RE-NEET 2026 | CHEMISTRY PAPER DISCUSSION | All India FULL SYLLABUS TEST (RE-FST-05) #newlight

Hinzugefügt: 2026-05-27

[00:00:01]Jai Hind children. So how are you all?

[00:00:04]Hopefully, you had a good experience in today's paper.

[00:00:11]We are going to discuss Re-NEET FST Test number five.

[00:00:14]Okay kids? So let's start.

[00:00:16]So let's start our pre with this mission of Dr. SP Singh Sir. Now everyone has the right to study and everyone has the right to become a doctor.

[00:00:23]So with this mission we are going to start this discussion.

[00:00:29]Okay kids? So we will discuss the questions one by one.

[00:00:34]So the first question is a very good question 46 and hopefully many children would have got this question wrong.

[00:00:41]Statement one says the limiting molar conductivity of KCl and it is a strong electrolyte is higher as compared to acetic acid which is a weak electrolyte. Is this statement true or false? So generally, if limiting molar conductivity was not considered then this statement would have been correct. But at limiting molar conductivity i.e. infinite dilution, every electrolyte, whether weak or strong, is strong.

[00:01:11]Is it clear? So which has higher limiting molar conductivity between KCL and CH3COOH?

[00:01:22]Acetic acid on infinite dilution. So what happened to this statement? Wrong. Why?

[00:01:27]Because whose molar conductance at infinite dilution is very high among all the ions? of H+. So, among all the other ions of H+, be it any cation or any anion, which ion has the highest molar conductance at infinite dilation, that of H+ is much higher. Because of this, acetic acid contains H+ ions, so what will be the molar conductance of acetic acid at infinite dilution due to the presence of H+ ions? will be more than KCl.

[00:02:04]And if we talk on the basis of data in NCERT, then the molar conductance of KCL at infinite division is given in NCERT as 149.8 moles per Simon centimeter square and that of acetic acid is given as 390.5 moles per Simon centimeter square. So this means that CH3COOH has a very high molar conductance compared to what? From KCl. So tell me what happened to this statement? Wrong. Next statement Molar conductivity decreases Molar conductivity decreases Molar conductivity decreases with increase in concentration of electrolyte. Now let's see here concentration kappa and concentration beta, these are directly. And if I talk about G conductance or molar conductance or equivalent conductance, all these three values ​​are inversely proportional to concentration. So tell me, Kappa is directly proportional to concentration. That is, kappa increases with increasing concentration. But the remaining three decrease with increasing concentration.

[00:03:18]Now look here, molar conductivity decreases with increase in concentration, so what is this, tell me correctly, statement one is false, statement two is true, statement one is incorrect but statement two is correct, answer will be third clear children, next question is, next question is, arrange following forces in decreasing order of the strength in 47, 47 and 47. So son, if we talk about it, many forces have been given. If we talk about decreasing order among these, who will be the highest? Ionic bond.

[00:04:05]Fewer hydrogen bonds than that.

[00:04:10]After this, Dapol Dapol.

[00:04:14]After this comes dipole induced dipole and at the end comes London force i.e. dispersion force. So which one will be at the back? London Force.

[00:04:27]Ok? London Dispersion Force. So tell me what is the order? Mostly B then A then D. So B A D then B A D after this. After this E then C E then C, this means what is the answer? BD and E are the first dipole induced dipole and who will come after it? Dispersion force i.e. C. The answer will be first.

[00:04:58]Next Question 48 Arrange the following in increasing order of Lewis acid character.

[00:05:03]Son, if we talk about Lewis acid character then BF3 BCL3 BBR3 and B3 B3 now look if I have boron and fluorine because here like this because what does it have son?

[00:05:27]Vacant P orbital is a vacant P orbital.

[00:05:30]What do they have? Loan pair. So in turn, sometimes it will donate it through coordinate bond. He will donate it someday. He will donate it someday.

[00:05:37]So tell me there is back bonding resonance here. Its deficiency is on the verge of being completed. So there are 2p2p 2p 2p orbitals. So back bonding will be better. Here 2p3p here 2p4p 2p5p so tell me here back bonding is the least. So the deficiency of boron will be highest here. That means the lowest acid character will be the highest here and the lowest here, so what will be the increase in Lewis acidic character from top to bottom, so tell me, the highest here will be BI3 and the lowest will be BF3, so B is more than A and D is more than B and C is more than C, so A B C D answer will be first, next question question number 49 is asking that tell the oxidation number of chromium.

[00:06:35]So look here CR NH3 Crnh36 and Cl3, here if we extract chromium then the one which has to be extracted, let us take son X and ammonia is neutral 0 and one Cl is -1, then what will happen to three? What does -3 = 0 equal to? +3 so tell me, in the first one there is +3, here everything has +3 so this will not be able to answer. Next, if I look over here, what happens between the two chromium two benzene rings? The sandwich is done.

[00:07:10]Sandwich compound okay? So this one will donate the pi electron here and it gets sandwiched in between.

[00:07:21]So it becomes a complex.

[00:07:23]Correct? So look here, this neutral, this neutral, what will be the value of the one that needs to be found, x + 0 = 0 x equal to? 0 means zero of chromium here, so tell me, if two options are exhausted like this then what two possibilities are left?

[00:07:38]Now look, the last one is important. Will have to check here. K2 Cr CN is twice O 2 O2 and NH3, now see if O is written inside the complex then it is always an oxide. There is no problem in this. But if O2 is written then it can be both oxide and superoxide and peroxide. So tell me, if this is O2 - 2 then it is super oxide and if it is O2 - 1 then it is peroxide. Now this will have to be checked taking both the conditions into consideration. Ok? So I'm going to take you through Perox Superoxide once. Let's go -1-1. Correct? Now look, if we take -1 then let's start.

[00:08:41]On one potassium there is +1 and on two potassium there is +2, the x of the one to be extracted is -1 on 1 CN and on 2 CN there is -2 oxide O -2 and tell me if there are two then how much will it become? -4 And if I take O2 -1 then this is -1 and ammonia is 0 so tell me what happened? So just look here -2 +2, this is over. So tell me how much will this be? If -5 goes there, what does x equal? +5 Now look, there is no option +5 here. So tell me what happened to this condition? It got rejected. Now what is the remaining condition? Oh 2 - 2 So now I just delete this right here. This one.

[00:09:25]So tell me, if I take copper oxide, will it become O2-2? And tell me what will happen? Plus will become six.

[00:09:33]So only one option is visible. The answer will be third. Clear kids?

[00:09:39]This next question 50 what is the structure of I3- son I3- if we look at this central atom how many valence electrons does it have? Seven.

[00:09:55]Plus how many monovalent atoms side atoms are there?

[00:09:58]Two. And if it is -1 then we will add it and what will this do? Half. So tell me, seven and 3 10 / 2 5, then five and five, what does hybridization mean, son? sp3d ok? And how many are five minus side atoms? Two. So how much loan will be paid? 5 - 2 Three lone pairs.

[00:10:17]Three loan pairs. Now look at the case of sp3d and there are three lone pairs, what does the shape become? Linear. And if there is a lone pair with sp3d then lead shape. And there are two lone pairs with sp3d, so the bent T shape is understandable, son, the answer to 50 will be which is the fourth, next question 51 in mixture A and B components so negative deviation, that is, if A and B are mixed then there is negative deviation. Negative deviation means that the vapor pressure is less than expected.

[00:10:52]And the decrease in vapor pressure means the force of attraction increased after mixing.

[00:10:56]That is, what is AB interaction? More Aa interaction and Bb interaction. Ok? And what will happen after adding volume? Will decrease.

[00:11:04]i.e. Delta H DeltaV Mixing Greater than 0 Not Less than 0 What happened after mixing volume? Decreased. So tell me DeltaV mixing less than zero. So, firstly, this condition will be correct. And the second condition is AB interaction is stronger than Aa and BB interaction. Then this will be correct. And here it is saying AB interaction is weaker.

[00:11:27]Son, if AB interaction becomes weaker then A and B will move away from each other, that is, the volume will increase, that is, there will be positive deviation. So tell me, the answer will be B & D.

[00:11:38]Which one will be B&D Answer? Third.

[00:11:42]Next Question 52. Phosphorus has the oxidation state +3.

[00:11:49]In which compound is phosphorus in the oxidation state of +3? So you should definitely remember these three compounds. H3PO2, H3PO3 and H3PO4.

[00:12:03]Son, if we extract phosphorus here, it will be +1 here, +3 here and +5 here, its name is hypophosphorus. Its name is phosphorus acid and its name is phosphoric acid. Hypo Phosphoric Acid. Phosphoric phosphorus, hypophosphorus, phosphorus and phosphoric. So what is this? Phosphoric acid. So the oxidation number of phosphorus in phosphoric acid is +3. Next question kids. The next question is 53, it should be minus here.

[00:12:34]x i- So overall I will do the balancing later. First I am writing down those whose oxidation state is changing.

[00:12:43]I- MN O4 -2 and this I2 and this MNO2 and the reaction has to be balanced in basic medium. Ok? Let's start.

[00:13:01]Tell me its oxidation number is -1, its +7 and its zero and its +2. Now tell me the mole change in oxidation state from -1 to zero. How much change? Of one. What is the change in mole oxidation state of manganese from +7 to +2?

[00:13:22]Sorry, what is the change of four from seven, it becomes two, so what is the change of five, okay, from +7 to +2, what is the change of five, now tell me, sorry, it is +4, here it is +4, son, now look, it is +4, son, okay, so tell me, now what is the change of four and this is seven, what is the change of three, now tell me, we will do their simple ratio, first the change in oxidation state and then cross multiply. So this is already simple.

[00:13:55]So cross multiply so three here and one here. So tell me, is this work finished now?

[00:14:03]Correct?

[00:14:05]Now look, first of all we have to balance those elements whose oxidation has changed.

[00:14:11]So tell me three. So here it has to be 3/2 and here mn is one and here also mn is one so this is balance. Now tell me, how much should I multiply this reaction by, since it's a fraction? From two. So tell me 3 * 2 = 6 this will become six. This will become two and if we multiply this by two then how much will it become? Three and this will multiply by two.

[00:14:39]Now tell me what do we do after this, first we balance the oxygen.

[00:14:43]So tell me how many oxygens are there on the reactant side? There are 4 * 2 = 8 oxygen. And if we look at the product side, 2 * 2 = 4, so it means there is four here. There are eight here. So there are four less here. So I'm going to add four waters and four waters. And in this process, if four waters are added then 4 * 2 = 8 hydrogens i.e. 8H+. In which of these is this son's reaction balanced? In acidic medium. But in what do you have to balance yourself? In basic medium.

[00:15:10]So give me as many H's here as there are OH's here and there. So tell me, give 8 OH here and 8 OH here.

[00:15:25]Now tell me, will four 8 H 8 OH become eight waters?

[00:15:33]Eight waters. Now look, if these four waters finish these four waters, then tell me how many will be left here in the net? Four waters will be saved.

[00:15:42]And all this water got finished. Now in which has the final reaction balance been achieved?

[00:15:47]In basic medium. Now look at this, here i minus the number is being called x. That means it became x.

[00:15:55]This is your y.

[00:15:58]Water's number is z.

[00:16:01]And the number of i2 is P and the number of MNO2 is Q and the number of OH is R. So tell me 6 2 4 6 2 4 3 2 8 3 2 8 the answer will be third.

[00:16:26]Next question 54 of the following molecule has allylic carbon. Who has allylic carbon? Son, just look at the carbon next to the doubly bonded carbon and how should it be sp3 allylic carbon. The carbon next to the doubly bonded carbon but how should it be sp3 allylic carbon. The carbon next to the doubly bonded carbon and how should it be sp3 allylic carbon. The answer will be fourth. Everyone has allylic carbon and what are all these halides? Allyl halide. Allylic halide sorry allylic halide.

[00:17:06]Ok? Next question. Next question 55.

[00:17:10]Number of visible lines when an electron returns from fourth orbit to ground.

[00:17:16]That means where are you coming from, son?

[00:17:18]On first. So tell me please. If it is coming from fourth to first then add the difference of four and one.

[00:17:22]How much? Three. And till where should you write? Up to one. So 3 + 2 + 1, now tell me where are you coming? On first. That means who has the first number? Lyman's.

[00:17:34]Second number is Balmer. And the third number is of the Westerner.

[00:17:37]Now tell me where does this happen?

[00:17:40]Ultraviolet. Where does this happen? Visible.

[00:17:43]And where does this happen? Infrared. So tell me what are you asking here? That number of visible lines. And what is the number of visible lines, son? Two. The answer will be fourth.

[00:17:55]Is it clear? So visible lines answer will be fourth.

[00:18:00]Next Question 56. Which of the following cannot be made by using Williamson synthesis?

[00:18:07]Son, what can we not create through Williamson synthesis? So let us look at it using the Williamson method.

[00:18:12]First of all, what is Williamson reaction? If we assume that 1 degree alkyl halide, with what should we react 1 degree alkyl halide? From RON RON. Now look at it, it becomes ether ROR.

[00:18:36]Ok? Now see, if I react 1° alkyl halide with RON here, then a nucleophilic substitution reaction takes place and what is this mechanism? SN2 and if I take aryl halide here then this cannot happen here because it does not give nucleophilic substitution.

[00:19:00]Ok? Now look, this R can be anything. It could be 3°, 2°, 1° or even aerial.

[00:19:09]Is it clear? So this R can be anything.

[00:19:12]But this cannot happen here.

[00:19:15]So now if I take this aryl halide here, then tell me if both aryl aryl halides will be there, then such an ether which has this kind of.

[00:19:27]Here also 3 degree alkyl halide and 3 degree alkyl halide cannot form such ether. Even if it happens on both sides, it cannot be made. And if there is vinyl like this on both sides and vinyl on this side also, then also it cannot be made. And if there is vinyl or phenyl, we cannot make it.

[00:19:58]So if there are 3 degrees on both sides then it cannot be made.

[00:20:00]So tell me here di tertiary butyl ether di tertiary butyl ether. So what is there on both sides here too? Tercy-butyl ether. So from what do we form this ether? Ca n't do it through Williamson method. Is it clear son?

[00:20:20]57 Identify product A in the following reaction. So son, look here there is OH as well as ketone. So what will Zinc Amalgam with Concentrated HCL do? In what will the son reduce the CO? In CH2. There is a climate change. So look, this has become useless here. And this one has reduced carbon.

[00:20:40]And both of these are fine. Now but here concentrated HCL with anhydrous zinc chloride if sorry concentrated HCL with anhydrous sorry zinc amalgam with concentrated HCL what is this? Clemmensen Clemmensen.

[00:21:05]Ok? So what will Clemmensen reduction do? It will reduce CO to CH2.

[00:21:11]But keep in mind this is HCL acid.

[00:21:14]What will it bring here by removing OH also? Cl will bring it.

[00:21:16]So what will be the answer? Fourth. Now if you don't want to remove OH, what would you take here? If WF Kishan had taken the reduction instead of Clemson, OH this would have been retained here.

[00:21:28]Is it clear? Then if VF Kishan had taken the seat here then who would have been the answer? Third.

[00:21:33]And if you are taking concentrated HCL i.e. Clementine along with Zinc Amalgam then what will be the answer here? Cl will replace OH and CO will also be reduced into which?

[00:21:43]In CH2.

[00:21:45]So this type of question has come many times, son. The next question is 58 a species and number of lone pairs. Ok? Let's see how many valence electrons does a beta xenon have? Eight.

[00:22:00]Two of these electrons went into bonding with fluorine.

[00:22:02]So tell me, if two electrons are used in forming the bond, then how many are left here?

[00:22:07]Six.

[00:22:09]Six electrons remain. So how many loan payers are left? Three. So tell me just one matching four.

[00:22:14]Tell me here, one oxygen will be attached to the double bond. That is, 2 2 4 and two halogens 4 2 6, that is, how many lone pairs? One. So tell me, what does it match with? From seconds.

[00:22:25]XCO3 3 * 2 = 6 and 2 8 here means zero lone pair. So tell me, with whom does it match? From this. So tell me, four halogens, four electrons have been used 5 6 7 8, that is, two lone pairs, that is, its matching with the third, so tell me, A's matching with the fourth, A's matching with the fourth, B's matching with the third second, B's matching with the second, C's matching with the first and D's matching with the third, the answer will be second.

[00:22:55]Next question 59, the molar conductance of HCl and NaCl and CH3C CO, what is their molar conductance given respectively? 500, 200 and 100 ohm inverse centimeter square mole inverse. Ok? And who are you asking about? So we're asking for acetic acid.

[00:23:13]Now tell me, in acetic acid we need two acetate ions and H plus ions and we will get acetate ion from here and H from here.

[00:23:22]So tell me, please add the first value and the third value. But what extra is there in this? Na also came and Cl came. So tell me we will reduce NaCl. That means we will decrease the second value. So tell me, add first and third. So 500 and one 600 and 600 - 200 600 - 200 how much will this cost? 400 ohm inverse centimeter square mole inverse.

[00:23:45]Next Question Question No. 60 Match the column List One with List Two. The cell constant has to be mentioned here. Beta cell constant g is star. g star is son l / a so tell me if this is centimeter and this is centimeter square then centimeter inverse or meter inverse so tell me here meter inverse now there is something wrong here because meter inverse is two. Son, let me check it in the paper once.

[00:24:17]Yes, that's the problem here. In this fourth option, son Om inverse meter inverse is written. Ok? Ohm Inverse Meter Inverse. Now let's check it. So tell me what did I tell you about cell constant matrix?

[00:24:35]Either centimeter inverse or meter inverse.

[00:24:37]So tell me, with whom does it match? There are two places where A matches third and A matches third. That means this and this option became useless.

[00:24:45]Next, molar conductivity Simon centimeter square mole inverse or Simon meter square mole inverse, so tell me, matching of b to the first, matching of b to the first, conductivity is kappa, what is kappa, son, the unit of 1 / rho and rho is ohm in ohm * meter, so tell me, what will be its unit, ohm inverse meter inverse, so its matching will be done with this and degree of dissociation of electrolyte degree, degree of dissociation alpha dissociated mole upon in initial mole, so tell me, mole above, mole below, so tell me, what will be dimensionless, so tell me, what will be the matching of d to second, so matching of d to second will be the answer to the third next question, question number 61, which one of the following is an excellent substrate for SN1, so son, what do we make in SN1, carbocation, so the more stable the carbocation, the higher will be its reactivity for SN1.

[00:25:44]So tell me here this is Bromi, if this lone pair is single double then it becomes partial double. Vinyl Halide: Here also if vinyl halide partial double and partial double character comes then it will not give nucleophilic substitution. So this is useless, this is useless, okay? These are vinyl halides. Ok? Now look, if I look here, this is benzyl halide. Benzylic halide. So tell me, if this is removed here, what will a carbocation become? Stable through resonance. And if you remove this, then tell me its resonance will not be able to stabilize.

[00:26:20]So tell me, will resonance stabilized carbocation be formed here? So tell me who will be more reactive for SN1? Answer will be second.

[00:26:31]Clear kids? Ok? Next 62. Which statement is incorrect? Tell me which one is incorrect here. So son, this 3 degree alcohol, this 2 degree alcohol can be differentiated by Lucas. You can definitely differentiate. Because if it is reacted with Lucas, it will give instant turbidity and if it is reacted with Lucas because it is 2° alcohol, it will take 5 minutes for turbidity to appear. So we can distinguish it or differentiate it from Lucas. Next, this is an alkene and this is an alkane. Can be differentiated by bromine water. So bromine water means test of unsaturation. So this is unsaturation. There is no unsaturation here. So you can easily differentiate them with bromine water. Correct?

[00:27:14]Look here, son, this is a terminal alkyne and this is a terminal alkyne and this is an alkene. So tell me, will sodamide react with terminal alkyne? And it will not react with alkene? So tell me, this will also become differentiated.

[00:27:27]And the next one is saying that we will be able to differentiate these two from carbon amine? Will not be able to do it. So tell me, who is the incorrect statement here? The answer will be four. Because who gives us the carbalamine test? They give 1° amines. And here this is 3° amine and this is 2°, both of them do not give the carbon amine test.

[00:27:50]So tell me, we will not be able to differentiate them. Next question number 63 a + 2b gives c. For this the rate constant rate law is given as r = k a to the power of 1 and b to the power of 2. Now see if the concentration of a is kept the same, that is, the concentration of a is kept the same and the concentration of b is doubled. So you double the concentration of b and what is its order? Order two is in respect of B. So if we square two times two then tell me what will be the rate? Four times. So tell me, what will be the rate-itself quadruple answer here? What will the rate be?

[00:28:35]Four times.

[00:28:37]Question No. 64 Given below are two statements. Statement One The Molarity.

[00:28:43]Pay attention son, what is written? Molarity of the solution because molarity is moles of solute up volume of solution in liters. So tell me, the term volume has come here, so it will change when the temperature is changed. The statement is absolutely correct.

[00:29:00]What is next molality son molarity? Moles of solute upon mass of solvent in kg.

[00:29:08]This is molality, son, how many moles of solute are there in 100, sorry, 1000 grams of solvent, or instead of saying 1000 grams, how many moles of solute are there in 1 kg of solvent? So look, the molality is expressed in the units of moles per 1000 grams of solvent. That is absolutely correct. So tell me what happened to these two statements? Correct. Answer will be Statement one is correct but both statement one and statement two are correct. The answer will be fourth.

[00:29:51]Correct?

[00:29:53]Next question. Question No. 65 Using the data given below find out the strongest reducing agent. Son, tell me who is the strongest reducing agent in this?

[00:30:07]Son, if I arrange them in increasing order of SRP, then the one with the lowest SRP will be the reducing agent. So tell me, the value of SRP is given here because look, here from higher oxidation state to lower, higher to lower, higher to lower, higher to lower.

[00:30:29]So tell me, all these are reduction potentials and the reduction potential is more negative, so tell me, if the SRP value is the lowest, then it means which of these will be the reducing agent? One of these will be Cr+3 or chromium. Now look, higher oxidation state is the oxidizing agent and lower oxidation state is the reducing agent.

[00:30:47]So tell me how will CR work?

[00:30:50]Like a reducing agent. Is it clear son?

[00:30:53]This will not work like reducing.

[00:30:55]Cr+3 is okay? The next question, question number 66, says that lactose is milk sugar.

[00:31:05]Between whom does the glycosidic linkage occur in this? So son, if I talk about all the diacids, muscle maltose, sucrose, lactose. If I talk about its linkage then here 1 4 here 1 here also 1 alpha alpha alpha beta beta beta glucose glucose glucose fructose galactose glucose now see everything is included in this so tell me one carbon of galactose and which form is beta and four carbon of glucose and which form is beta, between these we will form glycosidic linkage, so where is the glycosidic linkage formed between C1 of beta D glycactose and C4 of beta D glucose? In lactose.

[00:32:07]Next Question Question No. 67 Which of the following option is correct? Son, boron is a second period element. There is no vacant D.

[00:32:15]So how many bonds can this maximum create? Four. And this has been made BF6 -3 so this is not possible. So is this correct or incorrect?

[00:32:25]What do we need incorrectly? Correct. In diborane diborane is B2H6. So son making bonding is found here. This is it son B2H6 okay? And if I see the hybridization of both the boron, then tell me 1 2 3 4, then this four sigma will be hybridization, son, in hybridization we count this banana bond. So hybridisation will be sp3 and shape will be tetrahedral. So in diborane structure each boron atom has sp3 hybridization. That is absolutely correct. The Hydration Enthalpies of Alkali Earth Metals. Son, the size of alkali earth metal is alkali metal and if we talk about alkali earth metal of the same period, then if we talk about their size, then the size of alkali earth metal is more and that of alkali earth metal is less, then tell me, if its size is less and its charge is also more, then tell me, its overall charge density will be more and if its charge density is more, then its hydration will be more, that is, the hydration enthalpy of alkali earth metal will also be more. So the hydration enthalpy of alkaline earth metals are smaller but larger than those of alkali metals. So this thing went wrong. So tell me which one did we get correctly? Seconds. This also became incorrect.

[00:33:56]Diamond has low melting, not son. There is high melting, son. Region 3D network structure. Ok? Next question. 68 This is the son. If cumene is reacted with air or O2, then this O2 will get inserted between this benzylic hydrogen and what will be formed?

[00:34:18]Cumene hydroperoxide. So this is cumene hydroperoxide. I am making the structure.

[00:34:26]These and two methanes.

[00:34:29]This is cumene hydroperoxide. Now when we hydrolyze it, the bond breaks and it will get attached here, so one acetone and one phenol. So this phenol has been given, that means who is B? Acetone. CH3 CO CH3 ok answer will be which one will be the answer son?

[00:34:50]This is cumene hydroxide and this is acetone. The answer will be third.

[00:34:56]Clear kids?

[00:34:59]So the next question is question number 69.

[00:35:03]In question number 69, it is said that the standard electrode potential for the Daniel cell is given as 1.1 volt, so here E not cell is given as 1.1 volt, clear children, if you want to tell the approximate free energy of the standard given cell, then tell me delta g not, if you want to tell delta g not, then what is the relation between delta g not and e not cell, son? Deltag not is equal to -nf nfe not Deltag not is equal to -nf not and here son, what is minus n, how much total electron has changed here, so here from zinc to zinc +2, two electrons are lost and from copper 2 plus, two electrons are gained, that is, here the total n factor here is how much, what is this f, 96500 Faraday constant 96500 and what is E not, son? 1.1 And since this is kept in SI, this value will come in Joules and we have been given kilojoules in all four options, so to convert kilojoules, we will divide by 1000. Ok? And we will solve it.

[00:36:26]How much will this value be? of minus 212.27 kilojoules per mole. That means the answer will be third.

[00:36:36]Which will be the answer? Third.

[00:36:40]Question number 70. Very easy question.

[00:36:43]The boiling point of benzene, that is, pure benzene, that is, the boiling point of solvent, TB solvent, what is that, son? 300 53.23 Kelvin. Now see here, when we add 1.80 grams of non-volatile solute to it, then how much does the boiling point increase to? That means the TB solution comes and it becomes 354.11 Kelvin. Ok? This means what will the delta TB be? 0.88 Kelvin.

[00:37:29]This is delta TB, so we will convert TB solution minus TB solvent, how much will it be, 0.88 Kelvin, so okay, we know delta TB and here we know the mass of solute, and how is it? It is a non-volatile solute.

[00:37:42]What will be its n factor? One want of factor one will become.

[00:37:44]Dissolve in 90g benzene. So it is very easy. The mass of the solvent is given. The mass of the solute is given and the approximate molar mass of the solute is being asked. So very easy. Delta TB equals I * KB * M.

[00:38:04]What is the value of I that I have put here?

[00:38:07]One KB son, I have given you how much here?

[00:38:10]2.53 2 53 2.53 ok?

[00:38:19]Molality: Son, molality is moles of solute. Now son, did you give me moles of salt? If the mass of solute is given then 1.80 this mass in grams becomes divided by molar mass and I am writing that molar mass with capital m. Ok? This is the molar mass of the solute. Correct? into below weight of solvent in kg but weight of solvent in what? In the village. So round down 90 and divide by 1000 to convert grams to kg. So it will come up to 1000. Ok? And how much did this Delta TB give you, son? 0.88 And now what do I have to remove here? Molar mass of solute. So see what will be the molar mass of solute? 2.53 2.53 * 1 80 * 1000 divide by 90 and this 0.88 0.88 and if you solve this son, then how much will you get? 58 grams per mole.

[00:39:28]Clear son? If we divide this very easily by two zeros, it becomes 180 and see, if it is divided by two, this is fine son, then the answer will be 2.53 * 2 / 0.88, what will be the answer?

[00:39:43]58 Which son? 58 Which will be the answer?

[00:39:46]Fourth.

[00:39:48]Clear kids? Okay, next question. This is an easier question, son. Which has maximum number of atoms? So here we are asking about the maximum number of atoms. Now look, 0 point is talking about atom and what is it giving? Molecule. Ok? So what will happen if we multiply the atomicity of a molecule?

[00:40:08]Number of atoms. So let's see where to start? One gram molecule. And gram molecule is son of mole. So what does that mean, how much is it worth? Forest mole. And how much will the mole * Na become? If we look at the molecule and into atomicity, then tell me how much atomicity it has? 4 5 6 7 So this will happen. So tell me how much did you solve? 0.7 Na, this is how many atoms there are. Let's move ahead. Again gram molecule is again mole, if we convert this into one then it becomes number of molecules.

[00:40:46]In atomicity four and one 5 so this will become son 0.5 Na come here 5.6 liters if there is a gas at STP then how much is 1/4 mole how 5.6 / 22.4 how many times will this go? Four times.

[00:41:05]So tell me this is 1/4 and 1/4 son, 0.25 Na * atomicity 3 and one 4, so tell me if we do 0.25 * 4, what will this become, one means Na 36 ml, okay this is water, now son, what is the density for water, one so 36 ml means 36 grams and divide by molar mass 18, so what is this mole, son? If we add two moles and make it 2 * Na then the number of molecules in atomicity will be three, so 3 * 2 = 6 i.e. 6NA, so tell me which will have the maximum number of atoms? This 6NA, this Na, this 5 Na and this 7NA will be the answer, which son? Four. The next question is question number 72.

[00:41:58]Son, this is a question of Kjeldahl method and the percentage nitrogen has to be calculated here and the mass of organic compound has been given to you here, 0.50 grams. Now look, here the formula of percentage nitrogen in Jeddel method is 1.4 * milli equivalents of H2SO4 milli equivalents of H2SO4 but keep in mind that these are the milli equivalents of H2SO4 which have been used in neutralizing ammonia.

[00:42:30]Only those mill equivalents of H2SO4 which are used in neutralization? of ammonia divided by the mass of the organic compound in grams. Correct? So, this is the formula. Now look, you have 50 more centimeter cube, son, it is ml.

[00:42:51]So 50 ml 50 ml one normal H2SO4 So tell me how many millents were there in the beginning?

[00:43:02]How much do you get from 50 * 1 of H2SO4? 50 Now look at it, some part of it was used to neutralize ammonia and some to neutralize NaOH, so how much of 50 is used to neutralize NaOH and how much to neutralize ammonia, so look here it is saying that residual acid needed to 60 cm of N/2, so tell me how much will be required to neutralize it, i.e. how many NaOH solutions will be there? How much is the equivalent of 1 mil of NaOH? 60 * 1/2 means how much is it? 30 So tell me, 30 ml equivalent NaOH will neutralize its 30. So the remaining 20 will be used to neutralize the ammonia. So tell me, what is the mill equivalent of ammonia?

[00:43:53]20 milliards.

[00:43:56]Is it clear? So tell me, what will be the number of that milli equivalent of H2SO4 which is neutralizing ammonia? 20 millilitres. So tell me, 1.4 * 20 divided by mass of organic compound 0.2 0.5 0.5 and 0.5 means 1/2, so I am writing it as two above. Is it clear? This point from this zero point flew away so 14 2ni 28 28 2ni 56 answer is done which son? The fourth answer will be fourth.

[00:44:36]Next question question number 73, son, here we are getting benzene chlorinated in the presence of anide ClCl3. So son, this is Lewis acid. Halogen is the carrier.

[00:44:48]what will it be? Cl- will pull. So it became chloronium. And this electrophilic substitution, what will it become, son?

[00:44:54]Chlorobenzene. So this is A what happened? Chlorobenzene.

[00:45:00]This is chlorobenzene. Now if we react it with NaOH at high temperature and high pressure, then it will be aromatic nucleophilic substitution. And what will it become?

[00:45:14]OH here comes OH. But this is again acid base, that is, what will finally be formed here?

[00:45:20]Sodium phenoxide. That's why we had to do acidification. So what will it become after it is acidified? Phenol.

[00:45:29]Clear kids? So the answer will be phenol. Third. Is it okay son?

[00:45:38]Next question. Question number 74. Son, here you are given the toll and access chlorine in the presence of sunlight. So son, chlorination is taking place in the presence of sunlight.

[00:45:48]So this is free radical substitution and free radical substitution is not the reaction of the aromatic part. Whose is it?

[00:45:57]Who will react? of the alkane part. So this reaction will be here. So what will be installed in turn after all three hydrogen huts? Chlorine. So tell me this. This A is C and these three Cl i.e. CCL3. Now its reaction is concentrated HNO3 this is nitration and what will it provide? NO2+ is nitronium and it is electrophile. Now tell me what will happen? is working like -h.

[00:46:22]So -H what kind of director? Meta Director. So where will this incoming electrophile attach?

[00:46:27]on meta. So son, what will happen to this B?

[00:46:35]Nitro on CCL3 and these metas. Now son, when we make its reaction SN Zinc with HCL, Zinc with HCL, then it will also reduce CCL bond and it will also reduce NO2, in which? In NH2. So keep in mind son, Zn + HCl also reduces NO2 into what? In NH2. And this will also reduce the CL CCL bond.

[00:46:56]And this CH3 will come again. And here NH2 will be the answer, which one? Third. Clear child?

[00:47:05]This will become NH2 and this will become CH3.

[00:47:10]Next question 75 son, this is aniline and this is benzene diazonium chloride and son, this much part here, what is this?

[00:47:22]What will this electrophile and this benzene part give?

[00:47:24]Electrophilic Substitution. So tell me, what kind of director is this? This is +m. And +m how director? Ortho Para. So tell me which one will be the major? Paragraph. So where will this electrophile son connect? Paragraph on and what will it become?

[00:47:38]NH2 N double bed N, this will be called para amino azo benzene. Para amino azo benzene. And what kind of dye is this son? Yellow or orange? So the answer is it is yellow colored dye.

[00:48:02]Next question question number 76 son here is cyano benzene. Ok? Its hydrolysis with SnHCl to form SnHCl. So this CN change into CHO and this reaction is Stephen reduction. Is this what you call it? Stephen Reduction. So cyanide changed into aldehyde in the presence of Sn with HCL and also followed by hydrolysis. So the name of this reaction is Stefan reduction. The next question is question number 77. The pKa of acetic acid and pKb of ammonium hydroxide. So if this is an acid then its pK and if this is a base then its pKb.

[00:48:46]This is given to you. Ok? The PA is saying that he wants to tell you about ammonium acetate. Ammonium acetate. Ammonium acetate is the salt of a weak acid and a weak base. A weak acid is a salt of a weak base.

[00:48:58]So tell me what is the pH of the salt of a weak acid and a weak base? From plus 1/2 pk - 1/2 pkb it swelled up. Now tell me 7 plus 1/2 PKA and PKA son how much have I given you 4 76 minus 1/2 * PKB and how much is PKB son 4.75 4 75 now look this is son sorry what did I write 4.7 76 and this 4.75 so what is this small value? Is too much. This value is a little high and there is not much difference between the two. So tell me the answer is approximately what is seven? Is equal.

[00:49:56]So the answer will be around seven. But the value that is added is more. There is less that happens. So this answer will be slightly more than seven and if we solve it exactly then what will be the answer?

[00:50:09]7.005 7 5 clear kids answer will be second next question 78 there are three elements you have three elements a b and c having outer most electronic configuration this is given son ns2 tell me this is 3s2 and 3s2 means ns2 let's say this is your b element ns np4 4 3p4 so I am calling it NP4 and this is 3P5 i.e. NP5.

[00:50:51]NP5 right? Now look, they are saying which statement is incorrect?

[00:51:01]For this, we have to keep in mind which statement is correct for these ABC elements.

[00:51:04]Order of ionization energy. Now look, if I look at this, then tell me here NH2, after this NH2 NP1 NH2 NP3 NH2 NP4 NH2 NP5, then tell me here if I talk about NH2 NH2 NP1 NH2 NP 2 NH2 NP 3 NH2 NP P 4 NH2 NP 5, now tell me this is your alkali earth metal, this boron family, this carbon family, this oxygen sorry nitrogen, this oxygen and this halogen family, now look, if I look at the value of their ionization energy, then tell me what will be the value of ionization energy, there is just one exception, its ns2 is more than np1 and p3 is more than P4. Is it clear? So tell me what is the order? NH2 will be less than NP1. What more could there be than that?

[00:52:15]Whose ns2 will be more than that? NH2 NP2 will be more than that of NP4. And more than that will be NP 3 and more than that will be NP5.

[00:52:29]Now look, I have not even given myself everything here.

[00:52:32]ns2 np4 ns2 ns2 is np4 and np5 is there so look, the ion energy of this is more than this and this is more than this and here it means A is more than B, C is more than B, now see what is given here, A is more than B, C is more than B, so what is this, is the statement correct and he is asking us incorrect, so this is correct, delta H EG electron gain enthalpy, which one has the maximum in any period, son, halogen, so tell me, this is halogen, what is this? Halogen is less than that and chalcogen and this is chalcogen. So tell me about halogen and then chalcogen.

[00:53:09]And what is this electron gain enthalpy of alkali earth metal? Zero. So tell me what will happen here at the back? So tell me, who will have the most? of C.

[00:53:19]Then less than that B. Less than that. This is also correct. Come on. The compound form between A and C is ionic between A and C. That's absolutely right, son. What kind of bond will be formed between this alkali earth metal and this halogen and alkali earth metal, a metal and a non- metal?

[00:53:35]Ionic bond. This is also absolutely correct. And saying that no bond compound will be formed between B and C. A covalent bond will be formed between B and C.

[00:53:46]Which bond will be formed between B and C?

[00:53:48]Covalent. So tell me what happened? The statement was wrong. Which will be the answer? Four. Now see how it will be made?

[00:53:55]This is the oxygen family, son, and this is third period. If I talk about it, it is 3p, then tell me, it is 3p4, then tell me which third period is this and which chalcogen family member is it in the third period?

[00:54:08]Sulfur Now how many valence electrons does sulfur have, son?

[00:54:14]Six ns2 np4 right? Now look, it will form this bond with two halogens. Covalent bond will do the sharing of one by one. So tell me, covalent bond will form here. Ok? So this answer was wrong. So which statement is incorrect, son? Four. Next Question 79 Lanthanoids R Beta Lanthanoids are sixth period elements.

[00:54:37]First of all, from where to where? From 58 to 71. From atomic number 58 to 71. And which one is filled here? 4f. So look, there are 14 elements. That is absolutely correct. There are six periods. That is absolutely correct. And ranges from 58 to 71. And here the 4f subshell is filled. And what is this statement? It is correct.

[00:55:01]Next question Question number 80 The first order reaction The correct graph for first order is to state log concentration versus sorry log initial concentration upon final concentration versus time. So let's start. Son, you have an equation for first order CT = CT = concentration after time T C not initial concentration E to the power - KT. What is this equation for? The exponential graph is of CT concentration versus time. Now see, there is no concentration here. Is of log. So see, first of all I have AT'd this here and made it A not.

[00:55:47]Sorry R not this R and this R not same same E power - KT now look what more do I need? R0 / R So bring R0 / R R0 / R So it will come down. Ok? So 1 up to the power of e - kt So what happens if we bring this up?

[00:56:13]Now if I take its log form to the power of kt e, then tell me what will its ln form become?

[00:56:23]KT = lg r0 / r So look, sorry, ln, what should I write here? ln Now look, I am writing here ln r0 / r equals kt, so tell me what is yours here, now here it is in the form of log, so change ln to log, then tell me what will come out to be log log 2.303 log r0 upon r equals kt and if we remove this from here then it will come down to 2.303, okay now tell me, this is your y and this is your x, so what will be the graph of y = mx, straight line passing through the origin and its slope will be k / + k / 2.303 Next question 81 The Geometry and Magnetic Behavior of the Complex.

[00:57:36]Its geometry of magnetic behavior has to be explained. Ni CO4 Now tell me what is the niche here? Oxidation number zero and what is the meaning of oxidation number zero, son? 28 So 4s2 3d8 4s2 3d8 Now look always s we need empty. So where will these s electrons go? In n - 1d, tell me this will be empty, so this will become d10 and this will become s0.

[00:58:05]So I'm going to give this this is son d and this is 4s 0, okay? I transferred these two here.

[00:58:19]Now tell me, s is completely empty and what is your next? 4p0 and 4d0 coordination number is four so tell me this two these are empty three this further is empty so tell me what will be the hybridization sp3 unpaired electron unpaired electron zero and hybridization will be sp3 what will be the shape tetrahedral let's check for this square planar is wrong tetrahedral geometry will be diamagnetic absolutely correct unpaired electron is zero which will be the answer second next question next question is 82 vi of the following will favor maximum formation of product in the reaction maximum formation of product in the reaction so tell me what should be the condition for maximum formation of product tell me this reaction is delta H positive that means what kind of reaction is it endo and endothermic what temperature is required for maximum formation of product in the reaction?

[00:59:23]Hi. So these two options turned out to be wrong. Now tell me, if I look at the gaseous mole of the product here, it is less.

[00:59:32]What is the gaseous moles of the reactant? Is too much. So tell me, on increasing the pressure, the reaction goes towards the side where the gaseous moles are less. So tell me, the gaseous mole is less here, right? So where will the reaction go when the pressure is increased? Forward. So tell me, does this mean that pressure is also required? High pressure. So if you apply high pressure, where will the reaction go? Forward. That is, towards Les Moles. The answer will be first.

[00:59:54]Next Question 83. Match the species given in column one with the shape. So species are given here.

[01:00:02]Shapes are given here. First of all let's look at ammonium, which is easy. So tell me what will happen to its hybridization? sp3 zero lone pair and sp3 zero lone pair. What will the shape be like? Tetrahedral.

[01:00:14]So matching first of D to matching first of D has given son in the same place. The answer will be given in seconds. Let's tell the next Bro, please find the steric number half. The central atom has seven valence electrons plus the monovalent atoms have zero and -1, so what will we do with it? Plus seven and one 8/2, this will come to four. Four means hybridization has become sp3 and four minus side atoms three.

[01:00:40]That means how much is the loan payer? One. And if there is one lone pair with sp3 then what will be the shape? Pyramidal means C's matching second to C's matching second. Now come to C B.

[01:00:51]Brf3 again steric number half of Brf3 central atom has valence electron seven plus monovalent atom halogen three.

[01:01:01]Fluorine let's say 7 3 10 / 2 means five. Now tell me, what does five mean, hybridization? sp3d Now tell me 5 - 3 means if it is a case of two lone pairs and sp3d and there are two lone pairs then what will be the shape, bend shape means matching of B with four, matching of B with four, now tell me sf4 again find out the steric number, half central atom balance electron six plus monovalent atom four 6 4 10 / 2 five now five has come, that is hybridization has happened sp3d now look 5 - four means again one lone pair and if there is one lone pair then tell me what will be the shape of this? See Saw Shape See Sawsap Clear? So this will be A's matching third. Which one will be the answer from matching third of A?

[01:01:58]Seconds.

[01:02:01]Okay kids?

[01:02:04]Next question 84. Look here, there is magnetic moment manganese plus X ion.

[01:02:10]X tells us whether it is +2 or +3 or +4 or +6. Its magnetic moment in Bohr magneton is the magnetic moment of mx +3. Son, here it is missing 15 which was given in the question.

[01:02:28]15 bore magnetron is given. So saying what will x be? Son, if the magnetic moment μ is 15 Bohr magnetons, then it means you should know how many unpaired electrons should there be, son? Three.

[01:02:41]How? The unpaired electron sorry if n then the value of magnetic moment is n + 2 Bohr magnetons. Now look, if there are two unpaired ones, then tell me if I keep two here then 2 + 2 = 4 sorry not two, there will be three unpaired ones, so let's keep three, 3 + 2, so this is five and 5 * 3 = 15, so tell me if there are three unpaired ones, then what will be the magnetic moment, 15, and here the magnetic moment is given as 15, so the unpaired electrons are three, now tell me the unpaired electrons are three, and if I look at neutral manganese, then tell me, 4s2 4s2 3d5 4s2 3d5, now look, here, first of all, if I remove two electrons, if I remove two electrons, then this 3d5 will be left, this 1 2 3 4 5, after removing two electrons, this d5 5 and how many unpaired electrons do we have with d5? Five. So that means I have to take out two more. So tell me, if we take two more out of this, then tell me, these two have been taken out and if we take two more out, then tell me, it will become mn + 4. And if mn becomes +4 then what is this? This whole thing disappears and what will be left here? d3 d3 and if this becomes d3 then this will disappear this will disappear this will be saved.

[01:04:22]How much is unpaired three, so tell me which one here, it is given here mn + x, so tell me how much will x be equal to plus 4, it is clear, I understood the matter son, let's go to the next question 85 The basic character of the transition metal monoxide, son, transition metal monoxides are given, tell the order of their basic basic stand, son, what are the oxides of metals? Basic in nature. So if we talk about the most basic, then whose oxides are the most basic? Alkali metal. Less alkali earth metals. Then D Block. So tell me, the basic strength decreases as we go from left to right. So tell me, will the basic strength decrease if you go from left to right?

[01:05:08]So tell me who will be the most basic? Titanium, less vanadium, less chromium, less iron. So tell me the answer will be that the basic oxide will decrease when moving from left to right. The answer will be first.

[01:05:23]Basic strength increases from top to bottom and decreases from left to right.

[01:05:30]Next question 83 Whenever white phosphorus is reacted with aqueous NaOH, it is a decomposition reaction and what is formed, son? PH3 and NaH2PO2 you should know this. NaH2 PO2 NaH2 PO2 ok? And if we calculate the oxidation number of phosphorus and this phosphorus, then what will it be? Plus minus 3 and how much will this come to? Plus 1 is okay son?

[01:06:12]How to extract Hydro Sodium +1? +1 of one hydrogen is +1, then +2 of two hydrogens is the one to be removed, its x and -2 of one oxygen is -4 of two oxygens, if we make it equal to 0, then x will be equal to what? +1 Similarly, here x and one hydrogen non-metal, this one hydrogen, how much will this be? +1 so three hydrogens +3 = 0 x what will it equal?

[01:06:37]So tell me 3 of minus, here phosphorus is zero, so 0 to -3 and 0 to +4 means the same element is also getting oxidised. Same element reduction is also happening. So what kind of reaction is this? Disproposition.

[01:06:52]Whenever white phosphorus is boiled with aqueous NaOH then oxidation number of phosphorus in products will be So tell me, the oxidation number of phosphorus on the product side is -3 and +1, so -3 is common and +1 is at the same place. The answer will be second.

[01:07:10]Next question 87 In which of the following compounds pi d pi bonding is present. So saying which has p pi dπ.

[01:07:22]Where should the elements be to be beta p pi dπ?

[01:07:25]Where should at least one element be? The side atom or central atom must be in the third period or below. Third period or below. Because the second period has only p. D does not happen. P exists, D does not. Now tell me, here both the second periods i.e. the bonds that will be formed will be pi pi pi pi.

[01:07:48]If the bond of both the second periods is formed here, then whose will it be? Pi pi pi of.

[01:07:53]Look here, whose element is this below and this is the second period, so second period and third period, second period, third period, so tell me, here pi pi d pi pi pi d pi will be formed here, C and D, answer will be first, next question, question number 88, out of the various possible isomers of C7H7Cl, so C7HCl7, sorry, C7H7 Cl, containing a benzene, it has a benzene ring. So you are asking me to tell you who has the weakest CCL bond? So let's look at these options. Ortho chlorotolan this is tolan. These ortho chloro again para chlorotolan these tolan para pe chloro. Third is benzyl chloride CH2 this Cl this has become. Now look, here it is being said that in these isomers whose C7H7Cl has benzene ring, so among them it is being said that who has the weakest CCl bond, son, if I look here, lone pair single double, that is, what is there here, partial double resonance, here also there is partial double resonance and here, son, lone pair single single double is not in resonance, so it is complete single, so tell me here, complete single and here partial double partial double, so this bond, carbon chlorine bond, will be the weakest, the answer will be which third 89. Now look, this is a very easy question and a good question, here this is the substrate, if we are reacting it with reagent C, then this product is being formed, if we are reacting it with B, then this is being formed, if we are reacting it with A, then this is being formed, now see, if you look carefully at B and A, H and OH have reacted at this very place. H and OH have reacted at this place. That means the rearrangement that could have happened here because if this 1° or 2° or 3° was formed then it could have rearranged and reached here. So this means this is such a reaction ah if in both of these this is such a reaction in which what is the reagent? Where rearrangement is not possible. Correct? So that's the oxidation of Marconi or anti-Marconi. So tell me, here the negative part is connected to the place where there is more hydrogen. And look here, the negative part is connected where there is less hydrogen. So the negative part is where there is less hydrogen, that is, what rule is this? Where does Marconikoff and Marconik take place? OMDM oxymercuration demercuration. So tell me who is this here in BK? Reagent is third. Look here, antimarconic oxidation has taken place here.

[01:10:49]Ok?

[01:10:51]Antimarconic oxidation has occurred here.

[01:10:53]So tell me where does antimarconic oxy addition take place? Hydroporation Oxidation.

[01:10:57]And where is the hydroboration oxidation?

[01:10:59]In seconds. Now look here, first of all look here. OH given here is 3°.

[01:11:06]This means that carbodione must have been formed. There must have been rearrangement of carbodione. So this means this is acid catalytic hydration.

[01:11:12]So tell me who is Jara Si? It is first.

[01:11:15]Who is C? It is first. Ok? So tell me first of all how to tell it here?

[01:11:21]First tell A, then B, then C. So A means second third first. Second third first answer will be third.

[01:11:34]Next question. Question number 90. The correct order of acidity of the given acid.

[01:11:40]So look here, this monocarboxylic acid, this is dye, this is also dye, this is also dye, now tell me, as the number of carbons increases in dicarboxylic acid, then tell me, the distance of -I will increase and the acidic strength will decrease, so tell me, here between both the COH, there is zero carbon here, one carbon here, two carbons, four carbons, so tell me, as the number of carbons in between increases, the effect of -I will decrease and the acidic strength will decrease, so tell me, first of all, there will be more dicarboxylics and less monocarboxylics. Now tell me which one is the biggest among these? Second is first, third is less than that, fourth is less than that. So tell me which will be the answer 2 3 4 1 2 3 4 1?

[01:12:26]Third. Ok? So acidic strength in directly proportional to -i. And for -i we see, son, for inductive effect we see, what?

[01:12:39]Distance than number than power, so as the number of carbons increases, the distance will move away, so tell me, the effect of -i will decrease, the acidic strength will decrease, clear children, okay, so in this way your 90 questions have been discussed, okay children, thank you so much.