Transcript for:
Chemistry Essentials for VITEEE

hi everyone welcome to Vidantu J English I'm Dsha Koshel your chemistry master teacher So everybody let me know in the comment section in chat box if I'm clearly visible and audible to you so that we can start with a very very important session for all the students who are preparing for VITLE E right So uh here what I'm going to discuss today is the important concepts right along with that I'll discuss the questions based on that Okay So if you all are ready for that then let me know in the chat box Yes students if you all are ready for that then let me know in the chat box so that we can start with this session without wasting any time Yes Okay Am I clearly visible and audible to you is the screen clearly visible to you because after when I start teaching the concepts I might not be able to see the chat box continuously So if you have any problem with the voice or the video you can let me know in the chat box Right yes ma'am Uh for this session most welcome Most welcome students We are here to help you only Right okay So let us start this Firstly we are going to start with physical chemistry So what I told you I am not going to revise everything Let's be clear right everything I can't do in one or two hours right that's not possible right for that you obviously know one day two day that you know right but here we are going to do the most most important and most repeated concepts right so let us start with the first concept that is bohar model right so most repeated thing about bohar model is frequently they ask you question regarding radius velocity uh then energy questions they ask right now let us see how what kind of questions they can ask you based on this right so first is if you say that from radius right n everywhere n will define the orbit number and zed will define the uh atomic number right for hydrogen it is one for helium positive it is two for lithium 2 positive it is three right and n orbit number will be given to In the question only let's say if they give first orbit that means n is equals to 1 Now you tell me in the chart box if they give you first excited state first excited state then what is the value of n here what is the value of n for first excited state if they give you simple first state second state third state n is same If they give you first excited state second excited state that means you have to add one to this number So this is first So it will 1 + 1 is equals to 2 Second excited state is going to be 2 + 1 is equals to 3 So you will add one to that number That is going to be your answer for orbit Okay simple Now here if you see directly they can ask you question based on this is the orbit This is the uh atom you have to give the answer right of the radius from this formula that is radius is install I would prefer 0.5 to n² by zstro now they can also ask you question regarding the ratio two atoms will be given to you two orbits will be given to you they will ask you the ratio of this and this r1 by r2 right that they can ask you for that you don't need to remember this 52.9 or anything just remember that radius is directly proportional to n² by z directly only this thing right Similarly the question can be asked on velocity The formula of velocity is this and here the variability is velocity is directly proportional to z by n For time period also they can ask you the same question related to the ratios and this is the variability Time is directly proportional to n cq by z² For energy they can ask you question for first orbit what is the energy second orbit the difference in the energy delta all these things and the conversion part obviously you know what is what is conversion of electron volt to jewles but rather than you should also know the direct formula that is -3.6² 6 Z² by N² and - 2.18 into 10^ 18 Z² by N² J right now if you say your total energy is coming out to be let's say - XX2 then kinetic energy will be I'll just write it in a different form right okay if let's say this is X if let's say this is X then this will be minus of X this will be minus of X Okay this will be minus of X right okay and this will be minus of Wait a minute If I if I say this is X right then this is also going to be X This is going to be minus of minus of X right and this is going to be 2X Okay that is how also you can remember this energy conversions Okay Energy conversions of kinetic to potential potential to total energy the conversion of energies here right okay Now next thing we have is hydrogen atom Now a very important question based on riddberg equation Okay So firstly what is redber equation this is your redber equation right that is 1 by lambda equals to v bar is wave number equals to rh is the redber constant Z square is the atomic number right okay then you have 1 by N1 square minus 1 by N_sub_2 squ that is done right now here what kind of questions they can ask you first is you should know what is the value of N_sub_1 and N_sub_2 sometimes they will directly give you this is the value of N1 and N2 but I don't expect that in VITE E right rather they will give you something like N1 is this it belongs to Lyman series Balmer series so for that you need to know what are exactly these series So Lyman series Balmer series past bracket P1 that means N value is 1 2 3 4 5 Sometimes they can also give you that they will not give you that this belongs to LM series Rather they will say this belongs to UV region that means this is LM series They will say this belongs to visible region that means this belongs to Balmer series And infrared they will not give you actually they will rather give you directly past in bracket or preferred they will not give you this belongs to infrared because to infrared you have three things that are belonging so we can't justify that statement right okay so now that will be the value of n_sub_1 now second is ma'am what is going to be the value of n_sub_2 right for the redbug equation now for n_sub_2 what we are going to do here for n_sub_2 we should know these things here that is if you have let's say uh okay for this firstly understand this energy level diagram okay so when you put the value of n_sub_1 n2 n3 n4 in the energy formula here you will get the values as -3.6 then -3.4 4 - 1.51 minus.85 Now what you're observing here that as the value of n is increasing energy is also increasing here Right as the n value increases energy also increases But here the value of delta e that decreases That means the gap in the energies that will decrease Right here the gap is very large It is almost 10 units difference Then the difference is three units Then it is something 1 point something then it is 0 point something that means it is decreasing as n is increasing right now next is ma'am through energy level diagram see simple thing is for n_sub_2 either they will directly give you then this is the value of n_sub_2 this is coming from lman series 2 n= 4 directly you can put the value of n_sub_2 but what if they give you uh this is a first line this is alpha line this is beta line this is second line this is limiting line this is Last line Now what you have to do right so first is alpha line Okay Firstly let us understand last line Last line will be where it is n is infinity Right so last line or limiting line that where you have n_sub_2 is equals to infinity Right here why I am saying that it has maximum energy maximum frequency because it is the last line Okay so the energy difference is very large So it is maximum energy and we know that E is equals to H mu that means mu is also maximum frequency is also maximum We also know that E is equals to HC by lambda So lambda is inversely proportional that is why minimum wavelength is there Right next we have is if you have first line or alpha line first line means it is next to the line from which it is falling So if your n1 is let's say let's say this is n_sub_1 So n_sub_2 is going to be n_sub_1 + 1 right that is the first line or alpha line Beta line or second line is you have to add two here Gamma line or third line is you have to add three here Okay you have to add three here That is going to be a gamma line Now next is from energies how you will tell See we know that in this case first line means the if this is my nsub_1 this is n_sub_2 that means the energy gap is very less So it has minimum energy minimum frequency and wavelength is opposite So it has maximum wavelength From that they can complicate this simple equation of Rickburg equation Right by knowing this formula also you'll like you may get confused in the exam Right next is ma'am if they ask you wave number if they ask you wave number you can directly put the value of RH because in that way you don't have to take the reciprocal of RH but if they ask you the value of lambda right then you will have to solve 1 by 109 677 so rather than solving that what you do is put the value of 1 by RH as 912 Armstrong in that way you will get the value of lambda in Armstrong Okay Right Then you can easily convert it into centimeter meter nanometer whatever is given to you right okay So that is the inver energy conversions Next is a question based on number of spectral lines So let's say they have given you a line is falling from let's say an electron is falling from n= 6 to n= 2 So it will be 6 - 2 6 - 2 + 1 divided by 2 that is going to be the number of lines that are falling in between six and two right in between 6 and two the number of spectral lines that you have Now these are the total number of spectral lines What if they ask you specific spectral line that means they are saying you are coming from n= to 6 to n= to 2 right but you have to tell that how many lines are falling to n= to 3 Okay this is n= to 3 Now they have specified they are not asking you all the lines They have specified some lines So when they have specify drawing these lines all of this is not important for you You can't do that in the exam You will get confused and maybe after doing everything also you will end up losing marks You will end up marking the wrong option because you're in the panic state So directly do it from the formula right number of specific lines for specific orbit is n of height it is coming from n is equals to 6 it will be 6 minus n of specific is three so it will be three lines okay that is going to be three lines If that is clear to all of you then let me know in the chat box If that is clear to all of you then let me know in the chat box that yes ma'am this page is clear to us right Okay Now this also we have done right droi hypothesis right Now from here what you have to remember first is you should remember lambda is equals to h by mv right all these formulas Then you should remember that drogly hypothesis also prefers that also says that 2 pi r is equals to n lambda What is that 2 pi r because drogly hypothesis means dual nature So this is from particle this is from V nature and that's how you will get the boher angular momentum as well that is MBR is equals to NH by 2 pi So this formula this formula both are important for you right both are important for you Next is hence heenberg principle So in Hezenberg principle directly the uncertainty in position and momentum is greater than hx 4 pi and similarly you can put the value of momentum here Right Okay Now is the quantum numbers How they will ask you questions from the quantum numbers Right okay Now in this case first thing is quantum numbers You should know the principal quantum number that is n that is from 1 to infinity Right asal quantum number that is from L= 0 to N minus one Magnatic quantum number is from minus L2 + L Minus L2 + L Okay Let's say if my L value is 2 So this will be min -2 to +2 Okay So next is spin quantum number that will always be either plus half or minus half Now here what you have to remember is if you have zero L then it means it is S L is 1 then it is P L is 2 then it is D L is three then it is F Till F you should know right Next is what they represent from this theoretical question can be asked Principal quantum number that represents the size orbit right okay and as it increases right size and energy will all also increase As a mutal quantum number it represents the shape Shape of S P D F all are different right s has spherical shape P has dumbbell shape D has double dumbbell shape F has complicated shape Right magnetic quantum number that will tell you the orientation in the space Right like you have px py and pz right for p orbital you have for p subshell you have three orbitals px p y and pz So all these three orbitals are telling you the orientation in the space Right spin quantum number will tell you the rotation of that electron whether it's clockwise or it is anticlockwise Right now one more question they can ask you based on the formulas directly from here that is number of subshells If they ask you they have given you principle quantum number They ask you number of subshells that is simply n Number of orbitals when you have to represent this in the form of n that is n² number of orbitals in the form of l that is 2 l + 1 and in one orbital you have two electrons so number of electrons in that orbit will be double of this will be double of this right that is why we'll just you don't need to remember this formula or this formula that will be multiplied by two right because in orbit in one orbital you have just one uh you have just two electrons right next is the angular orbital momentum formula is under root l + 1 hx2 pi angular spin momentum is under root f s + 1 hx 2 pi directly these questions they can ask you in your exam last concept from this chapter which is very important for you that is n plus l rule okay so from here what kind of questions they ask you first kind of question is that they can ask you that uh Um one thing they can ask you that they will give you some uh different different orbitals and they will subshells they can give you different different subshells they will ask you that which of the following is high in energy So what you have to see that pursuable match if n + l right n + l increases energy also increases but if n + l is same then energy will also be no if n + l is same then as n increases energy will be increased right So for example you have 4s and 3d out of this you have 4 + 0 and this is 3 + 2 So here it is 5 here it is four So that means this has more energy as compared to 4s So we always fill the electrons in increasing order So you'll fill the electron in 4s first and after that you will fill the electron in 3d Right okay And if for example let's say your value of let's say you have 4 p and 3d right now here it is 4 + 1 that is 5 This is also 3 + 2 that is five If your value of n plus l is same in that case the value of n will define the more energy So this has more value of n So 4p is greater in energy That is why when you write the energy level diagram you write 4s then you write 3d uh uh then you write 3d then you write a minute 4s 3D is greater right uh yes then you write 4 here Okay this is how it is written Okay understood everybody now for hydrogen atom if n value is same if n value is same then value of energy is also same Energy is also same in that case there is no difference in the energy Okay 2sp 2p has same energy 3s 3p 3d has same energy in case of hydrogen atom Okay Is that clear to everybody yes students Now let us say the shortest wavelength of hydrogen atom in the Lyman series is lambda 1 The longest wavelength in the Balmer series is H E positive Now they are asking you what is the relation here Right so here they have asked you everything Okay So now here shortest wavelength means lambda is very small That means energy is maximum Energy is maximum when N_sub_2 you can say N_sub_2 is equals to infinity Lyman series means n_sub_1 is equals to 1 Right balmer series means n_sub_1 dash will be equals to 2 And longest wavelength in balmer series means longest wavelength means the uh wavelength is maximum that means energy is minimum that means the orbit the next orbit number it will be your n_sub_2 So if this is n_sub_1 dash is 2 then n_sub_2 dash is going to be n_sub_2 dash is going to be three here Right so let us uh just put the values in your redber equation and you will get your answer here Right so this is your redbug equation You just have to put the values in the redbug equation Right what is redbug equation wait a minute 1 by lambda is equ= to r hz² This is 1x n1²us 1x n_sub_2² Okay Now what do you see here is this is for hydrogen atom For hydrogen atom we know that zed value is equals to 1 For helium atom the value of zed will be equals to 2 So that is why here we have put 2 square Here we have put 1 square Here we know that n_sub_1 is coming out to be 1 n_sub_2 is coming out to be infinity here So when I put 1 by infinity that will come out to be zero Here it is 1x2 square 1x 3 square according to the n1 n2 values and you solve these two Now they are asking you the value of the longest radian in the palm series of this that they are representing by lambda is equals to how much amount of lambda 1 This is what you have to calculate that this lambda is how much times of this lambda that they have represented by lambda 1 So now you just need to take the ratio of both of them and solve it that directly you can answer Okay understood everybody yes Yes students let me know beta please focus here on this uh uh slides right rather than asking me questions regarding ma'am will I be able to do this or that you can do anything yes definitely if you give devote next 3 4 days totally into your preparation then definitely that will benefit you there is no harm in doing that right okay but I don't know you personally what is your level your basics are clear or not so I can't just generally say that yes you can do that No it depends that if your basics are really strong yes you can definitely do that If your basics are not clear then I then I have no comment on that right so I'll be very clear But if your basics are good you have done your class 12th board also in a good manner then definitely you can do that right okay then definitely uh these things are going to help you right and this session specially is definitely going to help you even if your basics are weak right because I'm focusing only on those topics where your basics might be weak but I'm focusing on those topics which are important so that I can at least lift you in those topics right now here is the quantum number of four electrons are given below the correct order of their increasing energies will be so two questions they have asked in this one right so what you have to firstly see is what is the orbit number what is this what is this orbital they are talking about subshell this is going to be 4 SPD 0 1 2 3 SP DF Okay So this is they are talking about 4D here This is 4 D This is 4 P Here they are talking about 3 uh LS 2D Right Here it talking about 3 P Okay So now here you have to solve these four things that is 4 D 4 P or you can say 3D then you have 4 P then you have 3D P right now calculate the value of N plus L here right so this is 4 + SP DD I'll just write it here only right SP DF so it is 2 that is 6 3 + 2 that is 5 this is 4 + 1 that is five This is 3 + 1 that is 2 So definitely they are going to have higher energies Right out of this which will have high energy 4 P will have high energy So that is why here what you can write 4 D then you will have 4 P then you will have 3D then you will have 3T That is going to be the order of energy here Now you have to put it over here Right okay So 4 D that was belonging to one that is going to have the maximum energy here right then 4 P that is the third that is going to have less energy as compared to that so order will be fourth is your 3P yes so answer will be C is the correct answer okay that is how you're going to answer these question is that clear to everybody or not right so two major important topics one is Bohar model redbug equation very very important Next is quantum numbers You can't afford to skip these topics from structure of atom Definitely you will see one or two questions from these two topics Very very important right okay Next is thermodynamics Okay thermodynamics You must have done this in physics also right uh so let me know in the chat box if sir has taught you in physics thermodynamics today If he he has taught you if he has taught you uh in physics also he has taught you all the things in detail for thermodynamics I have skipped this in chemistry right so that we can focus on other parts right everybody let me know in the chat box now uh by the time you tell me I'll start with some very basic things here right one thing for numerical that you should remember is at constant external pressure at constant external pressure what do What do you mean by that constant external pressure simply means that here we are talking about irreversible process Whenever you see variable external pressure in that case we are talking about reversible processes Right next thing is first law of thermodynamics in chemistry it is delta U is equals to Q + W right and W is minus V delta V One more formula to remember is QP is equals to delta H and QV is equals to delta U Very very important formula Now always remember the sign conventions in chemistry If you're using this formula then work done by the system by if you have to do work done you are negative If someone is doing work for you you are feeling very happy you are positive If heat energy is released right you have to give money to someone you are negative If someone is giving you the money you are happy you are positive right okay Then four processes here Isothermal where delta t is zero That's why delta u is also zero Isobaric delta p is zero Isoporic delta v is zero Adabatic where heat energy change is equals to zero Next we have is isothermal irreversible these are the direct formulas that you should know that in isothermal irreversible you will apply this formula which does not involve any integration But for isothermal reversible you will apply this formula which involves integration After integration we got this formula here It is log V2 by V1 This is log P1 by P2 Okay Whatever whether they have given you pressure or volume according to that you will apply any of these formulas for isothermal reversible and irreversible processes Right now before coming to adiabatic process firstly I would like to teach you a simple thing regarding heat capacity So heat capacity questions are also very important directly formula you should know heat capacity is Q byt If you talk about heat capacity at constant pressure you will put P and P here I have told you QP is nothing that is delta H Okay And QV is nothing but that is deltaU Right now here CP and CV are for N number of moles But if I calculate the molar heat capacity I'll divide this by N here Okay and it will be CPM This will be C VM M means molar heat capacity So that's why I divided it by N And from that you get this very very important formula that is delta H is equals to N CPM delta T This is very important for your VIT triple E We have seen direct question from these two formulas direct questions Okay So deltaU equals to n CVM deltat T right heat capacity is a very important topic for you right deltau is equals to N CVM deltaT Where this N is number of moles CVM is molar heat capacity at constant volume CPM is molar heat capacity at constant pressure Okay Now specific heat capacity is not at all important for you Right for specific heat capacity rather than dividing it by number of nodes you'll divide it by mass that is specific heat capacity right now next is you should remember is what is gamma that is poison's ratio that is CP by CV or you can also write it like this CPM by CVM where CVM is FR by 2 CVM is R + CV M right okay yes Okay what do you mean by that simple is when you're calculating the value of gamma you need CPM and CBM CBM you will solve from degree of freedom That means what if they give you they give you monotomic gas then degree of freedom is three Diatomic or linear polyatomic gases like CO2 you will have degree of freedom as five If they give you nonlinear polyatomic you will have SO2 like SO2 degree of freedom will be six According to that you will have the value of CBM when you put it here and when you add R into it you will have the value of CPM You divide it you will get your value of gamma that I have written already here These values are extremely important for you Right then this formula is very important Usually what students learn is CP minus CV is equals to R No CPM minus CVM is equals to R CP minus CV is equals to NR right in NCERTT it is it is written like this CP minus CV is equals to R but there they have written in the bracket that this is for one mole yes for one mole CV CP minus CV is equals to R but why can't we write for one mole CPM minus CVM right because we usually skip the things that are written in the bracket so rather learn this CPM minus CVM is equals to R right Okay Now I'll just go through the formulas again for adiabatic directly W is equals to N CVM T2 minus C T1 C CVM you can easily calculate from the gas that will be given to you right okay uh number of moles will be given delta T will be given they can also ask you two other things that is temperature right if they ask you irreversible adabetic process temperature you can solve from this equation where all the other terms will be given to you except for T2 Okay Or if they give you reversible adabatic process directly you can solve this this or this formula will be applied depending upon what they have given you whether pressure or volume or pressure or temperature or temperature or volume according to that you will use any of this formula but you should know the value of gamma here Okay Why you should know the value of gamma because gamma will be given to you whether it's diatomic monotomic according to that you will get to know which is the value of gamma here right okay then the next part is last part of this chapter that is this equation very very important for you that is if you want to be risk-f free you are thinking ma'am work done I find very difficult ma'am tell me what are the most important concepts here they say everything is very important they ask you very simple questions from thermodynamics Right but still if you want to see that uh the these two equations this equation and this equation are very simple yet very important Right okay So here directly all the values are given to you in the question You just have to put this this formula application is very simple Right So here if pressure and volume is given to you either this or this formula will be applied If they have given you any reaction if they have given you any reaction in that case you will apply this formula right where delta ng R value you already know that is 8.314 temperature will be given to you in the question either they will ask you delta H or they will ask you deltaU right now one twist they can do is instead of delta H they can write QP instead of deltau they can write QV they will ask you what is the heat energy G at constant volume that means they're asking the value of delta U here right then enthalpy positive means endo enthalpy negative means exo that we all know right not a big deal now let's come to gives hal mod equation very very important now delta g is equals to delta h minus c delta where this is entropy enthalpy this is gibs free energy now they can ask you like for a spontaneous reaction what is the temperature delta h Delta S will be given to you Right for a spontaneous reaction delta G should be less than zero According to that you can solve it or they can give you all the values They can ask you delta G They can ask you delta H Anything they can ask you from this gives Halmmont equation Right okay Now let us see some question based on this Right everybody five mole of an ideal gas at 100 Kelvin Everyone let us see this question Yes students VHA triple E Yes Now 5 moles of ideal gas at 100 Kelvin are allowed to undergo reversible compression till its uh temperature Wait a minute I forgot to tell you one important part that is these things Okay Iso isobaric process pressure is constant So directly this formula will be applied That is a general formula Isocoric process volume is constant So that is why work done will be zero because delta u delta v is going to be zero For cyclic process it is area under the curve But for clockwise it is work done is negative Anticlockwise work done is going to be positive Open vessel directly you will say open vessel work done is equals to minus delta ng RT directly you will solve this with this formula because PV is equals to NRT that's why instead of PV we have put delta NGRT in this case one reaction will be given to you in the exam okay free expansion all the terms work done deltaU q all the terms are going to be zero in this case okay yes students now let us See this question here Five moles of ideal gas at 100 kelvin allowed to undergo reversible compression till its temperature become this Okay Calculate deltaU delta PV for this process Two things they are asking you to calculate here Right okay Temperature will change definitely here Right Uh so five moles of ideal gas This they are asking auto reversible compression Right Okay Now CV is given to you here right cv is already given to you So from the formulas directly you will solve this right from all the formulas that we have now done Now I taught you that deltaU is equals to N CV delta T You will say ma'am you told NCVM ma'am here they have written NCV usually see what they have written if CV is equals to this per mole per mole means they are talking about CVM Okay these are the little little things that you need to see in the exam Right so it is CVM n CVM delta T delta T is given to you Number of mole is given to you CV is also given to you Right so you just have to put these values and get your answer here Right because the answer of deltaU is given in kiloj here mostly So you'll have to convert this into kilogjles and you will get 40 Right now we also know that PV is equals to NRT right they are asking you the value of delta PV right they're asking you the value of delta PV so delta put delta here this is PV number of moles is not changing this will be n r is not changing but temperature is changing so this will be n r deltat t right so it will be n r put the value of n put the value of r here had directly put it as 8 right okay this is basically 8.314 right and T2 T1 you can put the values here and again you will get your answer directly here okay so direct formula based questions right if you see it's not a difficult question okay it's not a these are the kind of questions that are focused in viol right so it's not like they will ask you very tough questions right they will ask you simple questions from thermodynamics you just need to know the formulas right if they are on your tips definitely you will be able to survive your thermodynamics a gas is allowed to expand in a wellinssulated container now insulated container means Q is equals to zero this is adabatic process right against the constant external pressure now you tell me constant external pressure whether it's reversible or it's irreversible yes students is it reversible or it's irreversible you tell me in the chat box of pressure is this pressure is given to you from initial volume V_sub_1 is also given to you right v_sub_2 is also given to you the change in the internal energy of the gas see first thing is delta u is equals to w + q if q is zero then internal energy is equals to work done now ma'am you have taught us that work done is equals to n cbm delta t but there is no mention see if in the question rather They given giving you pressure and volume They would have given you CBM They would have given you number of moles I would say calculate it this way only Calculate it this way only Apply this formula Calculate the value of delta U directly Right and that will be your work done But here they have not given you CDM or N Right so what you will do this formula work done is equals to minus P delta V Okay so pressure is given to you Right and volume you already know So directly you can put the val formula here Minus P Pressure is 2.5 right and this is delta V is this is 4.5 minus 2.5 You can put these values and you will get your answer But understand the answer is in jewles So you have to this will be in liter atmospheric So you have to multiply this by 101 Then you will get your answer in jewles here Okay Right The conversion part is very good Students have told ma'am constant external pressure Constant means a big external pressure you have used Right so a big external pressure that means we are talking about irreversible process Okay When you have variable external pressure when you have small small steps that is reversible process Okay So this is irreversible process Okay Next you have is a uh equilibrium Okay let's revise equilibrium The important concepts here of equilibrium See chemical equilibrium ionic equilibrium that is a very big chapter Very very big chapter So what are the most important things to do here right so first thing students in ionic equilibrium just focus on the pH formulas Right if you ask me ma'am what are the most important things ph formulas and solubility product pH formulas of everything If you're like ma'am right now I don't have time to understand the concept go through the pH formulas Okay mostly you will be able to answer the questions Okay first is for equilibrium We know that the KC is product raised to power the stoometry divided by reactant raised to power the stoometry Right this is not even a new formula to you So I'm not going to revise these chinu into formulas for you Right this we you have done already in boards right so many times These are the important things that you must do for your competitive exam that is let's say if the reaction is reversed right if reaction of A to B is K1 then B to A will be K2 then the equilibrium constant will also be inversely proportional to each other If you multiply the reaction with some digit with some number then that number will be in the power of the original equilibrium constant to get your new equilibrium constant Right if you divide it with the some number in that case your new equilibrium constant will be equals to the old equilibrium constant raised to power 1 by that number from which you divided it Right so it was basically divided by 1 by that's how you get K1 raised^ 1 by right next is if you add two reactions their equilibrium constants will be multiplied to generate new equilibrium constant If you subtract two reactions their equilibrium constant will be divided to generate new equilibrium constant Okay Is that clear is that clear is there any problem on this page any problem on this page students now I'll just pick up one question so that you're able to understand what exactly we are doing here Right okay I think not this question Uh this was the question Okay Right Because in equilibrium there are a lot of concepts Okay So that's why one by one I'll teach you what with the concept with the problem Right So that you are able to understand what I'm telling you Consider the following reversible chemical reactions Two reactions are given to you That might seem easy for you Now this might seem oh my god ma'am what they have asked Actually it's a very simple thing Let's say this is the first reaction which is K1 Now when you multiply this by three and then you reverse it When you multiply this by 3 that means K2 is equals to K1 raised to power 3 When you reverse it it will be 1 by K1 raised to power 3 that is K1 raised to power minus 3 So answer will be K2 is equals to K1 raised to power minus 3 So answer is going to be C is the correct answer Is that clear to everybody right okay C is the correct answer Those who have answered A they forgot that there is also reversibility Here you have A2 B2 on the reactant side Here it is on the product side So you have also reversed the reaction right So equilibrium constant will also be inverted here right so answer will be C is the correct answer So based on this concept a very simple question they will ask you but this has a very very high probability of asking right okay then next is this formula right okay this formula that is log K2 by K1 equals to delta H2.303 303 R T2US T1 by T1 into TS2 So if they have given you let's say at this temperature this is the equilibrium constant at this temperature this is the equilibrium constant whether this reaction will be endothermic or exothermic That means you have to calculate the value of delta H from here If that is positive that reaction is endothermic If that reaction is negative it is exothermic in reaction So very simple right then one more question they can ask you based on QC and KC right so they will not ask you this QC is greater than KC what whether this will be backward or forward no it's a very simple question they will not ask you that right if they ask you that that means they are giving you free numbers right so rather what they will say QC value is let's say X right let's say something I'm just randomly 1.8 date right KC value you have to calculate by this formula product raised to power concentration reactant raised to power concentration right so this KC value let's say it is coming out to be two in that case you have to tell okay KC is greater than QC if it is coming out to be 1.6 six then KC is less than QC right so according to that they will ask you if they want to ask a tough question okay so now here don't spam students don't spam it in the chat box don't spam okay right there is no benefit of you and me as well right when you spam things right okay I'll definitely see that what you are saying right so QC if it is here KC is fixed okay if QC is greater than KC you have to match this value while coming backwards so that's why it's a backward ward reaction If QC is equals to KC that is at equilibrium If QC is less than KC you have to move forward that means it is a forward reaction Okay understood everybody now next is a relation of delta G and K equilibrium Direct question based on this formula that is delta G is equals to minus 2.303 RT log K equilibrium Right directly numerical based questions they can ask you Also you should know that if KC is greater than one right then delta G is obviously negative If KC is less than one if this is less than one then delta G is going to be positive here because log of less than one is going to be negative Right okay That is why this log is going to be negative in that case and hence delta G will be greater than zero Right if KC is equals to 1 because log of anything is equals to 1 sorry log of 1 is equals to zero right that is why delta G will be zero and that's why this is at equilibrium right okay so now when delta G is negative it is spontaneous that is why reaction will itself proceed in the forward reaction if delta G is positive it is a non-spontaneous process which does not want to occur that is why it will be a backward process if delta G is zero it is at equilibrium there will be no shift here There will be no shift Then you have Lee chart layers principle right okay Lee chart layer principles principle How many of you find this topic difficult or easy let me know in the chat box Okay Knowledge B I'll take care of that right now Please focus here Right Lee chart layer principle So students now here mostly the question is asked on pressure and temperature right but you should know the all the cases here First is the concentration it will always go in the opposite direction whatever you do that is lee charter principle It's a stubborn kid your mom will say don't do this this will do that thing only which mom is saying not to do Right so it's a stubborn kid So here it is concentration opposite to the uh if you let's say you increase the concentration of reactant it will try to decrease it by moving it forward So it will be a forward reaction that's how it is going to change But if you add any solid component there is no effect on that solid component has no effect Then talking about pressure here pressure and moles are always opposite So if you increase the pressure it will go in the number of moles where number of moles are less When you decrease it will go in that direction where number of moles are more Right volume change volume is inversely proportional to pressure So when you increase the volume it will go in that direction where number of moles are more Right okay Because here when you increase the pressure it is going in that direction where number of moles are less So volume is opposite So when you increase the volume it will go in that direction where number of moles are more right Then talking about temperature here directly what you will see if temperature is exothermic or it is endothermic If temperature you see if the reaction is exothermic delta H is negative In that case when you increase the temperature in that case when you increase the temperature rate this equilibrium constant decreases So it is a backward reaction In this case when you increase the temperature equilibrium constant also increases So it is a forward reaction So it depends upon whether the reaction is exothermic or the reaction is endothermic Right okay Next is inert gas I feel that right now they can also ask you very simple questions like this At constant pressure when the pressure is constant number of moles are more Okay At constant volume there is no effect but at constant pressure it will go in that direction where number of moles are moved Okay Then you have volume This we have already done These are some of the examples where reactions are positive exothermic or endothermic in nature Like combustion is always exothermic Decomposition is always endothermic in nature Okay Next students we have everybody Next is strong acid Now I'll tell you directly here what are the questions that they might ask you based on pH Okay So pH is equals to firstly you should know the value of pH here I'm just going to write it here The value of pH will always be minus log of h positive of solution Right the value of p is going to be minus log of h negative of the solution Right and the value of pH plus p is going to be equals to 14 That you should remember right now What if they have given you simple strong acid in that case you will take the concentration of that strong acid If it is greater than 10^ - 6 you will ignore 10^ -7 here If it is less than 10^ - 6 you will add 10^ - 7 which is from water because water also has H positive right okay after putting the H positive value taking the log negative you will have the value of PH similar is for O negative if it is greater then you will have to ignore 10^ minus 7 if it is less then you can add 10^ minus 7 here now P is also you will take it the same way right next is if two two strong acids you If you have two strong acids directly what you will put n_sub_1 v_sub1 + n_sub_2 vs v_sub2 by v_sub_1 + v_sub_2 where this small n this small n are representing the n factors n factors like for there is one h positive so n factor is one for h2s4 there is two h positive so n factor is going to be two similarly for base if you have two strong bases it is n_sub_1 v1 plus n_sub_2 v2 divided by v_sub_1 plus V2 that is going to be right uh the correct answer right okay right next is strong acid now this is very very important that is strong acid and strong base if you have one strong acid and strong base let's say uh here there are choices whether you will use O negative or H positive let's say your O negative your base is greater Right or you can also write it like this Larger minus smaller value divided by V1 + V2 Right that will be the answer And if I say that uh for acid for acid N1 V1 is greater right then you are calculating H positive concentration If for base N_sub_2 V2 is greater then you are calculating then you are calculating is more then you are calculating O negative concentration So let's say if your N1 V1 that is for acid it is more then you will calculate H positive because acid is more in quantity right if M_sub_2 V2 that is for base it is more then you are left with only O negative that means you're calculating O negative ions here right for weak acids you can skip all these things right I'll tell you you can skip all these things because that is nothing but that is a ratio of the same thing but you should not skip is this formula K A is equals to C alpha squared This is for weak monoprotic acid right and pH is equals to/ pKa minus lo C These two are the important formulas for you Next is for buffer solution what you should remember is you have acidic buffer you have basic buffer you have salt buffer Here what you can understand is the formula for acidic buffer the formula for basic buffer and the formula for salt buffer Directly put these formulas here when you will calculate pH plus p is equals to 14 So you can calculate pH value is 14 minus p directly you can solve the value of pH as well here Right okay Now let us understand for salt hydrarolysis For salt hydraysis see you have three kind of things One is you have weak acid strong base strong acid weak base weak acid weak base In all these three things what you have to remember is the pH values here So the pH values you should remember from here Okay if you have weak acid plus strong base this is the pH value that is 7 +/ pKa plus If you have strong acid plus weak base this is the pH value right that you should remember directly remember the pH values right okay these are the tricks that I have already told you in physical chemistry formulas You can go through that But here I'm directly telling you the pH values here right for if you have weak acid weak base this is a direct formula of pH Direct pH formula is very very important for you Then last important topic from ionic equilibrium is solubility product which is very important So if they have given you some reaction like this Okay This is the stochometry This is the stochometry It will be X raised to power X Y raised to power Y solubility X + Y That is going to be the value of ESP That is the solubility product Right next if you have common ion effect let's say this is CF2 It is also giving us CA positive and 2F negative Right and NF is also giving you F negative It is giving you 0.1 and it is giving you this is s 0 0 this is zero for stochometry it is s for stochometry 2 it is 2s now here it is giving you 2s f negative from here it is.1 here it is 2s so you will add 2s here you will add.1 here and hence this is the final f negative concentration what you will have but solubility is very less as compared to 0.1 like we always ignore alpha value in place of one right so here we can ignore the value of solubility that's why answer will be 0.1 directly you can solve Ksp is equals to concentration or solubility of this that is Ka to positive that is solubility raised to power 1 this is.1 raised to power 2 that is the stoometry and you will get your answer for Ksp but remember one thing but remember one thing here that very important thing If you don't have common ion effect you will directly use this formula If you have common effect you have to go through this application You have to go through this proper steps Okay you have to follow all these steps here if you have common ion effect This is a very very important topic Okay understood everybody or not Now let us do some questions based on this Everybody write yes definitely you will get the session PDF Now let us see the uh questions based on this that is at 500 kelvin for a reaction this reaction is given to you KP is this much what is the value of KC So directly this is the relation of KP and KC This is the relation right kp is equals to KC RT delta NG Delta NG you will get to know from the equation that is 2 - 4 product minus reactants that is minus2 put this value and you will get your answer this formula is also very important next question is from leakart principle now this formula is given to you now you can see this equation is exothermic because heat is getting released now they're asking you which will favor forward reaction here forward reaction means Means let's say if I increase the temperature for an exothermic for an exothermic reaction temperature and equilibrium constant are inversely proportional so rate constant will decrease uh sorry for equilibrium constant will decrease that means that means if temperature increases equilibrium constant decreases that simply means one thing that it will go in backward direction which we don't want so that's why we need low temperature so that equilibrium constant increases that will go in the forward direction So low temperature is preferred Whether low pressure or high pressure that depends upon the moles So less moles are there on the product side That means if you increase the pressure because as pressure increases it will move towards less moles that is forward direction So answer will be low temperature and high pressure Right next is the incorrect match in the following incorrect match See simply K is equals to 1 We know that delta G is equals to -2.303 log of uh log of KC right so that means if you see that KC if it is one then delta G is equals to zero that is correct If KC is greater than one then delta G is negative that is also correct If KC is less than one then delta G this whole term will be negative Delta G will be positive that is also correct If KC is less than one delta G is negative that is incorrect So answer is D is the correct answer here Okay Yes students everybody let me know Next question is based on let me see this question is a very interesting question Okay Now let us solve this After that we are going to start with the physical chemistry of 12th class Okay H everybody in the chat box Everybody in the chat box So let us see that 20 ml of.1 molar H2SO4 solution is added to 30 ml of this much NH40 solution Now can you tell me what is the what is the value that basically you have to calculate the pH of the resultant mixture KP pb of NH40 is 4.7 okay so as you can see this is a strong acid this is a strong acid and NH4 let me know in the chat box whether it's a strong acid or weak base now if you see the reaction takes place when H2S04 is added to NH40 is as follows We know that this is a question of strong acid and a weak base Now it's not strong acid and strong base So you can't just directly apply N_sub_1 V_sub_1 minus N_sub_2 V2 divided by V_sub1 + V2 No that is for strong acid strong base If you have any of that is weak Now you have to identify whether it's a question of salt uh anal like whether this a question of salt hydrarolysis or buffer solution Okay that we will analyze here Now if you have strong acid and weak base the salt is of strong acid and weak base right strong acid and weak base Now tell me if you have strong acid and weak base that is nothing but resulting solution is a basic buffer right for example you had salt of strong base plus weak acid then it would be a acidic buffer Now this is a basic buffer right okay Now here directly what you will do you'll use the formula of basic buffer here that is P is equals to P KB plus log of that salt divided by the base If you remember this uh wait a minute let me just show you here basic buffer where you have okay wait a minute Yes Okay This is the basic buffer P is equals to PKB plus log of salt by base This is the formula that we are using there Okay This is the formula that we are using there So the main important part is to know that this is a question of buffer Right okay Then you will put this value of salt Everything is given to you in the question only 20 ml of this much mole is added to 30 ml of this much mole So concentration is already given to you that is 2 + 2 this is already given to you concentration right and pKb is also given to you in the question that is 4.7 so what you will do you'll just put this value here and you will get your answer here pH is equals to 14 minus p so 14 - 4.7 answer will be 9.3 is the correct answer if that is clear to everyone then let me know in the chat box because this is a very very important Important question Mark it Very very important question Okay Understood yes everybody Yes everyone If that is clear then let me know in the chat box Right Okay Yes One more thing in this case the concentration when you're adding you have to be clear on one thing that millles at temperature zero this is volume multiplied by this that is 2 here it is weak base is given to be 30 multiplied by2 that is six and in this it is zero so what will happen here all of this will be eliminated this will be all of this will be consumed here right here you will have this and here two of this will be consumed you will have two left and to go left So in the end you have weak base that is two and weak acid that is two because you have to look for the stochometry as well here right the stochometry is two here the stochometry is one here the stoometry is one okay for that you will put this value and you will get your answer here right next is solution chapters everybody let us see the solution chapter here now solution electrochemistry chemical kinetics I would say these are very simple chapters right uh basic questions are asked here from collleative properties nic equations so directly let's go through that RO's law firstly you should know Rous's law is vapor partial pressure is equals to vapor v pure vapor pressure multiplied by the mole fraction that is yours law okay next is see the one that follows rout's law are the ideal mixtures right these are the examples of ideal mixtures which are almost look alike They have just difference of one atom or maybe they are homologous of each other Right then one thing to remember in this case always remember delta H and delta V mix is zero But delta S and delta G okay always delta G is going to be negative Always delta S is going to be positive It doesn't matter whether it's a ideal solution or non ideal solution Right now let us see the examples of non ideal solution There are two kind of deviations that you have One is positive deviation other is negative deviation For positive deviation right it the pressure is greater than what we expect from Rout's law From negative deviation the pressure is less than what we expect from Rout's law So in positive deviation what happens the interactions are less the interactions of AB are less as compared to AA and BB interactions and in negative deviation the interactions of AB are greater than AA and BB interactions and here also because it's a positive deviation delta H and delta V is positive where delta H and delta V are negative delta S and delta G as I told you that is going to remain the same that is entropy is always positive and gives free enthalpy is always negative these are the examples is of the positive and the negative deviations which are very important You here what you will see interactions is decreasing Water is breaking the hydrogen bonds here Right here water is forming the hydrogen bonds between this So interactions are increasing here Right okay Then there is bond hall factor Right bond hall factor simply you can just ignore these formulas here You can ignore these formulas They are not useful for you Just remember if bond hall factor is greater than one it is dissociation If it is less than one it is association If it is equals to one that it is neither association nor dissociation That is in the case of non electrolytes That is in the case of nonelerolytes like glucose uh like you have ura right all these are nonelerolytes Now let us see the collleative properties here First is relative lowering in vapor pressure that is delta P minus pure vapor pressure is equals to I bond or factor will always be added into mole fraction of the solute Right then you have elevation and boiling point that is delta TB is equals to I into KB into marity Right what is marity marity is equals to number of moles of solute W2 by M_sub_2 m divided by weight of solvent multiplied by 1,000 What is molarity marity is number of moles divided by volume If the volume is in milll then it will be multiplied by 1,000 Okay this is your marity This is what we say is marity Right now that KB is elibiocopic constant KF is cryoscopic constant Right then next is osmotic pressure in which this C is nothing but it is marity It is nothing but it is marity Pi is equals to IC R Right okay Understood isotonic solutions are those in which P1 and PI2 that means the osmotic pressure is equal that is called isotonic solution Next is Henry's law in which we have pressure is equals to KH that is Henry's constant multiply by the mole fraction that is called as Henry's law that solubility and pressure are directly proportional to each other Okay understood now let us do one question based on this what we have A solution is prepared by dissolving 6 g of ura So how to visualize the collleative properties right how to understand this question right osmotic pressure they are asking you that means pi is equals to CRT you have to apply C is marity right okay uh now I'll see ura means I is equals to 1 bondoff factor is equals to 1 right this they have given us the weight of ura molar mass we know this much of glucose in 100 ml of water a solution is prepared by dissolving two things one is ura other is glucose in 100 ml of water the osmotic pressure of the solution is this now you have two solutions here one is ura and one is glucose both we are mixing here right so concentration of both will be added here for what we see we know that this is the relation given mass of ura is given to you molar mass is given to you mass of glucose is given to you mass is given to you right now here for What is n that is number of moles both the values will be added per ura and glucose both right so here number of moles is going to be6 by 60 here it is 1.8 8 by 180 right okay this is what you will add here this is going to be 01 this is going to be again 01 so here answer is going to be 02 that is 02 number of total moles that is n_sub_1 + n_sub_2 is going to be 02 right volume is already given to us in the question that is volume is given given to us as 100 ml Okay Volume is already given to us 100 ml Right now RT you have to multiply Now students ma'am why we have multiplied it by th000 See in this part if you see this equation here I told you whenever you are putting marity and you're putting volume in milll you have to multiply it by,000 That's why you multiplied it by,000 here Okay that's why it is multiplied by 1,000 because this is in millilit This is R value and this is T Now why the value of R is 0821 because whenever you're talking about volume pressure then the value of R will be 0821 that is 1 by 12 Okay understood this right Okay very good Very good Very good which does not depend on the quality instead the quantity very good that is polative property right the next one we have is chemical kinetics so from here two most important things are first order reaction and our heinous equation these are the two important topics here okay so first is these important things regarding the rate two things I'll tell you here one is if you calculate the rate from the reaction rate of the reaction In that case it will be divided by the stochometry Right so whenever I'm calculating the rate I'm dividing it by the stochometry I'm dividing it by the stoometry If you're talking about in terms of reactant you will put a minus sign If you're talking about products you will put a plus sign Right if you're talking about rate of appearance or disappearance in that case you will not divide it with the number of moles only the rate of reactions will be divided by the number of moles That is the important part Next important thing is let's say rate of reaction you have Okay Let's say rate of reaction you have Now students rate of reaction this is 1 by b db by dt Wait a minute Here you product is appearing and this is getting disappeared So it will be negative here Okay This will be negative here Wait a minute This will be negative here Okay This will be negative because reactant is getting disappeared So it will be minus 1 by a da by dt Right rate of appearance will be plus db by dt Right disappearance will be minus da by dt There is no division with the number of moles Next is what is order order does not depend on the reaction See it will depend on the rate equation If this is the equation then x + y is going to be the order of the reaction that depends on the stoometries in the rate law equation Okay Next we have is okay before that we will see the formulas of zero order first order and second order Now most important is the formula of first order reaction Here also remember that units of K units of K is equals to mole per liter 1 - N/ second and also remember generally where N is the order of the reaction T is directly proportional to A 1 - N These are the two important things to remember for your questions If they ask you general units right order 1 2 3 4 0 whatever is given to you put n value according to that you will get the units uh of k right k is the rate constant right t half is the halflife period the general equation is a not is the initial concentration raised to power 1 - 1 this is the general equation okay this is the general equation now these two formulas are very important for your these are most important formulas and I would say most important is first order then there is zero order order least important is second order very rarely they will ask you questions based on second order so first order is k= 2.33 log of initial by final t is 693 by k zero order is k= to r by rx t is r by 2k second order reaction is k= to 1x 1x t1 by r1 - 1x r2 t is 1x r into k right okay now let us do one question here right okay based on activation energy firstly we will do right here you have k is equals to a e raised to power minus ea by rt where a is a a is the arenous constant right and k is the rate constant e is the activation energy solving This right we also get log of K2 by K1 is equals to E by 2.03 R T1 minus T2 by T1 into TS2 Okay that also you get here Now if you have a question of our equation let's say for a reaction of H2 with I2 the rate constant is this The rate constant is given to you at this and rate constant this at this That means t_sub_1 this is k1 this is t_s_1 this is k2 and this is t_sub2 right the activation energy they are asking you so directly that formula you will put here right directly this formula you are going to put log of k2 by k1 is equals to ea 2303 ts2 minus t1 by t1 into ts2 from this k1 k2 you know t1 ts2 you know just you can solve the value of ea here right what is the value of R that you will put here because things we are solving in energies we are talking about energies here So whenever we are talking about energies we'll put the value of R as 8.314 Okay we'll put the value of R as 8.314 here Rest of the things will remain the same Is that clear to everybody or not yes students is that clear or not is that clear or not right Okay Very very good evening Very very good evening How many of you guys this is clear to how many of you right Directly it was a formula based question Okay Now we will see question on decomposition of X Decomposition of X exhibits a read constant This you might be like oh my god ma'am what is this say decomposition follows which order reaction that you should know right So remember that decomposition right uh either from the units also you should know that it follows zeroorder reaction Decomposition reactions are zeroorder reactions Now if you have no idea on that you can also find it from the units of K Exhibits a rate constant of this per year That means UG per year This is given to you rate constant of this How many years are required for the decomposition of 5 UG of X into this okay So this is your final concentration This is your initial concentration and this is your rate constant Right now from this we already know this is a zero order reaction For zero order reaction we have seen the value of t We have seen the value of t for zero order reaction that is r by 2k What is r r not is nothing but that is the initial concentration Okay R is nothing r a all these are representations of initial concentration So what what do you mean by that k is equals to k is equals to a by 2k or r by 2k A not or let's say r that is given to be five ug right You'll put it here The value of K is already given to you That is 05 Right 05 And then nothing you are left with You just have to put this and solve this value 55 will be cancel here You will have 100 This will be two This will be 50 So answer will be 50 years is the correct answer Okay Understood everybody 50 years is the correct answer Okay You want me to speed it up now we are towards the end of physical chemistry that is the last question right okay last qu uh no last question This is most important question from your electrochemistry First topic is n equation Then three topics majorly you have here Right so first is n equation that I would say the most most important thing here Right so for that you should know this equation that is e not is equals to e not cell minus.059 by n log of oxidation by reduction Right now next formulas are this these three formulas right delta G and E not cell delta G and log KC E not and KC these three equations directly the formula based questions can be asked here then this we have already done that if you have delta G is negative if this is negative right let's say if this is negative E not is going to be positive if this is positive E not is going to be negative that we have already done in equilibrium as Now next kind of question that is what is conversion conservation of energy Conservation of energy means if let's say you have a to a positive and a positive to a2 positive here you have e not value is e1 kn this is e2 kn and they're asking you for a to a2 positive what is the e not value So you can't just add e1 plus e2 No you can add the Gibbs free enthalpy here and that is how you will get by adding Gibs free enthalpy This is the equation that you will get from this formula because par will cancel each other and minus will cancel So it will be N3 number of electron exchange multiplied by E3 is equals to N1 into E1 plus N_sub_2 into E2 that is your final equation Okay right okay students Now next is very simple question based on this coro law and these formulas right one is uh this is the equation that you should remember from the graph that is uh molar conductivity is equals to molar conductivity at infinite dilution minus a into under root c that is the concentration now for strong electrolyte we can get the intercept here that's why for strong electrolyte we don't need cor law but for weak electrolyte there is no intercept So we are not able to find the molar conductivity at infinite dilusion That's why we needed coral law here Corlo simply states that if you want to find the infinite conductivity of a weak electrolyte that will be equals to the infinite conductivity of the corresponding ions here Right okay Now if you see ma'am how to find out the value of this uh uh molar conductivity molar conductivity at concentration C and infinite infinite you can easily calculate but how to find the molar conductivity of concentration at concentration C for that what we will do for that what are the formulas that you should know is that molar conductivity is equals to ka into th000 by marity equivalent conductivity is cap Cap into,000 by normality Kapa is the simple conductivity and here it should be in centm only then you can apply this formula If it is in meters then you will apply this formula But I would say mostly you will have copper in centm only So you'll apply this formula Right this is the molar conductivity at concentration C This is the molar conductivity at infinite concentration at infinite dilusion And hence what will happen from that alpha you can solve Alpha is the degree of dissociation And this formula for weak electrolyte we have studied in equilibrium that Ka is equals to C alpha squared Put the value of alpha alpha here and you will get the value of Ka here Right Okay Now other formulas to remember here is G into G star is equals to capa G is conductance G star is cell constant that is equals to length by area and kapa is the conductivity Okay this a very very important formula for you Then this is a general formula that you have studied in physics as well that r is equals to row l by a but this is very important formula in when we talk about chemistry right then okay then one more thing I'll just write it here right two formulas that also you should know the formula at least first law of parad first law of parad states that w is equals to z i that means molar mass mass by n factor into farad it that is going to be the weight deposited right if they ask you the weight deposited here right okay directly you will put this formula right that's a very very important formula and second law states that and second law states that here what you will have weight of let's say two electrol electrolytic cells are are in series right so weight and equivalent weight or you can say weight of 2 A and B will be equals to the ratio will be equals to equivalent weight of A and B Right so these are some of the important formulas for your uh chemistry section Right now let us do this simple thing here that is this is not related to any of the formulas Okay let me tell you because you know I need to save your time here as well So oxidizing power So these kind of questions repeatedly we see right in vit e So what you have to see here is for example any of three four options are given to you They are asking you the oxidizing power So always remember reduction potential is inversely proportional to reducing power Right is directly proportional to oxidizing power Right that means if someone is very good I want to get reduced then I will try to oxidize someone I'm a very good oxidizing agent So that's why if they're asking you oxidizing power that means you have to put it in the order of reduction potential Reduction potential means gain of electron everywhere you have gain of electron The one which has maximum reduction potential has a maximum oxidizing power So maximum is for cocobalt right that is having the maximum then you have PB then you have C4 then last one is BI so answer will be B is the correct answer okay so that is how you're going to answer these kind of questions students tell me if that is clear to all of you or not because after that I'm going to start with solid states right then I'll do chemical bonding and coordination chemistry right okay if that is clear to all of you then uh spam this chat box with green hearts right so that we can start with solid states here right okay yes next we have solid states so in solid states what if you have never done this ma'am I had no idea ma'am that was not also there in my board exams ma'am okay so what if uh see you may be from different different boards so what if that was not there in your board exam so what you have to do Right then go through this table here Right if you're like ma'am I find this table very difficult This is a second priority table This still you can skip here Right from the questions only I'll tell you two three kind of questions that can be there in your exam Right then there is cubic unit cell Right so this is your relation between R and A for simple cubic This is body centered BCC This is FCC Okay So I'll just write it here One is one is simple array array array One is simple cubic then you have BCC then you have FCC that is phase center Okay So I'll just put it here Simple This is BCC and this is FCC So you should know the relations of R and A here So what is R r is the radius of the atom right when there is a cube right the radius of the atom that you're putting it right so that is the radius R A is the edge length the edge length of all the sides of the cube that is your uh A okay so you have to know the relation between A and R here packing fraction directly you can just scram it also that uh FCC has a maximum packing fraction then you have BCC then least This one is for SEC right coordination number is six here is 8 here is 12 here if you understand ma'am how it is so see for FCC if you see there is the atom in the face right on the faces let's say first you talk about BCC there is atom on the body center right okay and then there are how many atoms on the corner 1 2 3 4 5 6 7 8 right so eight atoms are touching this body centered atoms So that is why coordination number is eight So to that body center eight atoms are occupying it Right if you talk about the face center right in that case what you will see if let's say it is on the face right it is on the face Face means let's say if this is the uh if I say this is the face means this length Okay If here you have one atom here you have one atom how it is touching different different atoms here Right so what you will see here right just imagine this in your room right you have this side wall you have one atom right that and other atoms are there on the corners of the wall So now this center atom will touch how many atoms here 1 2 3 4 Four corner atoms of this room Four corner atoms of other room Four corner atoms Uh okay Four corner atom of this room and the other room that is 4 + 4 8 Right now what you will see that corner atom it is touching how many things again it is touching four different atoms here Right so that corresponds to 4 + 4 + 4 that is 18 Effective number of particle is four Coordination number in case of simple cubic center is six So directly you can also remember this Okay Directly either the direct questions will be asked to you or numerical based questions will be asked to Next is density D is equals to z m n a cube This I'll tell you how to apply this That is effective number of particle according to this whether it's simple BCC FCC you will have your value of zed here right then you have mass is the molar mass that will be given in the question na is the ru's number a is the edge length if directly edge length is not given to you they will give you the radius from there you can calculate the edge length depending upon the formula depending upon which cubic cell you have whether it's simple whether it's FCC whether it's BCC now I'll tell you one question based on this what kind of questions they can ask you let's say at 100° C copper has FCC unit cell structure with cell edge length X what is the approximate density of copper at this temperature they're asking you the density okay copper we know the molar mass of copper right FCC unit they are asking you with edge length is also given to you that is X okay in terms of X you have to tell what is the density so directly what you can say that for FCC unit cells that is equals to four that we have seen mass atom that also we know number of we also know now edge length is there instrong you have to have this edge length in terms of cm because this formula has edg Ed length in centm right that is the major mistake that students might mistake might do here right so edge length is in cm here right now density is equals to z you have m you have na you have a you have now you solve this and you will get your answer right in g per cm cube is that clear to everybody this is the direct formula that will be applied here right very simple questions they will ask you Right Next is the arrangement of x negative Now it's a very interesting question Okay So every single question is telling you something else about the uh topic Right the arranger let's say they have given you uh this person from this arrangement what these are questions related to voids Okay So the arrangement of x negative around a positive this is x negative and this is a positive Solid ex is given in the figure not drawn to scale If radius of X negative is this radius of an first we have to see whether it's octahedral or a tetraedral void Now anybody in the chat box directly from the figure you can tell whether it's octahedral or tetrahedral void Okay So that is your octahedral void Okay These kind of figures are for octahedral void Three in front three in back that is the octahedral void Okay because you have uh like basically you have eight phases that that are occupying here So that is why it is octahedral void You don't need to get into the reason here Just see this is octa void This kind of position is octa void Now for octahedral void r of kion by r of anion is 401 That is the formula Right so r of kine r of kine you can just you have the value of r of nine here This you have this you have to calculate So directly you can put the values and calculate the answer here Right okay Okay students that is clear to everybody yes student is that clear now let us see four topics which are definitely will be there in your exam no matter what These are four those four topics that it's like no chemistry exam is formed without these topics right okay first is okay firstly to I'll just advise you guys that this I would suggest you to go through h I'll start with BBT right okay so first is your BBT that is from chemical bonding Right so here what you have to see is different different concepts here regarding the shapes Okay So are you all ready everybody if you guys are ready for to brush up your concepts then let me know in the chat box right let me know in the chat box if you guys are ready to brush up your concepts of EBT So now how you will identify the shapes first thing is if your sigma plus loan pair is equals to 2 it is sp it is three it is sp2 4 sp3 5 sp3d 6 sp3 d2 right for sp it is linear sp2 it is trional planer sp3 it is tetraedral sp3d it is trional bip pyramidal sp3d2 it is octahedral right now remember that whenever we are doing shapes we will only calculate bond pairs for geometry We will calculate both bond pairs and loan pairs both of them right now one by one we will do this let's say you have AB2 kind of molecule right let's say you have AB2 kind of molecule that is a simple molecule like this okay I'll just show you AB2 kind of molecule is a simple molecule like this right so if you have no lone pair AB2 is a linear right if you have one loan pair that means you have to ignore or the loan pairs in shapes So that will again be a linear shape only Right okay So that means shape is going to remain linear only in this case Next is a B3 A B3 Now here if there is no loan pair it is a trional planer geometry as well as shape But if let's say you have one lone pair that means this is a bent V shape Geometry will remain the same but shape has changed that is bent V Let's say you have two lone pairs In that case you will have linear shape here Right okay Next students next one is next one is your tetraedral This is a tetraedral Okay This is tetraedrral right so if you have all the four bond pairs it is tetraedral If let's say you have one lone pair in that case it is now a pyramidal shape right so that is why it is pyramid Let's say you have two lone pairs with two lone pairs Now the shape is bent to V Right okay Bend V As you can see it was like this So it is bent V Then if you have three lone pairs take out one bond pair and that is going to be your linear shape Right okay Next what we have is SP3D So what is SP3D everybody in the chat box What is SP3D what is SP3D everybody in the chat box here you have one trional trional This is the triangle that you will have and two pyramids here Okay two pyramids here So it is going to be trional This is trional bip pyramidal This is first pyramid and this is the next pyramid trional byparameal Okay trional bip pyramaral two pyramids you will have here Now very important thing is that if you have five bonds and trional biparam what if you have one lone pair loan pair see these two positions the pyramidal positions are the exial positions these positions are the equatorial positions So that is the reason here always in this position you will have the loan pair So if you have one lone pair that will be on the equatorial position that is why you have a seessaw shape Okay as you can see this is a seessaw shape Okay this is the seessaw shape Next is you let's say you have two lone pairs again second loan pair will also be on the equatorial So this is a T shape That's why it is a T shape Again if you take out one lone pair more that is from the you make one more loone pair on the equatorial position that will be a linear shape So hence it is linear This is a very very important thing Okay this is most important AB5 kind of molecules Right Next is octahedral So ready for octahedral Right everybody ready for octahedral Now I want to see you telling me in the chat box what is octahedral Right everybody I want to see you guys tell me in the chat box what is octahedral is this an octahedral position stop singing Okay What is octahedral yes everybody This is octahedral position Right now here as you can see this is one square two pyramids here Right you we call it octahedral only If you say you have all the six bonds pair this is octahedral If here you have one lone pair then you take out one lone pair This is square pyramidal That's why here it is square pyramidal If let's say you have two lone pairs that is called square planer That is why it is called square planer here Is that correct is that clear to everybody or not i can see you guys are not interested Oh ma'am we want VITE E session Ma'am you take the session we won't be interested at all Ma'am next theory is molecular orbital theory here what you have to remember I said before this I'll do the questions of uh let me see okay so I don't have questions related to that but I'll just solve one question here related to that let's say we talk about SO2 okay so you have this molecule they're asking you what is the hybridization here so how you will calculate firstly you will see central metal atom right then you will see that how many uh to which group it is belonging to Sulfur is belonging to group number 16 Okay So you will say six outermost electrons you have minus we know that oxygen always form double bond Oxygen always forms double bond So this is 1 2 3 4 electrons are occupied in making the bond pair So it is only two electrons you are left with That is one lone pair you are left One lone pair you are left with Now how what we will have here now students this is S double bond O double bond O how many sigma bonds you have two sigma bonds how many lone pairs one so it is three three means sp2 hybridization okay that means it is going to be something like this S1 23 okay so we know that you have two double bond O so I'll write two double bond O here and one lone pair so one lone pair will be over here that is how You will get the shape that is bent to V You will get the hybridization that is SP 2 Okay is that clear to everybody or not students is that clear to all of you or not then let me know in the chat box if that is clear Right now let's solve one more question like this Right let's say they are saying that solve for XCF4 Okay In this case xenon is belonging to which group it belongs to group number 18 That means eight outermost electron Four are already because florine always makes a single bond So it is already occupied in bond formation You have four electrons That means two lone pairs you have Now students for two lone pairs what you are going to do here very simple thing Two lone pairs ma'am we have here Two lone pairs ma'am we have here So this is going to be sigma bond is four two lone pair it is going to be six that is sp3d2 right So that means xenon 1 2 3 4 5 6 that is the octahedral gate right now here you have two loone pairs that I'll put here one here right one loone pair here and loan pair should always be opposite you can't put one loone pair here and one here that will cause a repulsion that is why theory states that loan pair loan pair repulsions are greater than bond pair bond pair uh are greater than sorry loan pair loan pair are greater than bond pair pair loan pair are greater than bond pair bond pair So we will not want two loan pairs to be together That was the same reason why loan pairs were on the equatorial position Okay So this is f This is f This is f This is f This is the square planer shape and octahedral geometry That is how you will answer the questions related to uh shapes and hybridization from chemical bonding If that is clear to all of you then let me know in the chat box Next very important concept is MOT molecular orbital theory Right here when two orbitals atomic orbitals overlap they have bonding molecular orbital and anti-bonding molecular orbital two things are formed right now here what you should remember that for bonding molecular orbital everybody over here right everybody over here right okay now here if you see bonding molecular orbital that is going to be lower in energy as compared to anti-bonding molecular orbital Now what I have written here is less than or equal to 14 electron it is 21 to1 If you have more than 14 electron it is going to be 1 2 1 Okay So simply what you will do here very simple thing we will do here First thing is always remember there will be a KK star that means it will be KK dash that KK dash will have 1 S2 or you can just write sigma 1 S2 sigma* 1 S2 sigma 2s and sigma* 2s So why we will not write it right because already we will write star which has eight electrons So you can just write KK 8 right that means already you have picked it because if you have atomic number more than 8 let's say you have more than eight electrons then already this whole shell is filled so you don't need to write it again and again so now we will use the tricks here how to solve this right we will use the tricks here right I'll explain you with the examples here so let's do questions which what is the from this question I'm going to explain you three four things so see this questions very very carefully Right what is the number of unpaired electrons in the highest occupied molecular orbital of the following species right so two things are first is you have to calculate for N2 Right now firstly you need to tell me they're asking you highest molecular orbital See here you can't use any trick You have to solve the diagram But diagram also we will do easily Right so N2 means 7 + 7 that means 14 electron As I told you KK shell you will already eliminate that is 8 electron is there in KK So you'll already eliminate the KK So that is uh that is going to be 4 6 electrons you are left with right when you have 14 electrons that is when you have equal to 14 you have 21 What is 21 211 is this 2 1 2 1 I'll tell you after this what exactly 21 means Now N2 positive means how many electrons you have this is 14 electrons minus one that is 13 electron from 13 8 electrons are already there in KK So this is minus 8 that corresponds to 5 electrons Right okay Now again what you have here is again what you have here is now here what you will see whether it's 2121 or 1 221 important thing is don't look for this N2 positive okay don't look into this N2 positive that it has 13 electron less than 14 no that is a long thinking N2 positive the neutral specy of this N2 positive positive is N2 which is having 14 electron which is equal to 14 that is why it will have 21 because N2 positive has 13 that is not why it will have 21 because the neutral specy has 14 electrons which is equals to 14 that is why it will have 21 2 1 here so you'll have to look for neutral species okay that is why again it will be 2 1 2 1 1 now electron filled is here you have six electrons 1 2 3 4 5 6 Here you'll have 1 2 1 2 3 4 and 5 Now they are asking you the number of unpaired electrons in the highest occupied orbital Now students see one thing here What is the highest occupied orbital this is the highest occupied orbital This is the highest occupied orbital This has one unpaired electron This has zero unpaired electron So this will be 0 and 1 This is going to be 0 and 1 0 1 0 1 So this is eliminated This is eliminated Right now one more thing here Now what is this dash dash dash dash so always remember sigma is z pi is x and y and x and y is always always going to remain the same So here how we will write this This is two same that means this is 2px this is 2p y because this is pi so it will be pi 2px pi 2p y which are equal these two are also pi 2px pi 2p y but they are antib-bonding this is anti-bonding this is anti-bonding this is anti-bonding this is anti-bonding this is anti-bonding this is antib-bonding right so here it is single one is z so it is sigma 2pz ma'am how to remember this See who is single person who is a single person who is sigma a sigma person is always a single person So sigma that is 2p that is sigma that is why it is single Right but this 2px and 2py are pi So that is why they are not able to remain single That is why they are 2px is equals to 2p y They have the same energy Okay So this is how you can remember this And same is thing for this 2p This is star This is pi 2p xstar Pi tp y star This is sigma tp zar Okay understood Now we'll do the same thing for O2 Okay we'll do the same thing for O2 here Okay for O2 also we will do the same thing Let's do that Now let me see how many of you are able to tell me the answer between A and C Right so very slowly I am doing this O2 how many electrons 8 into 2 16 electrons 8 electrons in KK you are left with eight electrons O2 positive how many electrons 15 electrons Eight electrons are occupied So this will have seven electrons Right now what you will see eight electrons and seven electrons because originally both are 16 electrons So it will follow more than 14 that is 1 2 1 So it will follow 1 2 2 1 That means it will be like this It will be like this That is 1 2 2 1 1 2 2 1 Okay Here it is 8 electrons 1 2 3 4 5 6 7 8 Here it is Seven electrons 1 2 3 4 5 6 and 7 So O2 has how many electrons unpaired electrons two This has one unpaired electron So it will be 2 1 So answer is going to be A is the correct answer Is that clear to everybody or not how you're going to solve the questions of MOT now one very important thing regarding MOT Let's say we ask you the bond order right so always remember for 14 it is going to be three For 14 it is going to be three Rest of it is if it is increasing 15 16 17 18 or it is decreasing 13 12 11 10 right what you are going to write here this will be 2.5 this is 2 this is 1.5 this is 1 this is 2.5 this is 2 this is 1.5 this is 1 okay that is how you will get your bond order otherwise the formula of bond order is number of bonding minus number of anti-bonding electrons dividing by two That is the formula of bond order here Okay But this trick will really help you to solve the questions easily and in a quick way Right okay students is that clear to everybody Now let us do the concept of coordination compounds Like coordination compound first are these exceptions that you should know when you're writing BBT questions Okay because if if you're just solving the questions as it is but you don't know the VBT the concept of uh like the exception then chances are you might do the answer wrong Okay So remember that coar 3 positive with H2O and oxilate then liant act as strong field liant Right Water and oxilate usually are weak fant but if they are bonded with phob 3 positive they will act as strong liant Right co2 positive when it is bonded with H2O it is acting as a strong filigant Water also is a weak filigan but with copper 2 positive cobbal 3 positive it is acting as a strong filigant Okay Nickel fluoride is a weak filigan but with fluoride with Ni4 positive it is acting as a strong filigan Okay Then iron and manganese right when they are bonded with NH3 right then it acts as a weak field liant Now we know that NHS3 is a strong field and it will cause the pairing right but with FO two positive and MN 2 positive it is acting as a weak field right next is this you don't need to know for H this you can ignore right for this concept you just need to know these important parts okay understood yes everyone done Okay Now I'll start with the VBT questions See from the question only I'll tell you how to answer that only from the questions right because theory is not important here The important part is steps Okay So the pair having same magnetic moment right so you have given you two pairs here So what we know here is that water see first you have to let's say I want to calculate CR2 Okay So firstly I'll calculate the uh oxidation state that is going to be x + 0 is equals to +2 So cobalt is in plus2 state So cobalt in plus2 state means water will act as a weak field again Cobalt in plus three state water will act as a strong So here it is weak field So here it is Sorry it's not cobalt it's chromium 2 positive Right so now here you should also know one more thing that I'll just write it here Pi Okay The electronic configuration this is 21 22 23 24 25 26 This is 27 28 29 and this is 30 Electronic configuration is 4s 4s You write 4s 4s 4s 4s This is 4S 4S 4S 4S you write 3D 3D 3D write 3D everywhere okay then I'll just directly tell you because if you know these electronic configuration you can answer things otherwise there is no use of doing this theory right okay then 21 so 2 1 22 22 23 23 but stop at 24 because of half filled electron it will 5 Okay Then you have 25 26 26 27 27 28 28 29 is again stop there with copper it is 2 No it is 1 10 That is how you have 29 Okay Then 30 is nothing but it is 2 and this is 10 So that is going to be your electronic configurations Okay If you know these electronic configurations directly you can answer it right now with chromium we know that chromium is what 24 is 4 s1 3d5 you take out two electrons it will be 3d4 now there is no pairing of electrons here because it is a beak liance that's why number of electrons number of unpaired electron is four right with cobalt now I'll solve for cobalt this is X - 4 is equ= to -2 again it is cobalt + 2 right cobalt is 27 that is 4 s2 3d7 you take out two electrons it will be 3d7 right 3d7 means 1 2 3 4 5 in five orbital 6 7 three electrons are unpaired that means here it is three so this does not have same magnetic moment for same magnetic moment number of unpaired electron Electron should be same right Uh here we have already calculated it was four Let us calculate for iron Iron here exist in iron 2 positive that means it is 3d6 because two electrons you have taken out from 4s Electrons will always be taken out from 4s firstly Okay So here no pairing of electron 1 2 3 4 5 and six That means total four unpaired electrons So answer is B is the correct answer Okay One more thing you should remember for this pairing and unpaired electron is new is equals to under root n +2 bohar magneton But you don't have to use this formula Okay you don't have to use this formula directly What you can do here is if number of unpad electron is 1 2 let's say three let's say four let's say five then the answer is going to be 1.7 2.7 or 8 like whatever you feel is more easy 3 let's say I'm writing it as 8 okay wait a minute okay 3.8 8 then you have 4.9 then you have 5.9 okay that this is what you will get your answer for unpaired electrons here right this is number of unpaired electron this is new directly you can answer this okay understood everybody h next is last topic that is CFT crystal field theory okay crystal crystal field theory because I think uh you guys are quite exhausted So what I'll do let me know in the chat box Uh so I'll do this I'll tell you this after this session only right crystal field splitting in octahedral complexes Okay in octahedral complexes what is the crystal peel splitting here So what you have to do this is dxy dy z in octahi head is flitting what will happen you will have three orbitals and then you will have two that means p2g then it will be 2g okay and whenever there is strong field means there is large splitting in that case electrons will be filled like this 1 2 3 4 5 okay but if there is a weak field legant which is which will not cause that much split In that case it will fill like this 1 2 3 4 5 6 7 8 9 10 That is how you're going to fill this Understood right now next is for tetrahedral splitting You have opposite This is E This is E and this is T2 Right and in tetraedral complexes usually the splitting is not very high Okay now one important part to remember here is one important part to remember here is see if you have CFST is very high in that case your gap is very high the gap is very high right in that case you will have pairing you will have no pairing okay it will try to cause it will not try to cause any pairing but the splitting is very uh wait a minute it will try to cause pairing because the splitting is very high Yes splitting is very high That is why it will not go to the further next level It will try to cause the pairing that is why you will have low spin complexes Here you will have low spin complexes and opposite here you will have if see if CFSC is high that means you have a strong field liant Okay Then opposite is for this If your CFSC is low gap is low you have weak field liant Weak field liant that will cause high spin because there will be no pairing That is why there will be high spin complexes will be formed here Right okay students understood Now we will see some questions here Right so there is a spectrochemical series given in your NCRT Okay follow your NCERTT for the spectrochemical series from here what kind of questions they can ask you identify the correct trend given below so they're asking you the correct trend here right this is delta O of this and this right okay so in this case you should know the CFSC trends okay so here chromium and monodinum you have here you have titanium and titanium so here uh your legant is changing in this case the oxidation state is changing So more is the oxidation state more will be the CFS Okay So that is why what you will see that more is the oxidation state more is the CFS3 This one and this one will come out to be the correct one This one will be eliminated So as oxidation number of the same atom increases CFS increases That you should remember I'll just write it here CFSA is directly proportional to oxidation number Right next is this is chromium This is molibid Right so chromium 3d 4 D So as you go down the group CFSC increases 3D then 4 D That is why here C is going to be the correct answer Got it understood as you go down the group C will increase Next is question based on how these are very very important questions right based on CMSC Uh usually these were ignored before but now they don't ignore these kind of questions Okay So I think that is the last or maybe the second last question Okay Now crystal field stabilization energy for high spin D4 octahedral complexes High spin D4 See you have octahedral complexes are like this Okay Now here D4 means high spin means high spin See what you have seen here for high spin what I told you for high spin for high spin your gap is going to be low Right that means there will be no pairing in this case Right so it will be like this 1 2 3 and this is four Now this is a berry center from here you have certain gap right so here this is 6 positive of 6 this is the gap it is minus of 04 in case of octahedral complexes so that is why 3 * minus of 04 plus 1 multi here you have one electron right so this is multiplied by 6 so this is -1.2 2 +.6 answer is -.6 is your correct answer So answer is -.6 delta O is for high spin D for octahedral complex This with this they can ask you anything high spin low spin If they ask you low spin in that case instead of this electron it would be here then you will have answer as this will vanish off you will have 2 into minus.4 then that would be your answer Okay So according to that you will have your answer right Next we have is which of the following statement is false four statements they have given you Let us see how many of you are able to answer this Let us see that First is in an octahedral crystal field the electrons on a metal ion occupy eg set of orbital before occupying t2g See uh no that is not true because in octahedral you have 1 2 3 1 2 this is T2G this is Eg so they occupy T2G first C because that is low in energy so that is the false statement that means this is the correct answer Next is diamagnetic metal ions cannot have odd number of electrons Right diiamagnetic means all of them are paired So pairing means if I say five couples so that means five into 10 10 persons will be there So that means they cannot be odd number Yes Low spin complexes can be paramagnetic Low spin means they uh that are pairing of energy basically is there can be paramagnetic is it can be it will not always be paramagnetic but it can be paramagnetic They're not saying always right in high spin octahedral complexes high spin octahedral is less than the electron pairing energy and is relatively very small Wait a minute In high spin octahedral complexes high spin means where your weak freant is occupied there Right weak fil octahedral energy is less than the electron pairing energy Yes delta o is less than pairing that is why in high spin complexes because delta o is less but pairing energy is very high that is why it will not cause any pairing So that is also the correct answer Okay So uh correct answer will be a but the correct statements are all of these are the correct statements Okay understood everybody Yes students Now to all my students here To all my students here Right Now for inorganic chemistry I have put some notes for you in this PDF In this PDF only you will get some notes here Right regarding the exceptional trends I have written everything here Right and then I also written because trends what is there to explain okay this is the trend right so that you have to do yourself right and these are the uh important trends right and these are some of the questions five six questions based on those trends and don't forget to do this table okay this table I'll just explain you what you have to do here you have to do the electronic configuration of lenthenum of serium of turbium of gdoleium europium then your turbium then utitium right your turbium ium is stable in plus2 oxidation state Turbium is stable in plus 7 oxid plus four oxidation state Gdoleium is stable in plus three oxidation state Europium is stable in plus two oxidation state Serium is stable in plus4 oxidation state That you should know Leninum is stable in plus three and dutium is stable in plus three So these are some of the very very important electronic configurations for you Right very very important But if you want to don't want to take any risk you can also do for neodyenium that also is one of the previous year questions right then where is samarium samarium okay this if you want to do you can do this as well if you don't want to take any risk otherwise these that I have marked with red this you can't skip okay in the oxidation state in which they are stable in okay understood right so now here they are which are The following is likely to deviate easily from plus three that is serium because it is more stable in C4 positive Rest of these three are stable in plus three oxidation state Lenthenum lutitium C If you see this here lenthamum gdoleium and lutitium these are stable in plus three oxidation state Okay So these kind of questions they are asking right Then students one more thing on this channel only for your preparation right here what you will see on the video section right physical chemistry formulas so I have already done but still if you want to see all the physical chemistry formulas that are there in this channel on this video section J 2020 April 10 physical chemistry formula solid state I have completed that was not there then next one is your live section I I know you guys are thinking about ma'am what about organic chemistry ma'am how you can skip organic chemistry so today only right before sleeping if you are a early bird then you can wake up at 6:00 p.m right sorry 6:00 a.m and start your organic chemistry because I don't want to repeat the same things again and again on this channel right that would be really wasted of your time and my time as well right so here if you see because I'm not going to change the reactions for vit E there are different reactions for J there are different reactions the reactions are same this labels are same right so go through this all naming reactions organic chemistry this one all naming reactions organic chemistry and also here I have done reagents this is just 1 1 hour 40 minutes lecture go through this lecture so that once you have complete once you have completed this lecture here then let me know in the comment section see only if you demand only then I'll take this class once you have gone through the naming reactions then I can take one thing for you that will be different for j means and that will be different for vit e That is how they ask you questions Okay So if you want me to take up question session specially of organic chemistry that will be entire organic chemistry right I'll do that right for your vit e especially for vit e only if I see let's say uh 70 80 comments in this video only then I'll take this on Friday okay on Friday then I can take take this in the morning in the morning because I'm free in the morning so I can take this in the morning at around let's say 12:00 p.m I can take okay so if you are actually willing to be there in the class around 70 80 comments I want if there are no 70 80 comments down on this video then I'll assume that okay you are feeling that ma'am no we are okay with naming reaction and important reagents we don't need any session for that right and if you want then definitely I can take that right okay so tomorrow students I have my class on need so I can't take that right otherwise I would have loved to taken that class right now Here this organic chemistry and then there is for inorganic exceptions also I didn't do it in this lecture Reason being because I have already done the same exception See all exceptions in one go So here also this is just 1.5 R 1 hour 30 minutes lecture So you can do that all exceptions in one go In detail I have completed that Right okay So these two things are very important for you to do it today itself right so that this lecture that we have done today that will be completed Okay Yes everybody Yes So that is all for this thing and one more thing that all the students who are like uh preparing for VIT E So here are some crucial in-depth topic that we have taken also for you Now if you're preparing for any other any other exam let's say estam B2 E comet K any other exam you are preparing for so these are the test series according to different different exams and if you're preparing for J advance I I can see some of the students they were asking for J advanc lectures the lectures are going there don't worry about that lectures are definitely going there right so if you're preparing for J advance so there is a very interesting book for you that G rank accelerator book that is just at 1665 right so in this book you have really interesting questions and if you want to see the sample of those questions you can go and watch my lecture of organic chemistry questions physical chemistry questions I have already taken up and soon I'll be taking up J advanced inorganic chemistry questions as well and if you're like ma'am my J advanc level is not so good then there is this batch that is ei batch going on where fullon revisions right full on revision Then previous question discussions right and then eab question discussion is also going on in this batch that is specially designed for J advanc students okay and that's it so now I'll take your leave students now let me know in the comment section if you want me to do questions on organic chemistry and surface chemistry because today I didn't take the questions of circ surface chemistry so if you want me to take the lecture like proper dedicated question based lecture on that no theory will be there theory I have already taught you where you are going to do if you want me to take that and you will be there in the live session then write down right now I'll end the stream after 2 three minutes you will be able to write down the comments I want at least 100 plus comments on this video so that I can take the session right okay because as you know that I'm already occupied with so many things but still I'm willing to teach you if I see the audience is there the students are willing to see the student If the students are willing to uh you know invest their time then definitely I would love to teach you but the condition is that definitely your motivation should also be there right okay so uh chats will definitely fade out but I'll you have one day still to think about that but yes 100 comments should be there before tomorrow's like morning or afternoon so that I can also prepare my lecture for organic chemistry it takes a lot of time to collect questions for you especially for VIT triple E right okay yes okay then bye-bye students bye-bye take care see you all in the tomorrow's class where I'm going to take up UG class right for your uh different different questions and really interesting awesome questions will be there so your J advanced students going to also watch that lecture okay till then bye-bye take care see you all in the next class bye-bye this is Dsha Koshal signing off bye-bye Okay