 Okay. Good morning, guys. Hello. Am I audible? Okay. Could you see my screen? Yeah. Okay. Okay. So last class we were discussing about group 15 nitrogen family, right? And we have discussed the properties. All the properties we have discussed about the various this thing, the different types of compounds found by the elements of group 15 that is hydrides and all oxides we were discussing. Okay. And there are a few things which was left in oxides that will continue with oxides today. Okay. So we have discussed about the preparation and properties of oxides. Now there is an important comparison of bond angle. Okay. So these bond angle as you know it is based on the VSEPR repulsion theory. Okay. Which is mainly based on again what which is mainly based on the bond pair, lone pair, lone pair, lone pair, bond pair, bond pair repulsion. So basically based on the electron pair repulsion, we can define that bond angle of any compound. And here also the bond angle in these oxides of nitrogen based upon the electron pair repulsion only. Okay. You see the molecule or ion is this have this bond angle comparison. So important you must keep this in mind. We have NO2 plus I'll write down the order here first of all. We have NO2 plus maximum, then we have NO2, then we have NO3 minus, then we have NO2 minus. Okay. This is that decreasing order of bond angle. Okay. The value of bond angle here, if I write down in NO2 plus it is 180 degree, you can find out the hybridization also here. For NO2 it is 134, 134 for NO3 minus it is 120 and for NO2 minus it is 115 degree. This is the bond angle we have. Okay. Like I said, it is based on the electron pair repulsion, if you find out the hybridization here, the hybridization of NO2 is SP. Okay. It is SP hybridized bond angle 180 degree. But in all the molecules, the bond angle is not because of hybridization. It is because of the presence of lone pair of electrons or electron pairs because you see NO2 is what? NO2 is an odd electron molecule, right? It has 6 into 2 is 12 plus 5, 17 electrons. Number of valence shift, valence electron we find out here, it is 5 plus 6 into 2, 17. So this is an odd electron molecule. Okay. It has 2 bond pair and 1 unpaired electron, single unpaired electron on this. Okay. That also we count into hybridization. This hybridization is coming out to be SP2 hybridized. This one is SP2 hybridized because of repulsion of electron pair, the bond angle increases here. It goes more than 120 degree, which is the ideal bond angle for SP2 geometry, right? But since because of electron pair repulsion, it is observed that the bond angle increases over here. But NO3 minus also if you find out the hybridization, 16 to 3, 18 plus 190, 19 plus 5, that is 24 by 83. So this is also SP2 hybridized and its bond angle is 120 degree. NO2 minus, if you see, it is 18, 18 by 8 is 2 plus 1, 3, so SP2 hybridized. But because of lone pair present onto this, the bond angle decreases. Okay. All three molecules you see these are SP2 hybridized but have different different bond angles. Okay. So you must remember this four different oxides of nitrogen which has bond angle. Okay. Copy this down. Okay. So we are almost done with this particular, you know, elements that is nitrogen. Only two more compounds we have to discuss of nitrogen. Most important part of nitrogen is oxide. Okay. So oxides property, you must remember. There is another important portion we have here, which is the reaction of HNO3. Okay. So what happens with concentrated HNO3 dilute or valley dilute HNO3? Three different conditions we have there that we'll see. Okay. But before that, we'll quickly finish this particular compound of nitrogen, which we call it as nitrous acid. Okay. So right on the heading, nitrous acid HNO2. Okay. How does it prepare? It is prepared by, right now it is prepared by the reaction of N2O3. N2O3, when you dissolve in water, it forms two moles of HNO2. This is one of the method of preparation of HNO2. We have nitrous acid. Okay. The another method is right down when metal nitrite dissolved in dilute acid. When metal nitrite dissolved in dilute acid. For example, HNO3 or H2O3, sorry, HCl or H2O4, anyone. Okay, dilute acid we can take. When metal nitrite dissolved in dilute acid HCl or H2O4, it forms nitrous acid HNO2. So metal nitrite, we can take any example. Suppose we have NANO2 plus HCl dilute. It converts into HNO2 plus NCl. This is another method of preparation of nitrous acid we have. Okay. There are a few important reactions, properties of this nitrous acid. So we'll see that properties of nitrous acid. Right now next, it shows oxidizing property since it evaporates, since it evolves nascent oxygen. Right? This HNO2 behaves as an oxidant actually. Oxidizing property means oxidant. Reducing property means reductant. Okay. So HNO2 evolves nascent hydrogen, H2O plus NO, and we get a nascent oxygen here. Hence, it shows oxidizing property. Correct? Okay. What it does, for example, you see a few examples of this particular thing. Like suppose we have Ki solution plus 2 HNO2 plus HCl. You don't have to memorize the balanced reaction over here. Just what is happening, you just keep in mind. Suppose HNO2 is an oxidizing agent, right? So it oxidizes Ki into this and evolves iodine gas. That is what you have to keep in mind. My entire reaction is not required. 2KCl plus H2O plus NO, and iodine evolves into this. I do. That is what you have to memorize. Okay. It also oxidizes SNCl2 to SNCl4. Means plus 2 oxidation state of 10 into the plus 4 oxidation state of 10. Okay. 2HCl. It converts into SNCl4 because we also know due to inert wear effect, SN plus 4 is the more stable form than SN plus 2, if you remember. Okay. This is the reaction we get here. Right? So this is the oxidizing behavior of HNO2. Apart from this, it also shows some reducing property. Okay. Since the oxidation state of nitrogen is what here? Plus 3. Right? It can increase its oxidation state or decrease also. That's why it shows both kinds of behavior. Okay. This is two examples for oxidizing behavior. Apart from this, it can also show some reducing nature. Reducing nature. For example, you see what it does? It accepts the, you know, this thing, the innocent oxygen and converts into HNO3, which is the next compound we have, nitric acid. Okay. So for example, you see this reaction you must have done. KMNO4, two molecules of this plus 5 HNO2. Again, I'm telling you, redox this reaction, the valence reaction, you don't have to memorize. Just you have to see how the product is formed, what the product we are getting. So HNO2, you see it is the acidic medium. Right? You know, in acidic medium, MN plus 7, this is plus 7 here. It converts into MN plus 2. Right? We have done this in redox reaction. So what it forms? It forms K2SO4, first of all, plus 7 to plus 2 conversion we have. So MNSO4 we get, plus 7 to plus 2, plus we get HNO3. Oxidation of HNO2 gives HNO3 and H2. Five moles of HNO3 we get. Okay. One important reaction we have. HNO2 also shows reducing behavior with urea. That is NH2, CO, NH2. And it evolves nitrogen gas in the reaction. And 2 plus CO2 plus H2. Nitrogen gas in the reaction. Okay. With ammonia, you see what happens with ammonia. All these reactions are important. With ammonia, it first converts into nitrite. Then this converts into N2 plus H2. The final result is nitrogen gas only. But via this, ammonium nitrite and then N2 to H2. Okay. This reaction you see HNO2 with enylene. Okay. Enylene is C6H5NH2 in presence of acid at CL. The temperature is, we kept low over here, less than 5 degrees Celsius. And this gives us C6H5N double bond NCl. That is the benzene disonium chloride to HCl to H2O3. And it is known as benzene disonium chloride. Okay. So these are the few reactions which shows the reducing behavior. All these reactions are important. What product we get into this? Okay. Reaction of benzene disonium chloride and evolution of nitrogen. Both are important reactions. Okay. One more compound of nitrogen we have. The last one is that is nitric acid. Nitric acid HNO3. Okay. This we also call it as aquafortis. Aquafortis given by the scientist, alchemist. Okay. Not that important. Name is this. How does it prepare nitric acid preparation right down? So there are different method of preparation we have. Okay. One is like on commercial scale and other one is laboratory method. So what is the laboratory method of preparation we have? Write it down first. Lab method. In laboratory what happens? KNO3 is allowed to react with H2SO4. And this sulfuric acid is concentrated. We take concentrated sulfuric acid so that it gives H plus here that H plus is taken up. This NO3 minus and it forms K2SO4 plus HNO3. So here we get this HNO3. We get in a vapor form which condenses and we get HNO3 finally in the reaction. Okay. After this we have commercial scale on commercial scale or we also call it as industrial preparation. Okay. Industrial preparation method. Write it down. Write down on commercial scale. On commercial scale it is prepared by the catalytic oxidation of ammonia, catalytic oxidation of ammonia by atmospheric oxygen. This process is known as Ostwald's process. Two different processes we have. This one is Ostwald's process. Okay. So what happens in Ostwald's process you see the reaction is the first step is the oxidation of ammonia and it's 3 plus O2 and it's 3 plus O2. We use platinum catalyst here around 700 to 800 degrees Celsius. High temperature it is. It converts into 2 NO. Right. 2 NO H2O also evolves here in the vapor form. Okay. Now this oxide of nitrogen, nitrogen monoxide. It is further oxidized into nitrogen dioxide in the step 2 NO2. NO2 when then dissolved in water and forms nitrous oxide first. It's a mixture of HNO2 and HNO3. That is what we get. Okay. So we get a mixture of 2 here nitrous and nitric oxide. So whatever nitrous oxide we have we can further convert this into HNO3 by dissolving this into water plus this plus H2O. In excess of water it converts into HNO3. Okay. This is what the method we have. Okay. HNO3 and HNO2 that we get. We get a mixture of this 2 here. HNO3 and HNO2 in the step 3 you see. This we call it as azeotropic mixture. I'll write down here. This we call it as azeotropic mixture. Azeotropic mixture will discuss this in a chapter called solution. Okay. It is there in the what we say in the in the syllabus but not that is extended. It's not that important only the basic understanding we should have what is an azeotropic mixture. Okay. And few examples of that we have to discuss. See I'll tell you it's just small things over here of azeotropic mixture. It is the component like the mixture has more than one component obviously. Right. So for a particular composition what happens the separation of the component of mixture is not possible. Right. Suppose if you want to separate this HNO2 and HNO3 it is not possible because it is an azeotropic mixture. The separation is not possible by distillation even because they have the same boiling point for azeotropic mixture separation is not possible by distillation distillation actually works on the boiling point of the component present in a mixture. Right. So for any mixture we can have a fixed composition and at that composition the boiling point of the components becomes equal. Right. And and that component that point of time the mixture is known as azeotropic mixture. Okay. And azeotropic mixture we cannot separate by distillation right because distillation is a process based upon the boiling point of the component present in the mixture. Okay. So azeotropic mixture of HNO3 and HNO2 we have 60% of HNO3 HNO3 and HNO2 it is the azeotropic mixture where we have 68% of this and 32% of this biomass. Right. But this composition this 1632 is a azeotropic mixture we have. Right. So obviously we have some amount of HNO2 present and we need to find we need to get this HNO3 so to increase the composition of HNO3 what we can do we can use water we can oxidize this further into HNO3 and then we get the 97 or 99 98% of HNO3 in the mixture. So that is the first method of preparation which we call it as host world process. What you have to keep in mind host world process we use ammonia first of all for this process ammonia is used. Okay. And it forms an azeotropic mixture of nitrous and nitric acid. Okay. Little bit idea I have given you. Excuse me sir. What? Sir if the mixture of HNO2 and HNO3 is is your dropping then how are we like separating HNO2 to oxidize? We are not separating it. We are oxidizing this. Suppose you have the mixture this won't get oxidized. Okay. This won't get oxidized because nitrogen is already plus five here. But for this oxidation is possible. So for the mixture we are trying to oxidize that so whatever HNO2 is there that will convert into HNO3. Sir so we are basically oxidizing the entire mixture for only HNO2. Yes but for oxidation of HNO3 is not possible because we can see if you if you if we can separate it right then fine we can take this HNO2 out and then then we can oxidize it separately. Yes. Right but since it is an azeotropic mixture so separation is not possible. That's why we have taken the entire mixture here and we are trying to oxidize it but what happens the oxidation of only HNO2 takes place that's why we reaction we have written for HNO2 only. Yes sir. Correct. So it's an azeotropic mixture at this particular concentration sir. Yes yes for any we will discuss this in detail in solution chapter but you must understand for any mixture we have a fixed composition right and that composition only the mixture is said to be an azeotropic mixture. So we have a graph also for this. Okay the graph you can probably understand it in a better way that how the you know the volume point becomes equal of the mixture of the component of the mixture. So I'm not going into that graph in all those things. Okay actually this particular thing it is it is there in your this thing the graduation book that you have there they have given this in detail azeotropic mixture. In RJH slapers they have just talked about what is an azeotropic mixture okay and few examples of that that is it. This definition and few examples but since it is there so try to understand the graph and some little more things about it how azeotropic mixture forms what are the various example we have what kind of deviation we have over there in the solution positive deviation or negative deviation those things will discuss in solution chapter but yes like I said in J also they have asked question on to this like two or three times before they have given the options right and which of these is an azeotropic composition like this they have asked. Okay so you have to memorize like this competition I've given you you cannot you know logically you cannot say the composition is this for azeotropic mixture it is a competition where the components of the mixture have equal volume point so this thing you cannot you know logically you cannot say the composition should be this for equal volume right so it is given in the book we'll discuss that in detail in solution chapter got it? Yes. So there are some properties of HNO3 and write down the physical properties first the first one HNO3 is a colorless liquid and have punzint order colorless liquid and have punzint order okay punzint order in presence of sunlight it dissociates it dissociates into NO2 nitrogen dioxide so four molecules of HNO3 in presence of sunlight it converts into four molecules of NO2 plus H2O and O2 so it forms nitrogen dioxide that is important here. Next write down nitric acid is generally stored in is generally stored in brown colored brown colored brown colored bottle to prevent photochemical decomposition of HNO3 the reaction I have given you photochemical decomposition means in presence of light only the decomposition which is takes place okay HNO3 HNO3 we all know it is a strong acid monobasic acid and it has very strong oxidizing nature okay since it releases nascent oxygen it has very strong oxidizing nature okay it oxidizes nonmetals okay it oxidizes metalloids okay there are many things nonmetals metalloids even compounds oxidize all those things okay so so the reaction over here you see and all these reactions the oxidation reaction it either involves NO2 that is nitrogen dioxide or it evolves N2O dinitrogen oxides or it evolves nitrogen monoxide okay is what I said oxidation on oxidation either it gives simple thing is you don't write this just for the understanding oxidation it either evolves NO2 or it evolves NO or it evolves N2O these three gas it evolves okay with some metal what happens with some metals it also evolves it also forms metal nitrate okay metal nitrate oxidation of nonmetals also we have to see right so how they frame questions over here yes but HNO3 cannot be oxidized it's already oxidized it oxidizes others that's what it means okay not on oxidation it provides oxidation reactions on this on oxidation of other things it converts into these oxides that what this means okay how they frame the questions actually they'll ask you HNO3 on reaction with this metal what oxides or nitrogen involves they'll give you all these options okay NO2 NO N2O like that okay so you have to know in which reaction NO2 is getting evolved NO is getting evolved or N2O is getting right there are many reactions for this okay I'll try to club those and give you the reaction so that you can memorize these things easily okay so write down first thing here and this HNO3 also we can take concentrated dilute and very dilute that is also very important what is the concentration of HNO3 we have if it is concentrated things will be different with the same metal if it is dilute with the same metal things will be different okay so first thing here you write down how does the brown color prevent the chemical decomposition of water it it absorbs the light actually okay it does not expose HNO3 to the to the photochemical reaction to the light actually so photon get absorbed and it won't interact with HNO3 okay yes yeah so write down the reaction with first one the reaction with concentrated HNO3 okay so first condition is concentrated HNO3 all these reactions are important and don't have to do anything okay you have to memorize this there's no other way okay concentrated HNO3 write down the point here metals like copper, AG, HG, PB, ZN, Zinc reacts with concentrated HNO3 reacts with concentrated HNO3 forms metal nitrate and N02 gas evolves and N02 gas evolves okay so metal nitrate it forms and forms and N02 gas evolves so obviously we we can write all the reactions with all the metals but that is not our concern okay you can easily understand what could be the metal nitrate here and you should keep this in mind that what gas evolves so when on reaction with this HNO3 concentrated so I'm not writing down all the reaction okay it gives metal nitrate and N02 evolves into this water is one of the product we always get in this kind of reaction whether it is dilute concentrated or very dilute are you getting it water always forms okay suppose we have zinc right copper suppose we have copper plus concentrated HNO3 what it forms CuNO3 whole twice plus N02 plus H2 you don't have to worry about the balanced chemical reaction it's not required in organic chemistry balanced thing is not required okay if you want you can balance this anytime but you should know what product we are getting understood guys okay so with concentrated HNO3 it forms metal nitrate N02 and this one example this one example I'll give you copper plus HNO3 concentrated forms CuNO3 whole twice plus N02 plus H2 H2O will always form in these reactions okay so concentrated HNO3 always gives N02 now if it is dilute dilute HNO3 dilute HNO3 right down in this metals like again you see the same metals Cu Ag Hg Pb reacts with dilute HNO3 reacts with dilute HNO3 forms and forms nitrogen monoxide nitrogen monoxide with metal nitrate again right so we get here metal nitrate plus nitrogen monoxide and like I said H2O always forms in this reaction right but if you have metal like zinc iron or tin this with dilute HNO3 it forms it forms metal nitrate plus N2 dinitrogen oxide and H2O evolves can we use the electrochemical series to tell you this difference you can you can use that to identify whether the reaction is possible or not okay but what oxide evolves here that you cannot say from this electrochemical series what it what it suggests it suggests whether you can displace hydrogen or not right depending upon the electrode potential correct yes so displacement of hydrogen gives you water that is fine we are getting water in all these reactions but what oxides of nitrogen you get that you do not have any idea from the electrochemical series right so because like Cu Ag Hg Pb are all below and certain F3s are above like that you can keep in mind yes so obviously it metal nitrate it forms nitrate it forms that's what the reaction is but see but what oxides of this you get in that way you can keep in mind that below hydrogen it will evolve N0 and then N2 or N2 like that that you can keep in mind but the point is you have to memorize this okay in terms of because they won't ask you whether this they can ask you one thing whether this reaction is possible or not that you can say from electrochemical series but if they ask you what oxides of nitrogen you get in that case you should know that what all metals gives N0 what all metal gives N2 or N2 depending on the concentration of HNO3 okay right so the third case we have when the acid is very dilute okay with very dilute HNO3 right on metals like metals like magnesium and manganese magnesium and manganese on reaction with very dilute HNO3 it evolves hydrogen gas H2 plus plus it forms again metal nitrates like Mg NO3 hold twice or it forms Mn NO3 hold twice like that gets but the thing is we get hydrogen gas over here if the metal is zinc iron tin right on reaction with very dilute HNO3 it forms ammonium nitrate it forms ammonium nitrate means one of the components will obviously be metal nitrate plus this is common NH4 NO3 we get always and H2O always forms into this one sir in the first case H2O is not formed huh first it's not formed first case H2O is not forming hydrogen gas evolves over there so when hydrogen gas evolves that is a specific reaction you can say water does not form over okay one important reaction you write down cane sugar that is C12 H22 O11 this question they have asked once O11 on oxidation with HNO3 forms oxalic acid this reaction is important forms oxalic acid NO2 evolves here and H2O but here what happens this I have given it here right but this we are using is concentrated acid not very dilute so again you see concentrated so NO2 evolves but acid is what oxalic acid so this also you must remember oxalic acid okay write down oxidation of non metals the last one here oxidation of non metal write down non metal does not react with dilute HNO3 non metal does not react with dilute HNO3 with concentrated HNO3 and metalloids with concentrated HNO3 and metalloids it forms corresponding oxy acids and nitrogen dioxide NO2 evolves so yeah yes sir can you repeat yeah I said non metals does not react with dilute HNO3 first thing but with concentrated HNO3 and concentrated HNO3 non metals as well as metalloids reacts and forms oxy acids and NO2 gas evolves oxy acids and oxo acids are not the same oxo acids it is defined for it is it is not the same we have different bondings over there in oxy acids and oxo acids the bonding is a bit different we'll discuss that okay we'll discuss that okay but they are not the same okay so in this we write down the non metals we have like carbon like carbon sulfur all these are non metals metalloids like we can take arsenic we can take antimony etc all these reactions what it gives I'll write down here suppose we have carbon we have you know sulfur we have iodine all these with concentrated HNO3 so again you see whenever you take concentrated NO2 forms H2O forms plus we get if carbon you are taking you'll get CO2 sulfur you are taking we get you know in case of sulfur we get what we get H2O4 iodine if you have then you'll get HIO3 like that okay so so that is the reaction here so I'll write down here either you get CO2 either you'll get H2O4 or you'll get HIO3 but this part like I said it is not that important but what gas evolves here it is important okay this is it for nitric acid these reactions are three different conditions you must remember now the last part of this chapter we'll discuss that is phosphorus sorry Sir when does H2O form? H2O form in H2O I did not give you Yes sir you told it can form but you didn't give any reaction where it's formed Oh just a second then I missed this reaction somewhere Sir you did under dilute HNO3 Yes under dilute HNO3 For ZN and FEN Yeah I have given this you see this Yes sir yes sir Under dilute there are two conditions C, U, A, G, P, B all those gives NO ZN, F, E gives N2O right Yes sir sorry sir Okay only OSTWAL process with it okay one more method of preparation we have HNO3 write down first that we call it as one is OSTWAL that we have done the another one is this write down second method of preparation I missed this write down it is again the industrial preparation method we have laboratory preparation method is KNO3 with H2SO4 this we call it as work line I process or we also call it as arc process industrial method it is apart from OSTWAL this is the another one in this what happens the first step is the reaction of N2O2 in an electric arc where we have a very high temperature 3000 degree Celsius around where we get monoxide this reaction is endothermic so minus heat further in the second step what happens simple reaction this NO by oxidation we convert this into NO2 nitrogen dioxide NO2 when you dissolve in water it gives HNO2 again the mixture of HNO2 and HNO3 then this HNO2 on oxidation gives HNO3 plus NO plus H2 sir so basically only the first step is different yes where we are using ammonia and here we have N2O2 sir even the mixture is aziotropic sir no here it is not the last thing you write down phosphorus there are different allotropes of phosphorus that we have to discuss it is not that important only the acids of phosphorus we have that is S3PO3, S3PO2, S3PO4, H4P2O7 there are 5, 6 different different acids those structure you should know is how many POH born in all these we have that is only the important part we have here so phosphorus write down the different allotropes of phosphorus the first one is we have white or yellow phosphorus why we call it as yellow phosphorus also because the white phosphorus on exposure to atmosphere it forms a thin layer which the color of the layer is red so red plus white it makes and it gives a yellow appearance so white or yellow phosphorus are the same thing the another one is red phosphorus, black phosphorus not important I am giving you this information that black phosphorus also has two different forms that is alpha black phosphorus alpha black phosphorus and beta black phosphorus beta black phosphorus only one thing you have to understand here or you have to memorize that this beta black phosphorus has graphite like structure graphite like structure layered structure we have here but beta black phosphorus that is the only thing you should keep in mind white phosphorus what happens white phosphorus write down it is prepared by it is prepared by heating it is prepared by heating phosphate rock that is CO3PO4 phosphate rock CO3PO4 with coke and sand with coke, carbon and sand CO2, silica also so the reaction overall is 2CA3PO4 2CA3PO4 PO4 whole-wise it reacts with SiO2 and the temperature we are using again is the high temperature 773 Kelvin it converts into calcium silicate CASIO3 plus P4O10 it forms now this goes under reduction with the coke at the same temperature and it converts into P4 plus CO10CO this is what the preparation we have for white phosphorus white phosphorus exists in tetrahedral structure like this phosphorus at the corners of a tetrahedron this is white phosphorus tetrahedral structure exists as P4 molecule the bond angle here is P bond angle is 60 degree and because of this strain only in the bond angle the phosphorus P4 is highly reactive bond angle is 60 and because of this high angle strain the P4 is highly reactive next point on exposure to light white phosphorus turns yellow white phosphorus turns yellow and it is also known as yellow phosphorus next point highly poisonous in nature one term we have here it is highly poisonous not important this value is not important at all but for an idea 0.15 gram of this is considered as the fatal dose okay a person can die with this amount only 0.15 gram is highly poisonous okay that's why the person that works with the phosphorus industry develops a disease right and this disease we call it as fossy jaw okay fossy jaw so what happens here in this disease the jaw bones decay the jaw bones that you have that decay with very slow rate the rate is not that fast but it is observed in all those workers or persons who are working at the ground level in phosphorus industry okay the labour class mainly this kind of disease they have like it has been observed the name of the disease is fossy jaw they have asked this question many times okay in the exam because the jaw bones starts decaying so this we call it as fossy jaw write down it is insoluble in water but easily soluble in organic solvents next point next point its ignition temperature its ignition temperature is very low around 25 to 30 degrees Celsius value is not important ignition temperature is very low and hence it catches fire easily catches fire easily this is the reason it is kept underwater to you know to stop this reaction with atmospheric oxygen right so it catches fire with the reaction with oxygen forms pentoxide P4O10 okay P2O5 or P4O10 you can say 2P2O5 or P4O10 but it is kept in the water since it is highly reactive because of low ignition temperature right down with caustic alkali solution with caustic alkali solution it forms phosphine gas this is important it forms phosphine gas I'll write down the reaction here P4 plus NaOH plus H2O it gives sodium hypophosphite it is NaH2PO2 plus pH3 you see it is minus 3 phosphorus here and here the phosphorus is plus 1 here it is 0 so it is a disproportionation reaction okay this compound we call it as sodium hypophosphite this is phosphine this reaction is again important it directly combines with halogens it directly combines with halogens and forms tri or pentahelite PCL3, PCL5 important reagents for organic chemistry right directly combines with halogen and forms tri or pentahelite with metals it forms phosphites for example suppose magnesium mg plus P4 it forms 2 mg3P2 metal phosphite magnesium phosphite Ca plus P4 calcium phosphite 2 Ca3P2 formula is important okay you must remember this formula Ca3P2 type formula it forms the next one you see red phosphorus and then we will see the compounds of phosphorus we have only oxides we will see over there red phosphorus just a second okay red phosphorus you see the reaction with metals right Sana the last one we are talking about yes Srishti the different types of phosphorus has different reactivity correct I saw this message now the last message you are talking about the first one first one is with NaOH okay NaOH the reaction is it dissolves with alkali solution I think this one you are talking about dissolves with alkali solution and forms and forms phosphine gas this one I think you are talking about dissolves with alkali solution NaOH and forms phosphine the reaction is P4 plus NaOH plus H2O gives NaH2PO2 plus pH3 oxidation is state of phosphorus you check it is plus 1 and minus 3 here it is 0 so it is getting oxidized and here it is getting reduced hence the reaction is what the reaction is disproportionation reaction okay now the next one is red phosphorus write down it is obtained by heating it is obtained by heating white phosphorus at 573K even this reaction is also important at what temperature white phosphorus converts into white phosphorus of halides we have done know first class pratham halides of nitrogen we have done the different properties I think only oxides halides of nitrogen we have not done first class we will see that in the last okay write down it is obtained by heating white phosphorus at 573K so the reaction is P4 white just you heat this at this temperature you will get P4 red the structure of these two are different okay white and red the structure is different the red one has the red one has a polymeric structure okay so it has a tetrahedron a retrahedron of phosphorus like this it is there but it has a polymeric structure right and this unit is repeating skipped on repeating in the structure like this and goes like this right so it has polymeric structure consisting of P4 tetrahedron linked together through a covalent bond this is a covalent bond here we have S8 not P8 shrunking okay quickly write down the few properties of this first thing it is non-poisonous insoluble in insoluble in water as well as insoluble in water as well as organic solvents insoluble in water as well as organic solvents next one more stable than more stable than the other allotropes of phosphorus more stable than the other allotropes of phosphorus next one less reactive than white phosphorus since its ignition temperature is high okay less reactive than white phosphorus since its ignition temperature is higher than the white phosphorus correct next one it does not react with caustic alkali this is very important okay unlike the first one okay finished it does not react with caustic alkali and that is how we can identify the two right in the reaction we can separate the two in the reaction one get dissolved it does not get dissolved into that okay so the another compounds we have so we have two three compounds of this we will just go through this quickly the first compounds of phosphorus we write down that is phosphine ph3 phosphine it is obtained by this reaction is important it is obtained by the hydrolysis of metal phosphates hydrolysis of metal phosphates so we can take metal phosphates for this purpose we can take ca3P2 or we can also take another 3P metal phosphates metals of first and second group okay so the reaction is ca3P2 this one they have asked many times in the exam ca3P2 when dissolved in water H2O it forms first of all hydroxide that is CaOH whole twice and pH3 forms similar kind of reaction you can write down with Na3P also apart from hydrolysis we can also allow this phosphates to react with dilute HCl dilute HCl so with dilute HCl also this can give phosphine gas that is 3 CaCl2 plus pH3 this is the first method of preparation write down in laboratory it is prepared by prepared by white phosphorus in laboratory we use white phosphorus for the preparation of red phosphorus so what we do the reaction of white phosphorus with alkali solution and this reaction we have already done 3 NaOH plus H2O it converts into pH3 phosphine plus sodium hypophosphide that is NaH2PO2 this is what we get further when you heat this this converts into Na3PO4 plus pH3 this is what the phosphine here also we get the phosphine but whatever is left here in the form of this hypophosphide we again heat that and we convert that into pH3 one more method we have not that important I'll write down with the help of S3PO3 4 molecules of S3PO3 if you heat this it forms H3PO4 and phosphine gas this is phosphorus acid this is phosphoric acid S3PO3 is phosphorous S3PO3 is phosphoric property is right down got a less highly poisonous the smell is like rotten fish smell how is the smell? right down to this next point phosphine decomposes on heating in absence of air phosphine decomposes on heating in absence of air so when you have phosphine gas there's a small difference here you see when you have a phosphine gas pH3 and you heat this in absence of air we get here P4 and hydrogen gas evolves but the aqueous form of this pH3 aqueous means in solution what happens in presence of light means the solution of pH3 in water decomposes in presence of light and it gives red phosphorus and H2 gas evolves here why? other thing is it reacts with halogens like chlorine and it forms a trihalide or pentahalide like that so we'll discuss the halides of phosphorus in that we'll see that pH3 plus Cl2 gives PCl3 excess of Cl2 gives PCl5 like that that is one reaction we'll discuss that later in halides of phosphorus with silver nitrate it forms a complex of silver phosphide first with silver nitrate it forms a phosphide of silver nitrate silver phosphide first which is reduced in presence of water which is reduced in presence of water to metallic silver and forms a black precipitate and forms a black precipitate the reaction you write down I'll repeat it again with silver nitrate it forms a complex of silver phosphide first which is reduced in presence of water to metallic silver and forms black precipitate so the reaction here is pH3 plus AGNO3 6 moles of this AGNO3 gives a complex which is AG3P dot 3 AGNO3 plus HNO3 we get it is an intermediate complex when this is allowed to react with water it converts into silver precipitate of this plus HNO3 plus H3PO3 black precipitate of this just a second I'm coming copy this down write down one next the important point here it is used in write down it is used in pH3 in HOMES signal in HOMES signal HOMES signal is basically used by the this thing by the various navy that we have which they use to send signal to the other ship or something in case of any urgency how they do this that is you need to understand this actually what happens we have a box in that box there are some chemical compounds present which is actually the mixture of calcium calcium carbide this thing they ask what is HOMES signal it is a mixture of calcium carbide and calcium phosphate so in that box there is an arrangement and there we have the mixture of these two compounds present calcium carbide and calcium phosphate and this is actually present like I said in a box or container where we have two holes in case of any urgency or any difficult situation if they want some help they throw this box into the sea ok and then the seawater enters into the box through that holes and reacts with the mixture ok so there is an opening the hole is there and then you can throw that box into the sea seawater enters into the box through those holes ok and react with this mixture and this mixture and this mixture provides phosphine gas it forms phosphine Ph3 right it forms acetylene C2H2 right this phosphine this phosphine catches fire highly reactive catches fire in presence of atmospheric oxygen and there is a small explosion you can see because of this fire right and that is the signal which is given to the other ship for any kind of in case of any urgency or difficult situation right so acetylene the purpose of this acetylene is what it produces a bright flame right and this catches fire so there is a small explosion there so explosion plus a bright it produces a bright flame right and that is the signal that they used to send to the other ship that's why we call it as Homes signal that's how it works but that working is obviously not required the reason behind this is the reactivity of phosphine with oxygen this question they have asked many times what is Homes signal ok calcium carbide plus calcium phosphate when you write down phosphorus halides phosphorus forms 2 types of halides trihalides and pentahalides ok trihalides and pentahalides so it's a very simple reaction for the reaction of phosphorus with limited amount of halogen gives you trihalide and with excess of halogen it gives pentahalide ok so write down the phosphorus forms 2 types of halide phosphorus trihalide and phosphorus pentahalide first one PCL3 phosphorus trichloride how does it prepare like I said phosphorus with chlorine limited amount and it forms 4 PCL3 it can also be obtained by the reaction of phosphorus with thionyl chloride which is SOCl2 thionyl chloride and it forms PCL3 plus sulphur dioxide evolves and we also get disulfuryl chloride S2Cl2 ok the structure of PCL3 is similar to you know this thing Ph3 which is this Ph3 structure I have in this I guess all of you know it is similar to NH3 like this trihedral geometry structure is this there's not much difference into this nitrogen in case of nitrogen what happens it forms only trihalides pentahalides it does not form ok because it cannot show you know usually like it can form to form pentahalides for nitrogen comparatively ok so it can form NF3, NCL3 and BR3, NIT nitrogen halides are not that important it is not there because the compounds of nitrogen with halides are very less ok how does it prepare again the same thing we have metal plus halogen you can take N2 plus Cl2 gives NCL3 structure is similar to this PCL3 like I told you right ok so this is about the trihalides same kind of pattern we have ok this behaves as a Lewis space PCL3, NCL3 also behaves as a Lewis space can donate electron right but obviously if you compare NF3 NCL3 and BR3 then NF3 has the minimum tendency to lose electron why could you tell me why? more electronegative fluorine is the most electronegative element so it won't you know it does not have the tendency to lose electron easily ok it's very similar actually it's not that you know much difference we have over here it's very similar nitrogen does not form pentahalides ok it's not the case at all we have ok so just 2-3 reactions I'll show you with trihalides of nitrogen that is it ok there is not much thing in the present it's very much similar to this so PCL3 forms by this tetrahedral geometry properties it is a colorless pungent oily liquid ok on hydrolysis it gives S3PO4 this is the one reaction is very important PCL3 on hydrolysis right down 3S2O gives S3PO4 you should know the structure of these acids the oxo acids of phosphorus S3PO4 I'll discuss that this part is important over here ok so this is one thing S3PO4 plus HCl goes out HCl forms in this reaction with water ok with concentrated S2S4 what happens you see if you are taking acid concentrated right so it forms sulphuril chloride SO2Cl2 it is sulphuril chloride sulphuril chloride plus it also forms HPO3 HCl HCl and SO2 also evolves into this ok these are the reactions I guess you all know the reaction of acid with PCL3 what happens with CS3COOH plus PCL3 could you tell me CS3COOH plus PCL3 we have done this in organic CS3COOH plus PCL3 it forms acyl chloride CS3COCl plus S3PO3 forms ok so S3PO3 forms over here right down it reacts with organic compound it reacts with organic compound containing OH group it reacts with organic compound containing OH group and the product is what the OH group is replaced by the chlorine for example this is plus PCL3 it gives CS3COCl plus S3PO3 even if you take an alcohol here CS3CH2OH plus PCL3 so the organic molecule which contains OH it replaces OH with CO S3PO3 is the product we get ok next write down phosphorus pentachloride phosphorus pentachloride it is again similar reaction we have P4 and this P4 is white ok this you must remember white phosphorus we use with 10 Cl2 it gives 4 Cl5 4 PCL5 ok phosphorus pentachloride we can also use Sulfurine chloride SO2Cl2 here with white phosphorus only forms PCL5 and sulfur dioxide SO2 could you tell me the structure of PCL5 trigonal hypermedicine diagonal bipyramid very good trigonal bipyramid the structure is this so it has two types of bond we all know this this is axial and we have three equatorial bond so two different bond lengths also we have here but this important point is what this structure is possible only in solid and liquid state sorry not solid gas in solid state and liquid state this structure is possible write down in solid state in solid state it exists as an ionic compound it exists as an ionic compound and that ionic compound contains this structure one is tetrahedral unit the cation part and an ion part is this square octahedral unit square bipyramid so you can easily draw the structure of this this is a tetrahedral part PCLClClCl CL and CL this is a cationic part so we have this positive charge and an ionic part will be we have chlorine, chlorine phosphorus here, chlorine, chlorine one on the top one on the bottom square bipyramid this is an ionic part has negative charge on it okay so tetrahedral unit plus octahedral unit we have into this okay so two types of hybridization we have here must remember this they ask this question in the exam so in solid state the structure is this we have two types of hybridization phosphorus here it is sp3 and here phosphorus is sp3d2 no it's not aditya it's not write down few properties of this quickly first point to write down PCL5 is a yellowish white crystalline solid PCL5 is a yellowish white crystalline solid with water it is hydrolyzed too PoCl3 which is one of the common product we get we see when we have this thing a reaction of PCL5 right just a second let me take the attendance take the screen shot so it forms PoCl3 what is the reaction we have reaction is reaction is PCL5 plus H2O it gives PoCl3 plus HCl further this PoCl3 if you put into water means if you are taking excess of water you end up getting H3PO4 phosphoric acid plus 3 HCl okay so we can say this reaction PCL5 plus excess of water gives you H3PO4 so two step reaction limited water gives you PoCl3 excess it converts into H3PO4 we all know PCL5 it easily converts into PCL3 and Cl2 we have this reaction also in organic chemistry how the reaction proceeds right PCL5 the reaction with the compound containing hydroxy group gives you the same kind of product PCL5 with this thing CS3 COOH it forms CS3 COCl plus H3PO4 plus HCl right so we will get acylene flow right here even alcohol also the same kind of reaction we have okay reaction with SO2 the last one SO2 plus PCL5 it forms thionyl chloride with PoCl3 SOCl2 thionyl chloride correct now the last thing and the most important part of this phosphorus is oxo acids oxo acids of phosphorus see one of you have asked what is the difference between oxo acids and oxy acids okay so oxo acids are those acids in which in which the molecule contains an oxygen at least right plus hydrogen oxygen should be there hydrogen should be there right and at least one hydrogen bonded with oxygen and one OH group should be there like this right oxygen hydrogen OH plus any other element one central metal atom so central atom I will write down here central atom like in some book they have also written like this the compounds which contains oxygen at least one hydrogen attached with one oxygen plus any other element apart from oxygen hydrogen those compounds are called oxy acids one simple example I will tell you S3PO4 you see it is an oxo acids of phosphorus right P the structure you see P double bond OH OH OH OH OH so either you can have this or this or both anything is possible right if you have another example you see if I write down H3PO2 then the structure will be P double don't write this now P double bond OH OH H and this is also oxo acids of phosphorus right so oxygen contains oxygen contains hydrogen which may be directly attached with the metal atom central metal atom or there is a hydroxy group does it any possibility is there we call it as oxo acids oxy acids are just the oxides of any compounds like nitrogen any elements like nitrogen I have taken so oxides of nitrogen are the oxy acids oxy acids are oxygen containing acids most of the covalent non-metallic oxides that we have for example NO N2O CO etc all these oxides are oxy acids because they react with water and forms acidic oxide did you understand this Anjali and oxo acids OH or OH should be present yeah any possibility should be there like this it is an oxo acid like this it is an oxo acid either OH means you see hydrogen either attached directly with the central atom or hydrogen attached with oxygen both possibilities oxy acids we should have at least three elements into this central element oxygen plus hydrogen three NO we do not have three elements it is only two oxides of nitrogen right ox oxy acids are those oxides of you know non-metallic oxides actually like CO is said NO N2O is it which reacts with water and it forms acidic oxides acidic compounds those are oxy acids if you want to write it down what there is no OH group at all that is still oxo acid yes you can differentiate this one simple thing in oxo acids you will have at least three elements more than like apart from oxygen and hydrogen we have one more element present oxo fine oxy we have oxides of non-metallic non-metallic oxides basically right on oxy acids are any oxygen containing acid oxy acids are any oxygen containing acids most covalent non-metallic oxides most covalent non-metallic oxides in bracket to write down CO N2O NO etc reacts with water to form acidic oxides and these acidic oxides we call it as oxy acids these acidic oxides we call it as oxy acids got it what at this point most of the last time you said I'll repeat it from the beginning oxy acids are any oxygen containing acid most of the covalent non-metallic oxides reacts with covalent non-metallic oxides reacts with water to form acidic oxides these acidic oxides are called oxy acids for all acidic oxides oxy acids yes non-metallic non-metallic okay sir yes sir thank you sir correct now this is not the point we have oxy oxy acids and all we are talking about oxo acids of phosphorus so oxo acid how do you find out you must have an oxygen an OH bond and and hydrogen attached with metal like this these two cases are possible now the various oxo acids and there are certain properties of oxo acids we have that you should know structure is important here because you know in mole concept 2 also rocks reaction where we find out the basicity or null to find out the equivalent mass we should know the structure right like what is the basicity of S3PO4 here basicity is 3 the hydrogen which is attached with oxygen here basicity is what only one so once you know the structure we can find out basicity and then equivalence equivalent mass other things right that is why the structure is important but before going into the structure directly we'll see some properties of these oxy acids oxo acids okay so write down two types of oxo acids we have basically I'll just try to classify this so that we can do this in you know now arranged manner you can memorize this so what I'm telling you oxo acids so we are classifying this into two types right one is us other one is ik us means phosphorus and other one is phosphoric two types we have phosphorus oxy acids are those oxy as oxo acids in which the oxidation state of phosphorus is plus one and plus three okay phosphoric one where the oxidation state is plus five this is one thing there are some important points we have here you write it down and this is common for both type okay depending upon the oxidation state we'll write down the name us or ik but these points the property I'm giving you now it is common for both write down all these at acid contains at least one P double bond O bond at least one of this bond and one P OH bond this must be there in all these oxo acids another important point oxo acids with plus three oxidation state with plus three oxidation state undergo disproportionation reaction undergo disproportionation reaction so suppose the molecule is this H3PO3 you see here plus six plus three so you have a plus three write four molecules of this right when you heat this it undergoes disproportionation reaction forms ps3 minus three oxidation state and three H3PO4 that is plus five oxidation state disproportionation reaction okay there are few oxo acids we have I'll write down the name and the structure you can draw this quickly the first one is go back to the previous quote for a second copy this down yes sir so the first one we have that is orthophosphoric acid the most important one this is again two three four five first one write down not ortho I'll just give you this in order write down the first one that is hypophosphorous acid hypophosphorous second one is orthophosphorous acid orthophosphorous acid the third one is peroxy monophosphoric acid peroxy monophosphoric acid then we have orthophosphoric acid phosphoric acid phosphoric acid phosphorous acid is H3PO2 orthophosphorous is H3PO3 peroxy monophosphorous is H3PO5 orthophosphoric is H3PO4 okay two three four and five orthophosphorous acid we also call it as phosphoenic acid the other name this is phosphoenic acid if you talk about the structure here right like I said all these oxo acids what you have to keep in mind they must have one phosphorus oxygen double bond and one OH correct P double bond O and OH now after this whatever H is present that you connect with the oxygen if it is there if oxygen is not there then directly you attach with the hydrogen like this similarly you see H3PO3 is what P double bond O OH then we are left with one oxygen into hydrogen it means one oxygen and one hydrogen S3PO5 it is a bit different because this you see peroxy means we have peroxy linkage here so in this what happens we have one P double bond O right one P double bond O and then we have two OH bond like this and we are left with two more oxygen here so here we have the peroxy linkage like this this is peroxy monophosphoric acid S3PO4 again you can draw P double bond O and one OH should be there they are left with two oxygen two hydrogen it means OH and OH the number of hydrogen attached with oxygen there we call it as the basicity of that particular compound like you see could you tell me this one is what is the basicity for this one mono basic acid die basic acid try basic acid try basic acid so once you know the structure you can find out the basicity and then equivalent number of equivalent equivalent mass anything can find out ok done can I go to the next board for us in any of these using your voice is too low I think you should increase the volume sir in any of these compounds can we calculate the oxidation number of phosphorus yes we can calculate without drawing structures you can do that but here you have to keep in mind where it is a peroxy linkage here without the structure what is the oxidation S3PO2 minus 2 minus 2 into 2 minus 4 plus 3 so it is plus 1 correct yes sir so without the structure you can do but I would suggest what you keep the structure in mind and do this with the help of structure only why because if only one phosphorus atom is there it's fine there's no any problem but I am going to give you 45 more structures where more than one phosphorus atom is present right then maybe with this x value maybe this x method that you are doing over here maybe from that you'll get the wrong answer because you know x values what it is the average right yes it is the average it won't give you the oxidation state of each phosphorus atom always we have discussed this in our redox reaction more concept 2 if you have your last year notes you go through once you will understand the difference between the x and the structure method yes okay so what I suggest the structure is not that difficult also I would suggest go by structure only like you see for example if I consider the structure we have double bond means it gives plus 2 here single bond then plus 1 right this this gives plus 1 here so minus 1 and minus 1 2 plus 1 3 minus 2 plus 1 because the phosphorus and hydrogen phosphorus will have minus 1 oxidation state hydrogen will have plus 1 always plus 1 so this is plus 1 so this is minus 1 oxygen minus 1 so phosphorus plus 1 oxygen minus 2 because we have 2 double bond so phosphorus plus 2 hence it is plus 1 did you get it yes I understood okay fine now you see few more structures I'll show you where we have more than 1 phosphorus atom present 4 more structure 5 more structure we have draw this quickly and then we'll take a break okay so fifth one we have metaphosphoric acid metaphosphoric acid the formula is HPO3 you can find out the oxidation state of phosphorus pyrophosphoric acid in net exam they ask this doubt this question actually what is the structure of pyrophosphoric acid what is the formula of pyrophosphoric acid like that it is H4P2O7 we have peroxo diphosphoric acid peroxo diaphosphoric acid it is H4P2O8 the eighth one we have pyrophosphorous acid pyrophosphorous acid is H4P2O5 and the last one we have is hypophosphoric acid hypophosphoric acid and this is H4P2O8 and this is H4P2O6 you can draw the structure I'll write down here the basicity on the right side okay you can draw the structure on your own the basicity of metaphosphoric acid it is mono basic H4P2O7 it is tetra basic H4P2O8 it is again tetra basic H4P2O5 it is diabasic could you draw the structure of this one eighth one H4P2O5 see the structure base question they ask in the exam so you must have the idea of it like for this one the structure will have one phosphorous phosphorous bond right four we have so one phosphorous oxygen double bond like this and OH OH OH this is the structure okay for this one the structure will be we have peroxo right so means between the two phosphorous atom we have one peroxy linkage so P O O P okay and then we have like three oxygen here and four oxygen here right so it means what we have we have one two P double bond O so we'll have over here two P double bond O and then what we can write H4P2O5 we are drawing this oh not this one that's what it is peroxo I am drawing no so this one is this peroxo is this one this is a second this one is peroxo so this is four and we have four oxygen left so it's fine OH OH OH OH OH okay pyrophosphorous acid is this pyrophosphorous acid we have I'll draw this way H4P2O5 the structure is we have phosphorous attached with one oxygen double bond O and then we are left with three okay so we have OH here double bond O we have OH here so we have used five oxygen so two hydrogen is like this it is present this is H4P2O5 similar kind of bonding we have in this H4P2O7 also that is pyrophosphoric acid okay they ask this how many POP bonds are present how many P double bond O bonds are present like that okay so H4P2O7 you see we'll try to draw the symmetrical structure this is common at least one OH one OH this must be present we again need what out of seven we have used five okay so what we can do next OH here three three plus six seven OH here you see H4P2O7 so just you need to know what you need to know that first of all symmetrical structure you have to draw if the name contains peroxy then you have to write down one peroxy linkage between the two phosphorous acid and then same like every phosphorous atom should have at least one double bond O and one OH group rest you can place as it is okay HPO3 the last one tell me draw this one also what is HPO3 we have one P double bond O we have H we have HPO3 so we have one more oxygen OH so it should be it won't I made a mistake so this won't be there HPO3 is this so one oxygen will be like this HPO3 two P double bond O and one OH that's why it is monogasic in nature okay so once if you can draw the structure again suppose if I ask you what is the oxidation state of phosphorous atom here there are symmetrical structures from the X value also you will get the same oxidation state okay here it is 2, 4, 5 and 6 plus 6 and plus 6 2, 4, 5, minus 1 that is plus 4 plus 4 oh it is plus 3 now why it is plus 3 because plus 2 plus 1 plus 3 plus 1 plus 4 and minus 1 because it is plus 1 and this is minus 1 so it is plus 3 plus 3 with X value also you will get the same value here because the structure is symmetrical you won't get any difference here it is 2 plus 2 plus 1 plus 5 the last one the previous one I will see this one is also you see all the structures but for this peroxy you have to take care of plus 2 plus 1 plus 1 right and for peroxy what happens 2, 3, 4 right and plus 1 because of this oxygen so here also it is 5 here also it is 5 here also 2, 3 plus 4 and plus 4 all these you can draw like this understood here guys 5 so like this phosphorous is done we don't have anything left in phosphorous okay we are done with it okay we will take a break now okay we will come back at 11.40 and then we will see the 2, 3 reactions of nitrogen highlights and then we will finish of this chapter okay next we will start with group 16 okay but again I am telling you the common properties like atomic radius ionization and healthy and all that is same in all the groups whatever the difference we have exceptions that we will see we won't waste our time into those things that we have already discussed right so quickly we will go through and then we will see the properties of compounds and the various different preparation methods we have okay take a break we will come back at 11.40 thank you hello guys can you hear me yes sir okay okay one last point you see just 2 reactions we have in trihalides of nitrogen first of all nitrogen nitrogen does not form pentahelites okay it only form trihalides so trihalides of nitrogen you see we can have NF3, NCL3, NBR3 and Ni3 as electronegativity increases its tendency to lose electron decreases and hence the Lewis basis is already if you write down the order is this the basic nature is this okay now just two reactions we have in fact one reaction we have here that is a hydrolysis of trihalides of nitrogen right so write down on hydrolysis on hydrolysis on hydrolysis it forms ammonia molecule ammonia molecule like NCL3 plus H2O gives NH3 plus 3 HOCl okay 3H2O we have here this is the reaction okay you can predict the reaction the product of the reaction this way NCL3 is this lone pair we have here and H2O is this so when the reaction takes place this OH takes this CL and forms HOCl one molecule of this similarly if you take three molecules of this we'll get three HOCl and this H comes over here so we'll get NH3 and HOCl right but this hydrolysis is not possible for NF3 so write down NF3 does not go under hydrolysis NF3 does not go under hydrolysis due to high bond strength of nitrogen and fluorine okay can you repeat that due to high bond strength of nitrogen and fluorine NF that's why NF3 does not go under hydrolysis finish so this is it for the chapter you do not have anything any other thing left in this chapter okay we have covered almost everything okay I won't say 100% because obviously in in organic chemistry there are many things given in the book right but we have covered all those things which has been or possibly asked in the exam right so the thing is NCRT you must do right Guru 15 I know there's not much thing given but if you have any other book okay for p-block like any other inorganic chemistry book you have after finishing NCRT you can read those books also okay but question you must go through once then only you'll understand what topic is important here okay so anyways we'll move on we'll start the next one that is elements of group 16 that is oxygen family right so like I said the common properties if you go down the group atomic radius increases ionic radius increases ionization enthalpy decreases okay and when ionization enthalpy decreases metallic nature increases metallic nature increases means electro positive character increases tendency to lose electron is more like many things you can assess from these properties right and these properties are common for for all the groups okay yes there are some exceptions here and there okay but more or less the properties varies in this order only talk to bottom and left to right okay so I am not going to discuss all those things again and again because many other things we have to cover okay and you must have that idea that how it goes what like what do you think do I need to tell all those things again or you can do this like atomic radius down the group increases ionization enthalpy decreases with ionization enthalpy you can predict metallic and non-metallic metallic character do you want me to do this no not required it's the same thing okay group 15 it's the same thing so you can go through so we'll start this the common thing I'm not going to discuss okay you can do this on your own if you have any doubt you can reach out to me anytime okay fine so the elements of this group we have we have oxygen sulfur then we have selenium tellurium and polonium okay atomic number we know it is eight then plus eight sixteen then plus thirty two what is the atomic number of selenium I think polonium is after this so it is it is eighty four eighty four minus thirty two is fifty two eight plus eight plus eighteen yeah plus eighteen forty four thirty four yes correct so here we have eight eighteen thirty two thirty two correct the magic number is eight differences eight then we have eighteen thirty two and thirty two okay that is what you need to keep in mind okay if you you know if you because these these I know of elements atomic number you're required in coordination compound somewhere not about this tellurium and polonium but the lower group elements you're required in coordination compound okay so the difference you must know eight eighteen thirty two and thirty two right okay so now as you know as you go down though I'll write down here only I won't like I said one by one I won't go down the group if you go what happens then atomic radius or ionic radius atomic radius or ionic radius increases see what happens usually what happens we know this atomic number of oxygen right eight for sulfur also we know sixteen right but we have a pattern or we have a difference in atomic number of these elements right suppose if you have to find out the atomic number of tellurium okay so if you know this pattern you can find out what is eight eighteen thirty two is here you see that difference in the atomic number of oxygen and sulfur is eight this difference is eighteen this difference is thirty two this difference is thirty two so once you know this difference you can find out the atomic number of the lower elements oxygen plus eight you'll get sixteen for sulfur sulfur you eight you add eighteen you'll get thirty four so if you add eighteen here you will get fifty two thirty two you'll get this these numbers we call it as magic number for this group okay similarly suppose the alkali metal we have hydrogen lithium sodium potassium rubidium cesium right so for these the atomic number is it is one then you add two it becomes three then you add eight it becomes eleven then you add eight it becomes nineteen then you add eighteen it becomes thirty seven then you add eighteen it becomes fifty five it is correct that's usually we don't remember the atomic number of the lower elements of any group right but if you know this difference here you see it is two eight eight eighteen eighteen and one more we have francium that is thirty two yes thank you right this difference this number is the you know it is the magic number of group one elements there's a magic number of group sixteen elements anyways so this is the one thing another one is what as we go down the group atomic you know number atomic radius increases ionic radius increases next what we can write what increases down the group what about electronegativity you know electronegativity decreases what about ionization energy ionization energy decreases why because size increases right size increases ionization energy decreases then what happens tendency to lose electron is more tendency to lose electron is nothing but a metallic character so metallic character what increases metallic character increases means non-metallic character decreases right metallic character increases means tendons to to lose electron is more means the metal has tendency to be to have a positive charge on it okay it means electro positive corrector electro positive corrector increases you can also understand this as electronegativity increases decreases so electro positive corrector increases two different things clear but when you talk about the elements of this thing group 15 and group 16 that comparison also you must do you see here if you take the example of group 15 element here and group 16 elements here suppose nitrogen and oxygen so if you look at the electronic configuration of nitrogen it is the outermost is 2s2 2p3 is 1s2 2s2 2p3 and this is what 1s2 2s2 2p4 correct so if you talk about the ionization energy as we know left to right you go and size decreases so ionization energy should increase and that is what the general trend we have left to right you are going in a periodic table the size decreases because of effective nuclear charge right size decreases so what happens ionization energy should increase right but here what happens the ionization energy i.e. one I am talking about i.e. one means what first ionization energy the i.e. one of nitrogen is more than oxygen because of what because of half field configuration 2p3 configuration is more stable and it is difficult to remove electron from this but here we have four electron right so from this we can remove electron easily so i.e. one of nitrogen is more than to that of oxygen but when you talk about i.e. 2 then what happens i.e. 2 of oxygen is more than i.e. 2 of nitrogen why because for this i.e. 2 we must understand or we must consider the ion which is O plus because you have removed already one electron so it becomes O plus and this becomes what N plus so O plus the electronic configuration is 1s2 2s2 2p3 any doubt and this one is 1s2 2s2 2p2 so here you see this becomes half field now this is not the stable configuration right this is not the stable configuration this becomes the half field so from here if you want to remove electron that is more difficult and hence the i.e. 2 order is other way and here we have this did you get it yes clear yes okay so these things are easy only it's not that difficult metallic character increases down in the group so one order is right down here if you talk about oxygen and sulphur right if you talk about selenium and tellurium selenium and tellurium okay and if you talk about polonium okay polonium is metal over here it is it behaves as a metal down the group you know metallic character increases these two are what these two are metalloids and these two are non-metals what selenium is non-metal see it is actually what happens it is something like that ionic and covalent character we cannot say any molecule is you know ionic or covalent when we say the molecule is ionic it means its ionic character dominates its covalent correct same thing we have here down the group if you go metallic character increases right so these two metals these two elements have tendency to behave as both non-metals and metals but precisely if you want to say then we'll use the term metalloids not non-metals for selenium even for tellurium also understood yes correct now the another thing here is the melting point and boiling point that you write down write down the melting point and boiling point the point the heading you write down write down as we go down the group the melting point increases the melting point increases with the exception of polonium with the exception of polonium so guys so when you are doing this you should focus on the exceptions only because general trend you know what happens down the group left to right and off so whatever change we are having over here that you must keep in mind focus on that one okay down the group you go melting point and boiling point increases right with the exception of polonium write down next line the melting point and boiling point of polonium are lower than to those of tellurium are lower than to those of tellurium okay now you see whenever you have to compare melting point and boiling point you can talk about when the wall forces because that only affects these melting points and boiling points obviously mass also has a factor over here right that also we can talk about but when the wall force you must consider right so that right on the reason next line as we go down the group the atomic size atomic mass and when the wall forces of attraction increases as we go down the group atomic size atomic mass van der Waal force of attraction also increases and more van der Waal force more melting and boiling point hence melting and boiling point increases right right on in case of polonium due to inert pair effect in case of polonium due to inert due to inert pair effect the valence electrons are less available for bonding are less available for interaction you right now are less available for interaction hence hence van der Waal forces van der Waal force of attraction decreases which decreases the melting point of polonium in comparison to tellurium so from oxygen to tellurium it increases and then for polonium it decreases yeah I'll repeat this again okay as we go down in case of polonium due to inert pair effect in case of polonium due to inert pair effect the valence electrons are less available for interaction the valence electron are less available for interaction hence hence van der Waal force of attraction decreases van der Waal force of attraction decreases which decreases the melting point and boiling point of polonium in comparison to tellurium okay so the reason is van der Waal force atomic mass of polonium obviously higher than with that of tellurium but van der Waal force is weak hence the melting boiling point is less okay so that is it guys we'll continue from this next class okay we'll start the last one our reaction mechanism portion now those who wants to leave we can leave now okay one more thing guys ncrt you must go through group 15 even you can go through group 16 also so that we can go a bit faster in the session okay so you must read out ncrt before coming for the next session okay if you want to do if you want to do some other stuff in this particular chapter you can you can go for the other books also like if you have opi tendon precise jd leaf you have that all those books you can take any any inorganic book you can no go through singed also it's fine okay fine guys see you in the next class okay thank you those who wants to leave you can go now okay tell me what have we done last class maybe i think we have to start pre radical reaction right is it yes okay so i think so i think only a few things are left we'll be finishing this in probably uh one or two sessions okay shall we start yes correct so write down free radical and its reaction the heading write down what is a free radical goc also we have discussed this free radical quickly i'll go through once carbon has three bond three bond x y and z and one unpaired electron unpaired electron right so the property is it has three sigma bond no lone pair one unpaired electron one unpaired electron neutral paramagnetic it is no rearrangement also stability we can compare directly proportional to electron donating group inversely proportional to electron withdrawing group so all these are property of this so what's the difference between lone pair and unpaired electron non-paired and lone pair and unpaired electron lone pair unpaired unpaired electron see lone pair means what suppose if you have this okay so this is lone pair of electron this is a pair and one single electron if it is present is unpaired electron there's no pairing of this electron it is unpaired electron both are unshared electrons but unshared electrons we can have one pair or one single electron so one single electron is unpaired electron you can also say see if you are not getting this term you can also say one electron present onto this but since we like the way we define carb anion one paired electron there one lone pair electron is present right so here also the electron is present but it is not paired so unpaired electron there's no pairing for this yes just a second like so these are properties we have okay now you see the important thing is since this free radical is neutral so it has tendency to get stabilized combines with the another radical and get stable right now the combination we have or the reaction of free radical we have two types of reaction okay so the combination of the radical takes place or free radical involves two types of reaction here the first one is combination combination reaction now what is combination reaction simply two radical combines and forms a compound okay for example for example we have a supposed CS3 radical combines with CS3 radical and forms ethane CS3 CS3 this is combination two radical combines now this kind of reaction the activation energy is zero easily possible this kind of reaction activation energy is zero the second type of reaction we have and we call it as disproportionation reaction disproportionation reaction what is this proportionation reaction means two types of reaction possible here i'll show you how this reaction goes one more thing you write down here then i'll go to the next page the combination reaction is possible mainly with one degree radical if you have like methyl smaller groups are there one degree radical generally combination reaction favors okay but we cannot eliminate the chances of disproportionation it is something like SN1 SN2 SN2 is there we cannot eliminate the possibility of SN1 but one is major so that gives the major product one is dominating that is the major so here also we have the same thing you one degree radical we have high chances of the radical to combines and goes under the combination reaction disproportionation reaction it is generally possible with three degree radical and it is easily observed in three degree radical this is one thing now you see what this disproportionation reaction we have suppose i have taken this example carbon with CS3 CS3 CS3 and we have a radical here yes correct correct fine so this is allowed to react with this is allowed to react with the same molecule which has this so this is another radical right so what happens here since it is three degree right so there are chances to go under disproportionation reaction is more and what is required for disproportionation reaction i'll write down here the condition is first of all its activation energy is not zero one of the properties this second property is what it is slower than the combination reaction slower than combination combination reactions are faster reaction but both are competitive both reactions are competitive like i said now right both compete with each other if the free radical containing carbon has large group attached with it then then disproportionation reaction is possible okay so what happens here in this reaction you see this radical this is the alpha hydrogen right so if you have alpha hydrogen present then what happens this hydrogen and this radical has some attraction here so we'll get a transition state which is something like this CH3 CS3 CS3 and this bond with hydrogen is about to form and this bond is about to break so we have here radical right this bond is about to form eventually what happened and all these happens on its own is spontaneous eventually what happens this homolysis takes place over here right so this radical this radical forms a double bond so one of the product that we get here is an alkane which is this CS3 and this hydrogen attached over here plus the another one is an alkene which is CH2C CS3 CS3 so disproportionation reaction gives you these two products alkene plus alkane clear so here what happens it is the addition of hydrogen so addition of hydrogen means what reduction so this is reduction gain hydrogen so it is reduction this loses hydrogen so this is oxidation since oxidation reduction takes place here hence the reaction is disproportionation reaction no doubts okay fine now what is the condition under which this kind of reaction is taking place whether it is combination or disproportionation so first point to write down here as crowding crowding at free radical means the carbon which contains free radical increases then percentage of disproportionation reaction increases more crowding more disproportionation nature will be there no in free radical dancing resonance is not there trippin dancing resonance is shifting off the sigma electrons two sigma electrons so free radical is not there it's not possible fine so as crowding at free radical increases disproportionation reaction increases so what we can conclude if you have one degree radical so it has major nature the major property of this is what it goes under combination reaction combination reaction if you have three degree radical then it goes under what we say disproportionation reaction if you have two degree radical so like we have both SN1 SN2 possible so here also combination then plus disproportionation possibly right we'll discuss just give me some time okay this is the one property any in doubt in this major minute I'm coming to that any doubt in this no right disproportionation reaction the activation energy is not zero so high temperature favors disproportionation reaction okay because the bond dissociates over there carbon hydrogen bond is dissociating right homolysis is taking place that's why as temperature increases as temperature increases percentage of disproportionation reaction increases and the first percentage of what we have the other one combination reaction decreases this is another property okay now apart from this we have one intramolecular reaction also possible right intramolecular reactions also possible what is this I'll tell you okay so first write down this if the molecule has two free radical in the molecule only if the molecule has two free radical then they combines combines and forms either a pi bond or ring pi bond or ring even a smaller ring also forms over here we don't consider the stability of that ring for example you see this if you have radical here this is the molecule just a second we have a radical here we have a radical here so these two combines together and forms cyclopropane this is intramolecular reaction intramolecular reaction and it is irreversible in nature intramolecular irreversible reaction won't go in backward direction okay since it is irreversible so it is rate controlled okay understand this carefully otherwise you'll ask me so three membering is not a stable why it is forming okay it is intramolecular irreversible so it is rate controlled reaction and if it is rate control so we can say it is kcp however this term we do not use okay so I am just I have just written here to make you understand this you don't write it okay don't write it rate controls so we can say it is we are not considering the stability of this product but we'll see how fast the product forms since free radicals are highly reactive so it forms very fast in this reaction right intramolecular irreversible so it goes in the forward direction with a very high rate correct this is rate control that's why we do not consider we do not consider the stability of product clear okay similarly you see the other example if you have this one radical here one radical here so it forms it forms a pi bond right this one you see these are the radicals present here it forms this one or this one which one is possible tell me first one there is a combination of a three degree radical and one degree radical then which reaction switch to justice again we'll discuss all those okay we'll discuss what reaction possible okay one by one we'll see that I saw your messages just now huh you see in this one what happens the reaction is not possible this is not possible this is also not possible because none of them are intramolecular reaction but the possible reaction product is this you get two molecules of this if you are combining these two these two it means it is not intramolecular right and sigma bond won't form because it forms what either a pi bond or ring so within the molecule and in the molecule we have two radical present so these two combines and these two combines understood so again how do we know if intermolecular or intramolecular reaction you see this this is what and this is the reaction it is within the molecule the radicals are combining so intermolecular yes which one tends to happen faster intermolecular or intramolecular intermolecular it is between the two molecule intramolecular oh did I say inter it is intramolecular all inter is not possible intramolecular I was saying inter right yes no no no it's not inter it's intra intermolecular is not possible not at all okay that's a slip of tongue intermolecular that's what you see these two are combined that's what the point is if the mod molecule has two free radical right molecule has two free radical one and two then they combined and forms either a pi bond or ring this is possible intermolecular is not possible right intermolecular we have discussed the first case that is a kind of combination reaction you can consider right another one so um few reactions few examples we'll see here we write down the possible product possible product of this one and we have it's not like only one molecule we have we have one mole of this suppose there are many such molecules what is the possible product we get again one moles of it radical one mole of it here see what happens since there are many molecules many such molecules and the radical is two degree correct radical is two degree so this will go under combination as well as what disproportionation so this is combination and disproportionation we know it gives two product one is alkene and other one is alkyne alkyne so these two are disproportionation and this is combination I want you to write down the steps into this you will understand this I think you won't have any confusion in combination disproportionation you see that steps tell me if you get it is it clear now this is one degree right so one degree will go under what only combination so the two ring combines with this this combination could you write down the product into this one here the product is what three degree so it will go under disproportionation reaction so we will have one hydrogen here and one alkene we get which is this clear what about this one so this is combination disproportionation this is combination this is disproportionation and this is intra intra and gives you what cycle of you are you getting it okay if you have this molecule where you can get the product in conjugation double bond in conjugation then disproportionation we won't consider into that you see this we have radical present here and here if you do the homolysis of this one electron comes over here another electron comes over here so this forms this product a cyclic diene so in this case we will form cyclic diene only not the intra molecule similar reactions of this you see means the radical is in conjugation see the two radical in conjugation so it forms diene deutadiene homolysis comes over here and this comes over here it forms a bond with this okay one very important one here suppose we have benzene ring and on the benzene ring we have two radical present here one is here another one is here and we have one mole of this so in this the product is this two such ring combines you won't get a triple bond here two such ring combines and you won't get a triple bond here like this because for triple bond for this intra molecular reaction for triple bond the ring must be at least eight membered right for triple bond the ring must be eight membered that is not possible over here hence it is combination reaction okay could you tell me the product in this one it goes under combination reaction and forms this yes or no so for the previous reaction isn't benzene also possible no benzene is an is an intermediate it cannot be a product of any reaction okay you never have seen carbocation is a product of any reaction right intermediate can never be any product okay so it's a benzene we won't form okay is this possible this one by combination it is possible but you see here this is anti aromatic yes or no it is highly unstable so this reaction is not possible here but what happens two such molecule combines and we get this structure all these are very specific reaction if you haven't done this before you won't get it in the exam that is for sure you should have all these informations any doubt guys sir all these reactions are irreversible sir what all these reactions are irreversible all these reactions are irreversible yes one more one more example you see this is also like you can take this as an information because again like I said if you haven't seen this you won't get it in the exam suppose you have this benzene ring and carbon also has two phenyl group attached to it like this and has one radical pre radical is this you see this radical is three degree yes or no it is three degree radical but it does not go under like combination is not possible here right even disproportionation is also not possible because there is no alpha higher you see there's no iron here there's no hydrogen no iron here correct are you getting it yes sir three degree the combination is not possible if you try to combine also then what happens the other molecule which is this a carbon has two phenyl group two phenyl group and a benzene ring another phenyl group is this if this radical and this radical you'll try to combine so this two will repel right phenyl and phenyl repel so whatever possibility of combination is there that is also that has also become zero because of the repulsion hindrance correct what we can say combination is not possible and there is no hydrogen present here at the alpha carbon so disproportionation is also not possible okay but in this reaction it's not like the product won't form into this product forms here but what happens this radical this radical will take hydrogen from here this one and you see how does this go this will take this so this forms a radical here this bond comes over here forms a double bond and this bombs comes over here forms a double bond here and this will take this hydrogen right so it forms a molecule which is this you see this ring I'm drawing here first carbon with pH pH and this carbon will take what in just a second it will come over here okay so it is like this this forms a radical here which is one hydrogen radical and this radical so this radical combines with this and it gives the ring okay one correction you do here I have by mistake I have drawn here the complete arrow it's in half arrow actually so this one electron comes over here this radical this radical forms a double bond one radical here one radical we have here so this form this forms a double bond here and this radical joins over here okay this hydrogen will be here only I have just made this thing here this comes over here this radical not on hydrogen did you understand this so it was in conjugation to this kind of reaction also possible and it is it is a combination reaction actually so these two three examples you must keep in mind okay very useful and important okay can I go to the next page yes okay now you see how free radical forms in any reaction so heading you write down formation of formation of free radical formation of free radical first of all when you have a peroxide it's very simple and easy you have a peroxide then what happens peroxide can easily forms free radical because they have tendency to go under homolysis because of lone pair lone pair repulsions two molecules of ro dot you get I'm not drawing the lone pair over here if you want you can draw the lone pair like this and the free radical like this okay this is one case right if you have halogen atom xx again the same reason lone pair repulsion it can easily go under homolysis and forms the radical third case is we can take rocl the same logic oxygen has two lone pair chlorine has three lone pair so lone pair lone pair repulsion it forms ro radical and chlorine radical okay fourth one in presence of metal all these three are of same category next one is suppose we have rcl here we not have no lone pair repulsion but alkyl lead in presence of metal it may form a radical and cl minus goes out easily you can understand this provides the free electron right it gives na plus plus electron right now this is delta positive this is delta negative because of high electronegativity of chlorine what happens next this electron gets attracted towards the alkyl group and makes this halogen atom to go outward that's why cl minus cl takes this two electron goes out as cl minus and this electron on this alkyl group forms a radical is it clear what I said you have rcl okay rcl with this electron electron forms the metal so delta positive and delta negative this electron gets attracted towards the alkyl group since this is delta positive so when it comes over here this makes this halogen atom to go out with this electron pair so it forms cl minus and one electron onto this alkyl group so radical yes sir correct now the fifth one is so one we have seen the repulsion of lone pair then alkyl halide in presence of metal like sodium the next one we have when we have this anodic oxidation reaction of carboxylate ion okay it is anode oxidation oxidation and anode what happens electron releases we know in oxidation electron goes out so we get one radical plus e minus right one more thing the free radical has property to eliminate neutral molecule to form a new radical which is this so what happens here in this one this one electron comes over here one radical one radical forms a double bond and this one radical comes over here so alkyl radical and co2 goes out so free radical has this property to eliminate neutral molecule to form a new radical which is the alkyl radical sir yeah sir full negative charge forms only when two electrons are lost right or a pair of electrons are lost where like suppose a molecule gains or an atom gains two electrons then only gets an additive charge right no very simple carbon ion example i'll take okay so carbon has four electron in the outer most shell correct so this is the three electron of carbon and this is the fourth electron of carbon so what is this it is a free radical what is carbon ion carbon ion is this the electrons of carbon are one two three and four right and this blue one that i have added onto this so this becomes carbon ion so why one negative charge because one electron i have added extra electron okay and one electron carbon already has awesome right so one electron if it goes out so you can't say if you remove one electron you'll get this radical only no that's what we get here awesome fine this is one thing one last possibility we have in case of azo compound okay ph n double bond n this is azo azo compound and it's nothing you have to do when you heat this it forms two phenyl radical and n two gas eliminates even in the reaction of this azo compound like di azonium chloride and then di azonium chloride right that is also azo compounds in all those reactions always nitrogen gas evolves okay and forms radical like this so this is the uh you know cases where where the free radical formation takes place now the last thing of this we have to discuss is what all the reactions we have which involves free radical mechanism most of the reaction we have done okay so we'll just go through it quickly reaction involves free radical mechanism any example can you tell me words reaction words reaction we have discussed free radical mechanism right but what reaction has actually two types of mechanism okay that we haven't discussed we have discussed only free radical we haven't discussed the another mechanism of words reaction that is ionic mechanism okay so sometimes they'll ask this question in the exam that how many products possible in this reaction words reaction so that thing you can answer now once you understand the free radical mechanism properly okay usually what we know write down this usually what we know what is the action is something like this two rx reacts with what n a drier and it forms r r plus nx this is what we know yes understood but what alkyl group will take here because you see what is the radical that forms if you look at the mechanism here quickly right so first step is what the first step is n a i'll take two n a gives two n a plus plus two electron then what happens rx the mechanism i told you in presence of this electron it forms r radical and x minus correct guys okay right now so what happens here actually we have this this this and this two modes of this so let's next what happens this r and r combines and we'll get this lk this is what happened right and we know this only till now correct yes how many of you are getting it i'm not getting any response from your side okay see just now we have discussed that depending upon this radical the reaction can be what the reaction can be either what combination or disproportionation yes so from here what we have discussed till now we have discussed only combination disproportionation we haven't discussed but when the radical forms both kind of reaction is possible here combination as well as disproportionation right so that's how we get the different different product here right if this radical is one degree then we'll get only combination if it is two degree then we'll compare when then we'll have both combination and disproportionation if it is three degree radical then we'll have only disproportionation reaction mainly right but since woods reaction is used for the preparation of alkane symmetrical alkane so ideally what we should take we should take primary halide here are you getting it primary halide gives you only symmetrical alkane right but the question that they frame in the exam they will give you secondary halide here like the question is this i'll write down the will finish this reaction so our radical our radical combines and it forms an alkane but you see how they frame the question in the exam they'll give you the reaction like this suppose we have CH3 CHCl CH3 and this reaction is taking place with Na dry ether okay and they'll ask you what is the number of products we get we get here that's very good and very important question usually we do not consider these the formation of alkane by a disproportionation reaction okay so number of products will be what it is not one product we won't get this only this is not the only product we get right but the product we get here is this plus this this and plus this these three products we get any other combination possible what happens here you see the first step is what we'll get the radical so i'm not writing down the whole mechanism here the radical is this is it right guys tell me this is the radical we get and the radical is what the radical is 2 degree and we have one more molecules of this right which is CH CH3 CS3 radical so when these two combines it forms this compound but the another possibility is what when this combines with H CH2 CH CH3 radical right so this gives you what this gets this hydrogen and we get this compound via disproportionation and one more compound is this via disproportionation so total number of product will be one two three did you get this sir yes sir can we get the isomers like cis and trans disproportionation depending upon the molecule what structure of alkyl halide you have taken obviously if it is possible then why can't we get we can get that if you get 2 degree possible no yes sir so then both will be found so so so if the question is total number of products if it is there then we have to consider the studio isomers also cis and trans okay if it is forming we will be depending upon the cases depends upon what structure of alkyl halide you have taken but always keep that in mind till now they haven't asked questions in which the studio isomers are involved okay similar this the example that we have done similar kind of question they have asked in the exam okay i am not very sure with the exact you know the molecule they have taken the reactant molecule but the concept is this okay similar kind of concept if a studio isomers possible then we'll count that also okay okay we use dry heater for this reaction so that water if it is present it gives you alcohol also one of the product to eliminate that we use dry heater one point okay so that's why for bourge reaction bourge reaction is mainly for the preparation of preparation of what preparation of symmetrical alkyl so ideal condition is what ideal condition that the alkyl halide should be what should be primary primary okay so if they ask you this question which of this alkyl halide it's suitable for bourge reaction you have to take primary alkyl halide there okay secondary gives you both product disproportionation combination tertiary gives you tertiary gives you what gives you disproportionation reaction you won't get alkene in that case okay understood one more question you'll see here tell me the number of product we get in this reaction cs3 cl plus cs3 cs2 steel with any dry one one mole of each we have number of product srishti what happened tell me the answer angeli what is the answer okay i'm getting five four three these are the three answers i'm getting five three and four srishti what about you so five five you're getting ripen what is your answer ajitya you're getting four ripen what is your answer swami i got four four i'm okay i'm also getting five okay now you see one product is what cs3 cs3 another one is butane another one is whatever radical we are getting those radicals combines with each other in all possible ways because since both are one degree so disproportionation we can eliminate easily it is only combination so combination of methyl ethyl combination of methyl ethyl combination of ethyl ethyl gives you these three products what happened what is what you're not getting i don't understand this this is not that tough any doubt in this tell me so a five on from before how far you won't get you won't get ethyne why we get ethyne here because uh it's one degree no so we won't consider eliminate disproportionation here ethyne forms by disproportionation or if two degree i desire then we'll consider that all are one degree no methyl it's a methyl radical and this one is one degree and if you are saying you'll get ethyne it means what you'll get methane also one of the product yes i know ethyne only forms in this case you see this h ch2 ch3 radical ethyne only forms in this case when you take this hydrogen and this one electron comes over here this one comes over here so you get methane also you'll get ethyne also all alkene and alkene you can form in this reaction we don't require any other methods understood yes right so we if the halide is one degree so simply we have to consider only combination reactions all possible combination we'll do and we'll get the number of products correct so what happens here you'll see that the problem is what now you get the mixture of alkene right so if you want to prepare this thing uh ethyne but you get propane and butane both then what you have to do you have to separate this which is also very difficult to separate because all have very close melting point and all ethane propane butane right separation is difficult over here that's why the preferred thing is what we always get or take symmetrical alkyl halide for this reaction so we'll get only one possible product that is the purpose of forage reaction so again one last thing i'm telling you you are not going to get tell me the product in this reaction forage reaction tell me the product you won't get that question they will ask you the number of product possible they'll ask you what is the possible or suitable alkyl halide we should use for forage reaction is it clear can i finish the another mechanism of this ionic mechanism i'll take two more minutes okay like i said there are two mechanism here free radical we have discussed one more mechanism we have we call it as ionic mechanism ionic mechanism is this we take some highly electropositive metal which is n a here right so n a releases electron right so first step here that two atoms of this sodium metal it gives two n a plus plus one electron two electrons okay now the second step is also fine that this has our x plus one electron on this so what happens if you if you heat this of it right then what happens here homolysis takes place right and this oh not like that see what happens in presence of this also fine it'll get a radical plus x minus right after this what happens one more electron that we have here this electron is taken up by this alkyl row and forms r minus so this is a carbon ion we get this carbon ion basically it is an ionic mechanism so in this ions are involved so we get carbon ion here now this carbon ion r minus behaves as a nucleophile and attacks on the substrate in s n two manner to gives you to give the product this comes over here this goes out so it forms r r plus x minus which these two combines with n a plus and forms n x right so ionic mechanism involves carbon ion intermediate and it has s n two path so it will be s n two like independent of whether it's three degree one degree two degree no no for this thing if it is three degree then elimination possible if ion forms then we can have elimination in this that is also one understanding you should have if they give you three degree an idea then the product you get by elimination not s n two so that's why the preferred thing is what for what's reaction we prefer that alkyl group should be one degree right right down over there alkyl group should be one degree any doubt no right so guys we are almost done with this there are few more reactions we have to discuss i can tell you some name coal bay electrolysis we have discussed that we have also discussed n b s and bromo succinamide okay alkyl light reaction and forming of formation of alkene from alkyl light sorry from alkene to allylic substitution of bromine then we have some anti marconic op reaction which for reaction which falls anti marconic op rule so there are basically four five reactions left okay chlorination halogenation reactions of alkene pinacol formations these two three four five reactions are left and then we are done with reaction mechanism okay so i hope you all have understood the concept of board's reaction both mechanism and what should be the alkyl light we take depending upon the alkyl light what should be the part possible product by a combination or disproportionation reaction okay so revise all these things once okay i'll share the assignment on this reaction mechanism with you you can solve those questions okay and now you can try the organic chemistry any chapter you can try the questions okay because the basic is clear now that we have done all the reactions how the reaction proceeds all the mechanism we have discussed so you can solve any uh you know chapter of organic chemistry if i will share the assignment with you but inorganic n crt plus one more book you must read out okay two pdfs i have shared last class have you finished that yes or no no you have done that okay those who are not solving you see if you look at this if you attend classes only that won't help you especially in chemistry you have to do some homework right after the class so fine guys i'll share some more assignments okay finished all those did you finished electrochemistry assignment yes sir all the all the all the pdfs if not then finish that okay the learners thing it is not till j advanced level i'm telling you frankly okay the learners assignment if you solve it is till j means ct bits at all those plus j means right but it is not up to j advanced level okay for j advanced level i'm sharing those pdfs okay so you must solve those pdfs anyways thank you guys thank you thank you thank you sir thank you sir thank you sir