you amide, right? CONH2 could convert into NH2, right? Only the thing is possible here. We don't have any bromine can attach over here. This is Hoffman bromide reaction. The name of this reaction is Hoffman bromide. We have already discussed this mechanism in the class. So here we are not discussing it. And we know the amide we have here, amide converts into amide, right? So CONH2 will convert into NH2. That's what the thing we have. CONH2 converts into NH2, right? This is also not possible because we are getting BR here. CONH2 attach to this also not possible. Only one option is true here. That is option 2 is correct in this. So we have discussed the further 15 questions here, okay? See you soon. Bye.