 Hello guys, good afternoon. Hello everyone, Aditya, Ranganath, Krishna. Are you getting me? Can you see the screen now? Hello everyone, please type in your name in the chat box. All of you, please type in your name in the chat box. Krishna, Ranganath, Shamik, Kirtana. Hello Kirtana, how are you? Sriramagiri, Aditya, Dhyansheyash, Bala Murugan, Ranganath. Guys, all of you, please type in your name in the comment box. We'll start the session very soon. Skanda, Sahana, Ananya, Anurag, Ravi Keden, Ronak. Okay guys, so this session is a YouTube live session. Okay, in case of any doubt, you can use your chat box for whatever doubt you have. Okay, so you cannot, no, I'm not hearing you. Right, so whatever doubt you have, you write it down in the comment box. Then we can discuss that. Right? Okay, so I guess most of you have joined already, Ronak, Vibhav, Sriramagiri, Vinaya, Vinaya, Ananya, Abhishek, Srishti, Aditya. Okay guys, so last class I guess we have discussed about the balancing of redox reaction. Yes or no? We have discussed about the balancing of redox reaction, guys. Okay, so any one of you have any doubt in the balancing of redox reaction? Any doubt if you have, you can ask me here. We'll discuss that first and then we move on to the class. All of you understood how to balance the reactions, redox reaction in different, different mediums, whether it is acidic or basic medium. Okay, so right. Okay, so the first thing that we are discussing this redox reaction to understand the calculation of N factor in reacting condition. Right, because N factor like I discussed already is nothing but the valency factor, valence factor. Right, so if you have to find out the equivalent weight, okay, and whatever guys I'm writing down here on the screen, you have to write it down on your notes. Okay, so copy the notes properly like you always do in the class, offline class. Okay, so like I said equivalent weight to calculate equivalent weight, this is equals to the molecular weight of the given compound and divided by the N factor or valence factor we call it as. This is the formula we have. Correct? Now N factor, valency factor in, in case of acid it is basicity, in case of base it is acidity, in case of salt it is either positive or negative charge, total positive or negative charge, all these things I have discussed. Okay, those who are not there in that particular class you can go through your notes. Okay, it's given there. Right, N factor is nothing but basicity in case of an acid for acidity in case of a base and when you have a salt it is nothing but the total positive or negative charge, correct? But I have also told you this thing that in reacting condition to understand the N factor you should know what is redox reaction, correct? Today in this session we are going to discuss the calculation of equivalent weight or N factor we can say because when you have N factor you can calculate equivalent weight. Okay, so basically in this session we are going to understand the calculation of equivalent weight or N factor in reacting condition. Okay, so first of all all of you write down the heading here that is calculation of, calculation of equivalent weight reacting condition, reacting condition, right? This is the first topic we have here. Guys, again I am telling you in case of any doubt, okay I am not hearing you, right? So with all honesty I expect in case of any doubt you use your chat box, write down your doubt, okay that only I can respond, okay? So in case of any doubt write it down in your chat box we will respond to that, okay? So first thing in case of when we have reacting condition what we write in reacting condition, in reacting condition all of you copy down this the equivalent weight, the equivalent weight in a compound for a compound, for a compound may change, may change as per the reacting condition, as per the reaction basically alright, okay? As per the reaction, may change as per the reaction, okay? So depending upon the reaction what is the change in equivalent mass and all those we have, change in oxidation number we have, according to that the equivalent weight will change or N factor will change, okay? And when this N factor changes, see here, when this N factor changes, right, equivalent weight will also change for the molecule, right? So first of all I will write down one compound here, one reaction here, right? And the reaction is this, suppose the first reaction I am writing it down, it is S2SO4, S2SO4 plus 2 NaOH, it gives Na2SO4 plus 2 S2O. Guys I know there was a noise in the background because of some work is going on in the office, okay? It will be fixed in few minutes, okay? So you have to understand that. So this is the reaction we have. Now we know one thing, when the two compounds reacts, they must have equal number of equivalents, correct? So this is the next point I will write down here, the next compound is, you write down the next note over here, that and I have discussed this, right? That when the two compound reacts, they must have, they must have equal number of equivalents, they must have equal number of equivalents. This thing we know, right? This is very important point we have. And because of this only, like if I give you a quick recall here, we have discussed about NH3, right? The equivalent mass of NH3 will be what? That will be the equivalent mass of nitrogen plus that of hydrogen. And how do we calculate that? That we know nitrogen and hydrogen are combining. So the number of equivalents of nitrogen must be equals to the number of equivalent of hydrogen, okay? Similarly here also since H2SO4 and NaOH is reacting, the number of equivalents of H2SO4 should be equals to the number of equivalent of NaOH, okay? N factor for H2SO4 we know, right? N factor for H2SO4, it is an acid and N factor for acid is nothing but its basicity, which is nothing but the number of replaceable H plus ion, N factor of H2SO4 is 2. N factor of NaOH we also know and that N factor is 1, right? What I said the number of equivalents of H2SO4 and NaOH must be same, right? This one we can understand from this. You see first of all this is again very important formula of calculation of number of equivalents, okay? So what I write the formula of number of equivalents is equals to what? Is equals to the given mass divided by equivalent mass. Anyone has any doubt in this particular formula? Let me know. If it is clear then please comment, then please comment CLR in the comment box. If it is clear then please comment CLR in the comment box. If you have any doubt in this particular formula, let me know otherwise please comment CLR in the comment box. All of you guys quickly. What do you mean by what is equivalence you are asking, right? Number of equivalents is nothing but see the number of equivalents I have discussed all this, okay? So I won't go into detail of all this. I'll just give you a quick revision of it. Number of equivalents is the amount that combines with 1 gram of hydrogen or 8 gram of oxygen or 35.5 gram of chlorine or 108 gram of silver. This one I have discussed in the first class Abhishek, okay? So if you are there then please revise your notes so you will understand what is one equivalence, okay? And equivalent mass is the mass of one equivalent, right? That's the thing. So you know this equivalent mass is nothing but the molecular mass divided by n factor. So this expression we can further write, this expression we can further write given mass, given mass divided by, equivalent mass is nothing but molecular mass, molecular mass and this is divided by the n, okay, in the denominator, okay? Further this expression becomes, this expression becomes, you see, we can write given mass divided by, given mass divided by molecular mass into n factor, right? So given mass by molecular mass is nothing but the number of moles, correct? Moles into n factor. All of you copy down this. So the formula of number of equivalence, formula of number of equivalence is equals to the given number of moles into n factor. Is it clear? This formula you understood? Please let me know. Yeah, it's look nice, right? So this is very important formula actually, number of moles, number of equivalence is equals to the moles into n factor. Now I am coming back to this reaction which I have written over here, right? Now you see the reaction is balanced, reaction is balanced, correct? So what is the number of moles of S2SO4 here? If you see the number of moles of S2SO4, here it is nothing but one mole, one mole of S2SO4 reacts with two moles of NaOH, right? So according to this formula, if you calculate the number of equivalence of S2SO4, that will be the number of moles into n factor. Two into one is nothing but two. So we have two equivalence of S2SO4 and here also two mole into n factor, that is one, two equivalence of NaOH. So again you see the number of equivalence, the number of equivalence of S2SO4 here it is two equivalence and similarly for NaOH also it is two equivalence. The same thing we have that when two compound reacts, they must have equal number of equivalence. Got it? Guys understood? Can we move on? Can we move on? Please let me know, guys. Anyone of you, if you feel any difficulty, please let me know. Yeah, hi Arpeetha. Yeah, I am going a bit slow today guys because this is your first online session, right? So I want all of you to be very comfortable with this. Okay, Sahana is there, Sahana are you there? Did you understand this because you are not there in that class when I discussed all these. Sahana if you are there, let me know please. Yeah, that's good. Okay guys, so two things again I want you to understand here. Yeah, okay. So you see this S2SO4, I have taken the n factor to here, right? Because, because both the hydrogen is taking part in the reaction and that is why we get Na2SO4 here plus 2S2O, right? That's why I wrote n factor for S2SO4 is 2. The basicity of S2SO4 is 2 but it is possible in a given reaction condition that its basicity can be one also, right? So that is the next thing that we are going to discuss here. So I am moving to the next slide here. Suppose the second reaction I will write down and which is this. Suppose the second reaction I am writing down here which is H2SO4 plus we have again NaOH, NaOH and this converts into NaHSO4 plus H2O. The reaction is balanced guys, please let me know this whether the reaction is balanced or not. Is it balanced? I guess yes, right? Quickly. So now the next thing can you tell me in this reaction, what is the n factor of S2SO4? What is the n factor of S2SO4? n factor of S2SO4 guys, Srishti, Aditya, Krishna, Kirtana, Dhyan, all of you are saying 2, Sharmik is saying 1, Areman is saying 2. Is it 1 or 2? Skanda is saying 1, Sahana 2, Ronak 2, Prajna 2. Okay, most of you are saying 2, few of you are saying 1. Okay guys, so answer for this thing is 1, right? Answer is 1. Why? Because out of the two hydrogen here, only one hydrogen is taking part in the reaction and that is why it is forming NaHSO4. The previous reaction you see, the previous reaction, the product is Na2SO4, which means both the hydrogen is taking part in the reaction and sodium displaces the hydrogen from here S2SO4 and forms Na2SO4 in the first reaction. But since we have one NaHSO4 here, it means this we can write what? H plus plus HSO4 minus C, other things you try to understand here. Okay, I will just write it down here. See H2SO4, H2SO4 has basicity 2. It means it can displace maximum two hydrogen. But depending upon the reaction condition, it can displace only one hydrogen also. Like for example, this H2SO4, one possibility is what? It will, you know, it dissociated as, it may dissociate as this. It gives you 2H plus and SO4 2 minus. This is one possibility, which was happening in the first reaction. Second possibility is what? It gives you H plus and HSO4 minus. That is also second possibility we have. So when the dissociation is like this, it means its basicity is 2. N factor is 2. Because 2H plus is coming out in the solution. But where, here we have only one H plus. So here the N factor is 1. Because only one H plus it is releasing. Correct. So when you look at the reaction, the first reaction, it is because of this kind of dissociation. And this reaction is because of this dissociation. So when you have H plus and HSO4 minus, this N A plus combines with HSO4 minus and forms N A HSO4. And H plus combines with OH minus, forms H2O. That is why we are getting N A HSO4 here and H2 here. This product is possible only when this H2SO4 releases one H plus ion. And that is why its N factor is 1 here. And for N A OH, its N factor is again 1. Because it is releasing OH minus ion. N factor is again 1. So if you see the number of equivalents of H2SO4, since it has one mole of H2SO4 reacts with one mole of N A OH. And that is why the number of equivalents of the number of equivalents of H2SO4 is one equivalent here and the number of equivalents of N A OH is again one equivalent here. Because they must have equal number of equivalents. The formula we have N factor into number of moles. So 1 into 1 is 1 and again 1 into 1 is 1. So both has one equivalence here in this reaction. Is it clear guys? I am sorry just a second. So I will write down the reaction was this. The second reaction was this. I am sorry. The reaction was this H2SO4 plus N A OH. It gives N A HSO4 plus H2O. The number of N factor of H2SO4 is one here. I have discussed this already. N factor is one. We have one mole of H2SO4 and one mole of N A OH. Hence when you calculate the number of equivalents for H2SO4, we have one equivalent and same for N A OH. We have one. So with these two examples, we can surely say this that in reacting condition the N factor or the number of equivalents for a given compound depends upon depends upon that what kind of reaction is taking place. Generally when you have H2SO4 present like this that its N factor is 2. But in reacting condition its N factor depends upon the type of reaction that we have. Is it clear? So far the formula of number of equivalents, we have discussed three different formulas. To sum up this, I will write down all those formula and you must write it down because all these formulas are very important. So three different formula of number of equivalents we have discussed. So what we can write the number of equivalent is equals to the first formula we have is the given mass divided by equivalent mass which is again equals to normality into volume in litre and the third one is the given number of moles into N factor. These are the three different formula of calculation of number of equivalents. This is the formula we have. Very important formula. So if you have normality, see this, if you have normality given and volume given then we use this formula N into V gives you number of equivalents. Given mass by equivalent mass we can find out number of equivalents. If number of moles you know and N factor if you know then also you can find out the number of equivalents. Now one more thing here you see when I write one mole and one mole, this you have to be a bit careful over here. To write the number of moles the reaction must be balanced. That is the condition we have. If the reaction is unbalanced, if it is not balanced then we cannot write down the number of moles here. The exact number of moles. You can write it down but you will get the wrong answer. To know what exact moles are reacting for that the reaction must be balanced. One or two more reactions will see. To understand this particular thing one more reaction will see. You see this reaction the third one I will write down. Suppose we have Na2CO3. You have to solve this guys all of you. Na2CO3 plus HCl 2HCl and this gives 2 NaCl plus H2O plus CO2. This is the reaction we have. Now you tell me what is the number of equivalents of Na2CO3 we have because both will have the same number of equivalents. Tell me this thing. All of you understood my question. Done guys. And number of equivalents you are saying two equivalents. All of you are getting two. I think all of you are getting two. Most of you are getting two. Swamig is getting one. Swamig is getting one. See again Na2CO3 we have. So it is a salt. It is a salt. So its n factor is what? For salt the n factor is a total positive or negative charge. So for this the n factor is 2 and for HCl the n factor is 1. So number of equivalents for this equals to what? We can write number of equivalents from this formula because we know n factor and we know the number of moles. The reaction is balanced. For Na2CO3 we have one mole. The number of equivalents is what? Moles into n factor which is 2. So we have two equivalents of both Na2CO3 and HCl. Answer will be 2. Next one is you write down the equivalent weight. Second condition we are talking about. This is one part. Now the second one you write down. We have to discuss this also. The equivalent weight of oxidizing and reducing agent. For different different cases we use different different method to find out the equivalent weight in a reaction. In this there are three methods basically. To find out the equivalent weight of an oxidizing or reducing agent there are three methods. So we will discuss all three. Whatever you feel easy you can use that. So you see suppose I write down the reaction and the reaction is this. We have K2Cr2O7 which is an oxidizing agent. K2Cr2O7 guys it is a strong oxidizing agent. Plus 4H2SO4. This is the reaction it will be given in that question. You don't have to worry with the reaction. It is K2SO4 plus Cr2SO4 whole thrice plus 4H2O and you will also get some nascent oxygen. This oxygen whenever you write oxygen like this it is we call it as nascent oxygen. This oxygen is what it is nascent oxygen. Now suppose this reaction is given. We have to find out the equivalent mass of K2Cr2O7 because this is the oxidizing agent. K2Cr2O7 is the oxidizing agent. We have to find out the equivalent mass of this. So what we can do you see whatever the molecular mass of K2Cr2O7 we have suppose that is M molecular mass of K2Cr2O7 is M and that we can find out M which is nothing but if you write down the exact value it is around 294.2 gram is the molecular mass of K2Cr2O7. Is it single oxygen atom? No it's not single. Also you calculate you count the number of oxygen atom. This side it would be 7 plus 16, 23 oxygen this side we have and this side we have 4 plus 12, 16 plus 4, 20 this must be 3 oxygen. I will discuss what is nascent oxygen. First of all you understood this. We should have 3 oxygen atom here. Now what is nascent oxygen is? See oxygen is a diatomic molecule. So in nature it exists in this form O2. It exists in O2. But how this oxygen molecule forms that's the question. So nascent oxygen is the oxygen atom when it is actually formed. What happens actually first we get one oxygen atom and then we get another oxygen atom. So whatever oxygen atom we are getting the time when it forms the moment when the oxygen atom is formed that's a second. So the moment oxygen atom forms at that particular time it is known as nascent oxygen. Clear? Now oxygen atom that's a second again. Now oxygen atom is actually not stable in its atomic form. That's why it converts into molecular form and it becomes diatomic molecule. Now the question is why oxygen atom is not stable. That is a different part we will discuss that later. But what is nascent oxygen? The answer is the moment oxygen atom forms at that moment it is known as nascent oxygen. And it combines with the other oxygen atom and forms a diatomic molecule which is O2. Correct? I hope it is clear now. Yes, 50. Now the next thing is what? Our purpose is to find out our purpose is to find out the equivalent mass of K2CR2 of 7. Correct? So you see whatever molecular mass of this that is 294 gram when it reacts in this condition it gives how many gram of oxygen atom, nascent oxygen atom. If you calculate the mass of this atom, oxygen atom here it will be 3 into 16. That will be 48 gram of oxygen atom we are getting. So what we can write here in the next line is 40 of oxygen, 48 gram of oxygen formed by m gram which is nothing but the molecular mass which is 294.2 gram m gram of K2CR2 of 7. Right? For 48 gram we require this is the mass for 1 gram we require m by 48 gram of K2CR2 of 7 and hence and hence for 8 gram. Why 8 gram? Can you tell me guys why I am taking 8 gram of oxygen atom? For 8 grams it requires m by 48 into 8 that is m by 6 K2CR2 of 7. Right? So this method is method one. This is method one with respect to the mass of the oxygen atom we are getting. Okay? Yes. Why we are taking 8 gram of oxygen? Because 8 gram is the equivalent mass of oxygen atom. It is the since one equivalence combines with one so obviously one equivalent of oxygen atom that produces with one equivalent of K2CR2 of 7. Now you see in this condition the K2CR2 of 7 the equivalent mass is what m by 6 and this becomes 6 becomes the n factor of K2CR2 of 7 under this reacting condition. Is it clear? Is it clear guys? Get down CLR. Did you write it down? All of you have copied this right? Now you see method two. That is oxidation number concept. This concept is oxidation number concept. Okay? Now you see the reaction we have here is nothing but this. In this reaction CR2 of 7 2 minus converts into CR3 plus. That is the conversion we have. Right? So I will write down the reaction first. And the reaction is this that CR2 of 7 2 minus converts into 2CR3 plus. That is the conversion we have. Is it clear? CR2 of 7 2 minus converts into 2CR3 plus. Yeah for the first method you must have nascent oxygen. If it is an oxidizing agent you must have nascent oxygen present on the other side. You must have that. Correct Vibhav? If you do not know the mass of nascent oxygen atom then you cannot find out from the first method. Correct? Vibhav? Okay. Now can you tell me guys what is the oxidation state of this chromium atom here? For this chromium atom what is the oxidation state? For this chromium it is plus 3 already. Tell me. No. Any other product we cannot use because for other product the equivalent mass we do not know. Again. Why we are using oxygen there? Because we know the equivalent mass of oxygen that is 8 grad. Understood? Yeah. Correct. So here the oxidation number of chromium is plus 6 and here it is plus 3. Now what is the total charge here on this chromium atom? That will be what? 2 into plus 6. The total charge here is 2 into plus 6. That will be plus 12. Oxidation number is this. Oxidation state is this. And here the total charge will be 2 into plus 3. That is plus 6. Right? Now according to this method the formula of equivalent weight is this. Equivalent mass is equals to the molecular mass divided by divided by the change in oxidation number in oxidation number of if it is oxidizing agent then oxidizing agent or reducing agent per molecule. Per molecule what is the change we have in oxidation number and of oxidizing agent and reducing agent. That will be the n factor in this case. You see molecular mass suppose that is m that is what we have assumed already divided by the change in oxidation number is what? We have plus 12 here and plus 6 here. Right? So if I write down here the change in oxidation number oxidation number per molecule we have only one molecule of CR 207 that will be equals to plus 12 minus plus 6 divided by 1 which is nothing but 6 and that is why the equivalent mass is equals to m by 6. Again you see you are getting the same answer. Got it guys? This is oxidation number concept. Any doubt till here? Any doubt till here? Didn't we separately calculate the oxidation number of chromium? We are calculating the oxidation number of chromium only. This is the oxidation number of chromium only. Swamy I am not getting you. Asha per molecule is what? We have only one molecule of CR 207 we have. That is why we have one here. If you have two CR 207 minus we will write here this minus this divided by 2. Understood Asha? Swamy I didn't get you. We are calculating the oxidation number of chromium itself. Is it clear Asha with you? Per molecule? Yeah it's potassium dichromate only because this CR 207 is coming from K2CR 207. So in K2CR 207 you see in K2CR 207 the oxidation number of oxygen and potassium is not changing. It is only for chromium. So N factor of this molecule will be the N factor which we calculate with respect to chromium. That's why it will be molecular mass of K2CR 207 divided by the change in oxidation number of that particular atom. Two molecules of CR we have this side Rajna. We have two molecules this side for this side if you write the equivalent mass of this particular salt that will be molecular mass of this divided by equivalent mass of this and there we take two here we don't take one over there because we have two CR 207. We are calculating the equivalent mass of either oxidizing agent or reducing agent. So K2CR 207 is the oxidizing agent. That's why we are taking one here because we have one molecule of K2CR 207. Why chromium we are taking? Only the oxidation number of chromium is changing. Is it fine? Is it fine? I want CLR from all of you quickly. M is the mass of K2CR 207. I will write down molecular mass of K2CR 207 Now again I will write down the same reaction which is nothing but this CR 207 2 minus and it converts into 2CR 3 plus. Now the job here we have to find out the number of electron exchanged in this reaction. The number of electron exchanged. Now for this method the formula of equivalent mass I will write down the formula first. The formula of equivalent mass whatever method you use will be M by 6 only for this particular reaction. And that will be again the molecular mass of K2CR 207 divided by divided by the number of electrons number of electrons exchanged that is also per molecule. Number of electron exchanged per molecule. This is the formula we have. Now molecular mass is M we are fine with it. The only thing we have to know is what how many electrons are getting exchanged here. To find out this you have to balance this reaction. The reaction you see it is given in acidic medium H plus. H plus the previous reaction you see there we have K2CR 207 plus H2SO4. So balance this reaction which is in acidic medium and tell me how many electrons are exchanged per molecule. Yeah it is almost same Swami that is why we will get the same answer. I want you to balance this reaction and tell me how many electrons are getting exchanged. Quickly we are getting 6 electrons right when you balance the reaction you are getting 6 electrons right. How do we balance the reaction? The reaction is in acidic medium. So first of all the chromium is balanced you have 2 chromium this side now next is what we will balance we will balance the oxygen atom and for that we have to add 7 H2O this side and 14 H plus this side ok. The total positive charge is plus 12 this side and plus 6 this side. So we have to add here 6 electrons correct we will get 6 electrons hence the number of electrons exchanged per molecule is 6. This 6 because we have 1 molecule 6 divided by 1 so the equivalent mass from this formula will be again m divided by 6. So from all these methods you see we are getting the same answer. Yes or no? Understood guys so for any oxidizing or reducing agent we can use any one of these methods to find out the n factor equivalent mass. Correct? So whatever method you feel easy you can use that method to solve the questions. Got it? Ok. Now moving to the next part here 2 reactions I will write down and we will see how this you know change is there how to find out the equivalent mass in all these Next thing you see all of you have copied it down can I move on to the next can I move on to the next slide again I am repeating whatever I am writing it down all of you must copy this ok so you see the next reaction how do you find out n factor here ok for example you see this reaction mnO4 minus we have mnO4 minus and this converts into the medium is acetic mn plus 2 we have to find out the n factor for this reaction n factor right now to find out n factor what we do you see first of all the medium is acetic so you can easily balance the reaction and then you can check the number of electron actions can you tell me the n factor ok try this one we will discuss then I will you will try this one and then we will discuss this n factor you tell me is it 1 swamig is getting 1 prajna is getting 1 kripan is getting 5 sivlela is getting 5 aditya, araman, afshish most of you are getting 1 thing is what we can find out the oxidation we can find out the change in oxidation number for molecule right so mn plus 2 we have this side oxidation state of magnet is plus 7 so what is n factor change in oxidation number for molecule that will be plus 7 minus plus 2 divided by 1 which is nothing but 5 n factor is 5 another way we will do it quickly you can balance this reaction suppose we have charath, charath when this work get over this daily thing when this daily gets over another 10 to 15 minutes 10 to 15 minutes because the class is getting this time yeah another 10 to 15 minutes after that they will go off this yeah so you see the reaction is this so I will write down the reaction quickly and then for this one I am doing this balancing one it is not required here but just to make you understand mn plus 2 now when you balance this quickly I will write down 4H2O this side then 8H plus this side 5 so we have 7 positive charge and 5 positive charge to do that we have to add 5 electrons this side and that is why the number of electron exchange per molecule is 5 and n factor is 5 got it so this is what another method we have this is the oxidation number concept ok another reaction you see suppose we have Cr2O7 2 minus and it is converting into 2Cr3 plus n factor we have already done for this C2O4 2 minus and it is converting into 2CO2 n factor the next one is this when this 2MnO4 minus converts into mn plus 2 plus mn plus 4 these are the reacting condition guys I am giving you this reaction and according to this you have to find out the n factor don't think about like how do we know this it would be given in the question one correction I will do here it is not plus 4 it is plus 6 ok in the last one the important one is Fe C2O4 Phidus oxalate it is and it is converting into Fe plus 3 plus 2CO2 tell me for all these reaction n factor first again third and fourth you are getting 2 for the first one Aditya Dhyan assume this is as the first question second third and fourth first one is 6 Dhyan is getting 6 as the first one first one guys I have already done this check your notes just few minutes back I have finished this CR2072 minus 2CR3 plus check your notes guys 6 2 and 10 the last one you are getting 10 second last one you are getting 10 Prajna third one is 0.5 ok now I will do this you just see first of all what we do we will find out the oxidation number ok so here it is plus 6 and plus 3 total charge is what 6 into 2 12 2 into 3 6 chain in oxidation number is 6 n factor is 2 into plus 6 minus 2 into plus 3 divided by 1 which is nothing but 6 am I clear for this one it is plus 3 for this carbon and plus 4 for this carbon right n factor will be for this one is 2 into 3 that is plus 6 minus 4 into 2 8 divided by 1 that is 2 n factor negative sign has no significance only 2 here minus 2 will consider it as 2 only correct this one for manganese it is plus 7 here plus 2 and plus 6 now you see n factor will be 2 into plus 7 plus 7 minus will have plus 2 and then minus will have plus 6 this whole divided by 2 which is 2 here that will be 14 minus 8 6 by 2 is 3 right here ion is plus 2 right so this you have to understand ion is converting into plus 2 to plus 3 the change in oxidation number for ion is 1 and this is nothing but if you see the anion part of it it is nothing but C2O4 2 minus is converting into 2 CO2 right and this one we already did n factor is 2 right so for this the n factor is 2 and for ion the n factor is 1 because the change in oxidation number is 1 overall n factor if you have to find out that will be 2 plus 1 for ion is equals to 3 so for ferrous oxalate FEC2O4 n factor is 3 this particular one is very important you can directly memorize this particular conclusion also third one using balancing yes you can do that but for that the reaction must be given which medium it is must be given this is actually disproportionation reaction you see manganese its oxidation only manganese is getting reduced ok the reaction overall its not given if it is given then you can do the balance thing and you can find out the answer the last one is what it is ferrous oxalate Fe2 FeC2O4 you see this we have FeC2O4 and it converts into Fe3 plus plus 2 CO2 so in this actually what happens you see there are 2 atoms present in this molecule and their oxidation number is changing because ion is converting into from plus 2 to plus 3 and at the same time C2O4 2 minus is converting into 2 CO2 you see the oxygen the carbon atom the oxidation number of this carbon atom is plus 3 here and for this it is plus 4 right so within one molecule only there are 2 atoms which is present one is ion other one is carbon their oxidation number is changing correct in this kind of situation when the oxidation number changes for more than one atom present in the molecule what we do we calculate this we will calculate the change in oxidation number separately for all those atoms whose oxidation number number is changing correct so we will calculate the change in oxidation number of ion which is plus 1 and the change in oxidation number of carbon which is plus 2 here like we did this one right so and we add the total change to find out the n factor plus 2 change we will have here in this reaction plus 1 we have here and that's why the n factor overall n factor will be 3 why we divide it by 2 because we have 2 molecule of MnO4 what is the formula change in oxidation number per molecule so we have 2 molecule that's why we have divided by 2 correct because there are 2 molecules that's why we have divided by 2 all of you understood this all of you understood guys tell me please write CLR if it is clear now the last thing in this we have to do and this part actually you have to memorize guys this you should keep in mind because in some questions you should know the change in oxidation number of different ions what happened first one you have doubt actually the medium would be given in the question then only you can use electron method because if the medium is not given then you cannot balance the reaction according to the medium that's what the thing is so it would be given in the question first one Sana you have doubt in this one CR2 or 7 2 minus 2 CR3 plus this one triple got it understood tell me I will move on then I don't know whether I have done it or in the class if I have done it please let me know so this actually you have to memorize write down the heading some important oxidizing agent some important oxidizing agent there change in oxidation number you should know actually right change in oxidation number you should know hello yeah guys are you there yeah are you there guys can you see the screen now give me just one minute it will come there was some issue with the network can you see the screen now okay guys I am sending you the link again okay join the new link all of you okay did you get me I am sending you the link again join 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