 The next one you write down, pentahelites, we already know nitrogen does not form Peltahelites, correct? We have discussed this in the beginning. These elements forms Peltahelites of Mx5 type. Mx5 type. Okay. Nitrogen next time, nitrogen does not form Peltahelites due to absence of d-orbital. Due to absence of d-orbital. Hence, it cannot expand its octet. The phosphorous pentachloride forms by these different methods. If you take P4 with excess of chlorine, with excess of chlorine, 10 Cl2, then it forms PCl5, four molecules of this. You can also take Cl2, sorry, you can also take SO2Cl2, SO2Cl2, thionyl chloride. This also gives four molecules of PCl5, plus we get SO2. If you balance this reaction, we have 20 chlorine, so 10 here and 10 here. The reaction is balanced. Okay. Phosphorous, if you allow this to react with chlorine, okay, in limited supply of it, first you get four PCl3, phosphorous pentahelite, trichloride, and when again it reacts with Cl2, you'll get PCl5, phosphorous pentahelite. So limited supply of first you get trihalides and then pentahelites. Okay. Right on next point into this, pentahelites are thermally less stable, are thermally less stable. On heating, they dissociate into trihalides and halogen. And that is why PCl5 act as a good chlorinating agent. Basically, I'm no organic chemistry, maybe use this reagent. Okay. That's why PCl5 behaves as a good chlorinating agent. Next point. All pentahelites behaves as Lewis acid due to presence of vacant orbital, due to presence of vacant deorbitant. Okay. So these are the, you know, few compounds which is formed by the elements of group 15 we have discussed, okay, halides, chlorides, sorry. I have a question. What's the hybridization of pentahelites? Pentahelites. Yeah. What is the hybridization of PCl5? PCl5, correct? Anyone? Hybridization of PCl5. Who is this? Who is this? Whose question is this? Pragwell. Sorry? It's mine, sir. Pragwell. Yes, Pragwell. How do you find hybridization? We'll find out first what? Valence electron, then a steric number. What is the valence electron for this? For phosphorate is 5, chlorine is 7. So 7 into 5. So what we get? 7 into 5, that is 35, 35 plus 5. 40. 40. Okay. This is the valence electron. So this is 40 divided by what? 8, 5, 40, and 0. This means what? We have 5 bond pair and 0 lone pair. So steric number is 5. So hybridization is what? SP3D. Wait, wait, wait, wait. What's the process? See, what do we have to do to find out hybridization? See, any molecule you have, suppose I'm taking the example of SO42 minus. Pragwell, SO42 minus. So the actual thing I'm not discussing because it takes time and that is not helpful in the case of when you get this question in the exam. So I'm not taking that now. How to get hybridized? That's what I'm trying to make you understand. What do you do? Any molecule or ion if it is given, you find out valence electron. Tell me the valence electron for this. For sulfur it is 6. Okay. Belongs to oxygen family. And oxygen family, all elements has 6 valence electron. Yeah. For oxygen also it is 6. 6 into 4. And if any charge is there, negative charge we have. So we have to add plus 2 for that. Positive then we have to subtract. So this is how we get the valence electron which is 24 plus 2, 26 plus 6, 32. Now this 32, whatever the number you are getting here, you have to divide this valence electron by 8. Okay. You'll get some quotient and you'll get some remainder. Correct? Right. This quotient is the number of, the number of bond pair. Okay. And this remainder is the number of, number of electrons in the lone pair. Got it? Okay. So if I ask you the number of lone pair, what is that? Number of lone pair is what? R by 2. R by 2. So with this, when you get the number of lone pair and number of bond pair, there's a number we calculate, we call it as steric number. And steric number is equals to the number of bond pair which is Q plus the number of lone pair which is R by 2. Clear? Okay. Okay. Now if this is steric number is 2, hybridization is SP. If it is 3, SP2. If it is 4, SP3. SP3 and so on. Yeah. Then SP3D. Okay. Like this week. So your question is PCL5. So that's what I did. For PCL5, the number of valence electron if you count, you'll get 40. Okay. So 40 divided by 8. So we'll have 5 bond pair and 0 lone pair. So 5 is a steric number. Hybridization is what? SP3D. Got it? Yeah. Thank you, sir. Okay. Okay. We'll start now. Okay. So we have to see now the preparation of nitrogen. Okay. So there are various method of preparation of nitrogen we have. Okay. The first method that we have from nitrogen compound. The first method you write down from nitrogen compound. Okay. Write on pure nitrogen can be obtained by pure nitrogen can be obtained by pure nitrogen can be obtained by passing ammonia vapors over heated COO. So reaction is 2 NH3. When passed over COO, it forms N2 plus 3 H2O. So by this method, the nitrogen that we obtain here, this is the pure nitrogen. You must remember this. Ammonium nitrate if you heat. Okay. That also gives you nitrogen. Okay. So the next method here. Ammonium nitrate by hitting ammonium nitrate, not nitrate, ammonium nitrate. So you see this reaction NH4Cl plus NaNO2. It forms ammonium nitrate, which is NH4NO2 plus NaCl. Now, when you heat this ammonium nitrate, we'll get N2 plus 2 H2O. Bottom water goes out. Okay. This is the second one. Ammonium compounds. Ammonium dichromate. Also, we can heat to get nitrogen. NH4Cl2O7. When you heat this, you'll get N2 plus CR2O3 plus H2O. So all these reactions, you see there's nothing to understand in this. You should know these reactions. Memorized. Usually you have to memorize this reaction. There are certain properties of nitrogen. The first one I'll write down here. It is a colorless, testless, orderless gas. First one is this. At very high temperature, at high temperature, and that to around 3000 degrees Celsius. It combines with, it combines with oxygen and forms, and forms nitric oxide. The reaction is N2 plus O2 at around 3000 degrees Celsius. It gives you 2 NaO nitric oxide. Nitrogen and oxygen are non-reacting at room temperature. And for this reaction, we require very high temperature. Okay? Another reaction here. Nitrogen when combines with hydrogen. N2 plus 3H3. All of you must know this reaction. The catalyst that you are using is iron. Temperature around 400 to 500 degrees Celsius. The process is known as Haber's process. Product is NH3 ammonia. Okay? The most important reaction we have here is with metals, with metals, nitrogen combines and forms with metals. It forms nitride. This question they have asked in the exam. 6 Li, lithium we are taking, plus N2. Temperature around 450 degrees Celsius. And you will get 2 moles of Li3 PN. Okay, lithium nitride. With magnesium, Mg plus N2. This is also metal reaction. Same temperature. You'll get Mg3 and magnesium nitride. Right? Okay? So with metals, it forms nitride. With metals, it forms nitrides. With non-metals right down, with non-metals, like you see this reaction boron, 2B plus N2. It forms PN. With silicon, 3Si plus 2N2. It forms Si3 and 4. Okay? The most important reaction here is the reaction with calcium carbide. The reaction with calcium carbide, CaC2. Calcium carbide CaC2 reacts with nitrogen N2. Around 1000 degrees Celsius. And it gives CaCN2 plus carbon. Okay? Now when this is dissolved in water, 2 moles of H2O, CaCN2, it gives CaCO3 plus NH3. 2 moles of NH3. You get calcium carbonate plus ammonia. Okay? This mixture, they've asked many times this question in the exam. This mixture, calcium, calcium cyanamide and nitrolym. Sorry, calcium cyanamide and this carbon. This mixture, we call it as nitrolym. This compound is calcium cyanamide. The mixture of calcium cyanamide and carbon, we call it as nitrolym.