 Hello guys, we are here to discuss the J main paper which has been asked in this few days back. So this paper which I chose today, it is of 9th of January 2nd shift. Before starting the discussion here, let me tell you first thing. These questions are memory based questions, so you may see some difference from the original paper. But more or less the question paper will be same. So here in this session we are going to discuss the first 10 questions of chemistry. The first question that has been asked is which is the following form nitride. So you see the question here is the formation of nitride. We know the general form because all these elements belong to group 1. So the elements of group 1 we have here. And the general formula of nitride if I write, it will be M3N type. Suppose if lithium it is there, the lithium will form Li3N, K3N. Like this we can have the formula of nitride. Now the question is which will form nitride in all this. You see as we go, if you talk about lithium, potassium, rubidium and cesium, they are going actually down the group. So as we go down the group, we know size increases. The size of all these elements increases as we go down the group. So lithium has the smallest size. And hence the charge density on these ions. If you find the charge density on lithium here, that will be maximum. So what we can write, the charge density of lithium will be maximum. And that is why the bond between lithium and nitrogen is the strongest over here. So what happens here when this lithium and nitride form, so this reaction at room temperature of lithium and nitrogen, this evolves a huge amount of heat and this is an exothermic reaction. Because of high lattice energy of this molecule, high lattice energy. So what we can say because of the high lattice energy, the nitride of lithium is stable but that is not found to be the same case with all the elements as we go down the group. So all these elements will not form stable nitride except lithium. The answer will be option 1. Now the next question is in which of the following bond order increases and magnetic behavior changes from paramagnetic to diamagnetic. So first of all, this question is based on molecular orbital theory. The concept that involves in this is molecular orbital. So according to that we know there are bonding and anti-bonding molecular orbital. Bonding molecular orbital and anti-bonding molecular orbital. And the bond order formula we have is equals to the number of electrons in bonding molecular orbital minus the number of electrons in anti-bonding molecular orbital. This is divided by 2. This is the formula we have. So we can draw the electronic configuration here according to these orbital which is sigma 1s, sigma star 1s, sigma 2s, sigma star 2s and so on. Now we can fill the electron into this and we can find out the number of electrons present in bonding molecular orbital and anti-bonding molecular orbital. So you see the question now coming back to this question, these are all theory or concepts required to solve this question. Now if you are coming back to this question, O2 molecule we know it is a known fact that oxygen is a paramagnetic molecule. So first option that you have is O2 to O2 plus. Now paramagnetic is because of what? Unpaired electron. So this is paramagnetic we have and in this the number of unpaired electron is 2. This is also paramagnetic. Number of unpaired electron is 2 here also. Similarly we can find out for NO and NO plus. We can fill the electron into these orbital and we can find out this bond order actually. So for paramagnetic behavior we must have unpaired electron and diamagnetic all the electrons must be paired. So the correct answer in this, if you find out the bond order of NO, it is found to be 2.5 or here also we have 2.5. Number of unpaired electron or bond order is 2.5, here the bond order is 2. It is not the number of unpaired electron. And NO converts into what? NO plus. So we are removing one electron into this and hence the bond order here you will get is 3. So this is paramagnetic and this is diamagnetic. So this is what the answer we have. NO converts into NO plus. The magnetic behavior changes from para to dia and bond order is also increasing here. Correct? Option 2 is right. Question number 3. M3 plus ions show blue, green, red color with ligands L1, L2, L3 respectively. Arrange the ligand in their strength. So in this the concept that is required here that we know the strength of any ligand, strength of ligand is directly proportional to the absorption energy or delta of absorption. Now you see here the question is the color is blue, green and red. So the wavelength of emission because it shows blue, green and red color, the wavelength that emitted in this process is we can write lambda for blue, then lambda for green and then lambda for red. Correct? So like this the wavelength increases. Red color will have highest wavelength. So this is the wavelength emitted. So the wavelength emitted is this. So this wavelength has to observe to emit this. So del of absorption if you write down, del of absorption it will be again minimum for the blue light. Then we have green and then we have red. Emission if you write del of emission that will be reverse of this, which is nothing but we have del V maximum, del R sorry del G here, del G and not lambda G and lambda R. This is the emission we have. Strength of wavelength, strength of ligand is that proportional to the absorption right here. So for B it is maximum then green and then for B it is minimum and then G and then red. So maximum is for red and for red it is L3. L3 is maximum we have only one of two options. These two we can eliminate after L3 we have L2. So L3, L2 and L1 option D is correct here. So strength of any ligand that you have is that proportional to the absorption value that you have. And more wavelength, more will be the absorption. So that's the concept we need to have to solve this question. Question number three, answer is option C. Which of the following transition elements will have enthalpy of atomization? Okay, will have enthalpy of atomization. One thing is missing here will have least enthalpy of atomization. The question is will have least enthalpy of atomization. Okay, this is actually a factual question we have right. And the elements are vanadium, first of all we have vanadium in 3D series. Then we have iron, copper and zinc. So first of all if you see the energy here or enthalpy required for atomization. For vanadium it is 515 kilojoule per mole. These data I am giving you iron it is 416 kilojoule per mole. For copper it is 339 and for zinc it is 126 kilojoule per mole. Okay, so clearly if you see the answer for this question will be option three. Enthalpy of atomization is minimum for zinc according to this data. Another thing is what this actually this facts depends on the size which is you know decreases as we go from left to right to a certain pair to a certain element. And then it keeps on increasing also. First it decreases correct and then it becomes constant in between and for last two elements it increases. Right, so zinc is the last element of 3D series we have. Correct, that is why from this the electron the removal of electron is easier and hence its enthalpy of atomization is minimum. Answer is option C. Unless you see by which iron arsenic sulphide is maximum coagulated. Right, this you see arsenic sulphide is maximum coagulated. Right, now the soul that you have here is arsenic sulphide AS2S3 it is the soul we have. Now if you know the nature of the soul what charge we have here we can easily do this question. Right, arsenic sulphide is a negative soul actually negative soul. Okay, so if it is a negative soul so it requires positive charge for coagulation for coagulation. Right, and we know more positive charge more will be the coagulation. Right, so coagulating power coagulating power is directly proportional to the magnitude of the magnitude of positive charge. Why positive charge because this is a negative soul. Okay, this is a negative soul. Correct, so you see here AG plus CL minus NA plus CL minus AL3 plus 3CL minus and this is 3NA plus PO43 minus. So hence we can easily say that this aluminium has maximum positive charge here which is AL3 plus. Right, this is AL3 plus and that's why the answer will be this because they are asking maximum coagulation. Right, if they ask you minimum coagulation then we can have any answer possible between 1, 2 and 4. So answer is option C. The only thing you should know to solve this question is the charge of this soul which is a negative soul. Then you can do this question. Next question, S2PO2 is a strong reducing agent. S2PO2 is a strong reducing agent. Okay, again I will do some correction into this. The question is S3PO4, S3PO2 sorry. S3PO2 is a strong reducing agent reason for this what we have. So for this question you should know the structure of S3PO2. Okay, so the structure will draw is P double bond O. We know in all these oxy acids of phosphorus one P double bond O is always present. Right, all other oxygen is attached with hydrogen if it is there. Right, so we have one oxygen only so one OH and other two hydrogen will be attached directly to this phosphorus. Right, now reducing agent is what? Reducing agent is because of, see if you take this OH bond here. Right, so this molecule has basicity, this molecule has basicity one. Why because we have only one OH bond? That's why basicity is one. Right, so oxidizing nature of this compound is because of this OH bond. Right, so reducing nature will be what? Because of this pH bond it will show reducing nature. Right, so this will oxidize other molecule because of this hydrogen present, but this will reduce others because of this hydrogen that you have here directly attached. Correct, so we have two pH bond and that's why it shows the reducing agent. So hence the answer will be option two. The only thing you require in this question is the structure of H3 CO2. Which is nothing but this. Next, copper crystallize in FCC lattice with unit cell edge length is given, calculate density. Right, so what is the formula of density? First of all I will write down the formula then we will see what is the data given here. So density D is equals to ZM divided by Avogadro number NA edge length Q. Right, where Z is the number of effective atoms, number of effective atoms. M is the atomic mass, if atom is there atomic mass. Do you see what all things are given here? Edge length is given X, FCC lattice we have. So FCC lattice we know we have eight corners and six phase centers. The value of Z from this if you calculate it is 8 into 1 by 8 plus 6 into 1 by 2, which is nothing but 4. Now this value we have to substitute here and we will get the answer. So density will be 4 into 63.5 6 into 10 to the power 23 into X angstrom it is given, X into 10 to the power minus 8 whole cube, right centimeter cube. So this will be equals to when you solve this you will get 421.71 by X cube gram per centimeter cube, okay. Option A is correct, right, direct formula we have to use here. Next question, okay, this reaction is given if at 300 Kelvin temperature E naught of the cell is also given 2 volt and find out the equilibrium constant, okay. So we know the relation of del G, del G is equals to del G naught plus RT ln cube, okay. Now it is given E naught of the cell is given and we know also know del G is equals to minus nfe this is the formula we have. If it is del G naught, then it is E naught is standard state, correct. And since we are talking about the equilibrium constant, so at equilibrium what we can write? Del G is equals to 0, right. At equilibrium del G is equals to 0 and Q is nothing but Kc equilibrium constant. These two formula if you apply here, so we will get del G naught is equals to minus RT ln Kc and del G naught is minus nfe naught is equals to minus RT ln Kc. The value of n is 2, right, one faraday it is given, right, this we have to assume. So n value is 2, right, 2 into f is 9600, E naught is also given 2, R is the value is given 8, temperature is 300 ln K, right. So when you solve this, you will get K is equals to e to the power 1, 6, 0. No theory, only calculation we have here. The formula is this and when we use this formula, the answer will be option 1, e to the power 1, 6, 0, right. Next question, the reaction is given 2A plus B gives C, if the initial rate of the reaction is 0.3 molar per second. On doubling the concentration of A and B, the rate becomes 2.4 molar per second. Doubling the concentration of A alone, it is 0.6 in order of the reaction. Okay, suppose the rate law we can write for this reaction is R is equals to K A to the power X and B to the power Y. This is the rate law I am taking. If you know this X and Y value, the order of this reaction will be what? X plus Y, we will add these two. So we have to find out this X and Y, correct. Now see, initial rate is this, correct. So first of all, when rate is 0.3 is equals to K, concentration of A to the power X, concentration of B to the power Y. When concentration of both increases twice, right, so the rate becomes 2.4. 2.4 is equals to K 2A to the power X and 2B to the power Y. When we double the concentration of A, rate becomes 0.6 is equals to K 2A concentration to the power X, B to the power Y. Okay, you see first and third equation if you compare, right. So rate becomes twice when we increase the concentration by 100%, right. Or when the concentration of A becomes twice, rate also becomes twice. You can easily solve these two equations and you will get the value of X as 1, right. Similarly, you can use these two equations and you will get the value of Y also that is 2. So basically, you need to solve this first, second and third equation. You will get X and Y value, the order will be with respect to A it is 1, with respect to B it is 2. The overall order, overall order is 3. What we can write, overall order X plus Y is equals to 3 with respect to B it is 2 with respect to A it is 1, right. It is 2, first one is wrong, second one is right. This is wrong, this is wrong. Option 2 is incorrect. And the mass of H2O formed, the combustion of this gamma, this takes place, right. So you see the molecule here, suppose if I take CX, write down the same thing. We will take the given molecule only. You see the molecule is C57H11046 and combustion of this. So we have O2, gives CO2. And if car number of carbon is 57, we will multiply here 57, carbon gets balanced. Hydrogen is 110, to balance hydrogen it should be 55 H2O. Now oxygen is 55, 114 plus 55, that is 169 approximately. You will have 163 by 263 plus 669. Number of moles of this molecule, moles of this will be equals to the mass taken divided by the molecular mass of this. You can calculate the molecular mass of this, 12 into 57, 1 into 110 plus 6 into 16. And when you solve this, you will get 445 divided by 890, which is 1 by 1 moles gives 55 moles of this. So half moles gives number of moles of H2O. So 1 gives 55, so half gives half of 55, that is 27.5, but you have to find out the mass of this. So mass of H2O will be half into 55 into 18, that will be 495 gram of H2O. Answer is option 2. These are the first 10 questions we have discussed. We will come back to another question very soon. Thank you.