9701 Chemistry Paper 4 2026 Feb/March Paper 42 | Full Solution & Explanation

Añadido: 2026-05-06

[00:00:01]Hi everyone. Okay, welcome to the channel. Yeah, today we are going to discuss about February March 20 uh 26 paper. Okay, paper 42. Yeah, so let's actually go directly to the paper. So the first question over here is chromium is a transition element. Okay, that forms uh colored compounds. Okay, so define transition elements. So transition elements are elements that can form at least one stable ion with incomplete d subshell electrons. Yeah.

[00:00:30]So complete the electrons in the boxes diagram to show the electronic configuration of an isolated gases chromium atom. So if you look at the periodic table okay the electronic structure from the periodic table you'll get 4 s2 3d4 but this is going to be an exception. Yeah, because uh chromium will be more uh stable if you follow uh the one electron go to the uh d orbital.

[00:00:58]So you will have 4s1 and 3d5. So when we draw it is going to be 4 s1 and 3d5.

[00:01:05]Yeah. So this is going to be how we draw. Yeah. Now sketch the shape of 3dz square orbital in figure 1.1. So how to sketch? So let's actually sketch together. So we wrap around Z. Okay. And then wrap around Z. And then we are going to do a some sort of a donut around. Okay. Here. So this is going to be 3DZ uh square. Just take note if you have uh let's say they ask us okay to draw for 3D XY for example. Now how we draw for 3D XY. So if you have X and Y rather than actually drawing on the axis itself. Yeah. So we will actually draw in between of the axis between X and Y.

[00:01:50]So X and Y. So you draw like that. So if you have let's say for example 3D uh YZ for example. So uh how to draw? So the way we draw is going to be we put Z over there. This is Y. So you just draw the same thing. Yeah. So something like this. Okay. Now let's move on. Okay. So state two reason that explain why chromium compounds can behave as catalyst. So transition elements they can actually behave as catalyst. One of the reason is going to be they can form variable okay they can form variable oxidation states. Okay oxidation states.

[00:02:30]So if they can form variable oxidation states okay uh this will enable okay this will enable uh them to get oxidized and reduced easily. Yeah. So that one is important when they behave as a catalyst. Now number two is going to be because that chromium you can see that they have empty orbitals over there.

[00:03:02]Yeah. So the presence of the empty orbital okay will uh allow okay the lians to form dative bond okay in which resulting in complexes. So we can just say presence okay of uh empty to be exact the empty orbitals if cr3 plus this will be empty. Yeah. So uh the presence of MTD orbitals okay MTD orbitals that allow okay uh lians to form dative bond to form complexes. Okay. So that will be the answer. Now let's actually move on. Now table 1.1 list uh standard electro potential for some of the reaction of chromium and other species you can see over there. Now here they ask us to define the electro potential. Standard electro potential including the description of standard condition. So let's actually define electro potential.

[00:04:10]So electro potential is going to be the wtage. Okay the wtage or you can say cell potential. Okay. Cell potential.

[00:04:18]Okay. cell potential of a half cell.

[00:04:24]Okay. Measured to SH. So what is SH? St standard hydrogen electrode half cell.

[00:04:42]Okay. Under standard condition.

[00:04:46]Now so they ask us to tell what is standard condition. So standard condition please write down standard condition is going to be uh at okay temperature 298 Kelvin okay and then the pressure is 101 kilo pascal concentration of the solution that we use okay must be 1 mole per d cube okay and then all the substances that we use substances must be India standard physical state.

[00:05:30]Okay, let's move on now. Explain fully with reference to relevant data in table 1.1. Why solution CR2+ need to be stored in atmosphere of argan instead of air. Okay. So, CR2+. Okay.

[00:05:48]This is uh anything that contains CR2 plus. You can see this one contains CR2+.

[00:05:54]This contains CR2+.

[00:05:56]And then they mention about something to do with the air. Yeah. Something to do with air is going to be the O2. Okay.

[00:06:03]You can see O2 there. Now, they are asking, yeah, why we need to store that in an atmosphere of argan instead of air. Now if you look at the um CR uh 2+ here okay uh if the E value is0.91 okay so let's say if I uh take this value okay 1.23 2 3. So what will happen over here is going to be this is more positive. This will move forward. Okay.

[00:06:38]This will move forward. This will move backward. So if you move backwards, it will react with CR. Yeah, it will react with CR. But we want it to react with uh Cr+. Okay. So uh the idea is to react with Crus. You see both of them are negative and here both of them are positive. Okay. Okay, we know that both of these equation will move forward.

[00:07:02]This one will move backward. So when it moves backward, this will become the reactant. This will become the reactant.

[00:07:09]And the reactant, we want the reactant to be CR2+ not CR. Yeah. So we will choose okay to focus on this equation and these two equation. Now if you look at that uh equation because why we want CR2 plus yeah. So if you look at this both of them are positive and this is negative. This will move forward. This will move backwards. So you can see that CR2+ is always going to be changed into CR3+. So it's going to be oxidized.

[00:07:40]Okay? It's going to be oxidized because this E value.41 is less than the E value of these two.

[00:07:48]Okay. So explain fully with reference.

[00:07:50]So what we can do over here is going to be we can tell that okay the uh the E value okay the E value of okay the both oxygen related uh half cell are positive 1.23 23 and 0.40 4.

[00:08:24]Okay.

[00:08:26]They are positive then.

[00:08:33]Okay. this value CR3+ plus electron to give you CR2+ E value 0.41 this will make >> oxygen easily oxidize CR2+ to CR3+ as the the E standard is more positive than0.41 41 for CR3 plus CR2 plus electron half cell.

[00:09:43]Okay, half cell not done. So therefore okay we need okay we need argan which is inert. Okay.

[00:09:52]Argan. Okay. Which is inert is used.

[00:09:58]Which is inert is used. Okay. So I'll just roughly make it slightly smaller.

[00:10:08]Yeah.

[00:10:12]Okay. Now a green precipitate. Okay. of CR3 forms when aquas sodium hydroxide is added to aquis cr.

[00:10:24]Construct an ionic equation for the formation of this uh green precipitate.

[00:10:29]Yeah. So we know that this one you have the cr okay plus okay3+ then you are going to have 3 minus. So we are going to get CR3.

[00:10:47]So just write down solid. This is aquas.

[00:10:50]This is aquas. Okay. So this is going to be uh formation of the green precipitate. So this is a green precipitate. Okay. Let's move on. Now green CR3 solid dissolve in alkali to form yellow solution containing this when it's left in air for several hours. Now use table 1.1 to construct an ionic equation for this reaction. Now from the table okay uh there are many things that you can uh look into okay V1 okay the CR3 okay dissolve in alkali to give you CR4 okay so CR4 okay is here okay and then you can see that something to do with alkali so this one you see so here I am going to choose Okay, I'm going to choose this equation.

[00:11:49]Okay, so I have actually chosen this equation. I rewrite for you. Okay, so this was 0 negative0.13.

[00:11:59]So in order Okay, we want it yeah to go left. Okay, why we want it to go left?

[00:12:05]Because we want okay this CR3 to give you C4. So we want this equation to go left. So in order for this to go left because this is negative we need to come up with another half equation which is more positive than0.13 okay then only okay this reaction will go backwards okay so what we need to do we need to look for an equation half equation which is going to be greater than0.13 for the E value okay and then they say that left in area so again we are looking for oxygen related. Okay, so oxygen related you can see between these two. Okay, between these two uh I need to choose okay one of it. So I have either O2 here and O2 here but I am going to choose 0.40 over here. Why?

[00:13:03]Because they have mentioned alkali.

[00:13:05]Okay, I do have O minus here. So I will actually use this rather than using this because this one got to do with acid.

[00:13:13]Okay. So I will try with this. Okay. So when I uh write the equation, this is the equation and 0.40.

[00:13:21]Now one important thing is this is positive and this is negative. We know that this reaction will move forward.

[00:13:28]This reaction will move backwards. So let's rearrange them. Yeah. So if you rearrange them because it's moving forward, so this stays as it is. But this this equation I need to reverse it.

[00:13:41]Yeah. Because negative. Yeah. So this will move forward. This will move backwards. So when I rearrange Okay. I need to change this into positive. So I have positive 0.13.

[00:13:53]And then I try to make sure the electrons between these two equations. I need to try to eliminate the electrons.

[00:14:01]So I have four electrons here. I have three electrons over here. Here I times by four. Here I times by three. Okay. So when I do that, okay, I will have all this the one that I've written in red ink. Okay. But here I can still make some adjustments. Okay. Adjustments such as I have six uh H2O here. I have 16 H2O here. Bring the 6 H2O to the right. You will be left with 10 H2O. Okay.

[00:14:30]Electrons and electrons will cancel.

[00:14:32]Okay. And then O minus. Okay. So we have O minus here 12 O H minus 20 O H minus I bring this to the front to the left left with 8 O minus. So if we do all that in the end we are going to get this ionic equation. Okay which will be our answer.

[00:14:56]Yeah. So you can actually have a check.

[00:14:58]Yeah. Now if it is not necessary yeah to write down this physical state but I think can since it's given there I think just write it yeah if you have time okay let's move forward now the equation for the decomposition of ammonium dromate is shown in the equation below so the entropy is going to be positive 915 joule per kelvin. Yeah. Now explain why standard entropy change is positive in reaction number one. Quite simple because this is solid and then the product is going to give you gas. So we can just write down okay the the reactant which was in solid gives products which are mixture of solid and gases.

[00:16:06]The presence of gases.

[00:16:11]Gases product indicates not just gases product. The presence of gases product from solid reactant indicates entropy change is positive.

[00:16:40]Now let's move on. Now table 1.2 gives some standard entropy values relevant to reaction number one. So the entropy is all given. Now use the information in the question to calculate the standard entropy. I have actually written the equation again. I took the equation and put it over here. So I just want to recalculate this. Okay. So when you calculate the thing is okay uh quite simple calculation. Okay. So we just need to find out the entropy of the reactant reactant. So it's going to be 114 which is this one is the reactant.

[00:17:16]And then we need to calculate the entropy of this. Make sure you times by four. Yeah. So you calculate okay so the four for water we don't know so we put as x okay and the formula is going to be the entropy change is going to be entropy of the product minus the reactant. So this is going to be entropy of the product and then we minus with the reactant. So in the end we try to solve and then we try to get the value of x. Okay. So the value of x okay is going to be positive 189. Yeah. So done.

[00:17:50]Now thermodynamic data for reaction one is shown. Now statement one they say physible at all temperature.

[00:17:59]Reaction one will not occur unless you heat them strongly. Now explain statement one and statement two. They didn't ask you to say which statement is correct, which statement is wrong. Okay?

[00:18:11]Both statements are correct and we need to explain why. Yeah. So number one the reaction is feasible at all temperature.

[00:18:19]Why? Because we know if we use the gives free energy the formula is going to be gives free energy equals to uh standard enthalpy minus the uh temperature okay times the change in entropy. So here we know that our enthalpy change is negative is given and then if your entropy change is positive yeah it's going to be minus. So this value is positive negative value minus okay a positive value in the end it's always going to be negative. Yeah if it is always going to be negative then your uh your um give free energy is negative it indicates that okay the reaction is feasible at any temperature. Yeah. So uh we can just write down yeah uh Gibbs okay Gibbs free energy okay is negative okay at any temperature okay since the enthalpy change is exothermic and entropy is positive.

[00:19:55]We know that the t delta s okay is always positive and g will be less than zero for any temperature.

[00:20:15]Okay, that explains statement number one. Statement number two, okay, is going to be um the reaction this uh the solid, okay, the solid uh NH4, okay, 2 CR27 has okay higher activation energy. Okay. If something is difficult to start or react at the beginning, they have higher activation energy making it necessary for strong heat energy provided at the beginning.

[00:21:07]Okay. Done. Now figure 1.2 shows bond hab diagram for the ionic solid CR23.

[00:21:13]Relevant data is given in table 1.3. So all these are given you can read them.

[00:21:19]So now when I look into this okay we can just label quickly. Okay that um this is going to be okay this is one this is two this is three this is four this is five and this is probably six. Okay. So here okay later we'll go into the calculation. They have given all this.

[00:21:40]Okay. But first the question is define first electron affinity. So first electron affinity is the energy change.

[00:21:47]Okay. Energy change when okay one mole of gases atom each. Yeah. Each of these atom gains electron okay to form okay gains uh one electron I think let's be specific gains one electron okay to form one mole of gases negative active ion.

[00:22:35]Okay. Now use figure 1.2 and data in 1.3 to calculate the latice energy. As I mentioned earlier, you just put the numbers easier for us. Okay. I'll just I've done it earlier. I'll just make it bigger. Okay. So here, okay, we just put okay clockwise equals to anticlockwise.

[00:22:55]All these are going clockwise. Okay.

[00:22:57]Clockwise, clockwise, clockwise, clockwise. This is anticlockwise. This is uh sorry all these are clockwise.

[00:23:05]Okay. Except for this is anticlockwise.

[00:23:08]So 1 + 2 + 3 + 4 plus let energy equals to 6. Okay. So important is to make sure we remember to times these numbers.

[00:23:20]Yeah. So I have times okay by 2* by 3* by 2 and so on. Okay. Equals to this value which is 128. All this value you can get it from here. Okay. So once you do that you solve we try to find out the latice energy. The latice energy you will get -13134.

[00:23:41]Yeah. Kiloj per mole. Okay. So all these are in kilogj per mole. Okay. So no need to change the unit. Now the ionic radius of C3+ is 0.069.

[00:23:52]The ionic radius of Al3+ is 0.050.

[00:23:56]Now uh suggest how the latest energy okay will differ. Okay we know that okay the charge density will be uh charge density is going to be uh written by okay radius. Okay radius. So charge divide by radius. Okay. So if your radius is smaller we expect the charge density to be greater. Okay. If charge density is going to be greater. Okay.

[00:24:23]when they bond with cation and an ion they will actually have stronger electrostatic forces of attraction they will release more energy. Yeah. So latice energy will be greater. So here okay the ionic radius is given as suggest how the latice energy of CR23 and Al23 will differ. Okay. So uh the latice energy we write down the latice energy of okay of um let's say um this one is smaller okay so the latice energy of aluminum is aluminium oxide is uh more ne uh exothermic okay more exothermic avoid to use greater and so on because latis energy is always exothermic. So just stay more exothermic than CR23. Yeah. So once you just mention now then you tell why okay the ionic radius okay of Al3+ is smaller than CR3+. Okay. When that is smaller what happens? charge density okay in Al3+ is greater.

[00:25:53]Now when the charge density is greater okay so we have more okay more energy is given out when Al3+ bonds with O2 minus as they form okay uh stronger electrostatic forces of attraction.

[00:26:32]M. So that will be the answer over here.

[00:26:36]Now next, okay, we read go through the question over here. The manganate for the manganese 7 ions is used in chemistry as an oxidizing agent. Three reactions are given. Reaction 2, three and four. Oxygen slowly oxidizes Fe2+.

[00:26:54]Reaction two can be used to determine the percentage of Fe2+ that is oxidized over time. A student prepares 250 cm cq okay of solution. Okay. After one week they titrate the portion okay and then we get that the average title is this.

[00:27:13]Okay. We want to calculate the percentage of FA2 plus Fe2 plus ion that were oxidized. Okay. So what we will do now is going to be first okay we will find out okay using this and this we will find out the number of moles of KM4 how to find quite simple okay 0.05 okay you times with um 23.80 80 divide by 1,000. So you get this number of moles. Okay. And then based on the equation, okay, we know that one mole will react with five moles. So let's actually find out the number of moles of Fe2+. So number of moles of F2 plus is five times okay five times this number of moles. So when you calculate you will get this value. Now this is going to be in 25 cm cube. So we can see here 25 cm cube. But earlier they prepared 250 cm cube. So what do you need to do? You need to times it by 10. So this will give rise to this number of moles. Yeah.

[00:28:18]So this number of moles of Fe2+. Okay.

[00:28:22]Is actually the same. Yeah. Because it's the same as the number of moles of this because this is the one that will give you Fe2+. So 1 mole and 1 mole. So the number of moles of this is going to be the same. So we can calculate okay the mass okay the mass which is going to be using this number of moles okay mass divide by relative molecular mass it's already given 277.9 okay once you do that you will get the mass okay what do we understand with that okay it means that earlier you use 17.70 okay now after the oxidation the one remaining is going to be 16.53505.

[00:29:08]What we need to do is we minus. Okay. So 17.70 you minus is 16.53505 which is this. Okay. Divided by the original okay of this value times 100% you can find out the percentage of this which is oxidized. Okay. Which is going to be okay. If you calculate okay if you calculate this will give you 6.58%.

[00:29:39]Yeah. So please make sure you calculate correctly. Yeah. Now explain why an indicator is not needed in this titration. Actually no need because okay you are going to use acidified potassium maganate. Okay. So KM O4 acidified.

[00:29:57]Okay. K MN O4 is a self indicator.

[00:30:04]Okay. What is self indicator? Okay. They change. Okay. They change color. Okay.

[00:30:13]from purple to colorless.

[00:30:20]when K MN when MN O4 minus ion changes to MN2 plus okay so self indicator let's move on reaction three uses hot alkaline KMO4 as a region the reaction mixture is then acidified draw the structure of the organic product okay of the reaction number three. Okay. So whenever you oxidize this okay do take note what is going to happen over here is going to be you have methyl uh methylbenzene. Okay. When you oxidize what happen you are going to change this into caroxilic acid benzoic acid. Okay.

[00:31:13]So then we are going to get the water and so on. So here okay when we say draw the structure the structure that we are talking about is going to be the benzoic acid. So uh I will actually make the answer bigger. Okay so this is the answer.

[00:31:33]Okay benzoic acid will be the answer.

[00:31:36]Now reaction four okay is called reverse disproportionation reaction. suggest the meaning of reverse disproportionation reaction. Use the oxidation number okay uh to explain your reasoning. Now if you look at this okay I have actually written down earlier also I'll just uh straight away uh use this to explain.

[00:32:00]Now if you look at this okay we know that okay uh the two okay the two reactant okay the two reactants okay uh of different oxidation number okay they get reduced get reduced and oxidized to give you product which is going to be the ox uh the pro to give you a product with one oxidation number. Yeah, if you want to answer like that. Okay, in this case you can actually explain further saying that okay the oxidation number okay one oxidation number the oxidation number of MN okay in here is + 7 and this is + two the oxidation number here is + three so we can say that it is reduced to MN3+ and then oxidized to MN3+ okay so here okay can we See in this case okay both okay from here okay they are going to be from okay + 7 to become three ox uh reduced here is going to be from +2 to become + three is going to be oxidized okay so in simple terms okay in simple terms we can say that two reactants okay with different oxidation number will get oxidized and reduced to a single oxidation number product. Yeah, that is going to be reverse disproportionation reaction.

[00:33:56]Yeah. Now the solubility okay is given.

[00:33:59]So please remember solubility we need to change into mole per dmq later before we continue. Now write an expression for the solubility product Ksp. So we know that okay your MN O2 okay it will give you MN2 plus and 2 O minus this will be aquas and this will be aquas too okay and this will be solid. So when we write down this is going to be MN2 plus and then this will be O minus but take note must put two no need to put two inside.

[00:34:37]Yeah. So this one we put it two over there. Now calculate the value. Yeah. In order to calculate the value earlier I mentioned that you need to change it into number of moles. So I'm going to take this this one is gram. So we are going to find out the number of moles.

[00:34:52]Okay. So if you calculate yeah if you look at the periodic table MN O2 is going to be if you put in the value this will be in mole. Okay. in mole per dm cq. Now this is our solubility. So what we do over here is going to be we know that ksp equals to this. Okay, we know that if it is uh solubility is x. So this one is going to give you x but this will be 2x. Okay. So here okay we will assume nothing happened here. So here we know mn2+ is x. Okay. Then O minus is 2x. Okay. 2x but to the power of 2. So to the^ 2. So this will be 4. Okay. So x to ^ 2 * x is going to be this. So this x we have already calculated here. So just put four and then this one will be 3.6. Okay. 10 to the^ of -5 and then to the power of 3. If you calculate that you will get 1.87 * 10 -13 and then the unit is mole ^ 3 dm9.

[00:36:12]Okay. So here is going to be 1.87 * 10 -13 and then mole to ^ 3 dm9.

[00:36:23]Yeah. Now next question they say that MN3+ ions are colored and we want to know why. Yeah. So you write down. So MN3+ okay has electronic configuration of okay configuration of argan. Okay 4S03D1.

[00:36:50]Okay. So this is going to be MN3+. Okay.

[00:36:56]The coordinate bonding. Okay. The coordinate bonding with like uh the lians. Okay. With the lians causes the Okay. It causes the d orbitals to split into two nondenerate d orbitals. So they are going to split with different energy something like this. Yeah. And then the difference.

[00:37:38]Okay. The difference of energy or rather than putting it as difference.

[00:37:51]Okay.

[00:37:54]Uh we can say that okay the visible light absorb. Okay. The visible light absorb when the electrons in lower energy level is promoted to higher energy level.

[00:38:32]Okay.

[00:38:34]So visible light is absorbed. Okay. When that happen, okay, complimentary color is seen.

[00:38:49]Okay. Uh based on base on the parts of visible light absorb.

[00:39:11]Okay.

[00:39:13]So important point over here is okay.

[00:39:16]You need to make sure to tell that coordinate bonding causes the orbitals to split into two nondenerate orbitals.

[00:39:24]Yeah, that will give you one mark. So when visible light is absorbed, okay, the electrons, okay, from lower energy will be promoted to higher energy level.

[00:39:34]Then once you do have that, okay, the complimentary color is seen based on the parts of the visible light absorbed during this process. Yeah, that will be the answer. Now MN3 plus ion forms complex ion that shows stereo isomearism. Okay. Use uh this uh to represent the lians in three-dimensional structures. Okay. Of this. Now the formula of the ligan. Yeah we did mention. Yeah they are going to be bidentic. Okay bidentic. So we are going to have okay three bente. Okay. By dentate which is going to be the formula is going to be the liant is C24 2 minus and because they have each of them okay each of them have two dative bonding.

[00:40:25]Yeah. So coordination number is going to be 6 3 * 2. So this will actually give you the octahedral shape coordination number is six. Okay. type of isomer is going to be optical isomeism. So we are going to draw. Okay. So how to draw?

[00:40:44]Okay. I will actually use this uh way to draw. Just put boxes there. Easier for us. Now we know that the one coming in front will be something like this. The one going behind okay is going to be maybe I just put like this. Okay. going behind I put like this. This one coming in front I put like this. Okay. So we know that okay uh we can represent like this. Okay. So here let's say O and O.

[00:41:20]So that's one.

[00:41:23]Okay. That's one. And then we have O from here and O from here.

[00:41:31]That's two. Okay. So side by side up to uh to the side and then side to the bottom. So O then O like this. Yeah. So this one we try to actually make the mirror. So this one here. So this O then this O and then from the top O. Okay. So it will come over here and then this one from here O. then here. Yeah. So, this is going to be my uh answer. Okay. Now, figure 3.1 shows a reaction scheme starting from four hydroxybutoninoic acid. Yeah. So, we have all the reactions that can happen. Now, under acidic condition, okay, it exists in equilibrium with compound G and uh all this water. But G does not react with sodium. So make sure okay when they react it means there is no caroxilic acid. There shouldn't be any alcohol because caroxilic acid and alcohol do react with sodium. So whenever you add acid most likely they are going to react with each other. Okay. So when you draw the structure of G and H how do I draw?

[00:42:54]So I know that okay the O will be gone.

[00:42:58]the H from here will be gone. Yeah. So we know that they are going to join together. So to be form a cyclic compound if they do that the oxygen here will stay. So I'll just put the oxygen stays. But how many carbon are we talking about? Okay. So we are going to talk about 1 2 3 4. Okay. So I need on top of this I need four more. So what I I'm doing is going to be just put like this carbon. I have another carbon. I have another carbon. Another carbon like this. Okay. So total four. Okay. You can see one 2 3 four. Okay. So the fifth which is going to be the oxygen. Okay.

[00:43:52]We know that oxygen then bonded to this carbon number one. content straight away will have the double bond O. So this one I will actually bring it down a bit.

[00:44:04]Okay, I'll bring it down a bit and then I put over here double bond O. Okay, now the they say also it can react to form condensation polymer. Okay, condensation polymer. So you will have the O. Okay, so I will have O.

[00:44:25]Okay. So, this one. Okay. I'll just put doctor. Okay. So, I have 1 2 3 4. Okay.

[00:44:36]So, 1 2 3 4 and then I will have double bond O double bond O and then continue. Okay.

[00:44:51]So, dotted.

[00:44:54]Okay. So this will be the answer. If let's say you draw it the other way around also still. Okay. Will be as long that you can show that one 2 three.

[00:45:07]Okay. There are three carbon there.

[00:45:09]Okay. So you see 1 2 3. Okay. Three carbon. And then another one more carbon here. Four carbon total. Yeah. Now construct an equation. Okay. With excess of PCR5.

[00:45:23]So do remember both this side and this side do react with uh PCL3. Okay. So if I want to write the equation I have done it earlier. I'll just bring it here and make it bigger.

[00:45:45]So if I write down okay it will look like this. Okay, this plus two moles of PCL5 because one mole PCL5 for this another mole of PCL5 over here you will get the Cl and Cl being added over there. the rest. Okay, just follow the PCL5. One mole is 1 mole of ACL and one moles of P3. But now because you use two moles of PCL5 because why two moles of PCL5? Do remember we are using alcohol, we are using caroxilic acid. So these two groups they are going to actually give us the Cl. So you balance and you write down accordingly. So four hydroxybutanoic acid is a weak acid with Ka value is given. So they are asking to calculate the pH of this solution at 25° C. So what we need to do is going to be just using this formula. We know that weak acid will actually have this. So we are going to use the Ka equals to this.

[00:46:45]Okay. Uh and calculate okay the concentration of hydrogen ion. Okay. And once you do that okay we are going to get 1.69 * 10 ^ -3. Then after that you can calculate using this formula log 1.69 * 10 3 and you will get the final answer which is going to be 2.77.

[00:47:11]Yeah. Now so when this uh solution is mixed with this conjugate base a buffer solution can form. Define buffer solution. Buffer solution is actually a solution okay that can resist changes in pH when small amount okay of acid and bases are added to it. Yeah. Now state the structural formula for the conjugate base of uh this. Okay. We know that okay when uh this is going to be added let's say for example uh when we have this okay added with H2O okay we know that okay the H+ okay will go here to give you this. Okay so we know that this is going to behave as an acid. So this is automatically going to become the conjugate uh base. Okay. So acid and this is conjugate base. So we just take this and we put it over there.

[00:48:09]Now a student makes a buffer solution by reacting this okay solution with this.

[00:48:15]Okay. Uh we will just use this uh calculate the pH of this buffer solution. Okay. We will just take it as we can just use the uh number of moles of sodium hydroxide. Okay. Sodium hydroxide you just use the formula we calculate. Okay. And then for this one okay the weak acid okay I will use okay 25 * 0.15 okay we get this and when we react them together okay according to this equation okay we know that uh this amount of moles will react with this so there is going to be excess over there so only this amount will be reacting okay to give you this. So the extra okay of the acid is going to be this amount.

[00:49:00]Okay. Now we know also the num uh the uh salt that is going to be produced is this. Okay. Total volume is 35 cmq because 25 cmq + 10 cmq. But this one when we use the formula it will cancel out. Yeah. So we use the formula of concentration of salt divide by acid. So this is just the number of moles divide by volume. Volume is 35. it will cancel out. So I didn't include that. Okay. But the formula is going to be pH is equals to pKa plus log 10 salt concentration of salt divide by acid. So it's 4.72 + log okay this value divide by this value. In the end once you do that okay we are going to get 4.78. Okay. Now four hydroxybutonic acid can be used to produce butin diuric acid. Okay. Suggest a suitable region. You can see the caroxilic acid is here. This is the alcohol. So if you look over here, this is going to be this entire part. If we count it as R group, okay, CH2, okay, and then O, this is actually a primary alcohol. Okay, so what we can do? Okay, if you want di if you want this part to have caroxilic acid group, okay, what we can do? We oxidize because primary alcohol we can oxidize to become alihide. and after that further oxidize to give you caroxilic acid. So the suitable region will be okay we use acidified potassium magnanate and then heated you can also use acidified uh potassium dromate okay uh and then also heated as well. Now a student okay does an experiment to determine the partition coefficient of butin di acid between two emissible solvers okay of etoxy ethane and water. What is meant by partition coefficient? Partition coefficient is the ratio of concentration of solute in two emissible solvents. That's very important. Yeah, two emissible solvents.

[00:51:02]Okay, next question. Student adds butin dioic acid to a flask containing all this. Okay, the mixture is taken and then left to separate into two layers.

[00:51:12]So we have one over here and here we just take it as two. So 10 cm cq requires this amount and then 10 cmq requires this amount. Yeah. For complete reaction equations are given. Calculate the concentration of acid in each solvent. So let's look at number one. So sodium hydroxide is 7.5 * 0.1 * 1ide by 1,000. So this is the number of moles.

[00:51:39]And for the butin diuric acid please remember uh we require okay uh if let's say butin dioic acid is like something a droic acid example of droic acid is like a sulfuric acid so sodium hyioxide react with sulfuric acid will give you something like this then you can see that the ratio is always 2:1 okay 2 moles and one so it means that the number of moles okay if it is you are going to have a droic acid okay the number of moles sodium hydroxide if you compare to the acid the acid is going to be half of it. So that's why okay for butin dioic acid so for when you get this number of moles this will be half of it. So you'll get this but do remember it is in 10 cm cq. So you need to change it into 25 cm cq. Yeah. So you need to times by 2.5 and then you will get this okay in 25 cm. So concentration you can actually calculate yeah number of moles. Okay. and then divide by uh this in dm cq okay you convert into dm cq and then you get this okay and then we do the same thing okay for number two this side okay so you calculate okay you will get same thing times by 2.5 okay you will get this in 225 cm cq so you can calculate the concentration which is going to be 0 2776 okay so the concentration of bin d acid in toxic you will get 0.0375 and butin dioic acid in water is this and then KPC because the KPC they mention partition coefficient but dio acid between etoxyane and water so atroxyan is at the top water is below so I'm going to actually take 0375ide by 0.2776 2776 and in the end you will get the answer which is going to be 0.14.

[00:53:42]Okay. So a student performs a series of experiments to investigate the chemistry of esther. Yeah. So we hydrayze it under acidic condition. The equation is given.

[00:53:55]Okay. So here okay they ask use the figure. Okay. to show that this is first order with respect to this. So uh the way that I use over here is I just find out okay from let's say 0.20 20 I bring it down and then I will get this value okay this value which is 1,500 okay 50% of this okay is going to be 010 so I bring it all the way here touch here bring it down okay 7,500 the difference is 6,000 now another half life okay half of it I will get somewhere here bring it down and I'm going to get 135 so from here and okay from here I will have another 6,000 if I minus now I can see that the halflife is constant m so I can actually show any working fully and explain your reasoning okay so I have done earlier so I'll Just make it bigger. Now the halflife okay is constant okay which is going to be uh 6,000 okay uh seconds okay from uh this to this and this to this okay which gives the conclusion that it is a first order with respect to this. Yeah so you just explain properly. Now use figure 4.1 to calculate the initial rate. Yeah.

[00:55:35]So to calculate initial rate what we need to do is going to be just okay uh use a ruler. Okay. And then we are going to uh adjust accordingly so that you can get the tangent. Yeah. The tangent at that particular point. Okay.

[00:55:56]So you adjust accordingly. Okay. So here when I see Okay. Uh I can draw. Okay.

[00:56:04]Actually I have done it earlier also.

[00:56:06]Okay. But let's say I try to draw it again.

[00:56:14]I draw like this. Okay. What I need to do is Okay. I just need to make sure Okay. I calculate. Okay. Uh the amount okay from here to here and then here to here I can find out the gradient. Okay.

[00:56:32]So here what I did. Okay, I use these values 0.25.

[00:56:40]Okay, from 0.25.

[00:56:44]So this is 0.25 is here. Okay, so actually this is we need to recalate again.

[00:56:55]So here 0.25 is here. So my value is not touching 0.25.

[00:57:05]This is 0.21 22 23 2 4. Yeah. So 0.24 0.24 divided by from here to here. So here I am going to get okay each one over here is 250.

[00:57:29]Yeah.

[00:57:30]250 500.

[00:57:34]So 500.

[00:57:36]Okay. Another uh sorry they here they have 2.5 okay boxes. So 2.5 boxes times 250 I have 625.

[00:57:54]625 I just add to 7,500.

[00:57:59]Okay, I will get Okay.

[00:58:08]8125.

[00:58:09]Okay. So, 8125.

[00:58:13]So, use the 8125.

[00:58:18]Okay. You can actually show the working also. Okay. If let's say here 0.24 24 from zero divided by Okay, this one is also from zero. Yes. So, uh 7,500.

[00:58:33]Okay. Each one is 250.

[00:58:36]So, you get this. Okay. So, 8125 8125 - 0. So, you'll get Okay. So, this one we calculate 0.24 divided by 8125.

[00:58:53]you will get okay 2 95 * 10 -5 so we will actually use this as our answer okay so I'll just change it okay that okay now reaction five is in first order with respect to H+ so you can write down because previously we found out that they are first order so you have this you have this okay write the equation so I have written the equation Construct the overall rate.

[00:59:24]State the unit. Okay, the unit. If you want to find the unit, we know that the K. If you want to find the K, rate divided by this. Okay, this is rate M per M cube per second. This one to the power of two because two times. Yeah. So to the power of two. Okay. So I cancel one. So left with per second divided by this. So mole -1 mole. This one go up d M. Okay. uh to the power of three. This is the uh negative. So go up positive3 and then per second. Okay. So this will be the answer. Now methyl metal ethaninoate okay forms this when they react with this. Yeah. So suggest why reaction six gives a higher yield. Yeah.

[01:00:10]Of course one this is reactive. Yeah.

[01:00:13]And then they give you HCL. But this one caroxilic acid with alcohol. Okay. is a uh reversible reaction and whereas this reaction is going to go to completion.

[01:00:26]Okay. So here suggest why okay if you look over here okay uh we can just write down okay CH uh CH3 CO is more reactive okay then it's more reactive than the CH3 CO reaction six will go to completion and HCl is given off and reaction Seven is a reversible and equilibrium is established. Therefore, okay, reaction six gives a higher yield. Okay. So, it gives you a higher yield of this. Okay.

[01:01:08]Because Okay. One because it goes to completion and it is not reversible.

[01:01:14]Yeah. So, next question. So, complete the figure uh 4.2. Describe the addition and elimination mechanism. Yeah. So here okay we want to include all the curly arrow charges and so on. Yeah. So first we know that this okay will give you delta negative this cl okay delta negative over here. This is going to make this to become partially positive. So this will go here. Okay. Once you have that okay they are going to form the intermediate.

[01:01:50]Okay. So the electron the single it will now have single bond the electron will go there so you have negative okay and then will form a bond okay with oxygen but this is going to be positive not delta plus yeah so it's positive so next is to keep on showing yeah the curly arrow so this one will go here this will come in again okay and then for plus so from this bond on it will go to the plus okay oxygen so what will happen you are going to get this okay during that process this CL and H will come out so you have ACL yeah now the relative rate of hydrarolysis okay is given investigated so we need to find out which one is easy slowest hydrarolysis and the fastest hydrarolysis yeah the one that is going to be difficult because we know that chlorobenzene will have a partial double bond uh character.

[01:02:53]Yeah. C double bond Cl kind of. Yeah.

[01:02:56]Because the electrons that you have over there, they will become deoized into the pi system in the benzil ring. So the uh the electron will will have the p orbitals uh of Cl and the p orbitals of carbon they will overlap slightly. Okay.

[01:03:16]Making the structure stronger. So it's quite difficult to break the bond between carbon and Cl okay but for this ethanol chloride okay it's going to be much more easier to be broken why because the oxygen and the CL they are going to be more electro negative so they will pull the electron over there leaving the carbon to become partial positive so this will attract okay any kind of nucleophile okay to it okay nucleophile In this case water they will go there and then the hydrarolysis can happen easily. Okay. Chlorane is like normal. Yeah they are not going to be compared to all this they will be in some sort in the in the middle ground.

[01:04:01]Okay. So uh state yeah so I have written the slowest is this the fastest will be this. Okay. The reason and so on I have written also I'll make it bigger here.

[01:04:15]Okay. So here ethanol chloride has a partial positive carbon okay which is partial positive carbon atom which is caused by both of these okay which makes it susceptible okay uh to be attacked by the nucleophile okay chlorobenzene is difficult to be hydrayed as a lone pair of electron in CL okay is deoized into the benzene ring and the p orbitals of carbon and Cl overlap partially to create a partial double bond character okay between this which is the bond is difficult to be broken. Okay. So this will actually help us okay to identify or to answer this question. Yeah. Uh chloroine will be somewhere in the middle. Yeah. So they are going to be uh not easily broken. Okay. At the same time not that strong as well. Okay. Next four methylphenol can be synthesized okay using three steps as shown. So in step one methile benzine benzene reacts with this. So write an equation. Writing the equation is going to be HNO3. Okay.

[01:05:29]When they react with H2SO4 they will give you NO2 plus H4.

[01:05:37]Okay. Because 1 H. Yeah. So it's going to be become negative. Okay. Plus H2O.

[01:05:45]Okay. Now draw the mechanism of the reaction. Okay. With methile benzene include all the similar like earlier include everything. So here okay when you have N O2 plus what will happen from here to here. Okay. And then what will happen? I'll just put okay this thing will become positive and then you will have H you will have N O2 so this one keep it as CH3 so this one will go in uh will come in okay so you will get this so this will be broken so H+ okay now one methile four nitro benzene is reduced in step two write the equation.

[01:06:36]So uh if you have uh CH3 C6 H4 NO2 so basically they will change it to CH3 C6 H4 NH2.

[01:06:53]So do take note the two oxygen yeah they will be taken out and you want another two hydrogen. So hydrogen and hydrogen and this thing must actually give you water. So we need another okay four hydrogen. So in total how many hydrogen we need? Four from here and another two.

[01:07:13]So total six. So plus six hydrogen.

[01:07:17]Okay. To give you this. So plus this two 2 H2O. Okay. Step three is two-stage process. Identify the region and so on.

[01:07:28]So first okay um step three okay we know that okay you are going to change the um uh NH2 okay the mine okay b uh we are going to change into phenol. So how we can actually do that okay is by first you react with this okay and then you get this uh benzin deninium salt and once you get the benzin denim salt we can actually do further warming with H2O then you can get the phenol so here okay uh rather than rewriting I'll just make it bigger so here okay we need okay we need the H NO2 okay or you can use this we dilute H2SO4 below 10° C then after that we need to further warm with water it will give you phenol a student suggest an alternative synthesis okay and decides to use the Fredal cross alkalation as the first step so here suggests the region yeah the region is going to be we use CH3 CL okay because we want put the CH3 there and if you use CH3 CL you must actually use okay Al C3 as the hogen carrier. So make sure heated okay very important. Now the explanation okay is going to be uh previously in this case we have the nitroenzene then when only we are going to put uh this this is coming from uh if this is one this is coming at position number four do remember that nitro group okay is going to be three and five directing so to get them into position number four is quite difficult so write down NO2 is electron withdrawing group and activates position three and five.

[01:09:45]Position four is not favored for CH3.

[01:09:59]And the withdrawal of electrons from benzene by NO2 causes causes the benzene less susceptible for attack of electroile.

[01:10:52]for electrofphilic substitution.

[01:11:02]Okay, done. Now next. Okay, the structure of three different phenols are shown. So this is just phenol and you have methylphenol and ethylphenol. Do remember okay when the number of carbon is increasing okay we assume that uh they are going to become more straight chain so this is going to be less polar okay compared to this yeah so a mixture of three phenol is analyzed okay the mobile phase is helium gas okay so you have this retention time and so on explain why helium gas is used as a mobile phase so if you want to use gas liquid chromatograph graphine you need to use unreactive gas. So unreactive gas helium is unreactive gas. Yeah. Suggest a stationary phase for the use in gas and liquid chromatography. Okay. Uh it's supposed to be a nonpolar liquid. Okay.

[01:11:58]with high boiling point but this is suggest so I will actually just say nonpolar liquid with high boiling point example okay I'll suggest example hexane okay so that will be one example now such as why two ethylphenol has the longest retention time as I mentioned to you earlier.

[01:12:34]Okay. Uh we know that the more carbon atoms that you have the alkal groups.

[01:12:40]Okay. When the number of carbon increases you will become less polar and we know that okay uh the stationary phase is non-polar also non-polar will be bonded more towards non-polar. Okay.

[01:12:54]So it will actually uh the interaction between non-polar and non-polar okay will actually make the retention time longer. Yeah. So here okay I'll just make it bigger here. So the ethyl group okay the ethyl group uh makes okay the molecule to be non-polar as the number of carbon increases. Okay. The interaction between the non-polar stationary phase and the non-polar uh molecule.

[01:13:25]Okay. Molecule makes the retention time longer. Yeah.

[01:13:30]Now the area underneath each pig is proportional to the mass. Okay. So straight away we use uh two active phenol. Two active phenol 22 22 so one you add. Okay. Times 100% you will get this. If you calculate now describe the relative acidities of phenol, water and ethanol. Yeah. The strongest acid is going to be phenol. Okay. And then the weakest acid is C2H5.

[01:14:03]Okay. Water is somewhere in middle.

[01:14:05]Okay. The explanation for it. Okay. I'll just make it bigger.

[01:14:14]Okay. When we look into ethanol, ethanol, okay, C2 C2H5, okay, they will give you C2 H5 minus plus H+. Okay. So in ethanol C2H5 is the electron donating group which increases the electron density of O in the ethoxide ion. Okay, because they will donate the electron. You can also include Okay, it gives you positive inductive effect or you can just include Yeah. So, I'll just say positive inductive effect.

[01:15:00]Okay. Making the position of the equilibrium, okay, shift to the left.

[01:15:05]Okay. So now okay it's going to be more it will become more negative so they can actually react so give you okay more of the undissolved ethanol yeah by reacting okay shift to the left by reacting with H+ to form ethanol.

[01:15:26]Therefore, okay, the amount of H form, okay, is less. Okay, phenol, okay, when they form phenox ion, so phenol, this one, okay, will give you O minus plus H+. Okay, they will give you phoxion. This is phoxion. Okay, the lone pair of the electron at the oxygen.

[01:15:54]Okay, the phoxide they deoize into the benziring they will come inside makes the phoxide ion to become more stable and the position of the equilibrium now because this is stable they will actually move to the right. Okay, so the position of the equilibrium is shifted to the right to form more H+ ion. Okay, so this will actually uh explain okay why these two are different. Yeah, so first maybe you will get one mark for this. Okay. And then mentioning this one is electron donating group. Okay.

[01:16:26]Positive inductive fact and uh uh all this explanation maybe this one. Okay.

[01:16:34]You will get uh one mark. Okay. And then this okay you will get another one mark.

[01:16:39]Okay. Now okay let's proceed. So glut ion. Okay.

[01:16:50]Gluten glution is naturally occurring compound. Okay. So it is hydrayed to form three amino acids. Okay. One of the amino acid is 16. Draw the structure of two other amino acids that are formed by this hydraysis. So from 16 I can see that SH is there. So NH2. Okay. Most probably we are talking at this combination. And then CO is around here.

[01:17:17]So this might be uh 16. Yeah. So the rest okay number one and number two will be here. Okay. The rest of the structure. So here if I want to draw okay I will draw like this. So just draw the using skeletal formula the one that they have given. So you have NH2 here double bond O then here O and from here okay N H double bond O H Okay. Now 16 has iso electric point.

[01:18:13]Okay. Of 5.0. Explain what is meant by iso electric point. Yeah. Isoelectric point is going to be the pH. Okay. When u when the molecule Okay. In this case the amino acid has no overall net charge. Okay. So this is going to be the isoelectric point. draw the structure of cyine as uh at its isoctric uh point. So this is 16. So when we want to draw the structure okay uh you can u you can uh draw maybe in skeletal formula you can also draw like normal no issues. Okay.

[01:18:52]So I will draw okay this h over here they will come over here. So you will get N H3.

[01:19:02]Okay. Plus then C H then C H S S H then C double bond O minus. Yeah. So this is going to be the iso electric point.

[01:19:32]Put it in the middle. Okay. Now, so they say that uh X, Y, and Z are three other amino acid that contains sulfur. Okay.

[01:19:43]So here, okay, let me erase some of it.

[01:19:47]Okay. A mixture of this amino acid is analyzed by electrophorosis.

[01:19:52]Okay. using a buffer of 5 okay five 5.5 yeah so we know that the iso electric point okay when they are going to be uh uh when this x y and z okay when they are going to be no net charge okay is going to be at these points okay so we know let's say at this point the iso electric point so we have NH3 H3 plus O minus NH3 plus O minus NH3 + minus Okay. Now have a look that they both of them are going to be having the same Mr. This is going to be slightly smaller Mr. Okay.

[01:20:43]And then generally okay if let's say 5.5 okay we can see that this is very close okay very close to 5 uh 5.7 and 5.5 is almost close so here at pH 5.5 this is going to be uh if you compare it's going to be slightly acidic over here now if it is acidic okay so it means that they contain H+ so what happen if they contain H+ this H+ will go here okay H+ will go here. So this part no longer O minus. So they are going to get O. Okay.

[01:21:20]So now this X okay the charge is going to be positive charge. Okay. If it is positive charge this will be attracted to the negative terminal. Okay negative terminal. Now same thing. Okay over here 5.7. Okay we know that. Okay. Uh 5.5 is almost the same. Okay, almost the same.

[01:21:43]Okay, but we uh in this case, okay, 5.5 is going to be more like acid compared to this. So what will happen the H+ will come over here similar okay you'll have O so that is going to be positive charge they will be attracted to negative terminal okay so now I need to check okay between X and Y okay which one will move further okay the one that will move further over here between X and Y you can see here these two okay one is going to be uh heavier okay compared to the other X is going to be lighter okay so because X is lighter okay and uh you can see this is generally almost the same I will actually put X will move slightly further okay slightly further than Y okay and also Y is similar 5.7 to 5.5 so maybe Y do not travel a lot now for Z okay we have 4.9 and this is 5.5 it means that this 5.5 means more they might have more O minus okay compared to this so what happen okay the H+ okay from here will go to here okay and then what happens this thing will become NH2 like normal but this will stay as minus Okay, negative ion. If it is negative ion, they will be attracted to positive terminal. So they will go to the positive terminal. But do remember that the uh they are going to be same okay charge uh sorry same mr as y. So I don't think so they will actually go far that far away. Okay. So I might bring it closer.

[01:23:46]Okay. similar to uh y.

[01:23:52]Okay. If we do like this, it's okay. Or if let's say we want to make it slightly further also. Still fine. So let's say bring it slightly further.

[01:24:07]Slightly further. So make sure that when we bring this is almost the same for X and Y.

[01:24:16]Okay. for x and y almost the same. Now explain your answer. Okay. So explanation for the answer is here.

[01:24:30]So for uh X okay at 5.5 which is acidic okay the H+ will go to O minus ion making the uh X to become positive okay and then move forward uh for uh towards negative terminal for Y okay this is acidic similar okay it will make Y is positive and move slightly okay to negative terminal and you need to also mention X is lighter and will move further then why okay and uh 4.9 which is basic okay which is basic for the buffer making H+ can leave making Z having negative negative charge and move towards positive terminal. And I can just write down the distance of Z from origin is same as Y since they have the same MR. Okay, you can just say that. Now define bient lian. Okay, biented lian is basically a lian. Yeah, that can form two coordinate bonding. Okay, per molecule of the ligan. Yeah, so two coordinate bonding. Now draw three dimen uh dimensional diagrams to show the two anthemas. Okay, so this one drawing.

[01:26:22]Okay. Uh okay.

[01:26:26]threedimensional diagrams to show two enantas of this. Okay. So let's identify the chyro center. This is the chyro center. Okay. So I will just put carbon.

[01:26:45]Carbon there is hydrogen.

[01:26:49]So there is hydrogen.

[01:26:51]So let's say here I am going to write down um carbon carbon carbon C bonded to CH3.

[01:27:23]C bonded to CH3. Then SH then this one NH2.

[01:27:42]This one will be C.

[01:27:53]double point O minus. Okay. So here, okay, we just need to draw similar uh just put CH3 2 SH one NH2 C double bon O minus okay so this the 3D structure suggest one reason why it's desirable to use single anthem. Okay. uh as a drug.

[01:28:52]Okay. Because only one has the biological effect and the other one the other anantis may have uh may cause side effects.

[01:29:31]Okay. Now, octahedral complexity forms when two moles of Z reacts with one mole of Cu2+ as shown. Okay. So, two moles do remember this is bent.

[01:29:45]Okay. So, they form octahhedral. So, they form two moles. Yeah. two moles. So complete the equation to to describe the formation of octahedral complex. Okay, we know that Cu and then when you have the Z uh 2 Z coming in. Okay, 2 Z coming in.

[01:30:10]Copper was 2 plus. Okay, 2 Z coming in.

[01:30:15]So they do not it will cancel. Okay, here is -1 -2 so it will cancel and then do remember it's bented earlier they have six so by dentate one molecule have two coordinate bonding so two molecule four coordinate bonding so we have replaced four waters here so four molecule of water will go out leaving the uh H2O okay to become uh remaining two only. Yeah. So, this will be the answer. Okay. Done. Okay. So, I will put the my jotted uh the note that I jotted down over here. And also for those of you who do not have this paper, I will also share the uh the blank paper for you. Yeah. Thank you everyone.