 Yeah, hello guys. Good evening. Good evening. So yeah. Okay. So today we have the last session means regular session and then next class onwards we have crash course. No, you have the info, right? Yes. So yeah. So you know how the crash course will go on how we have planned it. You know the info you have the information. Just a second Richard. So it will be a problem solving session. Okay, we'll have brief theory of the entire chapter will be theory and then problem then the theory and then problems. So theory will not go in detail. Okay, we'll just have a brief introduction of the entire thing. So you have to come prepared with the theory. Okay. So what is the first session? What is the first topic we have? You know that I think more concept more concept and atomic structure, I guess, right? Yes. Yes. So planner you have already you can revise chapters according to that. So it's going to be, you know, very hectic for you. Every subject you will be having two sessions in a week. Right. So six days is completely occupied and then Sunday you will have a test. Right. Okay. Any any info regarding notification from NTA? No. So we'll start it. See, first of all, the salt analysis chapter like I said last class, it has two, you know, two portions. One is that we have done. That is quantitative analysis detection of carbon, hydrogen, nitrogen, phosphorus, sulphur and all those hydrogens. Right. So like I said last class also three methods, Karius, Dumas, Jeldal. Okay. These three methods are important, very important in fact, if you are talking about these particular chapter, this particular chapter, correct, that you need to do. This portion, the second part that is qualitative analysis. Okay, salt and salt analysis, we have both parts over here qualitative and quantitative analysis. So we are going to do today qualitative analysis, qualitative analysis. So as far as the main is concerned, this chapter is not that important. This portion is not that important. Okay. I won't say they won't ask any question from this, but chance are very less. Okay, chance are very less. And the entire chapter is based on, it's theoretical completely, theoretical we can say, we have experiments and experiments based result we have then the chapter. Those results we need to study. Okay, like how to identify a given radicals. Radical means we can have acetic radicals, we can have basic radicals. So given an ion like suppose sulphate ion, how do we detect, how do we identify that this is the sulphate ion we have. What is the reactions for that? What is the test for that? Okay, so these things which is already done or all these experiments has been conducted in the lab and those results is there in this chapter. Okay, that is what we are going to study. So you need to mug up these things completely. If you look at the portions, the entire chapters, it's very big, the entire thing, because we have one theme. You will understand how these chapters we are going to study. Right, but suppose we have one ion, like suppose chloride ion. For chloride ion, we have many tests, many different reactions, five, six different tests we have. With that particular test, we'll get a particular result. If that result we get, then the ion is chloride ion. So like this we have test analysis results for all the ions, whether it is negative or negative ions or positive ions. Now you can understand that we study a lot of ions like chloride ion also you can see we have chloride ion, bromide ion, iodide ion, sulphide ion, nitrite ion. Okay, then some positive ions also we have as well. So for all these ions we have different different results, different different reactions. All these reactions we need to study. So obviously if you look at the chapters, it's too big. But the weightage is concerned, as far as the exam is concerned, the weightage is not that much in the proportion if you see. So that is what I would like to say, I would like to suggest. NCRT you can go through once for this chapter. Okay, for J mains point of view, the chapter is not that important. We have one complete session for this for our session. We'll try that we can cover up everything in this particular session. Okay, so we'll start with this chapter. See qualitative analysis is obviously if you try to analyze what ions are there, you need to what you need to understand what is the reaction of that particular ion, what is the property of that particular ion. So obviously to detect or to identify the ions present, this chapter is obviously it's important, like in order to understand which ion is there. So it is basically important part of chemistry. However, if you look at for a comparative exam point of view, it is not that important. But if you talk about chemistry, then this chapter is important. Okay, so what is the purpose of this particular chapter, the purpose is to identify the different constituents, constituent ions. Okay, the purpose is the objective here, the entire objective is to identify, identify the constituent ions, the constituent ions. Okay, so when you talk about ions, ions we can have of two types. Okay, cations and anions, right here and ions we also call it as here and ions we are calling it as acid radicals and ions are acid radicals and cations are basic radicals. This is how we classified anions, cations, anions are acid radicals or acid radicals, it is basic radicals. So first of all we can examine these ions, like preliminary examination we can do for these ions, like preliminary examination means what we can talk about color, we can talk about color, we can talk about heating effect. On heating what happens, heating effect, we can talk about a few tests, like we have flame test, borax bead test, etc. we will see that. These things we can talk about for preliminary examination. Okay, so what are the different techniques we use to examine these radicals that we are going to see, but before that we will see some physical characteristics. Okay, copy this down first, heading right down, physical characteristics. So what is the physical characteristics, so we can talk about like I said we can talk about the color, what are the color of the compounds we have and this obviously you need to memorize. Okay, so if you talk about the color here, the first factor we will talk about color. Yeah, if you talk about color, first thing you see, what are compounds which has black color, right, obviously this is something that you need to memorize. Okay, the black color compounds we have, we can have oxides, we can have sulphides and we can have some, you know, we call it blackish brown, okay, sulphides only compounds only, we can say. So you see some oxides of black color, if you look at the example, oxides of black color, we have MnO2, FeO, CuO, then we have cobalt oxide, CO3O4, Fe3O4 and nickel oxide, Ni2O3, Ni2O3. Some sulphides, if you see the example of sulphides of black color, we have Ag2S, black color, CuS, black color, FeS, Cu2S, all these are black color compounds. One compound we have, which we write in this one only, but these are blackish brown actually, exact color if you see, it is blackish brown. These are also sulphides like COS, cobalt sulphide, we have nickel, we have lead, PBS, we have HGS and we have then BI2S3. So obviously all these compounds have same color, right, so for sulphides we have two compounds of same color, Ag2S and CuS, so only color is not sufficient, right, only with color we cannot differentiate. After getting the color, we can obviously distinguish whether it is an oxide or sulphides, but then in that oxide or sulphides, which oxide or sulphide is there, for that we have different steps, like we have chemical reaction and other steps we have in this. But first step is this, that is why we are calling it as preliminary steps, right, we can figure it out whether any sulphide is there or any oxide is there. So first one is black, the second one you write down, that is blue, okay, blue color if you see, we have hydrated sulphide, for example, we have hydrated sulphide, that is hydrated CuSO4, hydrated CuSO4, we have anhydrous cobalt sulphide also in this category, anhydrous cobalt sulphide in this category. Third one is orange color, orange, we have CO2, some dichromate like potassium dichromate, K2CR2O7 and we have some ferric cyanides, and we have ferric cyanides. So you see all these compounds are there, but K2CR2O7 if you remember we use this in, this in redox, right, redox also we use this. Strong oxidizing agent. Sorry? Strong oxidizing agent. Yeah, so that's why this one since we use in other topics also, so this one is a bit important. Okay, color of this one, okay. So you have so many compounds, but few are important, which probably if you go through, I will also let you know which one is more important that definitely you need to keep in mind, I will tell you that. Third one is orange, fourth one we have green. We have nickel, the salt of nickel, simply write on nickel salts, nickel salts, then we have potassium magnet that is K2MnO4. Remember it is magnet stored per magnet, K2MnO4, some copper salts in which the copper oxidation state is 2, copper 2 salts. We can also have chromium the salt in which the chromium oxidation state is 3. Chromium 3 salt. Red is HgI2, lead oxide Pb3O4, yellow. Yellow is some sulfide in this also we have, CDS is yellow, don't get confused it is not black. CDS, AgBr you see it is important, AgBr is yellow. AgI is also yellow, okay, PbI2, okay, these are the examinables we have. Sir, it is HgI2 Nessler reagent. Sorry? Sir, it is HgI2 Nessler reagent. Nessler reagent is K2 HgI4. Okay. Not HgI2. Sir, for green we can write FESO4 also, right, hydrated. In which one? Silent green. Haha, that also we can take. Hydrated cobalt salt in which the oxidation state of cobalt is 2, that has reddish pink color, reddish pink. We have hydrated cobalt salt, oxidation state is 2. Hydrated cobalt, okay, one more important one you see. Purple color. Purple is a very important one. We have KMNO4, potassium permanganate, okay. FECL3 normally it is dark brown, write down one thing. FECL3 normally it is dark brown, but in solution it is yellow. FECL3, okay. Now there are a few compounds which changes its color on heating, right. So adding a write down. Color change on heating. So I'll write down here. We have compounds which is mainly oxides we have. We have compounds here. We have color here. This side I'll write down when it is cold. In cold means when the temperature is low and when the temperature is more that is on heating. What is the color change we have on heating. So let me just draw this. So if you look at compounds here, we can have ZNO, zinc oxide. When the temperature is low, its color is white. And on heating it changes into yellow, zinc oxide. SNO2, BI203, color is yellow. And when you heat this, it's changing into yellowish brown, okay. FE203, this is brown. And when you heat this, it becomes black red. Blackish red or black red we can say. Okay, so these are the few compounds we have which changes its color on heating. Few ions and its color. Okay, first you copy this down and I'll go to the next slide. There are a few ions and it's, you know, it's color of the solution. Heading write down. Color of the solution. Like if you talk about CR04, 2 minus, we have Fe3 plus, FeCN6, 4 minus, all these ions in solution. It is yellow in color, yellow color solution basically. If you have Cu2 plus, Ni2 plus, Fe2 plus, CR3 plus. And its solution color is green or blue. Okay, CO2 plus, MN2 plus. It is pink. Similarly, some salt also, they give characteristics smell. Okay, like you see, when you get smells like vinegar or acidic acid type, then it means it is acetate ion, we have CS3, CO minus. Ammonical swell means a monical ion present. Not much important as you can leave this particular part. Okay. Now, this is the preliminary examination. Okay, which we, which from which we'll get the idea that, okay, if the color is yellow, then these, these ions may present over there. Okay, green, these, these ions may present over there. Further, we have the next step that how do we detect each and every ion here. So for detection of each and every ion, this entire, this part only, the entire organic chemistry here is classified into three groups. Okay, that we discuss under analysis of ions. Sorry, anions, analysis of anions. Like I said, anions for analysis, it is classified into two groups. One is acidic radical and other one is basic radical, right? So if you talk about acidic radical here, which is nothing but anions, like ions are classified into two, acidic and basic radicals, acidic radicals are anions itself. Basic radicals are cations. So we are talking about anions here, which means we're talking about acidic radical, both are same thing. So for this also, we have three different groups in this, means under anions, we have three different groups, okay? So group one you see, in this group one, we have ions like carbonate ion, CO32 minus CH3 COO minus carbonate ion, sulfite ion, S2O32 minus, and we can also have nitrite ion NO2 minus. All these ions belongs into group one. It is kept into group one. And for this, for detection and other things, what we use here? We use dilute HCL, right? We use dilute HCL, H2SO4. Now, why these anions are kept here in this group one? We consider only those anions here, which gets decomposed by dilute HCL or H2SO4. These are the anions, which get decomposed by dilute HCL or H2SO4. Similarly, we have group two anions here. I'll write down first all the three groups, and then we'll discuss it one by one. Group two, the anion contains, it has sulfite ion as a 4, 2 minus, oxalate ion, C2O4, 2 minus. We can also have phosphate ion, and we also have BO3, 3 minus. But once again, once again, just a second. Group two is, I have written it in other way, not these one, these are the group three anions we have. Group two contains, group two are those compounds, which are decomposed by cong H2SO4, right? Concentrated H2SO4. So for that, we have halide ions we have. Generally, we have Cl minus, Br minus, I minus, NO3 minus, F minus also you can write. Mainly this, okay? So these are the anions, which are decomposed by concentrated H2SO4. Write down. These are the anions which are decomposed by concentrated H2SO4. Copy this down. Group three anions you see. Group three anions we have. Sulfate anion, C2O4, that is oxalate anion, phosphate and BO3, three minutes. In group three, we take all those anions, which are not covered in first two categories. This is what all anions are left. We'll drop into this category. That is group three over here. So here we cover, or we'll have all those anions, which are not covered in, or which are not present in group one or group two, which is extracted above, which is illustrated above. So when we have anion, no, mixture. So during the anion analysis of the mixture, we use two different types of extract here, right? So that is extract. We call it as water extract, or sodium carbonate extract. Okay, so write down this point first. An anion may interfere. An anion may interfere. During the detection of an anion may interfere during the detection of other anion. In such cases, in such cases, in such cases, water extract, water extract, or sodium extract can be used. Water extract or sodium extract can be used. So how do we prepare water extract, that you see, a small thing we have here? So, yes. So which one decomposes group three anions? Sorry? So what decomposes group three anions? No, in this, we don't have such classification. Like group one and group two, we have group one is valued at CNR as two as a four. Group two is concentrated as two as a four, right? In group three, we don't have such classification. It simply says that water anions are left after these two groups. That will drop into this category. Yes, sir. Okay. So water extract, whenever you use water extract after this, just to use the word WE, means water extract. Okay, we use this abbreviation for this. Water extract, obviously the preparation method is not important, but just for the understanding, you should know what is water extract. It is a small amount of, once again, so we'll take the mixture, the compound that we have, which we need to detect. And that we're very small amount. So I'll write down here, small amount of a small one is 40, 30, 40, 30 to 50 milligram in that range. Small amount, which can be less than 50 milligram. Of the mixture, of the mixture, is dissolved in, is dissolved in 225 ml, 225 ml of distilled water, distilled water, and boiled for, and boiled for like a minute, one or two minutes, and centrifuge. And centrifuge. Okay. The centrifuge that we get here, means after centrifuge, whatever the mixture we get, that we call it as water extract. One second. The centrifuges we get, we call it as water extract. When we said that there are two anions, so which is the second anion, which will interfere? See, that we don't have any control. You have a mixture of any organic substance, right? There are many substance present into that. It's like you can understand, it's like you have a mineral, right? In mineral, we have many impurities, no? So what impurities depends upon what, is the mineral you have, correct? And it depends upon the other condition also, like the geographical condition also. What are the impurities that present, right? So in any compound, we can have many such anions present. It's not like only one anion is present over there. Point is whenever you have more than one anion present, like suppose we have a sulphate anion present and phosphate anion is present into that. So to detect one of the anions, we need to understand what? We need to understand which anion is present over here. So if we have more, a mixture of anions present, if we have that condition, then one particular anion will also interfere in the reaction, and hence it will be difficult to detect that particular anion is present in this particular mixture. Are you getting my point? Yes, I understand, sir. Right, centrifuge, you understand what is centrifuge? Centrifuge is kind of, you know, are you ready? When you rotate it, right? Yes, sir, you rotate it at high speed. Washing machine, the concept is based on centrifuge only. Right, it rotates and very high centrifugal force acts on it. So basically when we have the difference in, you know, densities of the fluid, it is used to separate the two fluid, we can say. So different density we have, different force it will act and since it acts on the outward direction away from the centre. So in that way, it can accumulate the particles on the outer periphery and then when it settles down, you will get it to separate layer which we can easily separate. We can easily take it out. So in short, like in rough, you know, roughly I have given you the idea of what is a centrifuge and all. But again, in the college and all, you will need to do all these experiments. That is centrifuge. So the centrifuge is and the centrifuge that you get, the mixture that you get after centrifugation, that is called water extract over here, over here. Correct? Now similarly, we have one more, mixture that we use for detection purpose that is sodium carbonate extract. If you remember, we have done this in a metallurgy also a bit. We use sodium carbon extract over there. Sodium carbonate extract. SCE in short, we'll write. SCE. In this also, a small amount in short is right on a small amount of mixture is added, a small amount of mixture we have taken. So around 40 to 50, again, a milligram of the mixture. In that we add, I'll write down the composition. Suppose we have 40 gram of mixture. In that we'll add around 200 gram of milligram. I'm sorry. Here also it is milligram. This is also milligram. 200 milligram of Na2CO3, sodium carbonate we mix. Plus 5 milligram of distilled water. Distilled water. Boiled for five minutes and century fuse. We'll get sodium carbonate extract. Now one by one, we'll see all these groups. And I like group one water and I answer there. What are the methods? Then group two and then group three. So in group one, I'm taking carbonate iron. Carbonates. So obviously what is the reagent we're using here? I'm not giving you the theory over here. Like it reacts with value at zero. I'm not giving you that reaction. Simple use it. Whatever compound we have. And it was CO3. It reacts with H2SO4. Obviously we must have some carbonate. Reacts with S2SO4 forms. And it to SO4. We have valued H2SO4 here to remember that. We have any two SO4. Then H2O and then CO2. This is the. Reaction we have. Now CO2 evolves here with effervescence. CO2 evolves here with effervescence. And that confirms the presence of carbonate iron in the mixture. Okay. Further you see. This CO2. Reacts with lime water. That is COH whole twice. Converts into. C is CO3. Plus H2O. We have CO2. Plus H2O. We have here white precipitate of CO3. So these are the chemical reactions involved into this. Okay. Sodium calcium carbonate forms. It turns milky white precipitate and H2O evolves. One note you write down into this. One more thing you write down one more reaction we have here. The calcium carbonate that forms here. It also get dissolved in presence in water in presence of CO2. And it converts into CAH CO3. Whole twice, which is also insoluble. Sorry, which is also soluble. Biocarbate forms. Yeah, right. So this is a chemical reaction involved in this. One note you write down here. Obviously we have all these are reactions. It's not like all these reactions are true for all types of carbonates. Correct. So write down here. Carbonates of bismuth and barium. Bismuth and barium. Are not decomposed by are not decomposed by dilute H2SO4. Barium and bismuth carbonate are not decomposed by dilute H2SO4. Next point one more. PbCO3 reacts with dilute HCl. Or dilute H2SO4. So can you repeat this all name again? Pb. The second point. Yes, sir. Yeah, that's the second point right now. PbCO3 reacts with dilute HCl or H2SO4. But the reaction slows down. But the reaction slows down. Due to the formation of, due to the formation of protective insoluble layer, due to the formation of protective insoluble layer of PbCl2 and PbSO4. Okay. Protective insoluble layer of PbCl2 and PbSO4. This is for carbonates. So what you have to keep in mind that CO2 gas evolves, which is obviously a colorless and orderless gas. So it evolves with a brisk effervescence in cold, right? Effervescence with effervescence comes out. Okay. Next write down the second type of an ion in this group one. That is, that is carbonate. Sorry. Nitrite ion. Nitrite ion. Right now it reacts with dilute H2SO4. It reacts with dilute H2SO4. Hello, can you hear me? Yes, sir. I lost the connection, I guess. One second. Who is the host now? Which I have become the host anyways. Yeah, yeah, yeah. No, it's fine. Everything just a second. Yeah. So nitrous acid it forms, which in contact with air becomes brown due to the formation of nitrogen dioxide. This information you have to memorize. Okay. Can you repeat it again? Again, I'm repeating nitrous acid forms, which in contact with air, which in contact with air becomes brown due to the formation of nitrogen dioxide. Right. The reaction involved in this. We have an A NO2, nitrite ion NO2 minus plus H2SO4. It convert into an A2SO4 plus HNO2. This is called nitrous acid. HNO2 nitrous acid. Now this HNO2 decomposes into H2O, then NO and HNO3 nitric acid. Three moles of HNO2. Okay. And we have one nitrogen here. So two here. Two, three, three plus three, six. So six. Yeah, it's balanced. This is the reaction. This NO monoxide that we have. This combines with atmospheric oxygen and converts into nitrogen dioxide that is NO2. Brown gas. Okay. Now in this only we have a reaction. One more thing is there. This NO2 that evolves here. This may also react with a compound. So in any solution, reacts with KII basically and evolves iodine gas. Okay. So NO2 produce can react with KII which results into the formation of iodine gas. And this iodine reacts with the starch and forms the adsorption complex, which is blue is violet in color. Okay. should know that NO2 can also oxidize Ki into I2. That's the basic information we have. Now, in this, we have an important test. We call it as brown ring test, brown ring test. In this, what happens? In water extract, WE, in water extract, freshly prepared iron sulfate, FeSO4, in water extract, freshly prepared iron sulfate is added and we also add dilute S2SO4 in the mixture. In water extract, freshly prepared iron sulfate, FeSO4 is added, right? Freshly prepared iron sulfate is added. A small amount of dilute S2SO4 also added in the test tube, comma, a brown ring forms, a brown ring forms at the junction and it is due to the formation of, and it is due to the formation of this compound, FeH2O5NO, sorry, FeH2O5NO, bracket close, SO4. This complex forms and because of that only we have the brown ring forms over there. Okay. Now, what is important here? Remember this thing, what we say, coordination compound. We have discussed about ANO ligand, right? Okay. ANO is, if you go back and check your notes, ANO I have given in both type of ligands it is. It is neutral as well as positive charge ligand. Yes or no? Yes. If not then you keep this in mind, ANO has both kind of behavior. It can behave as a neutral compound and it can behave as a positive charge ligand, right? But in this brown ring test, this ANO is behaving as a positive charge ligand. Must keep this in mind, very important. It's a positive charge ligand, right? Write down this brown ring complex. This brown ring complex is paramagnetic in nature with three unpaired electrons. Brown ring complex is paramagnetic in nature with three unpaired electrons. So it's not like we have only one experiment. This one is a bit more important. In fact, that's why we have discussed this and brown color ring forms over here. But we have other kind of tests also. In order to detect nitrite ion, we can perform other tests also. Okay. So if you have this one by one, I'll write down two, three different tests over here. See, if you have the mixture that you have taken, in which we need to detect, that is the mixture we have here. In this mixture, we'll add dilute S2S4. Group one dilute S2S4, we are adding. If you add dilute S2S4, then we have the same thing. I'm just writing down in this equation manner. We have light brown, pungent fumes, light brown, pungent fumes evolved. So when this evolved, it confirms the presence of NO2 minus ion. This is another detection technique we have. Like I said, we have other methods also. Easier method is what? We'll take a paper. This paper will dip in KI and starch solution. In KI and starch solution, starch solution. And this paper, after dipping this into the solution, we place the paper at the mouth of the test tube. Place the paper at the mouth of the test tube. Of the test tube. What happens then? If this you will do, then the paper turns bluish violet. Blue also you can say or bluish violet. Paper turns bluish violet. This confirms the presence of again NO2 minus ion. These are the few more methods we have to detect nitrite ions. Obviously, the brown ring one is the most important, but these two also you can keep in mind. How is the brown ring paramagnetic if it's NO plus? Because this is an NO plus strongly there. Once again, I did not check the configuration. NO plus you have, right? H2 is neutral. So we have minus 2 plus 2, which is plus 2 and plus 1. So it will be. What is the option to stay for that of iron in this? So it comes plus 1. Iron is plus 1, right? So iron is plus 1. So Fe plus the configuration is argon 4s1 and 3d6, correct? Yes, sir. Okay. If you look at this orbital diagram, it is 3d4s4p. So we have 1 electron here and 6 we have here. 1, 2, 3, 4, 5, 6. Okay. So in this one, see, it is a factual statement. First of all, which I understood. Yes, sir. It is the factual thing. Remember, when you do this kind of question in VBT, there are some more information given that this compound has two different types of bond length out of all the bond lengths, four are equal, one is different. And then we realize that, okay, this is a square pyramidal geometry. An example I'm giving you. So along with the complex, you must require some information, right? It is true that when the ligand is strong, maybe you are thinking that all these electron will get paired and will have one unpaired electron, right? Yes, sir. Right. But that is not true here because it is a factual statement. Its magnetic moment you will get when the unpaired electron n value you are getting three. Then only its magnetic moment is getting satisfied. So what we conclude here, we conclude that out of this five unpaired electron, only one pairing is taking place. That is also possible in some cases. Okay. And the three electrons are unpaired only. That's why obviously number of unpaired electron gives you more or less magnetic moment. But if it is paramagnetic, that is true. So it's not like whatever number of unpaired electron is there, all the electrons will get paired just because the ligand is strong. That is not always true. Most of the time it is true. It is fine. But in some cases, like this one, you have to keep this in mind. This is actually this one is very important. Right. Magnetic behavior, I don't remember. But this information they have asked, like positive charge, you should know whether it is neutral or positive charge over here. Okay. Yes, sir. Okay. So this better you take it as an information. Tell me. Sir, how did you predict that only one will get paired another one? That is what you cannot predict. That's what Richard I explained. That we find out the number of unpaired electron or the geometry of the complex or the magnetic moment. It's not theoretical. Logically, we are not calculating it. It's experimentally, it is there. Right. But in some cases, we have exception. Like answer to your question, how do we predict only one electron will get paired? We don't have any control on this. Why are we saying three? Because if it is one, it has a root three magnetic moment. Right. But this is not the magnetic moment observed in the case of this. The kind of behavior it has in presence of magnetic or electric field, it is not satisfied by only one unpaired electron. Right. So these are the factual thing that you should memorize. You should keep this in mind. Even in VBT also, we have the same thing. But in most of the cases, we can understand this way that the ligand is strong. So pairing will take place. And like this, we have the vacant orbital, and then the hybridization is this. But in few cases, no, the entire thing goes on facts. So this one is one of those cases we have over here. Correct. Not much. This kind of cases, you won't get much. If you remember that time also, I'll tell you that in order to predict the geometry, they will give you the information. Right. Like suppose one example, if I go back and take, if you have trigonal bipyramidal, to understand this, we can also think of a square pyramidal geometry. Both has coordination number five, isn't it? Yes. Coordination number five. But how do we know that this is trigonal bipyramidal? It is a square pyramidal. For this, one information will be given in the question that it has two different types of bond length. Three bond lengths will be same and two will be different. Axial and equatorial, which is not in this case. And hence we predict that the geometry is trigonal bipyramidal. So this kind of information is given in the question. And that's how we can predict. Okay. Here we do not have. So this, you have to keep in mind. That's done. And retrieving one is we are done with it. We have sulfite next. Take care. Write down the sulfite ion next. So you can understand, there's so many reactions and so many, you know, tests we have for one particular ion. And like this, we have so many ions are there. So there's no end of this chapter, if I frankly, if I tell you, there's so many things you have to mug up simply. That's why I would say this, like for this one is very important. Okay. This one you have to keep in mind. Few things you just do. And that only you keep in mind. Because for J means this chapter is not that important. Third type of an ion you write down sulfite ion, that is SO3, two minus. Okay. In sulfite and what happens chemical reaction, I'll tell you what is the chemical reaction involved. The experiment by which we infer that sulfite ion is present is what? Okay. We'll take mixture, mixture and we'll drop few drops of dilute acid. Because it belongs to group one, we can take dilute Hc or dilute H2O4. Right. And we heat this. So on heating, what happens? We get what we observe here. The observation is a colorless gas, a colorless gas, pungent and suffocating order of pungent and suffocating order. We get this, which confirms the presence of sulfite ion. This order is similar to the order of burning sulfur, like the burning sulfur we have. This confirms the sulfite ion. Filter paper also we use for this. Another method is what the experimental process is. What we do? The filter paper we take and this will take in the environment of K2CR2O7. Dilute means we'll take the solution of K2CR2O7 also and this filter paper is moistened in this solution. Right. We'll have a moisture of this solution on this filter paper. Right. And when you add dilute H2O4 here, because this is the reagent we have for group one, dilute H2O4, then what happens here? The observation is what? This is the experiment. Observation is the color of the paper turns green. Color of the paper turns green. This also confirms the presence of sulfite ion, green color that we have. Copy this down. Now in advance, how they frame the question on this? They will give you one paragraph kind of questions, like that we have a mixture in which we add this particular H2O4, then the color changes to green. Filter paper was there of this. So this information, they will write in the question. The one that I'm telling you, they will write the question as it is and they don't mention the compound or they'll say, okay, it reacts with this and forms A, which is green in color. Right. So what is the ion present in the mixture which you have taken? So this information, if you remember, then you can answer those questions. Series-wise questions also they frame from this. Okay. This is one. Two more reaction we have. Both are important here for this one. One is SCSE, we use sodium carbonate extract we use and we add dilute HCl in this. Dilute HCl we add till the solution becomes acidic and we also add BACL2 into this. Barium chloride solution we add here. We get here white PPT of white precipitate of BASO4, white precipitate of BASO4, right, which is soluble in, which is soluble in concentrated HCl. So this will happens. It confirms the presence of SO3, 2 minus, sulphide ion. One more we have in this and that is in the mixture we add BR2 water, bromine water we add in the mixture. Simple one is this one. Bromine water we add in the mixture and it forms BASO4, white precipitate, BASO4, white precipitate, which is again insoluble in, like here also. Okay. Here I have written soluble by mistake. It is insoluble, which is insoluble in again concentrated HCl. This confirms the presence of SO3, 2 minus ion. One second. Then so it is insoluble in sulphide ion, which confirms the presence of sulphide ion over here. So these are the testing thing we have in the lab and do this. What are the chemical reactions involved? Chemical reactions we can write down, one, two chemical reactions I'll write down the important one. We have Na2SO3 because sulphide ion we have. So it can have in this form Na2SO3 and we add, you can add, we know it is either you can add H2SO4 dilute or HCl. If you add this, the reaction would be Na2SO4 plus H2O and you have to eliminate sulphide SO3, SO2 sulphur dioxide, sulphur dioxide evolving to this one. So the colourless gas with SO2? No, sulphur dioxide evolves here. No. So we say, because along with this sulphur ion is also forming. So we say, okay, in the mixture, the sample that we have in the sulphide ion is present. Correlous gases SO2 here we have. Okay. So it's not like that we have sulphide ion. So here in the product side we'll get sulphide ion. Right. Okay. Bacl2 that we have, the third one you see, the reaction here Na2SO3, SO3 plus HCl plus Bacl2 if it reacts sodium extract we are using here. Okay. So it forms NaCl plus H2O plus sulphide ion we get BasO3 precipitated forms. Okay. Just let me check one thing. I think I have done one mistake. Okay. Here there's a change. Just a second. This is, I think it was this. It forms white precipitate of BasO3 and BasO3 is soluble in consumption. That's fine. Not BasO4. I have written BasO4 initially. It is BasO3 it forms. Okay. Now in this only what happens we have all this solution and in this if you add bromine water then it converts into the insoluble sulphate that is BasO4. Right. So we say here what that in this solution in above solution in above solution if we add bromine water Br2 water it converts into insoluble BasO4, barium sulphate. One more reaction which is not obviously you know the detection of sulphide ion here but this you know one color that we get here a compound that color you must memorize if you have K2CR2O7 and it reacts with H2SO4 in presence of sulphur dioxide means basically you have this solution the first reaction you see we have sulphur dioxide and when we add K2CR2O7 into this solution then this reaction is possible right and it converts into sodium sulphate and into SO4 plus H2O plus sulphur dioxide SO2 is it this no we have CR2 which I've written it other thing what I'm doing I'm sorry this is one we have no it forms CR2SO4 whole thrice which is green in color so this question they have asked once CR2SO4 whole thrice it is green in color and along with this we get K2SO4 and H2O this color you must remember sulphur dioxide is the gas we have which has the suffocating smell just confirming all these things you get the suffocating smell okay so if you want to confirm this fine suffocating is SO2 you add K2CR2O7 to that solution then the color changes to green because of the formation of this chromium sulphate and hence it is confirmed okay next we'll see the test of sulphide ion S-2 S-2 okay so the first you know experiment what we do here again we'll take the mixture which we need to analyze and we'll add dilute H2SO4 dilute H2SO4 we get a colorless gas colorless gas with rotten egg smell rotten egg smell like H2S H2S gas smell be as I wrote that H2S and sulphide this confirms the presence of sulphide ion so basically H2S evolves in this reaction we'll see the reaction also another method is what we'll take sodium carbonate extract SCE and we'll add dilute H2SO4 this gives black precipitate black precipitate of PBS lead sulphide which confirms S-2 again right if you take sodium carbonate extract add dilute H2SO4 and along with this you are also adding CDCO3 solution then we'll get CDS cadmium sulphide you get and it is yellow in color here this confirms the presence of this sulphide ion so where is the PV coming from what CD no sir the PV in the second reaction okay here lead acetate solution we also add lead acetate along with this here we are adding CDCO3 here we are adding lead acetate solution and then this forms then okay so you see what are chemical reactions are involved here the chemical reactions we have Na2S plus H2SO4 gives H2S and Na2SO4 if you have lead acetate PV CH3 COO or twice plus H2S it gives lead sulphide PBS black and CS3COH this one is black okay so this is the test of the sulphide ion we have next we'll do the test of thiosulfide ion