 Okay, so these are the questions we are discussing, which has been asked in JEE main shift 1 9th of January 2019. This is the 11th question we are discussing and let me remind you, these questions are memory based question. Okay, so these are not the questions which has been declared officially by JEE. No, that these are the question has been asked. These are the JEE memory based question we have, which we are discussing. You see the 11th question here is from the chapter of polymers, right? And when these two combines, when the free radical polymerization takes place, which is usually takes place in presence of some peroxides or light, right? So what happens here, the hydrogen combines, okay, and comes out as HCl, right? But which one, which chlorine will combine, that is what the question we have, okay? So here you see, first of all, these because of this carbonyl group here and the electronegative difference, this bond is comparatively weaker than this bond, carbon-chlorine bond here. Obviously, this chlorine will get attached with the hydrogen over here, correct? So you see, according to the option also, you see one of the hydrogen combines with chlorine forms HCl and this carbon and this nitrogen will join, okay? So we have NH2, CH2, CH2, CO, NH2, right? So you see the option, we have this double bond obviously will break because it also attached with the other group, correct? So this we have carbon CCl, so CCl and this is attached with the carbon which contains carbonyl group, right? So we have this carbon attached with carbon with carbonyl group and the chlorine which is here is taken up by this hydrogen forms HCl and goes out, right? And this is this molecule we have from here to here, right? Now another thing, what we'll do into this one, we'll take the another molecule of this and the reaction proceeds in the similar way, right? So by looking at the option, you can easily see that the reaction which proceeds and the radical will end up polymer will get here is option A over here. It is easily observed from this, correct? This one is not possible here, you see it is, you don't have to draw the molecules here. Another thing is HCl will go out. So this H and this CCl will go out and the product will be this. This you see NH, but we don't have C double bond O present, do we have C double bond here? No. It only falls when this hydrogen will take this chlorine over here, right? But that is again not possible, right? Because this kind of reaction or the nitrogen-hydrogen bond that will break here is this one, nothing but this one, right? It is a free radical substitution reaction. The free radical forms at this nitrogen because of this carbonyl carbon, carbonyl oxygen here, the carbonyl group that you have, the molecule or this molecule that NH2 that is present this side, this will not form the free radical, okay? This bond will dissociate homolysis reaction takes place over here and hence the answer will be option A. So you may get confused with A and B, 1 and 2, right? These options are not possible at all. So you can eliminate these two options easily. You may get confused into this one, but since it is a free radical substitution reaction, okay, since it is a free radical substitution reaction, so the product will be this because of this hydrogen and this chlorine will join. So hence the option A is the answer here, right? Option A. Question number 12. The weight of Na plus in the solution of Na2SO4 is given, right? 92 gram, find the molality of Na plus, okay? So what is the formula of molality we have? It is the number of moles of solute divided by the mass of solvent in kg, right? So now you see in the question, number of moles we can find out easily because mass of Na plus it is given 92 gram, we know the atomic mass of Na plus also. So number of moles will be what, 92 divided by 23, this is the number of moles of Na plus divided by mass of solute, molality of Na plus per kg of water. So water is a solvent over here and the mass of solvent is given per kg, this is 1. So the molality will be what, 22 by 23, which is nothing but 4, option 2 is right. The question is which of the following alkaline earth metal nitrate does not have water of crystallization, okay? So all these nitrates belongs to which group? All these elements from which the nitrate has been formed belongs to group 2, correct? So first of all there are two things here you can see. Group 2, there is one fact that we have studied that the stability of nitrate or the tendency to form nitrate of the group 2 elements decreases as we go down the group. So for the group 2 elements, the tendency to form nitrate decreases as we go down the group, correct? So we have according to this logic you see the question is the following alkaline earth metal nitrate does not have, okay the question is for water of crystallization, so this is not actually useful here, let it be this fact, let it be this is not useful, okay? Since we have to find out the water of crystallization, okay? So it depends on the polarizing power, water of crystallization, polarizing power, it depends on the polarizing power, okay? As polarizing power Pp increases, water of crystallization increases, more will be the water of crystallization, okay? Because in water of crystallization, the water molecule surrounds the ions, okay, from all side, that's why it is the polarizing part, it is affected by the polarizing part. Now the point is what as we go down the group of this divalent cation that is Be 2 plus mg 2 plus and all, all this cation that we have, so as we go down the group the size increases, size increases and as the size of cation as the size of cation increases its polarizing part Pp decreases. So less polarizing part gives you less water of crystallization and hence according to this you see Ba NO 3 whole twice will have minimum water of crystallization or there is no water of crystallization for radium nitride, that is the answer we have here, option 4 or D. The following option is correct for the given curve, okay? This chapter is, this question is belongs to surface chemistry, surface chemistry and when I was teaching this I have discussed this graph, okay? So the relation we have here, we know what is x by m, it is the mass that is adsorbed at the surface, right? So x by m is the ratio of mass adsorbed at the surface of adsorbent, okay? x by m is that ratio, okay? And it is directly proportional to pressure p, pressure p to the power 1 by n, 1 by n, right? Where n is n integer, correct? So when I remove this proportionality, I will have x by m is equals to k p to the power 1 by n, right? Since the graph is for log of x by m and log of p, so we can write it as log of x by m is equals to 1 by n log p plus log of k, correct? Now here you see this is the graph here, so what is this 1 by n? That is what the question is, we have to find out the 1 by n value only, n value only. So 1 by n is nothing but we can write this equation, this equation actually is in the form of y is equals to m x plus, so 1 by n is nothing but the slope of the line given in the graph. So what is the slope of the line? That is y by x and that will be equals to 2 by 4, which is nothing but 1 by 2 and hence the option 1 is correct. Question number 15. Here the following rule contains iron and copper, okay? This is also, you know, memory-based question. Melisite, M-A-L-A-C-H-I-T. Melisite, the formula is it is a carbonate of copper C-U-C-O-3 dot C-U-OH whole twice. I have write down the formula of this. Next one we have as you write and its formula is it is the two molecules of C-U-C-O-3 dot again C-U-OH whole twice. Means it is surrounded by two molecules of copper carbonate. Copper pyrite you already know, copper pyrite it is the easy one, C-U-F-E-S-2, C-U-F-E-S-2. And the next one we have is none of these, fine. So the question is what which is the following over contains iron and copper? So you made it know, you may don't know these two, you know, these two formulas, okay? Because these are not the common pyrite you all know probably, right? They are very common or that we see it is given in all different books, correct? So copper pyrite is C-U-F-E-S-2 and this contains iron and copper. Hence the answer is a straightforward option three, correct? Next question number 16 we'll see. 20 ml of 0.1 molar S-2-S-4 is added to this calculate p-8 of the resultant solution, right? So first of all you see the reaction is what we have H-2-S-O-4 plus O-H and this gives you what NH-4, NH-4 whole twice SO-4 plus two moles of H-2-O. And if you back this here also we have two moles of H-4-4, right? This is the reaction. So how many millimoles of S-2-S-4 we have that is 0.1 into number of millimoles of H-2-S-4 0.1 into 20 which is two millimoles. Millimoles of NH-4-O-H 30 into 0.2 is equals to six millimoles, right? Six millimoles of this we have. Now in this reaction you see since the ratio is not same one is to two and number of millimoles those not same. So one combines with two so two combines with four, right? And this is the if you apply the concept of limiting reagent this is a limiting reagent we have because the acid will get consumed completely in this reaction and when the reaction completes we have this base present NH-4-O-H and the salt of this base weak base with strong acid. So first of all the another point you have to understand here the first point what we have discussed we'll have this salt and we'll have this base. What is the nature of this base? This is a weak base and this is a strong acid. So this salt is the salt of weak base plus strong acid. So when the solution contains a weak base and salt of weak base plus strong acid then in this mixture we call it as basic buffer, right? This is a basic buffer solution and for that we know the P-O-H value the formula of P-O-H is equals to PKB plus plus log of concentration of salt divided by concentration of base. PKB value is given in the question it is 4.7, okay? So we'll substitute this value 4.7 plus log of salt is what? How many moles of salt we'll get? What is the concentration of salt we have here? That is two moles of this, right? So log of salt, so it is 2 by, okay, so what is the concentration of salt we have? The reaction is balanced 8, 5, 7. This one is 4 plus 2, 6. Okay, reaction is balanced. So one mole combines with two. So two combines with four, so out of six, four reacts and two millimoles of this base will left, right? And one gives one, so two gives two millimoles of the salt we'll get. So two by two we'll get. So the answer here it will be what? The value of log one is zero. P-O-H is 4.7. H will be what? Is there any temperature thing it's not given? So we'll assume 14 only. So 14 minus 4.7 and you will approximately get 9 plus 4 is 13. So we are getting 9 plus 9.4 approximately we'll get according to the option I'm telling you. So option B is correct over here, right? Option B is right. Next question, question number 17. Consider the MOT comment on the stability of this, okay? So according to MOT we know from molecular orbital theory we'll have the concept of bond order, right? And from this theory only we'll come to know that the bond order can be, can have fractional value also, fractional value. And when it has the fractional value we consider that as a half bond, right? So if you compare stability here, so more bond order more bond order more will be the stability will be the stability, right? There are possibilities that for two molecule will have equal bond order. So when the equal bond order is there, in case of equal bond order, the molecule which has, the molecule which has less number of electrons, number of electrons in anti-bonding molecular orbital will be more stable. For all these molecule you have to draw the molecular orbital diagram, okay? That is energy order. Those bonding and anti-bonding molecular orbital will draw and then we'll calculate the bond order. If it is more then the stability will be more. If it is equal then we'll see how many electrons are present in anti-bonding molecular orbital, okay? So first option you see we have Li2 plus, Li2 plus will have five electrons, right? Li2 minus will have seven electrons, okay? So the order will be what? Sigma 1s, sigma star 1s, sigma 2s, sigma star 2s. This is enough for this two, you know, ions we have. So when we distribute the electron in all these orbital, sigma 1s will have two, here we'll have two, here we'll have only one because we have five electrons. This is for Li2 plus. So for Li2 plus if you compare the bond order, that will be equals to number of electron in bonding molecular orbital. So this is 2 plus 1, 3 minus 2 divided by 2, so that is half. Similarly, bond order for Li2 minus, that will be what? Seven. So two, four, six and seven, right? So three electrons will have in anti-bonding and four electrons will have in bonding by two. This is again one by two. So you see the number of bond, like the bond order for the two molecules will be same, but here the number of electrons in anti-bonding molecular orbital is one, sorry, two and here the electrons in number in anti-bonding molecular orbital is three, 2 plus 1, 3. So that's why Li2 minus will be less stable than Li2 plus, okay? So this is the option we have, Li2 plus stable, Li2 minus unstable. So that's how answer will be option one. Similarly, you can calculate this for other ions if it is given. We have these two ions only. Li2 plus unstable, wrong. Li2 plus unstable, unstable. This is wrong. Li2 plus stable, so this is wrong. So option one is correct. 17, now one. Option one is right. Question number 18, which is the following is not correct about Henry's law. On increasing temperature, the KH increases, right? On increasing temperature, value of KH increases, right? It's correct. The value of KH increases, solubility of gas increases, okay? So the formula we have here, that is the partial pressure of the gas, suppose P of any gas we have, it is directly proportional to its mole fraction, KH into X, mole fraction that has been dissolved. So the value of KH increases, right? Value of KH increases, the solubility of the gas increases, not possible, because we know this is the dissolution of gas is an, we can discuss about isothermic or endothermic nature of this also. If value of KH increases, that is possible when the temperature will increase, right? And when the temperature will increase, the dissolution of gas will be less into the water, okay? So this, we can also discuss over here that hot water, if you drink hot water, you'll feel like there is no test. It's very, you know, the test is very flat over there, correct? Or you can take the example of seawater also, the very, all the aquatic animals used to live deeper into the sea, not at the surface of water. The reason behind this is what? The surface water is comparatively warmer than the water in the deep level that you have. So on the surface water, water at the surface, right, it is warmer because of sunlight, right? The surface water, since it has comparatively high temperature, so dissolution of oxygen, the dissolved oxygen contained into the surface of the water, right? That is comparatively lesser. So what we can say the oxygen contained or the dissolved oxygen contained at the surface of the water in the sea is lesser than to that of in the deeper, like to that of the dissolved oxygen contained in the deep, deeper water in the bottom of the sea, you can say, correct? So that is how we can conclude that if the temperature of the solution increases, the amount of gas dissolved into the water decreases, correct? So the value of KH increases when temperature increases, correct? And when the temperature increases, the solubility of the gas will be less, okay? And hence the option two is wrong, okay? So correct one is option two. See the example I have taken now, you can easily compare from that for option one and two are contradicting each other. Suppose one is correct, if you assume, which is actually correct because KH increases as temperature increases, I have discussed this in the class, right? Now you see, suppose option one is correct here, right? So option one is what temperature increases, KH value increases. So when temperature increases, KH value increases, that is what the here we have, when value of KH increases, that is only possible when temperature increases. So it means in this option also, the temperature is increasing, so the solubility of gas should be less, right? It should decrease, but the option says as KH value increases, solubility of gas also increases and hence this option is wrong, hence the correct option is option two. The value of KH obviously it is different at the same temperature for two different gases, okay? So option two is right here. Question number 19, 0.05 Faraday charge is passed through a latest storage battery in the anodic reaction, what is the amount of PBSO precipitated, correct? Okay, this is question number 24, okay? If you have the same paper, it should be question number 24, by mistake, I have taken question number 24 here, you see we'll solve this one only, okay? So reaction here it is what PBSO4 we have, correct? So the lead solid combines with SO4 two minus and forms what PBSO4 and two electron goes out, okay? So the question is 0.05 Faraday charge we are taking, right? Now the molar mass of PBSO4 it is given in the question that is 303 gram, correct? Now since two electrons are releasing in this reaction, so what we can say if you apply the concept of this Faraday's law, two Faraday charge, if this mass has to deposit, right? For the deposition of 303 gram, why 303 gram we are taking? Because it is the molar mass of PBSO4. 303 gram of PBSO4 we need, we need two Faraday of charge because two electrons are going out, okay? So two Faraday the mass deposit is 303. So for 0.05 it will be 303 divided by 2 into 0.05. So when you solve this you'll get 303 divided by 40, right? So 303 divided by 40 it will be more than 75, right? 40. So 303 by 4 is more than 75 by 10 so 7.6 around. Option 3 is correct I think, okay? So when you solve this you'll get 7.6 gram approximately will deposit. Option 3 is right, okay? Very basic question, very basic application of Faraday's law. Whatever the values, whatever the charge is going out, equal Faraday will require for the deposition of the molar mass of that particular, you know, substance whatever you have, okay? So option 3 is right. Question number 20 you see, with the following property in a group decreases down the group and increases down the group. Very simple one. I think electronegativity decreases down the group and atomic radius increases. So option 1 is correct. We know, chlorine, chlorine, bromine, iodine. Fluorine is the most electronegative element, right? And atomic radius of iodine is more than to that of all other elements in the same group. So down the group, atomic radius increases, electronegativity decreases. So decreases down the group is electronegativity, increases down the group is atomic size or atomic radius. Option 1 is correct. These are the another 10 questions from 11 to 20 that we have discussed. The last 10 we will also discuss very soon. Thank you.