 Okay, good afternoon, everyone. Guys, please type in your name in the comment box. Let me know first who are out there. Then we can start with the session. Okay, I think some of you have joined. We'll join after some time. So we will start the session now. So today we are going to discuss the question of Redox reaction, okay? So here's the first question of Redox reaction. Which of this following is not a reducing agent? Tell me the answer. What is the answer of this question? Okay, you see, the reducing agent or what? Which reduces other and itself oxidize. Okay, so reducing agent reduce others oxidize itself. So when oxidation takes place, then the element should go under oxidation. The compound should go under oxidation. When oxidation takes place, what happens, oxidation state increases. So we have to check that which element in these compounds is at its maximum oxidation state. Because when the maximum oxidation state is already there, then further increase in oxidation state is not possible. And hence that compound won't act as a reducing agent. Okay, so you see here, H2O2, SO2, CO2 and NO2. First of all, SO2, the sulfur is in SO2, H2O2, CO2 and NO2. These are the four options we have. SO2, sulfur is in plus four oxidation state. H2O2, hydrogen is in plus one and oxygen is in minus one oxidation state. CO2, carbon plus four, NO2, nitrogen plus four. Right, we know for nitrogen, pentavalency is possible. Valency is possible. It means it can go under oxidation if the condition is there. For carbon you see, for plus four is a maximum oxidation state of carbon. Right, maximum oxidation state. Means further increase in oxidation number is not possible and hence it will not oxidize further. Oxidize any more. Because plus four is the maximum oxidation state of carbon. Okay, certainly O2 also, oxygen can go into minus two oxidation state that is also possible. And here sulfur is in plus four, but since sulfur contains deorbital, deorbital, it can show further more oxidation state and we know the compounds of sulfur exist as sulfur SF6. Okay, there are many compounds of sulfur we have which has more than, which sulfur has more than plus four oxidation number. So only CO2 can not go under oxidation. Right, since carbon is in plus four oxidation state which is the maximum oxidation state of carbon and this carbon does not contain deorbital also. That's why it cannot expand its octet, right? It does not contain orbital. Hence, the answer of this question is CO2 which will not oxidize further and it won't act as a reducing agent. Okay, answer is option C. Question number two, what is the answer of this question? We have to find out the equivalent weight of KBRO3 in the following equation. Okay, so we just have to find out, okay equivalent weight we have to find out and the reaction is already balanced. Okay, so we don't have to do anything into it since the reaction is balanced. A number of electron exchange is nothing but the N factor. What is N factor? C, equivalent weight of KBRO3 in the following equation. So the formula we have for equivalent weight is equals to molecular mass divided by N factor. Okay, molecular mass divided by N factor. So N factor in redox reaction in this case will be the number of electron exchanged per molecule, per molecule, okay? So here you see the number of electron exchange is the number of electron exchange is 10, right? And the number of molecules is two, right? So N factor if you calculate that will be equals to the number of electron exchanged divided by per molecule that is two, which is nothing but five, okay? So the equivalent weight of KBRO3 will be its molecular weight divided by N factor, which is five and hence the answer is option D, very simple one. Next question. What is the answer of this question? In the standardization of NA2S2O3 sodium thioculfate we have here using K2CR2O7 by idometry, the equivalent weight of K2CR2O7 you have to find out, okay? So in this question, what we'll do? How the reaction of K2CR2O7 takes place, okay? So in idometry we know this reaction takes place in acidic medium, okay? In idometry the reaction takes place in acidic medium. This information you should have idometric titration, okay? And we know in acidic medium, CR2O7 2 minus it converts into, it converts into CR3 plus in acidic medium H plus. This is the conversion we have. So what we have to do, we need to just balance this reaction and then we'll get the number of electron exchanged per molecule and then we'll get the equivalent weight, okay? So how do we balance this reaction? First of all, we'll balance the atom other than oxygen and hydrogen. So we have two chromium atom here. So two, two CR3 plus oxygen we have to balance and to balance oxygen we'll add H2O. How many molecules of H2O we add? Seven oxygen we have, so seven molecules of H2O, right? Seven molecules of H2O to balance this we'll add H plus on the left hand side. So 14 H, here we have, we should also add 14 H plus to the reactant side. Now we'll balance the charge and how do we check the charge here? Total charge here it is what? 14 positive and two negative, so 12 positive charge and here the total charge is what? H2O is neutral and this is six positive charge and 12 to six we have to go, it means we have to reduce the charge by six. So for that purpose we'll add six electron here, okay? So n factor for this reaction will be what? The number of electron exchanged which is six per unit molecule. We have only one molecule of this, so n factor is six. So equivalent mass or equivalent weight is equals to molecular mass divided by n factor which is six and hence the answer will be option B. Next question you see? Question number fourth, the equivalent weight of KMNSO4 in acidic, neutral and alkaline medium respectively or molecular weights of this is given, okay? So basically we have to find out the n factor in all these, one correction you make. Here they have given KMNSO4, right? So here also we should have KMNSO4, so do this correction. Okay, you see in this question we know KMNSO4 is nothing but MnO4 minus, MnO4 minus. So there are three different medium we have acidic, neutral and alkaline, acidic and then we have neutral and this is alkaline, okay? If you see the oxidation number of manganese here it is plus seven, so we know MnO4 minus in acidic medium converts into Mn plus two, right? In neutral medium it converts into Mn plus four and then in acidic medium it converts into Mn plus six. So here the n factor is what it is, the change in oxidation number, right? N factor is nothing but the change in oxidation state, oxidation number. So for this one in acidic medium the n factor will be five, for neutral medium the n factor will be three and for alkaline medium the n factor will be one, okay? So the equivalent mass in acidic medium that will be molecular mass by n factor which is five. So molecular mass is 158, n factor is five, 31.60. Okay? Equivalent mass in neutral medium that will be molecular mass by n factor is three. So 158 by three, 52.67 equivalent mass in alkaline that will be M divided by one which is nothing but 158. So answer is C, option C. Next question, so of this one the oxidation number of phosphorus in B in this compound and xenon in this compound, oxidation number. Okay, you see the molecule is BA H2O2 whole twice if you look closely. So the area we know it is in plus two oxidation state. So here the total charge on this part of the molecule is minus two. Hydrogen is always in plus one, right? Phosphorus we have to find out so we'll assume X oxygen always minus two, okay? So for this part you see for this part what we'll write only for this part taking for hydrogen two into one plus phosphorus is X plus two into minus two and this is multiplied by two because whole twice is also there and this is equals to minus two, okay? So when you solve this this two minus two becomes minus four. So we'll have two minus two plus X is equals to, is equals to minus two. So X will be equals to plus, yeah, right? Just to check the calculation if it is plus two. So next one you see NA4, XEO6, NA plus one, XEO6, NA plus one, XEI assume X and minus two. So it is four into one plus X plus six into minus two is equals to zero. So when you solve this you'll get X is equals to plus eight. X is plus eight and plus one, okay? So I'll just, let me check the calculation here. VA plus two, so it is minus two, so it is minus two, minus two plus two. Okay, so two plus two into, oh shit, I made a mistake here. You see this bracket is not here. This is actually here, it is present over here. You see this? Okay, so we'll have here also this bracket will not be there and we'll have a bracket over here. So basically we'll have two plus X minus four, right? So we'll have plus two also here and that will be equals to minus one this side. Okay, so when you solve this, so minus four, minus two, minus four, minus two, minus two, minus two will go this side, so we'll get plus one here, plus one. So answer will be plus one and plus eight. So option C is correct, yeah? Next question, in which of the following pairs there's the greatest difference in the oxidation number of the underlying elements. So you have to find out the oxidation number and then maximum difference you have to find it out. What is the answer? See the compound we have NO2 and N2O4. First pair is this, second pair is P2O5 and P4O10. Third pair is N2O and NO. The last pair is SO2 and SO3. N2O4, nitrogen is in plus four oxidation state and here nitrogen is in plus four oxidation state. So difference will be what? Difference here is zero because you have to find out the difference, greatest difference. Here, phosphorus is in, what is the structure of P2O5? Structure of P2O5 is this? Double bond O, double bond O, double bond O and double bond O. Each phosphorus atom you see, it is in plus five oxidation number. Here also it is plus five oxidation state. So here also plus five, here also plus five, right? So difference will be again zero. How here it is plus five? Nitrogen is in plus one, this is plus two, this is plus four and this is plus six. Here the difference is one and here the difference is two, right? Hence the answer for this question will be option D. Next question you see. Where are others I made? There's no one I guess. I was thinking of sending this as a test paper because none of you were there. I have taken almost half an hour, more than half an hour, it's almost first two minutes. Okay. I think only you are there. Tell me the answer to this question. A, you are getting A. A is ions having positive charge, ket ion, correct? These are some of the oxidation number of all atoms in a neutral molecule should be zero, right? So B is four. Oxidation number of hydrogen ion, it's plus one. So I think C is three, yeah. Oxidation number of fluorine in NF minus one, two, and E will be six, right? So A is correct. Next question you see. Tell me this one. Eight one is C. Nitric oxide act as a reducing agent in the reaction. Same as option C, you check the oxidation number of nitrogen atom, both side reactant side and product side. What is the oxidation number? If it is a reducing agent, so its oxidation number should increase, correct? Check that. Yeah, correct. B you can say easily because you see, first of all this iodine is getting reduced, okay? You see the reaction we have two NO plus three I2 plus H2O gives two NO three minus plus six I minus plus eight H plus. So since reducing agent they're asking, right? So you have to look at the compound which is getting reduced first, okay? So which you can easily see iodine from zero to minus one so it is getting reduced, right? So since this iodine is getting reduced, so this is acting as a reducing agent. It reduces the iodine. So oxidation number of NO here is plus two and that of NO three minus here we have plus five. Okay, so nitric oxide act as reducing in the reaction is option B, sorry, yeah, B, okay? Next question. The oxidation number of elements in a compound is evaluated on the basis of certain rules which in the final rule is not correct in this respect. Simon is getting D, Shreyash. Is it plus one in H2OF? You check once, first option is also wrong. First is not correct. And fluorine always have minus one oxidation number. D option is correct. Yes, did you check that? See in case of hydrides, yeah, correct. In case of hydrides, hydrogen also have minus one oxidation state, right? So the oxidation number of hydrogen is always plus one. That is not correct. Okay, it is wrong. But if you see in all its compounds, the oxidation number of fluorine is minus one. A is definitely not correct. Okay, just give me a second as I cross-check this option D. Fluorine, but there's a rule which says fluorine has since it is a most electronegative element. So fluorine always have minus one oxidation state. The fluorine is the most electronegative element. So it is always minus one oxidation state except F2 where it is zero. So in HOF, Hs plus one, O has zero, F has minus one oxidation state. There's no way F will never ever have oxidation state plus one, OF two, oxygen diaprodite, yeah, correct. That's what it is. Oxygen will have zero oxidation state here in HOF. Generally, this charge we only defines or we explains on the base of electronegativity. So in HOF, hydrogen is plus one. This is minus one and oxygen is zero, right? So you must keep this and this is the rule like we have done in assigning the oxidation number of any elements that fluorine being the most electronegative element, its oxidation number is always minus one, right? So if you have to choose between A and D, definitely A is wrong, okay? So we'll go with option A. In somewhere it is also explained this thing, HOF. That's why you are getting confused now. HOF in somewhere they have explained this that fluorine is plus one, oxygen is minus two and hydrogen is plus one. This is also written somewhere in the book, right? But we'll consider this only according to electronegativity. And the rule itself says because it is given, you see the question here, the oxidation number of element in a compound evaluated to the base of certain rules, okay? So rule itself says that fluorine being more electronegative element, its oxidation number is always minus one in all its compound. That's why we'll go with option A. Question number 10. A is correct, zero plus one and minus two. Why it is C? Stress, H2S you consider as H2O only. H2S, similar to H2O, it is minus two and it is plus one. That's why it is minus two, okay? 10th one is A. Oxidation number of oxygen arranged in increasing order. Okay, it is definitely maximum for OF2, okay? And that is given in only one option. See in BAO2, it is peroxide, okay? So here, the oxidation state of oxygen is minus one. It is a peroxide. What is KO2? KO2 is a superoxide, superoxide. An oxidation state is minus half for oxygen. O3 is ozone zero and OF2, it is oxygen diethrodide minus one, so plus two on oxygen. So obviously, the oxidation state of oxygen is maximum in OF2 and minimum in BAO2, right? Hence option A is correct, okay? This is the elements of group one, okay? So this forms oxides of this type. Peroxides of what? Na2O2 type. Na2O2 type peroxide, it forms. And it also forms normal oxides of Na2O type. Elements of group one forms normal oxide of this type, peroxide of this type and superoxide of MO2 type, okay? This you must remember. Next question, is it D? Bromine is getting reduced also and oxidized also, right? So bromine is getting reduced as well as oxidized, so hence it is option D. Can you tell me the oxidation number of bromine here, here and here? Here it is zero, minus one and here it is, plus five, minus one and zero, right? So oxidation number is getting reduced and increase also, that's why oxidation and reduction both takes place. Next one, yes. The largest oxidation number exhibited by an element depending on its outer electronic configuration with which of the following outer electronic configurations, the element will exhibit largest oxidation number. So it is when all these electrons, electrons present in D and S orbital takes part in the bonding, then the oxidation number will be maximum, right? So we have maximum electrons for option D. So when all these electrons takes part in bonding, the oxidation state will be plus seven, here it will be plus three, here it will be plus five and here it is plus six. So maximum is for this, okay? Hence answer will be option D. Yeah, that also you can say, manganese. Okay, so for phosphorus we have to write down, so outer most electronic configuration will be what? Three S two, three P three. This is the outer most electronic configuration. So the minimum oxidation number possible when it accepts three electron and maximum oxidation number possible when it releases all electrons well in shell, right? So minimum possible, this is for minimum, sorry, it is accepts A double C and this is for maximum oxidation number. So minimum will be what? When it accepts three electrons, since it requires three only to gain octet. So minus three will be minimum and maximum will be plus five. This is easily you can see from the electronic configuration. There's one formula of this also that I'll write down here that P block elements for P block elements, the highest negative oxidation number, highest negative oxidation number is equals to the number of electrons in the valence shell, number of electrons in the valence shell minus eight. So when you see these number of electrons in the valence shell is what? Three plus two, five. Five minus eight is equals to minus three and the highest positive oxidation number is equals to group number minus 10. So this phosphorus belongs to which group? 15th group minus 10, that is equals to five plus five. So minus three and plus five option C is correct. 15, what is this I think? With lithium it gives lithium iodide. I think something is wrong here. It should be lithium hydride. Why I'll tell you this? Yeah, there's a mistake in this question. First of all, you see the first option when iodine and hydrogen combines and it forms iodide. So here hydrogen is in plus one, here it is zero. Right? So it is getting oxidized, iodine is getting reduced. Right? Oxidized means what? It is a reducing agent. Third option, this is option A, third option. Nitrogen combines with H2 to gives NH3. We are not concerned with the balanced equation here. If you want to balance it, it would be two NH3 and three H2. It is zero again minus one, sorry, plus one. Zero plus one, correct? So again, you see hydrogen is getting oxidized. So here also it should, it is a reducing agent. Option D you see, sulfur plus hydrogen gives you hydrogen sulfide. Again, hydrogen here plus one. So oxidize, hence here also it is a reducing agent. So here in this option, option B, they should write here lithium hydride, not iodide. It should be lithium hydride. You see the reaction, LI plus H2. It forms what? Two LIH hydrides of first group that is alkali metals. It is minus one and here it is zero. So it is getting reduced, right? It means it is an oxidizing agent. Hence the answer will be option B. Yeah, okay, okay, fine. But the point should be this, okay? Option B is correct. Next question you see. 16th you are getting B. Tell me the oxidation state of nitrogen, both nitrogen atom. Six N minus four, is it? You check again. Is it six N minus four? Five N minus three, yeah, it's correct. It should be five N minus three. See, NH four NO three is an ionic compound. This is an ionic compound. So it actually exists in the form of ion, which is NH four plus and NO three minus. So if you find out the oxidation number here, it is plus five and here it is minus three. So plus five and minus three, not six N minus four. Answer is option B. What is the answer? 17th, sir, will it be in N three H? Okay, okay, we'll see the previous one you were asking, right? Okay, you see this 16th one first. N three H is what? The structure of N three H is this N double bond N, double bond N and then H. This nitrogen will have positive charge and this nitrogen will have negative charge. And this hydrogen is plus one, so this nitrogen will also have negative charge over here. Right charge if you want to see, if a whole this will have one negative charge, like this if I write, it is one negative charge onto this. So oxidation state on this nitrogen atom, if you try to find out, this one X is equals to minus one. So total charge on nitrogen atom is what? Is minus one. So charge on each nitrogen atom is minus one by three. So it is same. They are talking about oxidation state. So oxidation state is nothing but the average value. Okay, oxidation number is the, for oxidation number, we'll check each oxygen atom. Is it clear? So in N three H, the charge on each nitrogen atom is minus one by three. Understood? Okay, 17th one you are getting C. SO2 passed through an acidified solution of potassium dichromate. The oxidation state of sulfur. So we'll write down the reaction first and what is the reaction we have here? SO2 passed over an acidified solution of potassium dichromate. Okay, so potassium dichromate is K2CR207 and this with SO2 in acidified solution. So that suppose we have H2SO4. And the product we get here is what? K2SO4 plus CR2SO4 whole thrice plus SO3 plus H2. This is the product we get. The change in oxidation number is what? Here, the oxidation number of sulfur is plus four and here the oxidation number is plus six. The question is the oxidation state of sulfur changes from plus four to plus six. Hence option C is correct. That's what you also got. 18th one you see? See one thing you must keep in mind, they won't ask you this question directly. The question that we are solving today, they won't ask you these questions. But like questions like, you know, the questions of like electrochemistry, Faraday's law of electrolysis. And then the question of equivalent concept. There you require to find out the change in oxidation number. That's why these questions are important. They won't ask you these directly but application of this you must have, application of oxidation number you must have and other questions. That is why we are doing this. Okay, 18th one, in which of the compounds does mayonnaise exhibit highest oxidation number? Highest oxidation number of mayonnaise is what? It is there in? Oh, we have to find out then. So MnO2 it is plus four, right? Mn3O4 it is plus eight by three. K2 MnO4 plus six and MnSO4 it is plus two. So highest oxidation number if you have to find out that will be in KMnO4 which is two into one plus X minus eight is equals to zero. X is equals to plus six, option C, correct? In MnSO4 it will be plus two. Here plus eight by three, here plus four. So minimum value we'll have what? We'll have in MnSO4, minimum value. This one is off. 19th is D, stress is getting D. Here it is 10 ml of one molar. It is the concentration given, 10 ml of one molar. So we have to find out the oxidation state of cerium in the reduced product, okay? Now you see this Sn plus two releases four electron and goes into Sn plus four, right? So for this the N factor is what? The N factor is nothing but two. And for cerium we do not know the N factor, okay? So what we can write, what we can write that number of equivalents cerium is equals to the number of, okay, we have milliliters. So I'll write down the number of milli equivalent. Here also the number of milli equivalent of Sn, 10, right? So number of milli equivalent of cerium is what? It is the number of moles into N factor, right? And number of moles will be what? It is molarity, number of milli moles. So in molarity into volume, this volume should be in milliliters. Into N factor. This is for cerium is equals to the molarity into volume in ML into N factor of 10. So for cerium it is given that 40 ML volume we have into molarity is 0.5, right? N factor we do not know is equals to molarity is one. Volume is 10 and it's N factor is what? We have calculated two. N factor of cerium if you see will get 0.25. So it is 10, 20, oh, we'll get N factor, we'll get one. Okay, this is the N factor of cerium, right? Means what? One electron is required for the conversion. So C4 plus we have initially. For N factor to be one, we must have one electron. Like here you see two electrons are there, so N factor is what? Two. So for one, we must have one electron and it gives C3 plus. So hence the oxidation state of cerium in the product is plus three answer is option E, correct? Yeah, right, same here. And that should stress just number of equivalents we have to equate. Next question. Yes, that's why I always say, read the question carefully. Okay, don't assume always. Okay, usually what happens once we, if we study two, three lines of the question, half of the question, then we start assuming the question. Okay, the question is this. So don't do this. Finish the question first, read out carefully and then you try solving it. Question number 20, Simon is getting D. Suresh, tell me the answer. All of you are getting D. So I'll just write down HNO3NO and H4Cl. HNO3, the oxidation number is plus five. Here it is plus two. Nitrogen is zero and since it is an ionic compound, so it is minus three, right? So this is the decreasing order. You see it is decreasing plus five to plus two, plus two to zero and then zero to minus three. Hence the answer will be option B, decreasing order of oxidation state. Yeah, next question. Simon is getting D, Suresh, okay. You see this question? You need to draw a structure of this compound. Okay, S4062 minus, okay, S4062 minus. It is actually Na2S406 or you can also compare this with H2S406 when two hydrolyze replaced by Na, you'll get dissolved. The structure, if you see, the structure is like this. S double bond O, O minus double bond O, S, S, S, double bond O, O minus and double bond O. This is a structure. If you draw this for H2SO4, right, then here will be OH, OH. If it is Na2SO4, Na plus, Na plus, correct? So this is the structure we get here. So now in this structure you see because of this double bond, the charge on the sulphur is plus two. This single bond, the charge here plus one. This double bond, the charge here is plus two. So plus two, one, three and two. So here the oxidation state of this sulphur will be plus five. This is one type of sulphur atom here. Similarly, the oxidation state of this sulphur atom is again plus five, right, two, four and one, five. This sulphur atom, it is connected with sulphur itself, both sides. So there is no electronegativity difference. Oxidation state is zero. Here also with the same logic, it is zero. So there are two types of sulphur atom here we have. One atom in which we have plus five oxidation number and other type in which zero. So the difference in oxidation number, oxidation number is five. Yes, why it is zero? Because you see, both sides we have sulphur atom only. So there is no electronegativity difference. That's why it is zero, right? Why I'm taking a charge here because sulphur and oxygen will have electronegativity difference, right? So oxygen will be slightly negative. This will be slightly positive. Since two bonds are there, so these two contributes plus two charge. This double bond contributes plus two charge. This single bond contributes plus one charge. And this bond, since it is sulphur only, so this bond contributes zero here. So two, four, five and zeros plus five. Is it clear? Next one. How it is D stress? And how it is C, simul? Yeah, that's good. So this, actually this information you should have. You cannot do this question, okay? This you should know, you should have this information that when this potassium dichromate that is nothing but CR2O7 two minus in acidic medium combines with more salt. More salt has Fe plus two iron, plus two iron. So it gives chromium three plus. This is, we know already because this chromium is plus six over here. So this converts into CR3 plus in acidic medium. Plus we'll get H2O, seven moles of H2O we get here and then we'll get Fe three plus. And if you see this reaction, we'll get six moles of this and six moles of this. So the ratio should be one is to six, right? One is to six. And hence we can see the number of moles of more salt required per mole of dichromate is six mole answer is option D. This you have to memorize, okay? Because you don't have to write down the equation. That equation you cannot memorize. You can balance, you'll get this six balancing the equation. And for that, you should know what our products are there so that you can balance these equations, right? Or if you know this thing, that Fe two plus converts into Fe three plus like this you see, Fe plus two converts into Fe plus three. This you should remember, okay? So the number of electrons exchanged are one electron here. And if you take this CR two O seven two minus plus H plus gives two CR three plus, right? So if you balance this oxygen, you have to add seven H two of this side. Seven H two means 14 H plus, 14 H plus. The total charge here, it is what? Plus two, so plus 12, right? Here you see the total charge here, it is what? 14 plus and two minus. So 12 positive charge we have here. And here the total charge is what? Six positive charge we have. So to balance this charge, we have to add what? Six electron here, so 12 plus minus six gives you six. Okay? Now when you add these two equation, these two equation if you add, so we have to add in such a way that these electrons get cancelled. So this reaction we have to multiply by six. So we'll get six Fe two plus, six Fe three plus and six electron. And then when you add this electron gets cancelled, we'll get six moles of salt here, right? So that is why you're getting six moles of more salt. So ratio should be one is to six answer will be six. So what you have to keep in mind here, this reaction if you don't remember, you have to understand that in acidic medium more salt Fe two plus converts into Fe three plus. And then you can balance the equation according to the, you know, rule that we have for balancing our redox reaction in acidic medium. Follow that rule. You'll balance these equations and you'll get the mole ratio. You understood this? Any doubt in balancing of these reactions? Any doubt, tell me, understood? Okay guys, we'll do one thing. I'll take a break for 10 minutes. Okay, because I have to take my lunch because after this class, you know, this class we have till two 30, okay? So I have to leave for Rajinagar after two 30 quickly. Okay, so I can't get late. We have some, you know, a mock interview there for KV where we are conducting a mock interview there. Okay, today. So I have to leave at two 30 for Rajinagar. Okay, so I'll take my lunch quickly and then we'll resume the class at two o'clock. Will it be fine? Okay, so you can also take your lunch. We'll start at two o'clock. We'll take a break now. Okay guys, can we resume the class? Okay, next question you see is this. Can you tell me the formula of hematite and magnetite? What is the formula of hematite? See, again, this is nothing, but if you do not know the formula, you will have the doubt. Hematite is Fe2O3 and magnetite is Fe3O4, right? So in Fe2O3, we know the oxidation state of iron is, Fe is in plus three oxidation state, right? But this molecule actually exists in the form FeO.fe2O3, correct? So here it is plus two and here it is plus three. So in magnetite plus two and plus three, right? In magnetite plus two and plus three and plus three in hematite. So answer is option D. Understood? Nothing was there in this question, but the only thing you should know the formula. And this thing we have discussed many times. See, this kind of question, this is informative questions. You cannot put any logic into it. If you do not know this formula or if you have any doubt into this, there are all possible options. You cannot do this question. That's why chemistry at some point it is difficult. If you know, it will take hardly two, three, five seconds. If you do not know, you waste your time also and you do not know whether you will end up with long answer or right answer, okay? That's why these kinds of things you always, you should write down in a notebook, which you can revise in the last day of exam, last three, four days before. Those kinds of preparation actually it is required. Still you have time, okay? You can do that also from now onwards. That will help you a lot. Next question you see. All this question you see, I have chosen this question because these are the information-based question. Maybe they won't ask you this question directly, but this information, maybe it requires to solve questions in Faraday's law of electrolysis or calculation of equivalent weight and all. That's why I have chosen this question. Done? 24th one, you are getting A. Okay, why you took so much of time in this? We simply know that potassium dichromate is K2Cr2O7, right? Which we can write it as in ionic form, Cr2O7 2 minus and Cr2O7 2 minus in acidic medium. It converts into what? 2Cr3 plus, that is it. No, we know this already. You have this information for few molecules. The change in oxidation number you have to memorize. So it always converts into plus three oxidation state. Answer will be A, that is it. You don't have to write down the reaction and all. Understood? Okay, next question you see, this one. Is it B? MnO4 minus is, oxidation respectively are, so one, two, three, four. Okay, so from here to here, plus seven to plus two, change of five we have. So when you got five, these two you can eliminate easily, right? Then here it is plus two and here it is plus six. So plus four, so answer will be B, right? Okay, next question you see, I'll just give you this question. This is not the image I have. For this reaction you see, I'll write down the question. For this reaction, Cn minus, which converts into CnO minus. For this reaction, which of the following statements are true? Okay, so first statement, first option I'm giving. Carbon is losing two electrons per atom. The first option is carbon is losing two electrons, C, sorry, B. Oxidation number of carbon increased from, oxidation number of carbon increased from plus one, two plus three. Oxidation number of nitrogen remains unchanged. A and C, both are correct. Is it D? Okay, I'll do this, C. See Cn minus is nothing but this, C triple bond N, right? In Cn minus we have carbon negative sign bonded with triple bond N. And this convert into O, C triple bond N minus we have here. So minus charge we have an oxygen here and it is on carbon. So if you see the oxidation number of carbon here, it is minus three on this nitrogen. I'll change the color just a second. F minus three on nitrogen, so plus three on carbon, minus three plus three. Why minus here? Because it is more electronegative. So minus three on nitrogen, so plus three on carbon, minus sign we already have on this carbon. So carbon will have plus two oxidation number here. Similarly here, nitrogen will be at minus three only. Triple bond, so minus three, no change. So first of all, the oxidation number of nitrogen remains unchanged. So option C is correct, okay? And C is also there in option D, so we'll check option A also, whether it is right or wrong. Right? So you see here, the oxidation number of carbon here it is what? Minus three is the sides of plus three, minus one the sides of plus one, okay? So carbon here it is at plus four, right? So oxidation number of carbon is increasing from plus two to plus four. It means carbon is losing electrons. Then how many electrons? Four minus two, two electrons. It means what? Carbon is losing two electrons, right? So hence option A is also correct, right? Our answer will be option D. This one is obviously not correct. It is two to four, not one to three. Understood this? Clear? Okay, next question. You see this reaction I'll write down. XM NO four minus plus YC two O four two minus plus ZH plus YC two O four two minus plus ZH plus and it gives XMN three plus plus two YCO two plus by two H two O. You have to find out the value of X, Y and Z. Options I'll give you. It is two, comma five and eight, comma five and 16, five, comma two and eight, five, comma two and 16. Is it option B? Stress? Okay, so what have you done? You have checked the option, right? Tell me, did you check the options? Did you check the options? Did you check the option or you have solved it? Okay, yes. So you see in this kind of question, when they are asking X, Y and Z, okay? So it is the best way to cross check the options as you put all these options here and you check whether the elements are balanced or not. Or if elements are balanced, charge are balanced or not. That also you have to check, okay? So two things you must take care of. There are sometimes, you know, and when elements are balanced, charge will also be balanced, but then you have to keep that also in mind. But sometimes what happens, and when you put this two, five and 16 here, you'll see all these elements will be balanced, okay? That is the way to solve this question when they are asking X, Y and Z. But suppose the option is not this and the question is in this only, if they ask find the value of X into Y, into Z, product of this. Like this, they can form any expression. Like suppose they can also ask you Z into Y by X. They can also ask you Z plus X plus Y. Any combination they can, whatever they want, they can give you. But when you have questions like this and the options are given A, B, C and D, okay? Then it is difficult to cross check those because you don't have exact value of X, Y and Z, right? If you multiply product is some value you have, suppose we have 160, so the factor of this can be anything, right? And it is very difficult to solve those questions with the help of options, right? So the point here is what, when you have questions like this, then you have to balance this equation according to the rule. Yeah, they can ask you this question in the integer type, okay? So the point I'm trying, I knew this thing that you can put this option and you'll get the answer. That is not the big deal. But this is what I wanted to discuss, okay? When they ask you a question like this, this type of question usually they ask in integer type. And even in multiple choice, they can also ask. They'll give you four options, four number they'll give you, okay? So when the question is like this, you have to balance this equation. And to balance this equation, what are the information you should have? The information we should have is what? The reaction is in acidic medium, right? And acidic medium, this converts into this. Accordingly, you can have the rule. This converts into this. One is oxidizing part. Other one is reducing part. Write down the both half equation, balance that equation according to the rule. And then you add, so that the number of electrons gets canceled, okay? And then you will get the value of X, Y and Z. Accordingly, you will find out the expression. You understood this? Is it clear? Okay, guys, I'll wind up the class here only. It is like, it's already the two thirties there. I have to leave for Rajinagar. Okay, so it takes time also. So I'll wind up the class here only. Okay, I hope you understand all these things. These section that I have done today, right? Just you keep those things in mind that in which condition the reaction, what kind of reaction takes place, how the oxidation is state, oxidation number is changing. Like I'm repeating these things again and again, okay? You have application of these thing in other questions. Okay, so must do these questions, okay? So anyways, in the next class, we'll start probably, alkyl halide will do, okay? So we'll wind up the class here only. Hope you understand all these things, okay? Okay, thank you guys. See you soon.