 Can we start over there? Okay, so yesterday we were discussing some questions of atomic structure, right? So we'll continue with that. And we'll see today chemical equilibrium also. We discussed this one, just check once. The first one we have discussed. Yeah, we have discussed this, right? I think we have discussed this. This one is done also. Okay, we'll cancel this. What about this? This one we have. Okay, solve this question just once again. Question number 34. Question number 34. What is the answer? We've done this. This one we have done. Yes, this was the last thing we have discussed. You see this? This is the last page, fourth one, 25. Yeah, that's what. Solve this. We haven't done this. Question number 34. Siamese is getting C. What happened then? Tell me the answer. What is the answer? Tell me 34. Siamese is getting C. It is based on Heisenberg uncertainty principle, right? Heisenberg uncertainty principle, which says the uncertainty in position that is del X into del V is greater equal to or we can say equal to H divided by 4 pi M. So you see the velocity is given 600 meter per second and with an accuracy of 0.005 percent, right? So the uncertainty in velocity will be 0.005 of 600, right? So this will be equals to 0.03. Del X we have to find out and all other things are given, right? Just we have to substitute the value and the value of del X will be del X is equals to H is 6.6 into 10 to the power minus 34 divided by 4 pi is 3.14. Mass of the electron is 9.1 into 10 to the power minus 31 and then del V is 0.03. When you solve this, you will get 1.92 into 10 to the power minus 3 meter. In this kind of question, so option C is correct. In this kind of question, what you have to take care of is the unit. Unit, you must take care of, okay? There only you can make some mistake, okay? So take care of that particular thing, unit. Question number 34, option C is correct. Next one, question number 35, question number 35. Tell me, 35, Ramchandran is getting V, Samir is getting V. Where is Kondinia? Kondinia is getting V. Kondinia, what happened? Tell me the answer. Radius of the first Bohr orbit is X. De Broggy wavelength of electron and third Bohr orbit. Simple one, right? So 2 pi X 6 pi like this, it is given. So what we can write, that we can use the formula, okay? Or we can also write, since De Broggy wavelength, we have to find out, so that is nothing but lambda. And this lambda is equals to H by MV, right? This is what we have to find out, correct? So the, we can use this relation that angular momentum MV Rn, where R is the radius of nth orbit, is equals to the integral multiple of what? H by 2 pi. And H by 2 pi. This is the formula we have. And Rn, we can write MV, we'll write as it is. And Rn will be n square A0 by Z, right? And A0 is nothing but X. So n square X by Z is equals to n H by 2 pi. So we have to calculate H by MV. So H by MV will be what directly we can solve from here. First Bohr orbit is X. So nothing is given, so we consider Z as 1 for hydrogen atom. So we can write here, the square and this n will get cancelled. So 3X by 1, since it is 3rd orbit, so n is equals to 3 into 2 pi is equals to H by MV. So answer will be 6 pi X option B is correct, okay? Option B is correct. Now question number 36. What is the answer? 36? 36 is also B. All of you are getting B? Purvik is getting C. Okay. So you see, what is the formula of radius R? This is equals to an nth orbit 0.529 n square by Z, right? Thanks, Tom. Now the question is which hydrogen like species will have the same radius as that of Bohr orbit of hydrogen atom. So Bohr orbit of hydrogen atom means what? Which orbital we are taking? Bohr orbit means first Bohr orbit, right? First Bohr orbit. So for hydrogen or any write down as it is for hydrogen, this will be what? 0.529 n square by 1. Now if you see this option Li2 plus, Li2 plus Z value is 3, Be3 plus Z value is 4, He Z value is what? 2. Correct? So 1 by 1 if you see n value is 2, so 2 square 4 divided by 3 won't be equal to this, right? What we need if you substitute since they are talking about the Bohr orbit of hydrogen atom, so it means we are talking about the first Bohr orbit. If nothing is written, only Bohr orbit is written, it means we are talking about the first Bohr orbit. So radius is nothing but 0.529, n value is 1, Z value is 1 for hydrogen atom. This is for hydrogen, right? So we have to find out this for lithium if you see Z value will be 3, n value is 2, so 4 by 3 won't be equal to 0.529. For beryllium if you see Rn for beryllium plus 3, this will be equals to 0.529, n value is given which is 2 square and Z value is 4. So you see this and this will get cancelled and this will be equals to 0.529 which is equals to the first Bohr orbit of the hydrogen atom and that's why the answer b is correct, correct? Okay. Question number 37, 37a. Number of radial nodes of 3s and 2p orbital are respectively. What is the formula of radial node? The formula is n minus l minus 1. Number of radial node is n minus l minus 1. Number of angular node is l only, right? So n value is 3 for 3s orbital I'm talking about, n value is 3, l value is 0, minus 1, so it is 2 and for 2p orbital n value is 2, l value is 1, minus 1, so it is 0. So it is 2 and 0, option a is correct, easy one, right? Next question number 38, following quantum numbers, impossible set. All of you are getting A? Why it is A? A is impossible. Why A is impossible? The reason you tell me lies between minus l to l, right? So it cannot be less than minus 2 over here, right? That's why A is impossible. Question number 39, question number 39, 39 and 40 are true. The electron level which allows the hydrogen to absorbs photon but not to emit. Why it is 1s? Lowest energy level you can say and you can also say that for hydrogen atom, 1s orbital only will have electrons, right? 2s, 2p all are empty. Yeah, that is what ground state, lowest energy level, they both are the same thing in a different way you can say. Hence, option a is correct over here, right? 39 A is correct. 40, angular momentum in 2s orbital. Are you getting C? Diameter is getting C, Cymate is getting V. What is the formula of angular momentum? Angular momentum, the formula is what? l, l plus 1, root over of it, divided by 2pi. This is the formula of angular momentum, right? Since the orbital is 2s orbital, so l value is what? 0. And that's why the angular momentum is 0. Why are you getting C? How it is C? Lambshade improving. This is the formula we use. See, angular momentum is mvr also. mvr is equals to nh by 2pi. But this formula we use when the number of orbit is given, like in the second orbit, third orbit, like this if it is given. So n value will substitute accordingly. But if the orbital is given 2s orbital like this, then we use this formula, l into l plus 1, root over of it, h by 2pi, where l is the quantum number, okay? So l value for s orbital is 0. For 2p orbital, what will be the, for 2p orbital, what will be the orbital angular momentum? For 2p orbital, the l value is 1, right? So 1 into 1 plus 1, root over of it, h by 2pi. So it is root 2, h by 2pi, hence option d is correct in that case. The momentum, you will write this, right? Angular momentum will be this. Don't get confused with this. Orbital angular momentum will use the orbital here, l. This is important also. Maybe you will not get any question in JMAIN, but in all of the exams, BITSAT, CET, Manipal, you will get this kind of question. Very common question, every year they ask this kind of questions. You must remember this formula. 41, what about a? Does this violate any rule? Option a, option a, does this violate any rule? Yes, Hohn's rule is violated because what it says that the pairing of electron, what is Hohn's rule? Pairing of electron takes place once all the orbital is singly occupied, right? So here you see this orbital is vacant and the pairing takes place. So this option a violates Hohn's rule, right? Here, what is of the principle that the electron fills into the orbital according to their energy in increasing order, right? So lowest energy orbital fills first and then the higher one, right? So 2s orbital obviously has whatever this orbital, this obviously has the lowest energy than this. This should be filled first, right? So we should have two electrons here, but we have only one and here it is paired. This particular option violates oboe principle. Is there any violation of any rule in these two options, C and D? Is there violation of any rule in C and D option? No violation, okay? C and D is perfectly right. 42, all of you are getting A. What is this orbital? First one, what orbital we have here? First of all, increasing energy, we have to find it out. So we'll just add n plus l, right? So here the first one, it is 5, here it is 4, 5 and 4, right? So equal number of n plus l value, then we'll check the n value, right? So first one will be maximum, then we have third and then we have second and then we have fourth. First, third, second, fourth. Option a is correct. Right? So this is 4p, this is 4s, this is 3d and this is 3p. Correct? Option a is correct in this. Next. Now you see these questions have been asked in J exam, 50 and 51. In option a it is minus 34, option b it is minus 38, option c minus 31 and minus 30. ABCD in order I'll tell you, minus 34, minus 38, minus 31, minus 30. 10 to the power minus 38 you are getting. So it's option b. ABCD, the power of 10 is minus 34, minus 38, minus 31, minus 30. Rithvik is getting d and others are getting b. Ramchandran is also getting b. We have length of a car of mass, this velocity, this h value is here. So we have to only do the calculation in this, okay? So nothing we have to, there's no logic here. Lambda is equals to h by mv you have to use all the value you have to substitute and get. The answer option b is correct. 51, 51, how did you do? Purvik, how did you get the answer? b. So you got 6. Ramchandran how did you get b? See for the first one it is given n is equals to 5. So l can be equals to anything from 0 to 4, right? Which can be 0, 1, 2, 3 and 4. So if l is 0, m will be what? 0. If l is 1, m will be minus 1, 0 or plus 1. If l is 2, then again m will be minus, minus 1, 0, plus 1, plus 2. If l is equals to 3, m is equals to minus 3, minus 2, minus 1, 0, plus 1, plus 2, plus 3. l is equals to 4, m is equals to minus 4, minus 3, minus 2, minus 1, 0, plus 1, plus 2, plus 3 and plus 4. So where we have, the condition is m is equals to plus 1, right? So we have plus 1 here, right? So 2 electrons will have here, plus 1 here, 2 electrons will have here, plus 1 here, 2 electron here and plus 1 here, 2 electron here. So we can have only 8 electrons possible in the first case. Yes, correct. See you cannot, no, no, no, you cannot say p or d because p and d we can say when you have, when you know the l value, okay? So if l is equals to 1, then it is p. If l is equals to 2, then it is d. If l is equals to 3, then it is f and then g. So any of these orbital is possible, 5p, 5d, 5f, anything. But the point is whatever it is, we are talking about one particular orbital because it is plus 1. If suppose we have p, so we have px, py, pz, right? pz, px, py, like this. So out of p also, we are talking about only plus 1, correct? So this we can take only 2 electrons, right? 2, 2, 2, 8 electrons here. Is it clear? If you understood. So we have to only check for the orbital which has plus 1m value. So for all these possible value, we have to select only plus 1, right? So here we have plus 1, plus 1, plus 1, and plus 1. And each of these orbital can have 2 electrons, 2 into 4, 8 electrons possible, right? What about the second one similarly? n is equals to 2. So l value can be anything from 0 and 1. These two value possible? If l is equals to 0, m is nothing but 0. If l is equals to 1, m is nothing but minus 1, 0 and plus 1. So here in this question, n is equals to 2, l is equals to 1, m is equals to minus 1, and s is only 1 value minus half. So only 1 orbital possible, n, l, m, s, so 2 electrons. And then possible, so it is equals to 1. 1 is the answer. Why is 1, why 1 is the answer here, why not 2? One particular orbital we are talking about, right? So why 1 is the answer? Just a second. Okay, because s value is given, right? s is minus half and the other electrons in the same orbital will have s value s plus half, right? So if s value is not given, then the number of electrons, possible electrons will be 2. So in any orbital in this kind of question, if s value is given, like how many electrons can have these quantum numbers? If the question is this, with s value is given, then the answer will be always 1. One electron can have only one particular value of s, okay? Now question number 52, question number 52, number of the first emission line in the Obama series, number is nothing but 1 by lambda. So option a is correct, all of you are getting, that's why I'm not solving this. 53, in atom, how many orbitals will have the quantum number this? 3, 2, m plus 2. Is it c, only one orbital? c is correct. You see, these are the questions of J main, 54 and 55. 54 and 55. 54 is d, 55 is b. 55 is again some same question. 55 b is correct, 1 it is 5, is the same question actually it is. 54 is d, we use here, in the 54th one, will be this, e is equals to 2.178 into 10 to the power minus 18, z squared, 1 by ni squared, n1 squared, minus 1 by n2 squared. z value, it is given for hydrogen atom, z value is 1, n1, n2 is given, n1 is 1, n2 is 2. To substitute this value, you will get energy and that will be equals to hc by lambda. So this lambda we have to find out, h and c value it's also given. So lambda you will get around 1.214 into 10 to the power minus 7 meter, unit will be this. So option d is correct, again this is just the calculation. Solve this question. This is equilibrium now, c. What do we, I'm just going to continue, tell me the answer. So I'm here and Rithik is getting c, all of you are getting c. So in a chemical reaction, the rate constant for a backward reaction is this. So kb is given 7.5 into 10 to the power minus 4 and the equilibrium constant kc is also given, that is 1.5. So we have to find out the rate constant for the forward reaction. So we know this kc is equals to kf by kb, right? So kf is equals to what? kc is 1.5 and kb is 7.5 into 10 to the power minus. So when you solve this 1.125 into 10 to the power minus 3 you will get option a is correct. Question number 2. Question number 2. What is the answer? Question number 2. Still doing 0.118 you are getting. Okay, what about others? Amir is getting option b, is it? Tell me. Okay, see first of all we have moles of N2 that will be 56 divided by 28, 2 moles, the moles of H2 will be 6 divided by 2, 3 moles. The moles of NH3 we get here is 27.54 divided by 17. Okay, so once you divide this the moles of NH3 that we get here is 1.62 approximately. Now the reaction we have is N2 plus 3H2 gives 2NH3, correct? Initial number of moles of N2 we have 2, H2 we have 3 and NH3 we have 0 initially. So Amir is getting c. So it is 2 minus x, suppose it dissociates, 3 minus 3x and then 2 minus, sorry, this will be 2x. So the value of 2x is nothing but 1.62 because this is what we are getting NH3. So x is 0.81, right? So the equilibrium concentration of N2 will be 2 minus x, 2 minus 0.81 which is 1.19. Equilibrium concentration of H2 is 3 minus 3 into 0.81. Where it is 2 minus 2x? Where it should be 2 minus 2x? No, we have 1 mole only here, you see, 1 mole of N2. So 1, so it is x, 3, so it is 3x, 2, so it is 2x. What is the equilibrium concentration of H2 you are getting? Sime here, what is the equilibrium concentration of hydrogen? Tell me this value, 0.57. So this will be 0.57 and then NH3 we have already, 2 into 0.81. So Kc is equals to what? 1.19 into 0.57 cube divided by 1.62 square. Is it, the expression is correct? This expression is correct? 0.57 is. The answer you are getting here once you solve this, tell me the answer. Oh sorry, it should be product by reactant, correct? So I will write here 1 by Kc, okay. What is the answer of Kc you are getting? Kc you are getting 11.9, some mistake we have made or what? One mistake we made. Can you finally doubt what? What? The equilibrium constant that we calculate, there we make some mistake. Just a second. Just a second. Let me cross check the answer first. Equilibrium concentration of N2 is fine because x is reacting, x is nothing but 1.16. This gram of the 1 liter vessel, the equilibrium mixture contains this gram of NH3. This gram of NH3, equilibrium mixture contains this gram of NH3, means the mole of NH3 will be this. Can you check this value? What is 27.4 divided by 17? Is it correct? 1.62, 27.4 divided by NH3, is it 1.62? You check once, approximate value of Kc for our reaction in this, okay. Okay, it's 1.62, so it's fine. So number of moles of NH3 we are getting, volume is 1 liter, so whatever this is nothing but the number of moles. So 2x is 1.62, so x is 0.81. So when x is 0.81, so equilibrium concentration of N2 will be 2 minus 0.81, which is nothing but 1.19. This is correct? Equilibrium concentration of H2, 3 minus 3x only, which is 0.57 and 2 is 1.62. This is what we are getting. I think it's wrong then. Yeah, it's wrong. There's a mistake they have done. They have actually taken here, I saw it now, they have actually taken the x value 0.18 by mistake. What they have done? They have taken here 3 minus concentration of H2 what they have written, 3 minus 0.54, which is 2.46. Okay, so the Kc expression is this, 1.62 square divided by 1.19 into 2.46 cube. When you solve this, you'll get 0.148. You can check. Nostril, when you solve this, you'll get 0.148, which is the option we have. But what mistake they have made, they have instead of 0.81, they have taken 0.18. So that this 0.18 into 3 becomes 0.54. So 3 minus 0.54 is 2.46, that is what they have taken here, which we cannot take actually. So there is a mistake in this question. The correct answer will be whatever we are getting, 11.1. Is it clear? You can cross check whatever expression I have given you here. According to this, you are getting auction C. But in this, the concentration of H2, equal concentration is not right. It's wrong. Did you understand? Cross check this expression, this expression you'll get auction C. But in this expression, the concentration of H2 is not right. That is what I'm telling you. They have taken 0.54 instead of 0.81, they have taken 0.18 by mistake probably. That's why the answer Kc will be 11.9, according to our expression. Is it clear? Okay. Can we move on? Question number 3. Tell me question number 3. The concentration of CO2, which will be in equilibrium with this molar of CO at 180 Celsius for Kc is BLC. Kc is equals to concentration of CO2 by concentration of CO. Right? Just to show you have to take third one, B is correct. Just you have to write it down. Kc is equals to concentration of CO2 by concentration of CO. Right? Co is given, Kc is given, CO2 we can find it out, option B is correct. Question number 4. Just a second I'm coming. Okay. What is the answer question number 4? 4, D and C. You're getting C, PR, gas, LO, BR2 at equilibrium, pressure of BR2 is given and P is the total pressure, the ratio of Kp by P. It's easy. Ramchandran, what happened? All of other unwriting C, you are getting B. Expression for Kp, for this reaction will be what? Pressure of NO, the square, pressure of BR2 divided by NO, BR, the square. So pressure of BR2 is P by 9 at equilibrium. So pressure of NO will be what? Pressure of BR2 is P by 9. So pressure of NO will be 2P by 9. Okay. So we can write pressure of NO is 2P divided by 9 square. BR2 is P by 9 divided by pressure of NOBR. What is the pressure of NOBR? It is, okay. I'll write down the reaction here first. The reaction is NOBR gives NO plus BR2, all our gases, 2 and 2. So okay. Total pressure is given P, right? So for this will be what? Total pressure minus the pressure of NO and BR2, right? Which is nothing but 2P by 9 plus P by 9 will get the equilibrium pressure of NO directly. Total pressure minus the pressure of these two. Okay. So when you solve this, you'll get 3P P by 3. So 2P by 3, I guess. So it is 2P by 3. So we'll write here 2P by 3 square. This is equals to Kp. So when you solve this Kp by P at 1 by 81. Option C is correct. Yes? Question number five. Question number five. Question number five. You're getting C 1.6 into 10 to the power minus 23. You're getting 10 to the power minus 23. How it's possible? Check once, check your calculation. You must have missed the power. 6HCHO is there proving. So that will go into the power, right? So HCHO to the power 6 you will get in the expression. You must have missed that. The easiest to form glucose according to the reaction, the theoretical computed equilibrium constant for the reaction is this. Okay. So Kc for this reaction will be C6H12O6 H12O6 divided by HCHO to the power 6. So this Kc value is given, which is 610 to the power 22 is equals to 1 divided by HCHO concentration to the power 6. So when you solve this, you'll get the concentration of HCHO will be 6 into 10 to the power 6 into 10 to the power 22 to the power 1 by 6. This is what you have to solve. So answer you'll get here is option B 1.6 into 10 to the power minus 4. Option B is correct. Glucose will consider as solid. That's why we have taken 1 over there. Okay. Option B is correct. Next one. Tell me this A gives NB. The reaction is one more thing I'll tell you here. This is 0.2, which is not mentioned here. And this is 0.6. This is here. Kondinia is getting B. What happened? Tell me the answer? 6-1. You see in this question, according to the graph you see A gives NB. So the change in concentration of A del A minus of del A will be equals to what you see here from this 0.5, 0.5 to 0.3. This is for A, right? The change in concentration will be what? Minus of 0.3 minus 0.5. So that will be equals to 0.2. Similarly, the change in concentration of B will be what? From 2 to 0.6. So 0.6 minus 0.2. That will be 0.4. So you see A changes by 0.2 and B changes by twice of that. So obviously the reaction will have the ratio of 1 is to 2. So A gives 2B. So N value must be 2 here. So option B is correct. Now the equilibrium constant of KC, when you know this reaction, it will be KC is equals to option concentration of B square divided by A. So that will be 0.4 square divided by 0.2. That will be 0.8. This is again. 0.5, 0.3 is reacting. So concentration of B will be 0.6 here. 1.6 into 6 divided by concentration of A is, A gives 2B to 0.5 and 0. 5 minus 0.2 gives you 0.3, which is a concentration we have here. And this gives you 2 into 0.2, 0.4. So KC will be 0.4 square divided by 0.3, 0.4 into 0.4 divided by 0.3. So 16 divided by 30. So approximately 0.5 we are getting. Approximately 0.5, more than 0.5, 0.51 or something we get. Oh, just a second. Initial concentration of A is 0.6, right? So here it should be 0.6. 0.6 and final concentration of A is 0.3. So this should be 0.3. 0.3 into 2. Here we have 0.6. So we have 0.6 square divided by 0.3. 0.6 into 0.6. So we are getting 1.2. 1.2 option D is correct. See, actually this question is based on a graph. See, I didn't, see, actually where is the point of equilibrium? This is the point of equilibrium, right? Here the rate is not the concentration of B and A is constant after this point, correct? And that is what the condition of equilibrium, the concentration won't change of A and B. So this is the point of equilibrium where the equilibrium has been established, right? Now here you see these two values are not given. This is 0.2 and this is 0.6 that I have given you. It is not mentioned in the graph, but it is 0.1, 0.2, 0.3, 0.4, 0.5 and 0.6. Now from this graph you see the initial concentration of A is 0.6, which I have missed over here. I have corrected now. Initial concentration of A is, if I write down here, A0, initial concentration of A is 0.6 and when the equilibrium has been established, the concentration of A at that point of time is 0.3. Initial concentration of B is 0, right? And since we have to find out the changing concentration of B, so we required the concentration of B at any two point, okay? So one point is this and another point is this, where the equilibrium has been established. So the change in concentration of B is what? 0.6 minus 0.2, that is 0.4. Change in concentration of A is what? Any at two point, right? That will be 0.5 to 0.3, which is nothing but 0.2, right? So the difference in the concentration of A and B is in the ratio of 1 is to 2, right? 0.2 and 0.4. That's why this mole ratio should be 1 is to 2. That is what the first point, right? Now coming back to this graph again, the second question, the equilibrium constant Kc of the ever graphical study is. Now for this, we required the change in the equilibrium concentration of A and equilibrium concentration of B, so that we can find out Kc, okay? So initial concentration of A is 0.6 and from 0.6, the change is 0.3. So 0.6 minus 0.3 gives you final concentration of A is 0.3, which is easily understood from this graph. Now when 0.3 to get the final concentration of A is 0.3, from 0.6, 0.3 must react, right? So this is nothing but x we have in this case. 0.6 minus x gives you 2x here because the ratio is 1 is to 2. So x is 0.3, so 2x will be 0.6. This is the equilibrium concentration of B. So what we can write Kc, concentration of B square divided by concentration of A at equilibrium, that is what I have done. Understood? Yes, tell me, can we move on? Question number seven. Seven we have done, question number eight. Question number eight, you're getting A. How it is A? How did you get A? Poor thing.