RE-NEET 2026 | CHEMISTRY PAPER DISCUSSION | All India FULL SYLLABUS TEST (RE-FST-09) #newlight

Added: 2026-06-04

[00:00:00]Jai Hind dear children, how are you and today we have brought the discussion of REET full syllabus test number nine, I Anil Sanjeev Bhatia will discuss the first 23 questions with you with this mission of our Singh Sir, now everyone has the right to study and everyone has the right to become a doctor, so come children let's start question number 46, here we have been given 6.02 * 10 to the power 20 molecules of urea which is present in 100 ml of solution and we have to find the concentration i.e. molarity. So children, we all know the formula molarity equal to 1000 * number of moles up volume in ml, we will apply this formula here.

[00:00:46]Now 1000 now because the number of moles was not given to us. We are given the number of particles or molecules. So we will put n / na in place of mole.

[00:00:56]That is, 6 * 10 to the power 20 or we will first find out the mole and 6 * 10 to the power 23 inu volume is given to us in ml i.e. 100 ml. Now if you solve this, you will see that this 10 to the power 20 becomes 10 to the power 23, so it will cancel out. When it gets cancelled, this entire forest will come. If 100 goes up then it will come to 0101 molar. That means we will go here with option number second. Option number second will be correct for question number 46. Let's move on to the next question.

[00:01:35]What is question number 47 telling us?

[00:01:37]Which is not an example of a redox reaction? A reaction in which the oxidation number does not change will not be a redox reaction. The reaction in which the oxidation number changes will be a redox reaction. Now here the oxidation number of mn is +7. Look, here the oxidation number of mn is +4. That means the oxidation number is changing. So this is an example of a redox reaction. Here an arrow is given in the middle.

[00:02:01]Now look at this, here in the arrow, chlorine is reacting with water to form HCL and HCLO. There is 0 here. Here the oxidation number is -1. Here the oxidation number is +1. That means this was also redox. This is also redox. And in Cro2- the oxidation number of chromium is +6.

[00:02:22]Here also the oxidation number of chromium is +6. That means it is not redox. This option will be correct. Here the oxidation number of MN is +7. Here the oxidation number of Mn is +2. So this reaction is our example of a redox reaction. Therefore option number three will be correct here. Which of the following is incorrect? Look, the first one is incorrect because we have formula E note cell equal to 2.303 RT / NF log KC, there is no minus in it. There is a minus here, hence it is wrong.

[00:03:02]DeltaG Knot - NAP Knot is correct. DeltaG should have been minus in the knot. It is wrong because it did not feel right. So therefore both first and third will be considered correct. Both formulas are wrong. Where it should not have been minus, it was applied and where it should have been, it was not applied.

[00:03:21]The Correct Order of Energy for Hydrogen Atom In hydrogen atom, we have to keep in mind that the energy of the orbitals or subshells of the same shell is equal. So 1s will be greater than 2s and 2s and 2p will be equal.

[00:03:36]Here the above principle's n + rule will not apply. Here the energy of the orbitals you have will depend only on n. So this will be equal to 2s and 2p. Then 3s will be equal to 3p.

[00:03:50]So if we go according to this then option number four will be correct for us.

[00:03:54]Because there is more 2p than 2s. This went wrong. Here 1s is equal to 2s. This went wrong.

[00:04:01]Here also 2s = 2p. 2s = 2p, that's correct. But the 3s evened out. This went wrong. In this way the fourth option of 49 will be correct.

[00:04:15]Conversion is when ROH reacts with PCL3 to form a halide. Now the question is which halide will it be? Now Woods reaction is taking place with sodium. So what you do is break this bond symmetrically from here and give it a Cl.

[00:04:30]Give him a Cl. So its meaning is that our alkyl group will become CH3CH3CH i.e. the second option will be correct.

[00:04:42]See the next question. Which one of the following reactions wood produce secondary alcohol. Now look, this is OH- and Br2, that is, NaOH and Br2, so this is our haloform reaction and from this we will get CH3, this bond will break from here. CH3 COO- Now here if there is Na+ then sodium salt will be obtained and if there is K+ then potassium salt plus CHBr3 CHBr3 will be obtained. So right now we have n't got any secondary alcohol here. This LiH4 will reduce it. An H will come here.

[00:05:21]An H will come here. So I'm going to get C6H5 COH CH3 This is a secondary alcohol. So the second option became correct. This LH4 will reduce it to aldehyde and primary alcohol. And look at this CH3 MgBr this CH3 will get added here. If followed by hydrolysis it will be converted into tertiary alcohol. will convert to tertiary alcohol. Therefore, the correct option will be our second option of 51. Our second option will be correct. Let's move on to the next question. Now question number 52 is coming to us. So what will happen in 52 questions? Two statements are given. Lactose is a disaccharide made up of glucose and galactose. This is absolutely correct. You can read it in Bio Molecules.

[00:06:19]And so a linkage between two monosaccharides like glucose and galactose through an oxygen atom is called glycosidic linkage. This is also absolutely correct. The OH of glucose and OH of galactose will come closer to each other. Water will be released. And from the carbon O linkage of both, an ether linkage will come here.

[00:06:44]So this is going to be called our glycosidic linkage.

[00:06:46]Both statements are correct. So both statements first and second are correct i.e. should go with the third option.

[00:06:55]Here we will go with the third option in 52.

[00:06:59]See question number 53. It is Asan. Tertiary amines do not react with benzene sulfonyl chloride i.e. Husberg reagent. Absolutely correct.

[00:07:11]Because tertiary amines do not have H.

[00:07:14]And when there is no H then HCL will not be removed from benzene benzene sulfonyl chloride.

[00:07:20]So there will be no reaction. The Reaction of Benzene Sulfonyl Chloride. Now benzene sulfonyl chloride suppose we have C6 H5 H double BO double BD OCL. Now with what have we reacted this benzene sulfonyl chloride? From primary amines. Primary amine. So this HNH R will release HCL from here. So what we will get is C6H5 H double BO double BO NH R So now look here we have an alkyl group attached to S KN. R is a connected.

[00:08:07]If it is a primary amine then we will have an R attached to it, so we will call it N alkyl benzene sulphonamide N alkyl benzene sulphonamide, this N N di alkyl is wrong, yes if it was a secondary amine then N N di alkyl would have been accepted, hence here our assertion is wrong, tertiary amine does not react with benzene sulphone chloride, this is correct and because the reaction of benzene sulphonyl chloride with primary amine is wrong, so the assertion q R is false, that is, the first option will be 53. Let's move on to the next question.

[00:08:55]Question No. 54 Nessler reagent is used to detect. So what does the Nessler reagent detect? to the ammonium ion. Nessler reagent is a mixture of K2Hgi4 and KOH. That is, alkaline potassium mercuric iodide. It is a mixture of K2Hgi4 and KOH. From this we get brown colored iodide mixing base.

[00:09:26]This iodide combination will give us a brown coloured base.

[00:09:33]We'll have this Hg. On one side of this we will get NH2 and O Hg I. This iodide union base will become brown in colour.

[00:09:53]So we will use this here.

[00:09:56]Now let's see what comes next for us? Its formula is also made like this.

[00:10:01]However, he will not ask you for the formula. You can make it in this way also.

[00:10:15]We can make it like this also.

[00:10:19]It can be made in both ways. Or you can make it like this. One became oxygen free.

[00:10:24]This happened in an oxygen ring structure.

[00:10:26]Two Hg arrived. It became NH2+ and I-s added to it. Both can work in any structure. It will be brown in colour.

[00:10:34]Therefore the fourth option will be correct.

[00:10:39]Here we have to create the rate expression of kinetics.

[00:10:41]Rate reaction has to be created.

[00:10:44]So we know that here for N2O5 we will make - 1/2 DN2O5 / DT = + 1 / 4 DNO2 / DT = +1 / 1 DO2 / DT this is our main equation.

[00:11:12]Now we gave - DN2O5 DT in the first equation and multiplied O2 by 2.

[00:11:19]So this is fine. We want the incorrect one. - Gave 1/2 DN2 / DT and put DO2 / DT in plus as it is. Correct.

[00:11:30]Instead of DNO2 / DT 1/4, multiply 4 by O2.

[00:11:36]This is also correct.

[00:11:39]What did you do in this? Removed two from here and brought four here. So it went wrong. Therefore the fourth option will be incorrect.

[00:11:49]The given orbital which has electron density along the axis. So now see what happens in d? If we created dxy in d. Let's say this is the x axis. This is the y axis. This is the z axis. So now if we created dxy, one electron lobe would be between x and y and one would be here. That means it will keep touching z lightly.

[00:12:21]And this will just be the electron density in the middle of the plane.

[00:12:26]If we made this dxy, then this dxy came here, then this is wrong brother. dxy cannot move because its electron density is in the plane.

[00:12:41]Now the thing is that here also there is dxy so this also cannot happen. Now as far as yz is concerned, that too cannot happen. Now we are left with the option x² - y². Let's say we have the x axis. This is the y axis we have, so the electron density will be on the axis. There will be electron density on this.

[00:13:08]Electron density on axis. And the px is px, so let's assume this is our x axis.

[00:13:14]So this px will be created like this. So in this the electron density will be on the x axis. There is nothing to say. Therefore, we will have to go with option number three. If neon helium neon argon do not form bonds then the bond order will be zero.

[00:13:31]From S. N2 everyone knows. If N is a triple bond then the bond order will be three so it will be Q. O2O double bund O will form double bund then O2 will become R. If F2F forms a single bond F then the bond order will be one. So match it. S for first, Q for second, R for third, P for fourth, first option will be correct.

[00:13:58]Statement In cyclic process delta U and delta H are 0. So Delta U DeltaH is 0. So now here we have to know that in a cyclic process, the change of all the state dependent or state variables that we have is zero. The change for all of them is zero.

[00:14:23]Those with state change will have a value of zero. And the second statement is when the initial and final states of the system in a process are the same, the process is called a cyclic process. If it ends at the same place from where it started, then it will be called a cyclic process. This is also correct. So statement first and statement second both are correct. So both statement first and statement second are correct. That means the third option will be considered correct here.

[00:14:54]We got the third option correct for question number 58.

[00:14:58]Now we will move towards question number 59.

[00:15:00]What is question number 59 saying?

[00:15:03]What is the amount of CaCl2 dissolved in 4 litres of water such that its osmotic pressure is 1 atm. So brother, use the formula of osmotic pressure.

[00:15:12]ei = w / m rt up * i Now put up the value First of all how much pi is given 180m w I have to find w What is the amount w to find The value of r is 082 127 so Pi will be equal to 127 + 273 i.e. 400 * I We have been given the bond of factor three. The molar mass of calcium chloride is given as 111. The volume given to us is 4 liters.

[00:16:02]4 liters were given. Now if you solve all these values and find out W, you will get W around 4.5 grams. So the second option will be correct.

[00:16:16]4.5 grams. The second option will be correct here.

[00:16:22]The energies of E1 and E2 radiation are given.

[00:16:26]And I want the relation between λ1 and λ2. So, you should know this formula. E1 / E2 =λ2 /λ1 Given E1 ≥ 75 75. E2 is given 300 300 is placed. The relation of λ2λ1 has to be found. Cancel 75 for 300, it will come 4 times. So 1/4 =λ2 /λ1 is to be expressed in terms of λ1, then λ1 = 4λ2.

[00:17:02]So λ1 = 4λ2 is matching the first option.

[00:17:07]So the first option of 60 will be correct.

[00:17:09]In question number 61 we have been given a conversion. When phenol reacts with zinc, benzene is formed and zinc oxide is released.

[00:17:22]When you nitrate benzene with zinc oxide in a base of concentrated HNO3H2SO4 at 60°C, you will get nitrobenzene.

[00:17:36]Nitrobenzene arrived. If you reduce nitrobenzene with SNCl, you will get aniline.

[00:17:44]Got Analine here.

[00:17:46]So who are the ABCs? A is benzene, B is nitrobenzene, C is aniline. So correct option number four. We have to match the lists to see how the things given in our List One match with List Two.

[00:18:03]So look R m gx is Gragnod Regent. All the children know this.

[00:18:08]So A will become third.

[00:18:12]A's third is in the same option, so if you want, you can kill it directly. There is nothing simpler than this.

[00:18:20]But if you still want to match then look this is propene. When its hydroboration oxidation takes place, we will get OH on the corner. You will get 1 degree alcohol. So therefore our B match will be first. So here B's first match took place. This NaOI means it's showing your haloform reaction. So if acetone has a haloform reaction, that is, if it is NaOI then iodoform will be formed. So C's second will be. Here also C is second and if you heat CH3COH with dilute NaOH, aldol condensation will take place. So this will become four.

[00:19:02]A hydrocarbon C5H12 does not react with chlorine in dark. Alkanes you know they do not give halogenation in dark. Give it in the presence of sunlight. Only one mono chloro compound in bright sunlight. Everyone knows that you have read many times that neopentane is the only five carbon isomer which gives only one monochlorinated product. As it is now in this either Cl will be connected here or will be connected here or will be connected here or will be connected here. It will be the same thing. There is no H in the middle.

[00:19:36]Well, see, this is normal pentane, so three of it will be formed. Cl can connect here once, can connect here once, can connect here once. This is our double bond given in it. So the double bond will show addition reaction at room temperature or at dark.

[00:19:52]Dichloro will be formed. This cannot happen.

[00:19:55]And there are one, two, three and four possibilities in this. So four monochlorinated products will be formed from it. So the hydrocarbon will be neopentane.

[00:20:10]Look, decarboxylation is happening. The intermediate of decarboxylation is a carbanion. How? RCOO- Na+ This will change to RC = O- Now this will rearrange. How will the rearrangement happen?

[00:20:32]This minus will come here. This electron will go here. And from this we will get R-, so from this we will get R minus.

[00:20:49]R minus will be found from this. That means we will get the option could not be printed here. Our slide maker brother got under pressure while working and forgot to show the options. So let me give you the option here.

[00:21:05]We have one option in this, our first option is in which R+ is there and the second option is in which R is free radical, the third option is R- and then the fourth is all of the.

[00:21:19]This means that we have to create a carbanion.

[00:21:23]So our third option will be correct.

[00:21:28]Now here the basic nature has to be compared.

[00:21:31]You know the basic nature, as the electron density on the donor atom increases, the basic strength will increase and as the electron density decreases, the basic strength will decrease.

[00:21:41]If it is an aromatic compound then the ones with +r at the ortho para position will increase the electron density and increase the basic strength and the ones with -r will decrease the electron density and hence decrease the basic strength.

[00:21:55]Now here you first check that Cl will decrease the electron density due to our -I effect.

[00:22:06]So this partial plus will be created on this.

[00:22:09]This will create a plus due to NO2 - R. Will greatly reduce donor capability. The CH3 that is there will increase the minus density on it because of +H and will increase the basic strain.

[00:22:26]Due to CH3O +r +H +, the electron density on it will increase a lot. So the most would be four because +r is strong. It will be less than that of the third because +h is weaker than +r. The first key will be less than that because -i will reduce the electron density less and -r will reduce it a lot. So option number second is matching here as 65.

[00:22:59]Option number 65 is matching second. Now let us talk about question number 66. V is a polar molecule. So let's look at polar molecules.

[00:23:08]SF4 is symmetrical and is tetrahedral. Its four dipole bonds will neutralize each other.

[00:23:20]Therefore, its total dipole moment or permanent dipole moment will be zero. Therefore it will be non-polar. So Siff4 is non polar.

[00:23:31]XCF4 has a square planar shape and all the vectors neutralize each other.

[00:23:38]So this will also be our non-polar.

[00:23:44]BF3 is also a symmetrical shape. The trigonal planar and dipole bond dipoles of all three polar bonds will neutralize each other.

[00:23:53]Here also the total dipole moment was zero. Here also the total dipole moment will be zero.

[00:23:59]So BF3 will also be non polar. But in SF4, the lone pair on S is not able to neutralize our bond dipole.

[00:24:13]So this will be our polar. So μp will not equal 0. So the fourth option is the correct option.

[00:24:26]Geometry and Hybridization of XCF4 Look, XCF4 was formed and it had two lone pairs, so there were four sigma bonds and two lone pairs, so the total was six and if six is ​​being formed then it is sp3d2 hybridization and it is octahedral geometry. The shape is square planar but the geometry is octahedral.

[00:24:51]The calculated spin-only magnetic moment. So, now you have to write down the configuration of CR 2+.

[00:24:58]When you write this configuration, 3d4 will appear in the last shell. That means there are four unpaired electrons present.

[00:25:09]1,2, 3, 4. So there are four unpaired electrons present. So if n = 4 then μ is equal to nn + 2 when you solve it, you will get 4 9 Bohr magneton. So the second option will be correct. In this way son, we discussed the questions of our part. I hope you guys have understood. Thank you. Hail India.

[00:25:36]Jai Hind children. Hello. Today your online REFST was held, in which respected Sanjeev Sir and I are in front of you to discuss Chemistry, so you discussed till question number 68 with respected Sanjeev Sir.

[00:25:54]Now after this let us start with question number 69.

[00:26:00]Which of the following examples is saying mainly +4 oxidation state? See, when we talk about actinoids in NCERT, when you are given the table of oxidation states in actinoids, then this thing is mentioned in the table of oxidation states.

[00:26:21]It is mentioned in the table of oxidation states that there are a few things given there which you have to remember. As it is said that uranium shows from +3 to +6.

[00:26:35]Ok? You can see. Thorium, which is specially mentioned, shows +4 there. Ok? If you look at that table, it clearly mentions thorium.

[00:26:50]AC +3 shows. So what was asked of us here? Mainly +4 oxidation state. So he's doing thorium, son.

[00:26:59]Okay, right? So son, this second option will be absolutely correct. And you must remember this table. Isn't it?

[00:27:08]In this you have to remember and tell us by commenting.

[00:27:12]I am saying one thing. Which actinoids show +7 oxidation state?

[00:27:16]You have to tell me this by commenting. It is given in this table. Ok? So your question has been answered and we have also got its answer. Which actinoids show +7 oxidation state? I have to tell you son. Ok? Let's move on to question number 68. 6 Sorry it's 69.

[00:27:35]Now let us move on to question number 70. It is being said that the rate of reaction is doubling by increasing the temperature by 10 degrees Celsius. The rate of reaction doubles at 10°C temperature. If there is an increase of 50° in temperature, how many times the rate of reaction?

[00:27:58]If it increases, it is saying 50 degrees. So son, if t is t degrees Celsius then it becomes t degrees + 50. Isn't it? So if we look, let's say it was 10 degrees.

[00:28:15]So how much did it become? You can also do it with the 20° 30° formula. 40 degrees became 50 degrees and 60 degrees. So let's say R here, then it becomes 2 R here. There are 4 R's here. 8 R 16 R and son how much is it here? 32 R 32 Times Done. It's been 32 times. How much is it? It's been 32 times.

[00:28:41]We know that at every 10 degrees the rate of reaction doubles, minimum, so you could have done it like this and you also know the formula, we could have done it with that too, see carefully, we could have taken out the delta t to the power of 2 from there, we could have done it with 10 also, okay see, there is no problem and like this because it is not a very far value. 50 is the value increase. So, if R is 10 then it is 2 R at 20. It doubled again at 30. Then it doubled again and then it doubled again and it became 32 times. How much is it son? It's been 32 times. Ok? It's been 32 times.

[00:29:27]And we could also do it using the formula, if we're talking about delta t by 10 to the power of two, what would it be?

[00:29:42]So this is a change of 50 degrees, so 2 to the power becomes 5. If it was 50/10 then 2 to the power of 5 means multiply it five times, so five times 4 becomes 8 * 4, so it becomes 32 times. It's been 32 times. It's been 32 times. So the second answer got correct here. Ok? You can also do it with a formula. You can do it like this also.

[00:30:09]Question number 71 See, at which of the following temperatures the reaction will occur spontaneously, what should be delta g? It should be negative.

[00:30:22]What must happen for a spontaneous reaction to occur? It should be negative. Now listen carefully. We know that Delta G is equal to Delta H - T Delta SH, we know this son. Do you know or not? Everyone knows. So we know that at equilibrium delta g is zero and what is t equilibrium called? Delta H/Delta S. Is it okay son? So first of all find the T equilibrium. So if we calculate t equilibrium, how much delta does it give us? 170 kilojoules. It became 170. So how much did T equilibrium come from here? 1000 has arrived. Isn't it? Meaning if we are taking t equal to 000 Kelvin then the value of delta g is zero and delta g is equal to delta h minus t delta s. So now you understand if the value of t is 1000. If the value of t is 1000 then deltag is zero. We want to bring delta g to negative. What do we have to bring to Deltag? We have to bring negative. So to make deltaG negative, the value of T deltaS should be high. The value of T deltaS should be greater than deltaH. And for T deltaS to be high, T must be high for T deltaS to be high.

[00:31:44]Now that too should be more than what? It should be more than 1000.

[00:31:49]Because there is zero on 1000. So the t which will be greater than 1000 Kelvin, you see if any option is given to you, so see son which is the second option, the second option of 71 is given to you, what is given, the second option of 71 is given to you here, okay, the second option of 71 is given to you, okay, for the azimuthal quantum number, the total number of magnetic quantum number is given by, so it is a very simple formula. We all know that if we talk about m equal to m total, if we talk about m total then it becomes 2l + 1, so from here m = 2l + 1.

[00:32:41]Hence 2l = m - 1. So l = m - 1/2.

[00:32:47]m - 1 / l = m - 1 / 2.

[00:32:55]So l = m - 1/2. So the second option will be absolutely correct. Very simple question related to quantum number and surely you all must have solved it.

[00:33:06]This was a very fundamental thing and it just had to be related.

[00:33:16]Here is List One and List Two. Here is the conversion. Here is the number of Faradays. So see, we told you what is the number of Faraday equal to? Mole in valence factor.

[00:33:30]What happened to Faraday's number? Mol in va balance factor. What happened? Mol in va balance factor. So the price has been given to everyone. Let's figure out the valence factor. From H2O to O2, from -2 to zero, its valence factor is coming to two. So for this case A, the number of Faradays will be mole in balance factor so two, so it becomes 2 Faradays. For A it became two Faraday. So A is the second of A. Now come here.

[00:34:05]MNO4 minus MN2+ means this plus to 2 plus valence factor becomes five so for this if we see how many Faraday equivalent moles inu va balance factor moles for B? How much is one valence factor? Five going from seven to two. So it becomes 5 Farades. So look what is coming for answer B? Where is 5 Faraday? B's four. A 's two B's four. Say A ka two B ka four son. A is two B is four, right? A's two, B's four, is clearly given to you here. Now come over here.

[00:34:44]1.5 moles of Ca from CaCl2.

[00:34:49]How many moles is there? 1.5 inu va balance factor ho gaya do.

[00:34:52]So this is three Faradays. Meaning, the answer of C became first. And the D from FeO is 2 plus in FeO and this is six, this is 3 plus, so this is a change of one.

[00:35:08]So for this it becomes 1 * 1. For this it became 1 Faraday. Ok sir? So the D has become three of D. D's got three. A's second, B's four. A's second and B's fourth. There is only one option.

[00:35:23]C has one and D has three. So son, the fourth answer is absolutely correct. The fourth answer is absolutely correct. Look carefully.

[00:35:30]Look carefully. Such questions may come son. So you all have to remember all these things. Ok? Let go.

[00:35:41]How many chiral carbons are present in open chain and cyclic? So absolutely correct.

[00:35:46]Son, we have told you this many times.

[00:35:49]We have told you this many times that when we talk about open structure, the chiral carbons here are four and in the case of cyclic, there are five chiral carbons. And we are talking about glucose. Who are you talking about? of glucose. We have told you many times that you should remember all this directly.

[00:36:11]Creating structures again and again. Isn't it?

[00:36:15]You should remember all this.

[00:36:19]You have to remember all the important things about it.

[00:36:21]Ok? There is no need to build any structure there. You must remember that there are fours in the open chain. Cyclic has five. Let me ask you a question related to glucose. If we have to tell you how to do prolonged heating of glucose with HI. Isn't it? So what happens?

[00:36:42]You have to quickly tell its answer in the comment box and tell what is formed when it is reacted with bromine water? Please tell me all this in the comment box. I will wait for your comments.

[00:36:51]Ok? So what is the answer to this? Four and five. Fourth answer got correct in question number 74.

[00:37:01]See question number 75. CH3CHOCN ok? So this is becoming A. So son, if we write here first CH3CHO + HCN, then if we look here, we are asked directly A and B, so if here he is asking us what will be A? So let us understand this, first you make it by opening it CH3 CH double bed O, this is it. Now break it, break it. So this is minus and this is plus.

[00:37:44]This H plus and CN- came. So this will become cyanohydrin, son. CH3 is CH, right? Neither it became OH nor CN was added. OH also joined here and CN also came. OH also joined and CN also came. Okay, right? OH also joined and CN also came. Now what about it? Acidic hydrolysis is taking place.

[00:38:11]So when its acidic hydrolysis takes place, son, the CN will be converted into COOH.

[00:38:19]What will it turn into? will change to COOH.

[00:38:22]So if you look at what it turned into? In COOH. Now look at this carbon carefully. There are different groups or atoms all around. So how did this happen? He became cowardly. Meaning it became optically active.

[00:38:34]Ok? So compound A was a cyanohydrin. Absolutely correct. The compound was also optically active. Absolutely correct. Formation of a taxon by nucleophilic addition. It is absolutely a nucleophilic addition.

[00:38:47]Who went here first? So CN- was the first one to be attacked by it. Ok? So what happened? Nucleophilic addition took place. So first, second, third all three are correct. So we'll go with Fourth Answer in 75 with all of these. Who will you go with? Will go with All of These. Is it okay son? Let's look at question number 76. CH3 CH2Cl reacts with KCN so it becomes CH3 CH2CN right? Now after this, in this reaction of SnCl2Cl, what will it turn into? This will convert to CHO. What will CN change to? will be converted to CHO. What will CN change to? will be converted to CHO. Ok? So CH3 CH2 CHO, now after this it will be reduced with Li AlH4, so after reduction it will form alcohol, son. So will this finally happen? CH3CH2 CH2OH ok? CH3CH3 CH2OH So son, look carefully at the third option here. This is what is given.

[00:40:07]Look carefully at the third option. This is what is given. Look carefully at the third answer of 76.

[00:40:13]is given. Ok? The third option of 76 is given to you here. Let's move ahead to 77. So most reactive towards electrophilic attack. So son, in this you have to see +m -m. So who will do it?

[00:40:31]+m will do. + Whoever has it. So you can see this. This is son -m. Here it will be -i.

[00:40:39]Here it will be -i. So the one having +m will be most reactive towards electrophilic attack.

[00:40:45]So the first answer got correct. If we look at question number 78, question number 78 if we are talking about the geometry and magnetic behavior of NiCO4.

[00:40:58]So NiCO4 you remember that this is its sp3 hybridization.

[00:41:03]You must remember this. It is tetrahedral. You should remember this. It is diamagnetic. These are all familiar examples from NCERT which should be on everyone's lips.

[00:41:16]Nickel's is 4s2 3d8, so since CO is a strong field ligand, there will be pairing in everything, son. You have to keep in mind, see it this way, sp3 hybridization, tetrahedral and all of them diamagnetic, how are all of them diamagnetic, so you can see this, it is a very simple thing here, so in this way we will go here with tetrahedral and diamagnetic, okay, with whom will we go, with tetrahedral and diamagnetic, we will go with the second answer in 78. There is a very famous example of NCERT which is given. Ok?

[00:42:09]79 Now here it is saying geometrical isomerism of so look this is MA2B2 type complex.

[00:42:19]So it will have two GIs. One will be cis and one will be trans. One will be cis and one will be trans. So look carefully. This is it, it will also be made into cis platen and trans platen. If it is made then it will also have only two GIs. So there will be only two GIs. So its fourth answer will be son. This will be the fourth answer. What will happen? Fourth Answer. Son, you have to answer me, in whose treatment is this Cis Platin used? You have to answer this to me in the comment box.

[00:42:55]What is cisplatin used to treat?

[00:42:58]Ok? Let's say the pair of lanthanide ions are diamagnetic. So if we talk about which one is diamagnetic. So let us take a look at this also.

[00:43:15]See if we talk about cerium, right?

[00:43:21]So it goes something like this. Correct? And if we're talking about cerium four plus, then if this 4F becomes zero, it will become diamagnetic. So look at Cerium Four Plus, it is given as an option. Ok? Well, we will take out the rest.

[00:43:38]What happens in Eu2 plus it becomes 4F7.

[00:43:41]So here there will be unpaired electrons.

[00:43:43]Now when there are unpaired electrons it will become paramagnetic. Is it okay son? And how about Cerium 3+? will remain paramagnetic. After this, son, if we talk about EU3+ then this will also become 4F6. This too, son, will become paramagnetic. What will happen to this also? This will also become paramagnetic. There will definitely be unpaired electrons. There will always be unpaired electrons. Now after this we talk about PM or now if we know yb then y b two plus is 414. It is 4f14, son. What happens? It is 4f14.

[00:44:31]So if you look, it will be 414. So from here we can directly say yes and PM3+ will become 4F4. So what is that even? It is also paramagnetic. If you asked about diamagnetic then this fourth option would be absolutely correct. The fourth answer in question number 80 will be absolutely correct. Is it okay son? Look carefully. Is there any problem? Is it clear? Move ahead.

[00:45:09]81 Among the oxides of given below vs below are acidic vs below are acidic so chromium is in its highest oxidation state means it will become acidic MN2O7 is MN in its highest oxidation state. This means it will also become acidic. COO This is in +2. Isn't it? So this will not be acidic. CO son will become neutral. Isn't it?

[00:45:36]SO2 How are oxides of non metals formed? Are acidic. How are they? Are acidic.

[00:45:42]So CO2, SO2, SO3, all these are acidic. So what happened to SO2? It became acidic.

[00:45:51]So what happened here with Cr3, Mn2O7 and SO2? The second answer is absolutely correct. The second answer is absolutely correct. What happened? The second answer is absolutely correct.

[00:46:03]Ok?

[00:46:09]Correct? So here you can see that we have told you everything. Cro3 is acidic. MN2O7 is acidic. CuO is basic. CO is neutral. SO2 is acidic. So there will be a second option here.

[00:46:26]The General Configuration of Lanthanoids. So the general configuration of lanthanoids is N -2, F1 to 14 n - 1 d0 or 1 ns2. So look carefully. The second option will be absolutely correct. The second option of 82 will be absolutely correct. The second option of 82 will be absolutely correct. Look carefully. Ok? Let us consider the question of Kjeldahl method, that is, the question of estimation of nitrogen in Kjeldahl method.

[00:47:00]So the percentage nitrogen is 1.4 normality in volume up W.

[00:47:08]So 1.4 * normality 2 molar H2SO4 is given sir. We said normality equals molarity in the balance factor. The valence factor for H2SO4 became two. So normality equals 2 * 2 i.e. 4n. So this is four. What is the volume sir?

[00:47:27]How much is 20 upon W? 1.5 grams. So this is 14 * 4 * 20/ 50, okay, so if you look carefully, if we write this as 14 * 4 * 4 up 3, then this is 56 * 4 up 3, meaning 1124 up 3, so 224 up 3, let us calculate 224 up 3 224/3, then it will come to 74.6%. The third option is absolutely correct. The third option is absolutely correct.

[00:48:18]Calculate the delta g not for the conversion of given this and KP given this.

[00:48:24]So if we know, son, that DeltaG not equals -2.303, right?

[00:48:37]RT log K, so we have to keep this value for it.

[00:48:44]Ok? Now if you see, this question is only about calculation, son. There is nothing else in the question. Ok? There is nothing else in the question. So you have to keep this in mind.

[00:48:59]Now we have applied this formula in it. Now all the values ​​have to be kept. Ok? If we have set the values ​​then definitely we will get the answer. Like what is -2.303 * 8.314 * t here? Is 298. And if we look at the log off, it's 2.47 * 10 to the power -29. So if we look at this, if we solve it, then the first value we should know is 2.47 * 10 to the power of -29. So if we look at its value, son, it comes out to be log 2.47 plus log of 10 to the power -29.

[00:49:59]Now its value will become slightly greater than log two. Isn't it? So, it will become 0.39 and we know its value, this -29 will come first. If log becomes 10 then it becomes -29. So the value that comes after solving here will be -28.6.

[00:50:21]Approximately how much will -28.6 be?

[00:50:27]How much will it cost -28.6?

[00:50:30]Its value will come to -28.6. So now put it in the formula here. - 2.303 * r 8.31 * t 298 how much is it inu?

[00:50:46]- 28.6 So minus minus became plus. So all the values ​​are plus. Now let us calculate this, son. Let me calculate it and show it to you. 2.303 * 8.314 * 298 * 28.6 so son 100 63 is coming to 187.2 joules per mole. So what can we write about this? 163.1 kilojoules can be written as moles. So the first answer will be absolutely correct. Look carefully. The first answer will be absolutely correct for this, son. Ok? Let us consider a solution containing 10 grams per decimeter of urea. The molecular weight is given to you. That's right. Is isotonic with so isotonic. Pi one equals Pi 2 or meaning if we see that both of them here are non-volatile solutes. So we have given this condition here that C1 = C2.

[00:51:54]Ok? So now what will be the concentration for this? 10 and what is its molar mass?

[00:51:59]Decimeter cube is already given on 10/60 brother.

[00:52:02]So volume is mentioned in this.

[00:52:06]5% solution means 5 grams of solution in 100 ml.

[00:52:14]5 grams of solute in 100 ml of solution. That's what it means. So if we look and what do we need? It is required in litres.

[00:52:26]So 5 upon is its molar mass, right? And there is 100 and we need this in which one?

[00:52:35]Required in litres. So 60 = 5 * 1000 up to m * 100.

[00:52:42]So what is this now? This is 10, so from here 1/6 is equal to 50 / m, so from here m is equal to how much it came, 300 m is equal to how much it came, 300 second option is how much it came, son 300 300, okay, see carefully, we could have seen this with the formula of osmotic pressure, here for isotonic solution, put C1 = C2. Both were non-volatile.

[00:53:22]What current is to be passed for 0.5 seconds for deposition of certain weight of metal which is equal to electrochemical equivalent?

[00:53:32]So the question of Faraday's Law is W = ZIT w = ZT son. Ok? So here i is asking you and w = z is given to you. Isn't it? And how much has t given you? 0.5 seconds.

[00:53:53]So w = z in here now w = wi * t okay?

[00:54:01]So i = 1 / t. Meaning 1 / 0.50 means 1 / 1/2 means 2 amperes.

[00:54:10]Tell me brother.

[00:54:12]Okay, right?

[00:54:16]Ok? 2 amperes. So the fourth answer will be absolutely correct.

[00:54:22]Even in REET, the question has come from Faraday's law, son. The question has come from Faraday's law only.

[00:54:29]Hypothetical electrochemical cell is given to you. The EMF measured is given to you.

[00:54:34]Cell Reaction is asking you. So son, we know that A B C means anode on the left, cathode on the right and this B in B means bridge salt bridge. So you look carefully. What is happening to A? A plus oxidation. What is happening to B?

[00:54:49]Reduction. B+ is being reduced to B.

[00:54:51]A is getting oxidised to A plus. So this is going to a plus a. This is wrong.

[00:54:56]This is going to a plus a. It went wrong.

[00:54:58]This is going from a to a plus. Correct.

[00:55:01]This is going to b plus b. There is reduction.

[00:55:03]Correct. So we will go with the third option.

[00:55:05]87 will become third. What will happen to 87? It will be third. Ok? Let's move ahead. Let's go ahead son. Let go.

[00:55:17]Reduction of aldehyde and ketone into hydrocarbon using zinc amalgam and concentrated HCL is called so if we see here then CO gets converted into CH2.

[00:55:32]Ok? And all this has been given to you.

[00:55:34]So this reduction is called claim reduction. What is it called?

[00:55:40]Claims Reduction. What is it called, son? This is called claim reduction. Ok?

[00:55:48]Identify the correct reagent that should be the reaction. So what is given here.

[00:55:53]Ok? So first of all, if we pass this out BH3H2O2OH- then this is son HBO reaction. Here the anti-Marconian rule will apply. Isn't it? Which will do hydration of water according to the antimarconic rule. According to the antimarconic rule. So from here, I will make whatever is possible.

[00:56:22]This will become CH2. This will become CH2 and this will become OH.

[00:56:25]Correct? Now if we look further, what if we take PCC? So this will oxidize it into what? In CHO.

[00:56:40]In which will he do it? Just wait and see what will happen in CHO.

[00:56:49]CH2 CH2 CHO Is it okay son? So wherever BH3 PCC etc. has been given, look, we have already got it. Already got it.

[00:57:01]So firstly its answer will be absolutely correct.

[00:57:03]First will be the correct answer. Firstly, it will be absolutely correct. Just look.

[00:57:08]This is what we wanted and this is what we got.

[00:57:11]Question number 90 Question number 90 see what it is saying that reaction of carbonyl compound with so reaction of carbonyl compound if we look at HCA then in one question we had seen from HCA whether cyanohydrin was being formed or not, if it was being formed then A will become four, okay son, A will become four, if we react any carbonyl compound with NH2 2OH then it forms oxane.

[00:57:43]What forms with NH2OH?

[00:57:46]Produces auxin. You must remember all these reactions, son. Okay, right? Like you saw the reaction with glucose. Ok? And if you see, when carbonyl reacts with alcohol, beta acetal is formed. Okay, right? And if we look at the reaction of carbonyl with amine, what is formed when the carbonyl compound reacts with an amine? This becomes Ceph Base. What happens? Siv becomes the base. Ok? So here if we see A is four then what happened to B? It will be third. I just told you that it will be C's second, son. and Dee's Forest. So, we could have told the one of D directly through one option, son. And we had already told after seeing cyanohydrin HCN.

[00:58:39]So this will become A of four. So A's four. So look at this second option, B is three, C is two. So the second option is absolutely correct here.

[00:58:48]So son, this was your today's online FST discussion and in this way, son, we discussed the paper.

[00:58:59]Keep giving tests regularly. We will meet again in some next paper discussion. For now, thank you so much for today.