[0:10]Hi students, in this session I'm going to teach you eight tricks from P-block elements. This topic is from Class 12 Chemistry. Let's see the trick number one. N2O By the end of this video, you are going to learn so many, eight different tricks you are going to learn, so watch up to the end of the video. Coming to the trick number one, N2O, NO, N2O3, N2O4, N2O5, they are asking us to find the acidic order strength. What you need to do here is, you know, first calculate the oxidation state of of N2O, +1, and it is +2, +3, +4, and +5. As the oxidation state of the central atom increases, as oxidation state increases, acidic strength also increases. This is the easiest way. What is that? As the oxidation state of central atom increases, acidic character also increases. So which is having more acidic strength? N2O5 is having more oxidation state, so it is more acidic than this. Coming to this, N2O5, P2O5, As2O5, Sb2O5. If you observe this question, here central atom is different. This is Nitrogen, Phosphorus, Arsenic, Antimony, Bismuth. Whenever central atom is different, if electronegativity of atom increases, acidic strength increases. So which is having, which is more electronegative? Top to bottom of the group electronegativity decreases, so acidic strength also decreases. Which is more acidic in nature now? N2O5. So the summary of this trick is, when same element, you need to see oxidation state. When the, when there is a different element, you need to see the electronegativity. Let's see trick number two. Now see trick number two, P2O3, actually it exists as a dimer. P2, P2, so P4. O3, O6, it's a dimer. P2O5 also exists as a dimer. The question is calculate the number of sigma bonds and pi bonds. What you need to do is first draw the structure of P4O6. The easiest way is first draw one phosphorus here, one phosphorus here, one phosphorus, one phosphorus. Just keep oxygen, connect oxygen in this way, connect oxygen, connect oxygen. And then at the center, connect one oxygen. Don't draw the oxygen exactly at the middle because if you draw the middle, then these both will intersect here. That's why you draw little top here and then here. So this is P4O6. That's it. What do you need to do? Just count the number of sigma bonds. Single bond is a sigma bond. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 sigma bonds. And there is no pi bond. So answer is 12 sigma and 0 pi. When they are asking us to draw the structure of P4O10, how you need to draw is same structure, how you drew for this, same thing you draw again and then, see same structure. And then now just keep a double bonds O. Same structure. And double bond O, you are keeping. So here how many 12 sigma bonds already you calculated. 13, 14, 15, 16. 16 sigma bonds and how many pi bonds? 16 sigma and 1, 2, 3, 4, 4 pi bonds. This is your answer. Now let's see trick number three. In this, I'm going to teach you the bleaching action of O3 and SO2. Bleaching action of ozone is permanent and SO2 is temporary, and here it takes place through oxidation, and it takes place through reduction. Let's see this. Ozone upon decomposition it gives a nascent oxygen. This nascent oxygen, if any colored matter is there, colored matter. This nascent oxygen, you know, addition takes place here. It oxidizes, ozone oxidizes that and we get the colorless. So, this is, you know, this is a permanent. It gives us nascent oxygen through oxidation, it takes place. And coming to this, SO2, SO2 if it reacts with the water, then it gives the nascent hydrogen. Along with H2SO4, it gives nascent hydrogen. This hydrogen, if any colored matter is there, this colored matter, it undergoes, you know, reduction and it becomes colorless. But why it is temporary? Because this colorless matter which absorbs the oxygen from atmosphere again, and it can convert into again colored matter, that's why it is, you know, it is not permanent.
[5:39]It is temporary. And now coming to the trick number four. The reaction between ozone and uh with ozone with a dry iodine and moist iodine, when you are doing a self-study, uh you may not observe this clear difference. When you when we treat ozone with a dry iodine, we get an yellow color solid I4O9. This is yellow solid. Yellow solid will be getting. And this if you observe, it it is having two different units, one is I3+, and another one is IO3-. So to balance this, I'm keeping three before it, I3+, IO3-. It contains two units, important. And coming to the moist iodine which upon reaction with ozone, we will be getting iodic acid IO3. Clear? Trick number four is easy. See the just the difference you are learning here. And coming to the trick number six. HF, HCl, HBr, HI. They are asking us thermal stability. Cl2O, ClO2, Cl2O6, Cl2O7. They are asking the acidic order strength. What is this acidic order strength? What do you need to do? First, you find the oxidation state of central atom. Once you find the oxidation state of central atom, 14 by 2 is 7. Okay. Just, you know, it is +4. Once you find the oxidation state, here I'm my topic is not to calculate, not how to calculate oxidation state. So, I uploaded another video how to find oxidation number. If you don't know how to calculate, just go through that. I'll give in a description. Cl2O, ClO2, Cl2O6, Cl2O7, as oxidation state of central atom increases, what I told you? Acidic strength also increases. So, which is having more acidic strength? Easily you can say Cl2O7, then Cl2O6 and then this is the order. If suppose, central atom is same like this and side atoms are different. Side atom, this is a side atom. If side atom, you know, size increases, side atom is different. Side atom is different and size of side atom increases. Size increases then bond angle also increases. Bond angle also increases. So which is having more bond angle? Can you tell me now? Can you tell me? OF2 is having more bond angle, then OBr2, then OCl2 because this is oxygen. And this is one iodine and another iodine. Due to the bigger size, you know, repulsions will be there. They both go away from each other and more is the bond angle.
[8:24]Coming to the trick number six. HF, HCl, HBr, HI. Coming to the thermal stability, if suppose, observe here iodine is larger in size.
[8:40]Whenever size of the atom increases, the bond length increases. Bond length increases, if bond length increases, between hydrogen and iodine, because size increases, bond length increases. Then we can break it very easily. So when you can break it very easily, thermal stability decreases. This is having least thermal stability. HF is having more thermal stability. And when thermal stability decreases, reducing character increases. Both are inversely related. Thermal stability increases, reducing character increases because we can break the bond easily, which means that we can give the, it gives the hydrogen easily. That's why HI bond length is more and its thermal stability is less, it dissociates easily and it gives hydrogen easily. That's why HI is having more reducing character than HBr, than HCl, than HF. Is it clear? This is trick number six. Here you learned, you know, relation between thermal stability and reducing character. Coming to the trick number seven. XeF2 reactions with hydrogen. This is very, very easy. They ask you like, they give the reactants and they ask by products by giving the dash like this. What do you need to do is, how many hydrogens are present? First you remove that. How many HFs I can remove? I can remove 2 HFs. So keep 2 here. Is it clear students?
[10:19]And remaining what is left, you just write here. And see the next one. HF, how many HFs you can remove? Tell me. Yes, correct. You are right. 6 HF and xenon is your product.
[10:44]So this is these are the products you learned in a trick number seven. And coming to the trick number eight, hydrolysis reactions of, you know, xenon compounds like xenon difluoride, tetrafluoride, and hexafluoride. XeF2 when we treat with water, I mean, if it undergoes hydrolysis, we will be getting, you know, the products like xenon + HF + O2. These are the three products. And same way you will be getting when, when XeF4 reacts with H2O also we get the same product. But here the simple trick I'll tell you to remember this reaction is along with this three, these three are same, same product. Along with this three, we will be getting XeO3 also, xenon trioxide also as a product we'll be getting. And coming to the XeF6, what happens is sequence hydrolysis takes place. It is very explosive reaction. We will be getting XeO3 as a main product along with the HF. So what are, what we are getting here? XeO3. This is a slow reaction, and this is a little fast reaction. You can say, and this is very fast, rapid reaction, which is this reaction is explosive in nature. Explosive in nature. These are the eight tricks I covered from P-block Class 12 Chemistry. Just write in your books and practice one time. You feel, you know, this video actually this video really helps you. My name is Komali, I'm your chemistry mentor. Thank you so much for watching my videos.
