 Okay guys So what is this Rosinman reaction? This is what in this it is a formation of what aldehyde? Right when which carbonyl chloride? Reacts with PDB SO4 H2PDB SO4 and it converts into aldehyde Okay, we've always done this the name reaction in which the aldehyde formation takes place formation of aldehyde Okay, so reaction if I write down here only just one single line carbonyl chloride RCOCl when heated in presence of PD with BASO4 Poisoned with BASO4 it forms aldehyde which is RCH Okay, this is the reaction we have equal number of carbon atom will have your in heroes Can you start a reaction in this reaction? We know that we take two molecules of aldehyde like suppose I'm taking at CHO and We use 50% of any OH which concentrated base will take here one of the molecule Will get reduced forms alcohol and another molecule gets oxidized and forms the sodium salt of it So this is the canizaro reaction we have one will get reduced other one will get oxidized Okay, out of this this which one gives I do form test many times. They have asked this question if you remember methyl ketone if it is present in any compound CH3CO methyl ketone gives positive Idoform test Right answer is this one Next question you see this is the 29th one these reactions you have to write down the product See if the reactant molecules are different if formaldehyde if one of the product you have for big it always undergoes Oxidation see what happens? I have done this in detail in class. You can refer your notes You will understand okay We cannot discuss all those things here again But I'll give you just one thing if you go through the mechanism you will understand Which one which one of the molecule will oxidize and which one of the molecule will get reduced, right? There's a no transfer of Hydrogen and then the reactions the mechanism is given over there one thing Just you have to keep in mind here if you have one and he had you have and other one is formaldehyde So formaldehyde always undergoes oxidation and the other molecule will go on to reduction Right and the height with reduction gives alcohol on oxidation gives Acid we'll get here salt of that acid. Okay, so which one will go under a Reduction or oxidation that depends what molecule you have but one thing you can Assure or you can keep in mind always that formaldehyde if one of the molecule is there then this will always undergo Oxidation and the other one will go on the reduction this you should know Detail mechanism. I have done this in the class you go through with your notes. You will understand What is the product in the first reaction? This is getting cyanohydrin Hydrazone second one the first reaction is very common. We have done this many times CS3 CHO C double bond O edge When the reacts with at see H plus and CN minus so this OH will get and see and we'll come over here to the product here. It will be CH3 see CN OH This is the cyanohydrin product we have With NH2 OH what happens is NS2 H will have oxyme. It forms oxyme, right? So the product will be what? CH3 CH double bond O When reacts with H2 and OH if you remember this Oh and H2 forms H2O and this carbon attached with the nitrogen with double bond the product here It will be CS3 CH double bond and OH Oxyme formation, okay CS3 CHO in presence of dilute NaOH CH3 CHO what is this reaction? Tell me can you tell me the name of this reaction and all condensation? What is the product we get here? Do we have condensation takes place? Do we have condensation here in this reaction product here? It will be what? CH3 CH OH CH2 Sorry CH2 CHO Right after this if you heat this which is not the question the answer is this only for this question, but if I Go forward one more step If here also if you heat this then the product will be what one of this hydrogen and this H comes out as H2O and it forms CS3 CH double bond CH CHO Right, but since there is no heating so condensation won't takes place here Condensation won't take place, but only aldol forms here. So product will be aldol Test to distinguish the following pair of compounds benzoic acid and phenol see phenol. I'll just write down here with neutral FECL 3 It forms violet color with neutral FECL 3 it forms violet color, but benzoic acid Does not react with with FECL 3 There is no reaction With benzoic acid. So with neutral FECL 3 we can distinguish benzoic acid and phenol What about propanol and propanone? This is easy right? Idoform test we can use here in the second reaction The keto group gives you Idoform positive Idoform test and The aldehyde won't give Idoform test, right? So that's the reaction we have for the second one We have Idoform test and the first one we use neutral FECL 3 Yeah, we can use FECL 3 NAHCO 3 test. We can also use like any one of these you can use Okay, so this is the answer Question number 28 done this with reference to structural Variability and chemical reactivity. Let me check this question. I think something is missing here Question number 28 With reference to structural variability that and chemical reactivity write the difference between lanthanoids and actinoids Okay, so actually here Let me correct this name of the member. Okay, so this is not the second one Okay, this whole thing is the first question till here So with reference to structural variability and chemical reactivity write the difference between lanthanoids and actinoids actually it is second question is name the member of lanthanoid series, which is well known to exhibit plus four oxidation state correct complete the following reactions Okay, fine question number one. Yeah, it's actinoids. Correct So what is the difference basically we have here? We can also talk about here about the oxidation state of lanthanides and octanides and One more thing actinides all the elements are radioactive in nature. Okay, there's two points. You tell me then Greater range of oxidation is just yeah, fine actinoids these two three points. You can write or whatever. I'm telling you actinoids Generally shows plus three oxidation state, but they can show oxidation state up to plus seven right lanthanoids also generally shows plus three oxidation state But they can also show oxidation state up to plus four plus seven is not possible in case of lanthanides Okay, one important difference here. It is lanthanides are generally a non radioactive lanthanoids are not Radioactive with one exception here that is except a Element called promethium PM is the symbol of that promethium Promethium Yeah, right that also you can say lanthanides are not radioactive, but actinoids are generally radioactive radioactive Okay, oxidation is that I've already told you this can show plus three or Up to plus four generally shows plus three But can go up to plus seven right these are the two, you know Major difference we have apart from this you can also say one or two difference between these two You know the series of elements we have that is for this lanthanides So this is also, you know Experimental like it's magnetic behavior can easily explain for lanthanoids its complexity is less over here comparatively So it's magnetic behavior is explained easily Magnetic behavior explained easily But here is still the study is going on magnetic behavior does not easily explained over here Right with in comparison to actinoids lanthanoids physical and chemical properties. We know a bit more right Actinoids is still the you know research and study is going on So there are a few two three points first two points are more important which you can write Okay, now next question or the numb name of The member of lanthanoid series, which is well known to exhibit plus four oxidation state tell me Which lanthanoid elements shows plus four oxidation state serium. Yeah, it's correct. So for the second one The answer is Serium next question you have to complete the reaction and Next question is out of mn3 plus ncr3 position is more paramagnetic and why can you balance this reaction? Tell me Mn2 plus Then how you balance this equation? See I'll write down this in the next page You see first of all the question that is given here is mn O4 minus Plus how many h plus we have 8 h plus and 5 electron. Okay, so 8 h plus and 5 electron. This is the question. We have to complete this reaction Okay, so you can do this question. It is actually a redox reaction Okay, so you can solve this question with the concept that you have studied in 11 plus redox reaction Okay. Now, what is the meaning of this thing? H plus here first of all this iron comes from what? Comes from KmnO4. This means h plus means the medium is acidic and Since it is a half reactions. So there are electrons also into this Okay, so now you see in acidic medium in KmnO4, which is nothing but will write here mn O4 minus and this converts into mn Plus 2 this information we should have in Acidic medium the conversion is this Okay, here the manganese is in plus 7 oxidation state and this is from plus 7 to plus 2 If you see this this change is there in acidic medium, right? It will go to mn plus 4 in Neutral medium and then it will go to mn plus 6 in Alkaline medium. This is the Change you have to memorize 2, 4 and 6 Right now the question is this mnO4 minus 2 mn plus 2 that will Will focus on that only here. So you see The question is what mn why I'm explaining this to you in so detail Because if you remember, this is the second question we have seen in board exam last class Also, we have seen one question based on this balance question. Okay, mn plus 2. So this is the change we have Now the point is the first rule if you see what is the first rule to balance a redox reaction half reactions in Acidic medium, what is the first rule? Can you tell me just you tell me the first rule? How do we balance this? Okay, so first we'll check what first of all you see this reaction means mnO4 minus or KmO4 is reacting in acidic medium. That is it. Okay, and we know in acidic medium mn O4 minus goes to mn plus 2. So I have written this now We have to balance all atom first other than oxygen and hydrogen Okay, we'll balance oxygen and hydrogen on the basis of what medium we have Correct. So since other than oxygen hydrogen, we don't have here. We have only manganese here Right. So manganese is already balanced. We don't do anything into it Now in acidic medium, we know the medium is acidic In acidic medium will balance oxygen by adding water molecule on the Other side. Here we have oxygen. So we'll add water molecule on the opposite side But how many water molecule? That depends on how many oxygen atom we have here. So we have four oxygen atom So one H2O molecule has one oxygen so multiply with four here. So we'll add four H2O molecule on the other side So if you add this H2O, you see the oxygen is balanced both side Right now to balance hydrogen will add H plus on the other side again But how many H plus it depends on how many hydrogen we have here So we have eight hydrogen. So we'll add eight H plus But if you see this Reaction all the atoms have been balanced manganese oxygen hydrogen all the atoms have been balanced Now we'll balance the charge balanced charge What we'll do we'll just calculate the charge On both side of the half reaction Like you see here The total charge here is it is what? Eight positive one negative. So we have seven positive charge here plus seven What is the total charge here plus two and zero? So we have Two positive charge from seven to two you have to go It means you have to decrease this charge by five. So for that we'll add five electrons on the left hand side You see seven minus five gives you two. So here the charge is also balanced atoms. We have already balanced, right? So that's why you see the left hand side of the reaction is given in the question Mn o four minus eight h plus and five electron So when you do this you will get mn plus two and four h two o which you have already given the answer So your answer is correct Understood Okay, last question you solve Out of mn three plus and cr three plus which one is more paramagnetic and why tell me Correct wishes mn three plus is more paramagnetic since it has four unpaired electron You can draw the configuration. You will get the same Guys all of you done mn three plus Yes, more paramagnetic with four unpaired electron next question Complete the following reaction again. You see this reaction is again The same thing that I have discussed now Can you do this now? How did you get this? again, you see See first of all This cr two or seven again There are two possibilities it can go to two cr three plus when the medium is acidic Or it can convert it into cr o four two minus If the medium is basic, right? So again this thing you have to know, right? Cr o four two minus in basic medium Two cr three plus in acidic medium Now you see the reaction is what cr two o seven two minus Since the medium is basic here basic medium So this will convert into Cr o four two minus this you have to memorize. There are a few things here You have to memorize like in the previous example. I've given you for k mn o four that is mn o four minus This is now for cr two o seven two minus acidic medium cr three plus basic medium cr o four two minus Now you see again, you have to balance this Right, so medium is basic. So what I am doing here. You just try to you know look at this example very carefully You must have done the balancing of reaction and there are two set of rules One each for acidic and basic medium. Yes or no Yes, you must have seen one set of rule for basic medium and one set of rule for acidic medium, correct? For balancing of redox reaction Yes, so what I'm asking you to do here. You don't you just forget those rules Which you have started for basic medium. Okay, just you let it be those rules basic medium rule You let it be only you memorize the rule of acidic medium We are going to use the same rule here And you take care of one thing here the medium is basic But we are going to apply the rule for acidic medium And once we balance all this Just one change will do in the last and the whole question will convert into basic medium So the benefit of this thing is what you don't have to memorize the rules for basic medium separately Okay, so let me do this first you will understand So first of all you see we'll balance all the atoms other than oxygen and hydrogen So again, you see here we have two chromium. So we'll add two here Right chromium is balanced now. We have seven oxygen here eight oxygen here four ox Sorry eight oxygen here. So to balance oxygen will add H2O molecule only one H2O molecule will add So that's seven plus one eight oxygen eight oxygen this side Now we have two hydrogen to balance this hydrogen here will add H plus on the other side And how many H plus since two hydrogen we have here. So we'll add two H plus now all these things we do in acidic medium also Right next thing is what we have to balance charge. What is the total charge two negative here? What is the total charge two positive and four negative? Yes So we have two into two four negative and two positive So we're getting two negative is it So charge is balanced here. So we don't have to do anything into it Now the point is all these things we have done According to acidic medium Right according to acidic medium. But since the question is in basic medium So what we'll do here what change we'll do in the last this you take care of But what change we'll do here that the number of h plus you are getting here Equal number of OH minus iron you have to add both side in the reaction You will convert the whole reaction into basic medium. So what we are doing here till here We have followed the rule of acidic medium now to convert this into basic since we have two H plus we'll add Two OH minus here Right since we have two H plus we'll add two OH minus If you have eight H plus we'll add eight OH minus two OH minus we have added this side So to balance this reaction to to maintain the you know the number of atoms constant both side Two OH minus will add on the left hand side also, right? Now what happens we'll write down the net reaction now this two H plus and two OH minus gives you two H2O Right and this two H2O and left hand side. We have one H2O This will get cancelled and we are left with one H2O molecule on the right hand side So the whole reaction becomes it was if you have done everything correct You'll you'll get the left hand reaction left hand molecule here. Exactly. It is you see the molecule is CR 2072 minus plus two OH minus and it gives Two CR 042 minus plus H2O. This is what the answer you were getting Right, you see this is the question we have CR 2072 minus plus two OH minus The answer will be two CR 042 minus plus H2O. So nothing you have to memorize here Did you understand this? That's why we do not have any free electron here wishes you see Since the oxidation number is same There is no free electron present here You see there's no electron present. So what is the See you only have to memorize the rule for acidic medium To convert this into basic medium the number of H plus you are getting Equal number of OH minus you have to add both side and then you have to write down the net reaction All of you understood this? Okay Now the second question is what again you see the kmn of four minus acidic medium converts into what? converts into mn plus two So here the product will be what? mn o2 Plus four h plus we have so two h two will get here two h two Two plus two four Yes, got it. Okay. What is this answer? Zn is not considered as a transition element. What is a transition element? Can you tell me the definition of transition element difference between the transition element and d block elements? transition element are those elements which has partially filled d orbitals When d orbitals are completely filled it is d block elements Right partially filled d orbital If you draw the electronic configuration of z n you will get the completely filled d orbital is not partially filled That's why it is not in transition element. It is a d block elements Why transition elements forms a large number of complexes? Why transition elements forms a large number of complexes answer? Yeah, right High nuclear charge and availability of d orbitals. Correct. That's why it forms complexes third one the enought value of mn three plus two mn plus two couple is much much more than much more positive than cr three plus two cr two plus couple the reason behind this is what? The third question Okay, you draw the electronic configuration of mn three plus and two plus and see which one is more stable. Okay, draw the electronic Yeah, that's what when you draw the electronic configuration, you will understand mn two plus is highly stable because of half filled electronic configuration Okay, that's why it's enought value is much more positive Got it Okay, next question. We'll see Give reasons Right Since nitrogen does not have d orbital. It cannot show pentavalency So the second molecule does not exist phosphorus has d orbital it can expand its octet correct Yeah, right second one. You tell me 27 2 Electron can help is low because small size. Yeah The small size it has high electron densities. So it can produce more repulsion to the coming electron. That's why it's electron getting help is less with negative side Yeah, right third one s 3 p o 2 is stronger acid stronger reducing agent than s 3 p o 3 uh in s 3 p o 2 the oxidation state of phosphorus is plus one and that of phosphorus in s 3 p o 3 is plus 3 So in s 3 p o 2 it has more tendency to increase this oxidation state Right and can easily get oxidized and act as a reducing agent. So reduction tendency or the reducing behavior of s 3 p o 2 Since it has lower oxidation state is more to that of s 3 p o 3 next question What are emulsion? What are their different types? Give one example of each Emulsions are liquid liquid quality solution. So we have actually there are two types of emulsion Okay, just write it down There are two types of emulsion water in Oil and oil in water You see first one we have oil in water That we also write like this o slash w oil in water Okay, example of this is milk And in this milk the dispersion medium is water dispersion medium Is water dispersed phase Dispersed phase Is oil So this is oil in water emulsion example is milk water in oil water in oil given in n c i We write as this Example Is butter and in butter the dispersion medium is oil. It's reverse actually Dispersion medium Is oil Dispersed phase Is water these are the two types of emulsion we have in general With example you must remember always try to write down support your you know theory with some example. So always good Next question you see Let me check this question. The following data were obtained during the first order The first order thermo the constant volume associated to This and this the following data Were obtained during the first order thermal decomposition of s o 2 c l 2 at a constant volume Reaction is also given two experiments And data are given here time and you know pressure You have to find out it's not written here You have to find out the rate constant of this reaction rate constant which is k Is equals to what for this reaction? So all this Done 1.38 Into 10 to the power minus 2 Okay Yeah, it's right All of you have done Perfect. You got the answer Okay So don't need to solve this right just write down the expression for first order of k which is uh 2.303 divided by t log of Initial pressure p not By p not minus p Okay, so the time is given k you have to find out What is p not if I see this question? I'll just write down here Initially, it is p not 00 And then it becomes p not minus p p and p so total pressure will be what I'll just write down this very quickly Total pressure will be p not minus p plus p plus p. So p not plus p And this is equals to what at 100 seconds point seven P not is point four So p is point three P not is Point four P is point three you substitute here. You'll get k value as 1.38 10 to the power Minus two. What is the unit of k? What is the unit of k tell me? Second inverse, right? So don't forget to write down the unit also Always second inverse first order reaction Second inverse is the unit. Oh, sorry Next question you see question number 24 This is the abstract question This one Yes Account for the following primary a mine have higher boiling point and tertiary a mine The reason for this Hydrogen bonding in primary amines. Yeah, so the number of hydrogen bonding even in tertiary amine hydrogen bonding is not possible in primary amines possible So due to intermolecular hydrogen bonding Primary amines has higher boiling point. Correct. Second one. Aniline does not undergo fritill craft reaction. Why? Yeah, see see what happens the basic nature of aniline reacts with the acid which is used in fritill craft reaction Okay, that's why it does not go under fritill craft reaction. It reacts with acid and forms a salt Right alc l3 is what alc l3 is a louis acid, right? Do you do? Electron deficiency So it can accept the lone pair of nitrogen atoms. So it reacts with that louis acid alc l3 and forms a salt. That's why it does not go under fritill craft reaction Right, so that's the reason what is the third one is more basic than CS3 whole-trice n What is the conjugate acid of CS3 whole twice NH? Okay, CS3 NH2 plus. So this will get stable You see one more thing you can understand here that See the question is in aqueous medium. I'll just give you the You know detail of thing of it when methyl is there and when ethyl is there then what happens? Okay, so you see first of all First of all, there are Two things here. If you remember, I have done this in the class also. Let's give you the quick, you know A division of this so that you can understand. This is an important question as far as you know any competitive exam or board exam is concerned See when gaseous Phase is there in case of gaseous phase like purvic, uh, this vicious said Said that because of plus i It gets stable. So that is possible in gaseous phase only In gaseous phase we only it is only defined by the plus i nature of the alkyl group plus i nature Okay, and that is why the order of You know, basicity of three degree amines will be maximum than two degree than one degree and then NH three, this is the order in gaseous phase. But when we have aqueous phase Aqueous phase then there are two three things which comes into the picture Okay, like suppose if I take the primary amine r NH two in aqueous phase, which is H2O it forms R NH three plus Plus OH minus Right Now this shows hydrogen bonding Right and the reaction has tendency to go in forward direction Right, but the number of hydrogen bonding you see here. We have three hydrogen. So three hydrogen bonding will be there if it is secondary Then the number of hydrogen bonding will be what? NH two plus it is less than the previous one and with tertiary. I'm not writing down the whole equation This will be r3 NH plus even the number of hydrogen bonding will get reduced Right, so the number of hydrogen bonding will be maximum in case of primary amine then secondary and then tertiary But this is not the only you know a factor we have here Right second thing what happens because of hydrogen bonding Because of hydrogen bonding that stability will be Uh more over here more salvation more hydration energy will be there. So because of hydrogen bonding this one is most stable Then this and then this But according to eye effect if you see Three alkyl group this one is maximum stability will have maximum stability then this and then this so we have the reverse order So two factors one is hydrogen bonding and other one is eye effect. Both are contradictory to each other Right. So in this case uh when when when more than one factors are there and Those factors are contradictory to each other in this kind of cases The you know result becomes experimental in these kind of cases Okay, it depends on which factor is dominating when according to that the result will be there Okay, so what happens here? It has been observed that if the r is cs3 if r is the alkyl methyl group If r is the methyl group cs3 Right, then the order of basicity is found to be secondary amine. We have maximum then one degree and then three degree But if r is any other molecule Other than methyl higher molecule like ethyl propyl butyl whatever it is if methyl is not there any higher alkyl group is there Then the order is found to be secondary will have maximum then tertiary and then one degree So all these things you have to write down in this reasoning Okay that in aqueous phase there are two factors hydrogen bonding and eye effect Both are contradictory to each other. So in this case is the you know The result becomes experimental and it has been observed that according to you know kb value that secondary amine is more basic than Then primary and then tertiary in case of cs3 and r is any other group other than cs3 like c2h5 or you know ethyl propyl butyl whatever it is Then the order is found to be this because when the you know alkyl group increases So we have steric hindrance also around it Right that steric hindrance will also come into the picture. So hydrogen bonding eye effect steric hindrance Steric crowding around the nitrogen atom Got it the keyword here. It is what hydrogen bonding you can say Steric hindrance you can say and the eye effect you can say but in case of gaseous phase eye effect is the dominant factor So we'll only decide according to the eye effect more alkyl group more will be the eye effect and hence more will be the basicity Understood. So this is very important one. Okay next question. We'll see Tell me the answer The first one is cs3 i second one is cs3 cl and why is it so why cs3 i reason Both are primary halide Okay In first case both are primary halide why cs3 i is the answer answer is correct. Your answer is correct. But why cs3 i Yes, primary alkyl net go under s and 2 correct But first question if you see We have cs3 br and cs3 i Both are primary halide only Guys, what is the answer? Tell me Yeah, it's simple I iodine is the better living group Right iodine is the better living group. So in case of cs3 br and cs3 i Removal of iodine is easier. That's why it's it easily goes under s and 2 reaction or faster. The reaction rate will be the will be more Right, so the keyword is what The iodine being a better living group. It can react faster in s and 2 reaction cs3 i Okay, in second case what happened? What is the reason of cs3 cl to be the answer? Why cs3 cl is the answer in the second one? Because the first one is tertiary halide and second one is primary So tertiary halide has more tendency to go under s and 1 reaction Okay, formation of carbocation is easy over there because tertiary carbocation is more stable Okay, so cs3 whole thrice cl goes under s and 1 reaction, but cs3 cl goes under s and 2 reaction Okay Next one you see question number 20 a The reagent is pcl 5 This is pentachloride. It is pcl 5 phosphorus pentachloride Instead of over which we'll get cl over there. Correct What is the name of this compound the first compound that you are getting? What is the iopsc name of this compound? What is iopsc name of this compound the first one that you have got? Is it benzyled just the ring is not benzene ring you see that and this is you cannot write down you cannot write space in between Cyclohexyl methyl chloride name is correct, but space you don't write cyclohexyl methyl chloride Cyclohexyl methyl should be together and then chloride What is the second one? What is the answer? What is the product for the second one? Suppose I'll write down The numbering suppose if I do first carbon Second carbon and third carbon obviously bromine will attach so at which carbon the bromine will attach At which carbon bromine will attach Markovnikov any kind of shifting here So that we can get more stable carbocation. Is it possible? See first of all The h plus will attach at first carbon. Okay, and we'll have positive charge on second carbon, right? Okay, then we can have one two hydride shift right hydrogen from the third carbon Will attach onto the second carbon and then we'll have carbocation on the third carbon That is benzyl carbocation resonance stabilized And then the br minus will attach on the third carbon finally the answer will be what the bromine Will attach at the third carbon and we don't have any double bond present in the molecule. Did you get this? See Here the product will be this ch2 cl Now the another molecule the second one The molecule is this hbr So what happens this molecule will take this h plus from here And it forms ch2 ch positive ch3 And then what happens 1 comma 2 hydride shift Which makes the carbocation even more stable ch positive ch2 ch3 And then the br minus Will attach onto this carbon atom and the final product Will be this chbr ch2 ch3 this is the Answer in this question Did you get this understood? Okay Yes, now you tell me the iopc name of this product the final product you get what is the iopc name of this One bromo one phenyl propane correct Okay, the name will be one bromo one phenyl propane Must remember this rule that whenever carbon chain and benzene ring is attached Okay, and if the carbon chain Contains any functional group Right halogen or also functional group then we consider carbon chain as the As the primary chain parent chain And we write down phenyl for benzene ring in that in the name of that particular compound Okay, so one bromo one phenyl propane is the right answer Next question you see Tell me Is it dichloro? Is it dichloro? Vesist and Purvik Right, Saim Premnath don't write any space between chromium and then oxidation state Saim Premnath chromium no space three chloride Don't write any space do not put any space over there oxidation state is not too fresh So dichlorido it will change into Ido it will change into a to I it will change into I to Okay, that you must take care of don't I know do these kind of mistake You know the answer, but you won't get marks if you do this kind of mistake. Okay, the whole question will get wrong Take care of this small small thing Okay, coen whole price three plus what kind of isomerism? Is it optical? Right What is the density of this ligand ethane one two diamine optical isomerism is correct It is by dented ligand. Okay. There are two donor atom by dented ligand Coordination number of cobalt is what over here. What is the coordination number? The coordination number is what? Yes Coordination number is six Because one ligand has two donor site ethane one two diamine is what? See again. I am again repeating this this thing CH2 CH2 NH2 right, this is the Ligan we have NH2 and the donor atom is what? The donor atom is nitrogen here. This is this will donate electron to the metal Right, so there are there are two donor atoms in one Ligan, right? So similarly we have two more donor atoms in the another ligand Right and two more in the third ligand You see so density is what you can do like this like this you can write density is what one two three four five six Okay, so sorry the coordination number is six here Density is what density of ligand is two because you have two donor atoms Okay, so coordination number is equals to what we can write number of ligand Into the density of the ligand roughly we can say this coordination number Right second thing also see they can ask you this question They'll give you one formula like this They'll give you one compound and they can ask you two three questions on to that Suppose they can ask you write down the structure of that Okay, you have to write make that structure like this Okay, then what is the density of the ligand you have to write down the density What is the coordination number that that also they can ask what are the donor atoms in each ligand? They can also ask you this so donor atoms here it will be what nitrogen and nitrogen Are the donor atoms? Okay, so like this they can again ask you hybridization geometric anything so with one one molecule one complex compound They can ask you two three four questions this kind of questions you can you can get in three marks Okay, three questions they can frame one mark each so Take care of this thing also oxidation number. They can ask you Understood do this one Okay What is the product we get in rhymer time and reaction that is williamson synthesis Okay, so williamson synthesis will get ether Right the action is correct. We just what we get in rhymer time and reaction. What is the product we get? Can you tell me so the reaction if you write down here I'll write down the reaction. There are two products actually possible here both reactions. I'll write it down Okay You see the rhymer time and reaction. We have done this in phenol chapter alcohol chapter alcohol phenol ether so When you have phenol And this is allowed to react with chloroform c at cl3 In presence of a base, which is koH The product we get here is this Mechanism we have done this in the class But here you don't require mechanism. You don't have to write down the mechanism here This compound we call it as celly cell dehyd Right celly cell dehyd now second reaction again in rhymer time Which we can write with again phenol But we can also use here carbon tetra chloride instead of chloroform Plus koH This gives a product as cellycylic acid OH And then cOOH here This compound is cellycylic acid So both reaction is possible in rhymer time and reaction This reaction we call it as rhymer time and formulation Formulation and this one is rhymer time and carboxylation Carboxylation rhymer time and formulation Remember with carbon tetra chloride you'll get acid with chloroform you'll get aldehyde Understood So this reaction you have to write down over there Next reaction you see Tell me slope is nothing but rate constant Yes order the reaction is zero zero And slope actually you can easily do this Suppose the concentration of a dca by dt minus of dca by dt is equals to k Concentration of a that is ca Right If it is one Then we'll get a log term here Right since it's a straight line here and if it is zero you can cross check like this So it will be what? You'll get y is equals to mx plus c form so it will be a straight line So zero order reaction it is Now for zero order reaction the slope of the line will be what? Concentration and time concentration slope is nothing but minus of k So it is nothing but the rate constant of the reaction Write down c the rate expression whenever you write down It will always be de concentration of The concentration by dt Minus dc by dt is nothing but the rate is equals to rate constant k Concentration of the reactant species to the power the order of the reaction Now whenever you get this kind of question you write down this expression and then you analyze if it is zero Then what could be the graph if it is one then what could be the graph and generally in this kind of question You'll get either zero order or first order second order you will you can you may get but Less that's chances are less for second order if it is second order also you can easily find out That if you have one by concentration term then you can get a straight line graph over there But point is whenever you write down this relation After this zero one and two you can easily cross check it Okay Next question explain the mechanism done finished How many steps are there? protonation formation of carbocation nucleophilic attack. Okay, so three steps are there so And in this kind of since they have asked the mechanism So you have to write down step by step properly Okay step one What happens in step one? We have primary alcohol ch3 ch2 OH Or it would be better if you You know show the lone pair of electron here This is again. I'll just write down this step one step two And then we have step three. We have a lone pair of electron here on the oxygen atom two lone pair this lone pair will attack on to this h plus and This is nothing but the protonation ch3 ch2 OH2 positive charge On it. Okay protonation of this now next step. What happens? This ch3 ch2 OH2 plus To stabilize this carbocation positive. Sorry this positive charge on this oxygen atom It will take this one pair of electron And we get a carbocation here, which is ch3 ch2 positive charge on it And h2 a molecule will go out Okay Last step is what this ch3 ch2? positive charge on it And then the nucleophile, which is br minus here This nucleophile will attack on to this positive charge carbocation. There's no rearrangement possible here and the product we get here is ch3 ch2 Br So this is the three step mechanism one thing you must take care of the direction of arrow is also important Lone pair will attack negative charge will attack always negative charge attacks on to the positive charge Okay, positive charge will not attack. Okay, so that's so I know arrow The direction of arrow is also important in this mechanism part understood The first step is protonation Second step is formation of carbocation. If there is any possibility of rearrangement, we'll do that also here and then We'll get the attack of nucleophile in the last step Understood What is the slowest step we have here in the mining step? What is the slowest step step two? Yes, the formation of carbocation is the slowest step here Once this carbocation forms the nucleophile will all of a sudden attack on to the carbocation Okay, so this is the slowest step or we can say rd s rate data mining step next question Define these term What is molar conductivity and secondary batteries? Just a second you try this Okay, then what is molar conductivity? Yeah, secondary batteries can be recharged Yeah, what is molar conductivity is the conductivity power of all ions furnished by one mole of an electron Present in a fixed volume of solution. Yeah, it's correct Both definition is correct. Then you after this you write down the formula for more conductivity also, okay And secondary batteries are those batteries. What is the example of secondary batteries? Any example? Can you tell me Secondary batteries any example any definition if they ask you in the exam Okay, write down the definition and then you write down one example Okay, like secondary batteries are the lead storage battery Which can be recharged again and again Example a lead storage battery next this one Write the type of magnetism observed when the magnetic moments are aligned in parallel and anti-parallel direction in unequal number It is unequal number. You see Unequal number is anti-ferromagnetic in unequal number. It is anti-ferromagnetic Short key is correct But it is unequal number. So it is anti-ferromagnetic Due to short key density decreases What happens in frankle defect as far as density is concerned? Frankle effect does not affect the density density Yes anti-ferromagnetic will have equal number of equal Magnetic moments are aligned in parallel and anti-parallel direction Second one is short key. Yes Next one you see numerical See this type of question If you observe we have seen in every question paper every year One density related question What is the answer Just to use the formula of density that is it Take care of unit always like I said always in this kind of question. They must change Sometimes they change the unit also So you must take care of unit over there 26.9. Yeah, it's correct. 26.97 It's better to write exact value 26.97. Don't write it 27 it won't be wrong, but it's better to write always 26.97 like you get Yes unit is also Gramp and liter wishes or is it mole? Next question you see all of you have done right. Do you do I need to solve this? Do I need to solve this? Okay. So just formula you have to use Next question you see Write the name of monomers What monomers we use for Teflon Tetraflodoethene it is Flodoethene it is or tetraflodoethene. Yeah, it's correct So for Teflon it is allied on the formula cf2 double bond cf2 Tetraflodoethene tetraflodoethylene For bunion what will write 1 3 butadiene which is nothing but ch2 double bond ch Single bond ch double bond ch2 1 3 butadiene and acrylonitrile Right acrylonitrile is this ch2 double bond ch And then cn here. So this is acrylonitrile And this is 1 3 butadiene. This is tetraflodoethene ethylene Next one you see structure Done Okay So how do we what we calculate to find out the structure of this first of all you tell me What we'll find out this one is easier like you know this s double bond o Double bond o OH And OH this is a structure of h2o4 Dibasic acid What we'll find calculate to find out the structure of this number of valence electron, right? So how many valence electron we have here 8 plus 18 That is 26. So now the central item is x e And we have three oxygen o O O We'll put what one bond pair of electron here I'll just draw it simply for for for We have two lone pair here Two lone pair here and two lone pair here So we'll have 24 total and one lone pair will have here in this orbital The structure of this will be what double bond o Double bond o And double bond o with one lone pair here Understood Understood. Okay. So we'll see one last question for today Done this Right on these reactions Which one is trigonal bipyramid? Both are sp3 hybridized Right geometries tetrahedral. Yeah shape is trigon. Yeah, it's not See the shape is trigonal pyramidal Geometry is tetrahedral. It's not bipyramidal Okay x e o3 So what is the product in the first reaction carbon with concentrated h2so4 That gives co2 and so2 H2o right on the balance reaction here. So it will be 2so2 2h2o x e f2 We'll have 2h2so4 here x e f2 plus H2o It gives xenon x e That is 2 x e plus o2 plus hf We have 2 here And 2h2 This is the product we get in this reaction So we'll wind up the class here only right we have solved today like Many questions. I'm a small questions in numeric numeric as we haven't done today because it would not there in the last part Okay, anyways, so you keep on realizing these things and Like I said the small small things that you are making mistake try to uh Try not to repeat the same mistake in the exam like the unit Whenever you write calculation mistake. Okay, if you have got it wrong in first attempt, you don't know what there that it is right or wrong Okay, so while you are doing some calculation take care of that also Okay, small small things you take care of you'll get good marks. Okay It's not like you do not know, you know like almost everything But don't make those silly mistakes in the exam. That's the only key thing you should keep in mind Anyway, so we'll see you soon. Okay, we'll wind up the class only. Thank you all for joining all the best