 Okay, hello everyone. Please type your name. Who are all there? Hi, Ritwik. Hello, Sanjana. Khushal, Purwik, Sanjana and Ritwik. In today's class, we are going to cover three, four different chapters of chemistry and we'll mainly solve questions that has been asked previous year and then some other questions also which are important. So basically, we'll not discuss theory today. The problem-related concept that we have to use in the solution of that particular problem. Only that, those theory or those concepts we'll discuss otherwise we'll mainly focus on problems today. Okay, so like few questions I'm just going to give you. You have to solve these questions and then we'll see the solution and we'll discuss the solution of those questions also. You see this one, this is a chapter that belongs to inorganic chemistry which is coordination compound. These questions you try to solve first and then we'll do the solution of this. We'll discuss the solution of this. Question number 13. What is the answer? Question number 13. Sandeira is getting A. Okay, solve these questions quickly. One, two, three, four, five questions I've given you. Tell me the answer of this 13, 14, 15, 16 and 17 quickly. Then we'll discuss one by one. We'll see question number 13 first. Question number 13 you see the complex which is given here. The complex which is given is COCl3.3 NH3. So if you see this complex, this complex has written in this simple form. It's not in a complex form but the actual formula or actual molecule is this. CONH3 whole thrice Cl3. This is the actual compound. The actual molecular formula is this for this particular compound. So all these are complex parts. This is nothing but the complex part we have. This is the complex part. And we know that complex part does not dissociate into its ionic form, into its ions. So there is no ion possible into this. So answer will be no Cl minus ion present. Right, because the complex part is why this is the formula of this compound because we know the coordination number of cobalt. We know this, the coordination number of cobalt is six. The coordination number of cobalt is six. So the possible formula is this only. You see three, all these NH3 and Cl3 are monodentate ligand. NH3 and Cl both are monodentate ligand. Right, so coordination number is nothing but the number of ligand with which it is attached in a given molecule. Right, so three here, three here, six coordination number of cobalt we have, which is also possible. So the molecular formula or the exact formula is this. There is no ion present outside this ionization sphere. And that's why there's no Cl minus ion present in the form of ions. Right, hence the answer will be option D. Is it correct? So I understood this. Question number 14, tell me what is the answer. Okay, 14, most of you are saying C. Okay, if the question is the axis of AGNO3 is added to 100 ml of this smaller solution of dichloro bis ethylene diamine cobalt 3 chloride. How many AGCl forms? So obviously we are having this AGNO3. Right, and then some complex compound also forms and that is not our concern here. Okay, the only thing is what this AG plus and NO3 minus and this AG plus combines with chloride ion from this complex. So first of all, we should know or we should find out that we have to find out that how many Cl minus this complex compound can eliminate. Right, or how many Cl minus can exist in the form of ions of this complex compound. Right. So first of all, what we have to do, we have to write down the structure or the molecular formula of this complex compound. Ethylene diamine is nothing but En. Right, where NH3 is the donor atom we have. Nitrogen is the donor atom we have. Right. So and cobalt is in three oxidants in the state. Right. So it's possible formula is COEN since it is saying bis ethylene diamine. So bis stands for 2 dichloro. Dichloro means Cl2 close bracket and then we have chloride. So chloride is Cl here outside. Right, so you see if you cross check the oxidation instead of cobalt here, chlorine is minus one, minus two is what? Minus two is chlorine here. It is zero. So minus two will go this side, one is on this side, so it will plus three here. So oxidation instead of cobalt is plus three in this compound. So the possible molecular formula of the given complex is this. And it gives obviously only one Cl minus. Right, that's why it's Ag plus and Cl minus combines and gives Ag Cl here. You see one mole of this reacts with one mole of this. Now coming back to the second part of it. This complex has molarity, 0.024 molar and 100 ml. So number of millimoles if you find out that will be this into this. Right, 0.024 into the volume. Right, or if you write this in the form of liter that will be 0.1 liter. So 0.1 into this, if you multiply, this gives you 0.0024 moles. Okay, 0.0024 moles of this complex we have. So one mole of Ag NO3 reacts with one mole. So 0.0024 mole of this reacts with 0.0024 mole of Ag NO3. And also it gives 0.0024 mole of Ag Cl because the ratio is one is to one is to one. Okay, hence the answer will be C. Okay, next one. Question number 15. 15A, what about others? What is the answer of question number 15? Electrical conductance in aqueous solution. Coordination number of platinum is 4. Which on the L cross take question number 16? First let me check question number 15. Yeah, it's 4 Sanjana. Coordination number platinum is 4. Sino complex. The color of the coordination compound depends on the crystal. What will be the color? Order of adsorption of wavelength of light. Yeah, I think something is missing. Let me cross check this. NH3 H2O, CO, H2O, CO, NH3, 6, 3 plus. Okay, let me CO, NH3. Here we should have this 3 plus. This is plus and CO, CN6, 3 minus. And this is nothing but CO, CN6, 3 minus. This is CO, CN6, 3 minus. The first one is this. Let me write down. CO, CN6, 3 minus. What is the answer? Question number 15. See, in question number 15, you just have to check the number of ions in the complex. You see question number 15. We have K2, PT, CL6. You see the oxidation number or the coordination number of platinum here. It is 6. Oxidation number is 4. So in this case, when it dissociates, it forms 2K plus plus PT, CL6, 2 minus. Okay. Total ions produced are 3 ions produced in this. Similarly, if you cross check the second one, PT, NH3, CL. Okay. So this is nothing but the second compound is nothing but this PT, NH3. 2 and then CL4. Oxidation number is plus 4. Third compound will be PT, NH3, NH3 whole thrice. CL, NH3 whole thrice, CL3. CL4, we have CL3. And here it is CL. And the fourth compound will be PT, NH3 whole 5, CL. And outside we have CL4. Okay. We have CL4 we have here. So here it will be CL3. Here we have CL3. So number of ions produced here are 3 CL minus and 1 this. 4. Here number of ions produces 2 and here number of ions produces 0. Okay. So maximum electrical conductance is for the last one. This is the fourth one. The order will be this. Maximum 4. Then we have what? 3 ions. So 1. Then we have 3 and then we have 2. So maximum is for fourth. So for 2 it will be 0 because there is no ion present here. So for the second one it will be 0. For this one it will be 0. The electrical conduct. Because there is no free ion present into this. So for second one it is 0. So this is not possible. This is not possible. Any one of these two can be the answer. You see the fourth one has the maximum value. Right. The maximum value we have. So the answer will be option A. Which one? Okay. Let's wait. Question number 15. The coordination number of platinum. Oh you are asking Sanjana. You are asking coordination number of platinum. It's not coordination number. I thought you are asking oxidation number. The oxidation number of platinum is 4. Its coordination number is 6. Okay. That's why you see the first compound K2PT CL6. In all this compound you see the coordination number of platinum is 6. 4 plus 2 is 6. Here it is 6. Right. K2PT CL6 is 6 only. Here also you see 3 plus 3 is 6. So coordination number of platinum is 6. But oxidation number is 4 here. Right Sanjana. Oxidation number is 4. Coordination number is 6. So this thing you should know. Now the point here is what? The question is we have to find out the electrical conductance. Electrical conductance we know it is because of the movement of ions. Right. More number of ions more will be the electrical conductance. Right. So we have to find out how many ions are possible in these compounds. Right. So first of all you have to write down the exact molecular formula of this. Since the platinum coordination number is 6. So this 2 potassium will be outside and the formula will be K2PT CL6. Okay. So the number of ions produced here it is what? 2 potassium ion from here and 1 is this complex ion. So 2 plus 1 there are 3 ions present in this. Right. Similarly you see question number 2. Option 2. NH3 CL4. So all will be inside the square bracket. CL4 NH3 4 plus 2 6. So there's no ions present into this. So number of ions will be 0. Okay. Option 3rd you see PT CL4 NH3. NH3 you see neutral molecule cannot, can never be outside the square bracket. One thing you must keep in mind. Wherever you have neutral molecule present like NH3 is neutral. There is no charge on NH3 molecule. So all NH3 molecule must be present within the square bracket. Right. So this NH3 should be present in the square bracket. And since 3 NH3 we have so I have written PT NH3 whole thrice. And we know the coordination number is 6 for platinum. So we require 3 more chlorine atom inside the square bracket. So CL3 for that. So that 3 plus 3 becomes 6 coordination number of this platinum. Now out of 4, 3 is inside. So one will be outside, right? One chlorine ion will be outside. Now when this molecule dissociate it gets one CL minus and one this complex. So 1 plus 1 total 2 ions present into this. Similarly for the last one NH3 5 will be within the square bracket. And since the coordination number is 6, one chlorine requires one CL and all other chlorine will be outside. So 3 CL will be outside CL3. So number of ions present here it is what? 3 plus 1 4, right? So more number of ions, more will be the electrical conductance. So for the fourth option the electrical conductance will be maximum because it contains maximum number of ions, right? So and for the second one it is 0 because there is no free ion present into this. So first of all in the option you check where is 0 is the answer for the second option. You see in the option A and B for second one it is 0 it is given. But C and D there is no 0 present for the second one. C and D is not possible. The answer will be any one of these two. A and B option, right? First information is for PTCL4 NH3 second option the electrical conductance is 0. So from this we have eliminated C and D option. Now we are left with A and B. Second information what we have? The electrical conductance of the fourth option is maximum because it contains maximum number of ions. Now you check these two. Here the fourth option has maximum electrical conductance and here the fourth option does not have maximum electrical conductance, right? Hence the answer will be option A. Is it clear now? Question number 16. Question number 16 you have done. See question number 16 depends on the properties of ligand, okay? We know what is the concept we have here. Strong ligand means more splitting. And when the splitting will be more energy required for the transition of electron. And we know energy is nothing but E is equals to Hc by lambda. So more energy means what? Less wavelength or low wavelength. So basically if you see a strong ligand corresponds to low wavelength, right? So when you see these three ligands, ligands we have is Cn minus, right? Then we have what? NH3 and then we have H2O. If you see the splitting of these ligands Cn minus is the strongest ligand we have here, right? The crystal-filled splitting energy, the order is this. It's found to be this. CfSe is this. Maximum for Cn minus and minimum for H2O, right? And hence the wavelength required here will be what? Minimums, order of wavelength will be minimum suppose it is 1, it is 2 and it is 3 we have. Lambda 1 is minimum and lambda 3 is maximum, right? So wavelength in the case of H2O as a ligand is maximum and Cn minus is the minimum. So option C is correct, which all of you have got, I guess. Understood? Number of D electrons in CR H2O 6, 3 plus. 17, how many of you have done? H2O is a weak ligand. 17, option B is correct. All of you have done. Okay, so I'm not going to solve this. Okay, can we move on? Tell me fast. Question number 18, hybridization. Poor weak is getting A. Plus 3 is, it's plus 3 poor weak. Charge is plus 3 everywhere. Plus 3, minus 3, plus 3 and plus 3. All the four options. No, it's plus 3 poor weak. Check once again. Okay, you see. See, we have to find out which of the following complexes does not have D2-SP3 hybridization. Okay, so 2-SP3 hybridization. Means what? It is inertial complex, still complex. Or we also call it as low spin complex. Low spin complex. Right, inertial complex or low spin complex. Low spin complex, why it is low spin complex? Because inertial orbital is involved in hybridization. Right? It's not often involved. Then what we write? Then we write SP3D2. When we write SP3D2, it means it is high spin complex. Right? But when you write D2 first, it is low spin complex. Means inertial orbital involved in hybridization. Now, when this low spin complex possible, when the ligand is weak, if ligand is weak, then only pairing of electron is not possible. Electron is not possible. And then only the lower understood. So low spin complex, we are getting this now. No spin complex means ligand is strong. I'm talking about this. Ligand must be strong. Then pairing of electron is possible. Then only lowest spin complex forms. If I take one example of suppose this one, the first option I'm taking. CR NH3, 6, 3 plus. Right? Chromium oxidation number is plus 3. And chromium, we know it is what? It is 24th atomic number. So we can write AR argon 4S1 3D5. This is what the configuration we have. Now, three electrons we have to remove. So chromium plus 3, if you write, it will be argon 4S0 and 3D3. Right? 4S0, 3D3. So when you see this configuration, when this ligand is strong, then pairing of electron takes place. Okay? And then the bonding will take place and the complex molecule forms. But if this ligand is weak, then pairing in this orbital against the Hund's rule is not possible. Right? And in that case, high spin complex forms, the outer D orbital involves. So the only thing you have to do here it is what? One thing, you just find out the hybridization of one by one and you cross check whether it is D2SP3 or not. But other thing is what? D2SP3, since it is a low spin complex, which is only possible when the ligand is strong. So you just have to check whether the ligand is weak or strong. Now you see NH3 is a strong ligand. So strong is strong. CN is also a strong ligand, because it is not a strong ligand. Right? When the ligand is not strong, so then pairing against the Hund's rule is not possible and hence the low spin complex does not form. Right? So in this case, D2SP3 hybridization is not possible. You can easily eliminate all the options since all these A, B and C contains a strong ligand. Right? Or in this case, you can do this. If all the ligands are strong, then you have to go with the hybridization thing. You have to find out the hybridization one by one. Okay? And then you can do this. So option will be D, correct for this question. Now question number 19. Question number 19, CUSF4 reacts with KCN solution and forms. CUSF4 with KCN. This is just a reaction which you have to memorize. Okay? Sometimes this kind of reaction also they ask in the question. So question number 19, some of you are saying B, but here you see the reaction. 2CUSF4. This you must remember. 2CUSF4 plus KCN. It forms 2K3CU CN4 plus one more complex we get that is CN2 plus 2K2SO4. 2K2SO4. So K3CU CN4 is the complex we get here. Right? And hence the answer will be option C. Okay? This you just remember. Okay. Don't... There's no logic of this. You have to memorize this reaction. Okay? One more thing you can keep in mind this gives a complex compound. So either C and D can be the answer. Since these are the complex compound we have. Okay? These two not possible. See Simon, it's just an experimental thing we have. CUSF4 copper sulfate solution reacts to potassium cyanide. It gives a complex compound where the central metal atom is copper. Right? Complex compound of potassium where the metal is copper. So it gets K3CU CN4. Like I said, it's just a reaction. It is the fact we have. You will not get copper cyanide into this or cyanide into this. One thing you can keep here in mind that potassium and cyanide KCN bond that we have it won't dissociates completely in this reaction. If you get CUCN 2 and CUCN it means there is no potassium and CN bond we have here. But here in this case potassium is bonded with copper sorry, one thing. Potassium is bonded with copper and copper is bonded with CN. Right? So CUCN bond forms but this won't detach from this CN completely. Somehow it is related with this complex. Okay? Like surrounded this potassium ion surrounds this complex part and then it is attached like this K3CU CN4. So basically it's just a fact we have in which the copper sulphate ion reacts with potassium cyanides and gives a complex compound of potassium. Right? Where the copper is the central metal atom. Okay? This you have to memorize. Okay? There's no logic of this. Not exactly. But if you get this kind of question one thing you just keep in mind with it that you will get some complex compound. In this chapter only they ask this kind of questions. So if you are having these complex compound as an option then you should go with that. If you are going with some you know if you are trying to you know if you are trying some if you are going with some hunch and all if you are like if you want to attempt this question then the best way is to attempt the complex compound that you have because these kind of questions they ask only in complex compounds. Okay? So there's not much questions on this. Okay? But yes. And they haven't asked this kind of question before in JEE at least. But yes in BITSAT or in Manipal they may ask. Okay? Not in CET. If I get some other question on this or reaction based question I'll give you. I'll send you. Okay? Question number 20. I hope you all have done this. Question number 20. Just now we have discussed this kind of questions. How many CL minus is outside? Then how many AGCL we get? Then mole ratio you have to take. Tell me. All of you are getting B. So Andrea is getting B. Purvik is getting B. All of you are getting B. I think you got this. This concept you understood, right? Yeah. Yeah, it's B. 20 is B. Since all of you have done so I'm not solving this one. 20 is B. What about 21? Octahedral complex. 21. What is the answer? 21 is B. B and C. What is the hybridization for complex? SP3D2. Both possible? Right. Now when you write D4 with low spin. Low spin wins what? Strongly Gantt. Right? So D4 strongly Gantt we have. So configuration is this. One, two, three, four and five. Maybe we'll get this. This. And one more if you check. D8 high spin. D8 high spin. Now high spin means what? Weekly Gantt. Now when it is weekly Gantt. So what we can write? One, two, three, four. Five, six, seven, eight. And the last one we have D6 high spin. Again high spin means weekly Gantt. High spin weekly Gantt. Again D6 we have. So one, two, three, four, five. So one, two, three, four, five. SP and then again D. One, two, three, four, five. Okay. Now in this you tell me where this D2 SP3 or SP3 D2 is possible. That is what we have to think. D6, D8 both. Is it true? D6, D8 I'm getting. Yeah, right. Because when high spin is there Ligandt is weak. So SP3, D2 may take part in hybridization. Here also same thing. SP3, D2, D2 may take part in configuration. So D8 is also possible. And D6 is also possible here. Right. For the formation of octahedral complex and for octahedral complex we require either this or this configuration. Okay. So B and C both are right into this. What is the answer of 22? Question number 22. This question is based on one relation of splitting in tetrahedral and octahedral complex. Tell me the answer. 22. 22 you are getting C. What is the relation? Delta T is equals to 4 by 9 delta O splitting in octahedral complex. Delta T is equals to 4 by 9 delta O. Delta O is given. The value of CFSE for complex and this is this CFSE for CCF for complex and will be. So why I am using this relation because here you see this is Cucl6. This is octahedral complex tetrahedral. But this one is tetrahedral. So this formula we can apply for the same compound like no, the same ligands and metal we have here. One forms gives you octahedral and other gives you tetrahedral. Then only we can apply this. When the constituents elements are same metals are same ligands are same then we can apply this. Delta O is given and why it is for octahedral because this complex is octahedral complex. So we can use 4 by 9 into delta O is 18,000. So when you solve this you will get 18,000 per centimeter. Answer is option C formula based this one. Question number 23 Question number 23 10 by 23, Sondaria is getting C Simon is getting B. Only CFSE you have to calculate okay. Minus sign we don't consider over there because that represents the splitting okay. So minus sign we don't include in magnitude. Just you find out CFSE octahedral complex right? So delta O or calculation of CFSE we have for octahedral complex is equals to minus 0.4 into number of electron present in T2G orbital plus 0.6 into number of electron present in EG orbital and T2G and EG is the number of electrons right? Now for various different configuration D6, D4, D5 and D7 we can calculate the number of electron present in T2G and EG orbital okay? So for D6 configuration octahedral complex the splitting takes place like this and which orbital will have the higher T2G or EG? Tell me that one first. This is the splitting in octahedral complex. High spin complex it is given I am talking about first option. High spin complex means weak ligand right? EG will have the higher energy. So now in this if you have first option if I am taking D6 so the first three electron T2G orbital one two three four, five and six right? So T2G has four electron EG has two electron so O if you calculate for D6 I am just solving the first one you can do the same way for the other three BCND it is minus 0.4 into four electron present in T2G plus 0.6 into two electron present in EG. So when you solve this you will get minus 0.4 like this only you find out del O for option B del O for C, del O for D okay? And then you solve this you will get the value of del O this minus sign we do not consider here right? It just represents the splitting below this average line okay minus or plus where this side splitting is more or this side splitting is more okay? So minus sign we do not consider we just consider the magnitude of this that we are getting and we calculate the value of del O you just check the maximum value of del O okay? And that comes for option C in this case which is nearly two answer will be option C when you solve the similar way solve these questions in the similar way T2G and EG orbital you always remember low spin means low spin D4 means strong ligand high spin means weak ligand low spin means strong ligand now D4 if you have low spin means the ligand is strong so 1, 2, 3 and 4 right? So for option B there are 4 electrons present in T2G and 0 present in EG like that we will calculate understood this option C is correct this one 28, 29, 30 direction based question pass it solve this one No negative sign we do not consider see we are just when the splitting is there you see this the plus and negative sign that we have here it is because of the average energy that you have here negative means the splitting takes below this line positive means above this line right? So overall when you calculate this this negative sign means what the shifting or the splitting electrons orbital is more below this line than above this line we have ok that's the simple meaning we have ok in calculation of magnitude we don't consider the negative sign we just check the magnitude here understood here it is this one is this complex is also FeCl6 3- 3- this complex FeCl6 3- FeCl6 so the information we know we have already what is what Cl- is a weak ligand and Cn- is a strong ligand ok we have Fe Cl6 3- oxidation state of iron here is plus 3 26 electrons the electronic configuration is argon 4s2 3d6 if we take Fe3 plus its electronic configuration will be argon 4s0 3d5 5 ok so 5 electrons we have in d orbital and since the octahedral complex we have splitting will be like this eG will have the more energy and t2g will have the lesser energy 5 electrons we have so 5 electrons will distribute to what kind of ligand is Cl- ligand means less splitting 2 electrons will be here less splitting means it follows one's rule 1 2 3 4 5 so t2g has 3 electrons but eG has 2 electrons this is the configuration we have option A is low spin complex FCNC also the same thing but here we have what is strong the configuration will be t2g5 it will have any electron present in eG so its configuration is this and the electrons of Fe3 and this are respectively here 5 electrons present in Fe3 and 1 electrons so 5 and 1 on BZ then can we move on now you see these are the few questions which has been asked in J exam now you solve this question 49 48 49 50 50 with this 6 question 6 question and I will give you 8 minutes so its 5 12 now 5 20 you can answer me all of you try this one by one 6 question all these questions has been asked in J exam is it visible now is it visible are you getting my voice ok I will do one thing I will send you these questions on whatsapp you solve and just send me the answers ok lets not ok I will send you the question on whatsapp and then after half an hour or 45 minutes I will send you the answer key and solution both you can cross check and you can ask your doubt if you have any fine ok and next class I will be there I will be available so that we can have physical classes there right