 Hello guys. Good evening. Few minutes. We'll start the session. Few minutes more. Yes, Anuram. Yes. Okay. So I think the last class we have discussed about electronic configuration. Unstool we have discussed. Yes or no? Tell me the last thing that we discussed last class. Where we stopped? Yes, sir. Sir. Excuse me, sir. Can you help me? Unstool. Fine. Okay. Is it fine? The last few things we need to discuss and then we'll move on to the next chapter in this. Okay. Poly exclusion of both we have discussed. Have we discussed the examples of, just a second mother, have we discussed the example of electronic configuration, chromium copper? Okay. Fine. Fine. I got it now. Yeah. Suppose electronic configuration we need to draw for, draw the electronic configuration for the first one is hydrogen, helium, lithium, perilium, boron, carbon, nitrogen, chlorine, sodium. For all this draw the electronic configuration. So we know the energy order. The energy order is this. Last class I have given you already this thing. My audible to all of you. Arnav is not getting me. What about others? Arnav, you can rejoin. Okay. I think there's a problem from your side. Because all other can hear me. So you can rejoin. Yeah. You don't know my name. My name is this. It's Gaurav Sinha. Then what? Okay. This one's name. Post name, right? Yeah. I have changed that. I have changed that. Okay. Not an issue. I'll rename it in the next session. Let it be. So electronic configuration. The first thing we should know is the atomic number, right? Which is one for hydrogen. Two. Three. Four. Five. Six. Seven. What is the atomic number we have here? Nothing. This is the first thing we are discussing. Okay. So electronic configuration. We discussed last class that the electron is the, it is a distribution of electron in the orbital. Right. And electron fills into the orbital. According to. The energy of the orbital from lower energy to higher energy. Right. One orbital can have maximum two electrons. Right. So if you see this, we have S sub shell. S sub shell has only one electron. And it can have maximum two electrons. If you talk about P sub shell. P sub shell has three orbitals. And each orbitals can have two electrons. So six electrons. D sub shell. Has five orbitals. Maximum 10. Maximum 10. Maximum 10. Maximum 10. 10 electrons f sub shell can have 14 electrons maximum possible right so with this reference only will distribute the electron in the what we say in the atom correct so hydrogen has only one electron so you see we'll start from here 1s one electron we have so 1s1 the configuration is for this one two electron so 1s2 leave him as three electron so 1s can have maximum two then the third electron goes into next higher energy orbital 2s1 fourth electron will go into 2s again 2s2 it is 1s2 then 2s2 that is for fifth electron will go into 2p and 2p can have maximum six electron one will have here if you talk about for carbon so 1s2 2s2 2p2 nitrogen 1s2 2s2 then 2p3 chlorine 1s2 2s2 2p5 and sodium 1s2 2s2 2p6 10 electrons then after two picks the next higher is 3s so it is 3s1 electron left I'll give you some more mother finish it first they are the very basics one no I didn't forget intensely I did not give sir yeah sir how how do you represent the like the core like the noble gas core only the violent shell in terms of noble gas you're saying yeah argon and then 4s2 3s2 4p like on Google I saw the electronic configuration it has the noble gas and then the noble gas we can use I'll show you we've just started copied this one did you understand this just you have to follow this part like this and whatever cells orbitals are coming in the part you have to give electron into that shells okay depending on upon their capacity spdf this is the capacity we have maximum finish all of you okay now we have one more rule in this and that is write down that is hunts rule hunts rule write down the statement and then we'll see the rule how it what it means statement write down in an orbital in an orbital I'll also write down wait the statement is in an orbital bearing of electron is not done is not done unless all orbitals belongs to unless our orbitals belongs to the same sub shell sub shell are singly occupied what is the meaning of this statement we have copied this down I'll explain this you see these are p sub shell I am assuming and this box is represents the orbital in the sub shell because it says what when all the orbitals belongs to the same sub shell means the orbital belongs to d sub shells are dx y dy z dz x dz square d x square y square okay orbitals belongs to p sub shell is px py pz right so it says in an orbital the pairing of electron is not done unless all the orbitals belongs to the same sub shell are singly occupied means orbital belongs to the same sub shell must have at least one must have one one electron then pairing is possible pairing of electron is possible suppose this one here this one suppose we have it is 2p1 configuration p sub shell 2p1 configuration this is 2p2 configuration this is 2p3 configuration this one is 2p4 this one is 2p5 excuse me this one is 2p6 and these are what these are the three orbitals belongs to the cell sub shell p that is px py pz here also we have px py pz and same thing we here we have for the other three okay so when you have 2p1 configuration first of all you see if there is no electron present in this orbital the orbit is empty right at this stage when the orbital is empty I want you to write down these things in your own language into your notebook just in short you write down so that you can understand what this cousin was going on why I said that when the orbitals are empty then the energy of these orbitals are same and it is known as degenerate orbitals right degenerate orbitals all these orbitals belongs to the sub shell p has equal energy now when electrons we have 2p1 configuration it means one electron one electron is trying to fill into this orbit right it is not is trying to occupy the space in this orbital so this orbit this electron can go into any one of these orbitals that we do not know right we can it can go into px py and pz because all three orbitals has equal energy right it can go in any one of these orbitals right and the first electron that enters into any orbital that electron we assume to have clockwise orientation and that we represent upheaded arrow like this so for you know just an assumption we write first x y and z so first electron goes into px then p1 and then pz that is the assumption we take but it can go into this also and this also because all three orbitals has same energy now suppose you have two electrons like in this case two electrons trying to fill into this orbital then one electron comes over here the next electron goes into which orbital according to hunts rule this will go into the next orbital which is here pairing is not possible now if it is three electron then one one electron will go into the each orbital first i'm sorry one one and one like this the distribution of electron takes place in the orbital okay so pairing is not possible unless this stage is there i'm talking about p sub shield if you have d then first one two three four five and then pairing takes place okay so pairing is possible when we have this particular configuration then we have two p four configurations of four electrons you are trying is trying to enter into this orbital so out of four we have one then two three when we have this all the orbitals belongs to same sub shell are singly occupied and clockwise orientation so fourth electron will go into any one of these three orbitals having anti-clockwise orientation right two p five when five electron enters into this orbital so first all the orbitals are singly occupied like this and then the pairing takes place fourth and fifth anti-clockwise orientation no no it's not i'll talk about the direction first let me explain this if you have six electron then what happens will distribute the electron like this one two three four five and so this is what it means okay pairing is not possible unless all the orbitals belongs to the same sub shell are singly occupied did you understand the rule first of all tell me right now since electrons are rotating about its own axis this rotation we do not know actually okay again i repeat this one second for see in any sub shell you have a given number of orbitals correct so according to hunts rule what he suggests that all the orbitals belongs to the same sub shell must have one one electron singly occupied one one electron present over here after that only the pairing takes place like suppose if you have two electron just a second the point here is suppose this is a p sub shell we have right and it has three orbitals that is px py and pz px py and pz so if you have suppose two p two p four configuration suppose you have right so you cannot do like this because you know one orbital can have maximum two electrons so you cannot directly do this this is not possible it is not right it is wrong according to hunts hunts rule what you need to do you have to do this you have to do this one again i am doing this just to make you understand okay you have orbitals like this two p four configuration so what happens first all the orbitals takes one one electron like this and then we'll start the pairing like this this is hunts rule read the statement in an orbital the pairing of electron is not done unless all all all orbitals belong to the same sub shell are singly occupied this is what the rule is correct fine right so as far as the orientation of electron is concerned the electron can be clockwise can have clockwise orientation or anticlockwise orientation that we do not know again and we don't want to get into this discussion also actually why because first electron we assume to have clockwise orientation means if you have up headed arrow like this if it is written up headed arrow whenever we have it means the clockwise orientation clockwise orientation no it's not like yeah it's relative it's relative you can see but but the electron that enters into the orbital the first electron is assumed to have clockwise orientation yes correct correct and i can say that but the first electron is clockwise that's why we always write up headed arrow the first electron now since it is clockwise right and when the second electron tries to enter the same orbital if the second electron has second or technically i should say fourth electron because second electrons will come over here third electrons will come over here fourth electrons then pairing takes to the piece of if the fourth electron tries to enter into this orbital first electron has already clockwise orientation if the fourth electron also has clockwise orientation then there would be repulsion and hence that is not possible right to minimize the repulsion when electron is trying to approach here into this orbital this clockwise repels this clockwise orientation so this electron changes its orientation from clockwise to anticlockwise or this fourth electron prefers to have anticlockwise orientation in order to have minimum repulsion that's why the fourth electron we place like this and it is anticlockwise orientation did you get it tell me no doubt that is what han sol means if you have supposed d seven configuration i'm assuming configuration here suppose we have d seven configuration this is one three four and five configuration is d seven d seven configuration the d seven is what all the orbitals will be singly occupied first so one two three four five six seven this is the d seven clear no doubt okay this is han sol okay write down the electronic configuration anyone anyone like i said we do not have any control in which orbital the electron goes right so it can go into any orbital can be py can be pz can be px but can be like like convenience like what what the you know assumption we have or by convention what we'll write we always write x first then y and then z that's why it is written in that order otherwise the electron can go in any one of the orbitals write down the electronic configuration for argon electronic configuration for calcium then scandium phosphorus chromium ion then copper yes box diagram i'll tell you what it is see actually no no box diagram you don't have to write down just try to 1 s o 2 s o 2 p 6 like that one more thing i'll tell you here this box actually represents one orbital because every time it is difficult to draw this dumbbell ship right so we usually take the reference of this box this means it is an orbital one box orbital and that is what the box diagram we have okay but for this configuration you don't have to draw the orbital the box diagram you can write down the order like that one is two two three six and so on what is the atomic number of argon could you tell me the atomic number of argon 18 atomic number of calcium 20 20 scandium 21 phosphorus 15 chromium 24 ion 26 copper 29 from hydrogen to zinc at least better if you memorize still krypton but to zinc you must memorize the atomic number now you write down the electronic configuration here okay one more guys one small request to all of you i would request all of you to write down the nodes properly during the class only you don't uh you know expect me to send up pdf nodes after that class because we are you know going to change this particular thing because we got to know that some of you some of the students maybe from not from your class maybe from the other batch they don't write down the nodes properly that's why from the last two class probably if you see past one or ten days one week i'm not sending the pdf nodes on the group however i'll send it for today right but you never know maybe for some class maybe next class i won't send so i request all of you write down the nodes during the class only okay last week many of you have asked me the nodes for the last class okay and i have already i had deleted that i could not give you that time i request all of you so write down the nodes during the class only done okay so the electronic configuration here for argon is 1s2 2s2 2p6 3s2 3p6 for calcium it is 1s2 2s2 2p6 3s2 3p6 4s2 right this one you see here from 1s to 3p6 here this represents the argon configuration right and argon is the inert gas that's why we can also write the electronic configuration in terms of argon and till here i'll write down ar simply that ar represents 18 electron and then we have two more so 4s2 that is what it means okay now in this one it would be 1s2 2s2 2p6 3s2 3p6 4s2 and then 3d1 okay 3d1 if you write down in terms of argon this electronic configuration would be argon then 4s2 3d1 okay 15 this is no we don't we cannot we cannot write that strong as a neon you are writing it down neon neon 3s2 that that it is not wrong if you write this but we write in this way only so that we can have see till here it is same no 18 electrons till here it is same that's why we write till here there's no point of writing down neon here and then there's all these things here the purpose of this is if you have a very long electronic configuration then we can write in a shorter way when we use argon over here what extra ever 1s2 2s2 2p6 3s2 3s2 3p3 okay you can write neon here that's not wrong neon then 3s2 3p3 it is not wrong you can write like this okay this would be 1s2 2s2 2p6 3s2 3p6 4s2 3d4 have some patience I know you are going to object over here have some patience okay I intensely I have written it 1s2 2s2 2p6 3s2 3p6 4s2 3d6 this one 1s2 2s2 2p6 3s2 3p6 4s2 3d9 okay all these configuration we have correct but these two are not right here this one is not right this one is not right and this one is also not right these two are wrong the correct one is what the correct one is in presence of this two will have 4s1 and 3d5 and this one would be 4s1 3d10 why it happens because this 3d5 has exactly half filled configuration exactly half filled configuration half filled configuration means all the orbitals are singly occupied 1 1 electron each present in the d orbitals half filled configuration and symmetrical also you can say symmetrical arrangement sir yes so then why does this not hold true for silicon sir because for silicon also it is 3s2 3p2 right sir so wouldn't 3s1 and 3p3 be more stable more symmetrical see energy gap is the factor i'll discuss that let me finish this i'll discuss that why it is not true for s and p transition okay here what happens you see this one is exactly or fully filled configuration fully filled configuration and again symmetrical arrangement symmetrical arrangement means symmetrical arrangement of electrons in the orbital like if you draw 3d10 configuration all the orbitals are fully occupied completely filled 3d5 configuration is this all the orbitals are fully occupied completely filled one two three and then four one two three and then four right if you draw the orbital diagram you'll get one two three four and five here we get one two three four five and then pairing six seven eight nine ten you see all the orbitals are one have one one electron that's why it is symmetrical arrangement and this gives extra stability to the atom symmetrical arrangement gives extra stability extra stability and hence it is and hence it is more stable here also we have extra stability hence it is more stable and electron forms this arrangement okay so now if you see what happens if you look at this why it does not take place in s and p like 3s2 3p2 that we have the question that you asked okay so why it is not happening here because you see the orbitals one factor is the energy factor here suppose it is this is the third shell we have okay and in the third shell first we have s then p and then d okay i'll just write a bit small and we have a fourth shell which is this which is this right so this energy gap it is a transition of electron action from 4s to 3d there is one transition of electron and hence the configuration becomes this okay why this transition takes place because the energy gap here is not that great the energy gap between 4s and 3d 4s and 3d orbital is not that great that's why the electron delta e is not large here and hence the electron can jump into this transition is possible here but s and p orbital if you see 3s and 3p the energy gap is some is relatively large and that energy gap is sufficient enough which prevents the the excitation of electron from s to p orbital that's why this thing is true only for s and d orbital like 4s and 3d we have here understood this all the atoms tries to have try to have lesser energy state and the distribution of electrons in orbitals affects the overall energy of an atom right so when we have symmetrical distribution the overall energy is less the electron is the atom is more stable in that case that's why whenever it is possible obviously depending upon the energy difference whenever it is possible there's a transition of electron in order to have more symmetrical arrangement and lesser energy state understood this tell me did you get it that's what i said uh anura that's what i said that whether it is fulfilled configuration or half filled configuration it's a relative thing if you have only one electron here you see this if you have only one electron here then there will be some energy right if you have one electron like this one here and other one is here it will have some energy means unsymmetrical arrangement so it has been observed that when the here in this state the energy that we have and in this state the energy that we have this has more lesser energy more stability over here because of the symmetrical distribution of electron all the orbitals has one one electron each right so this arrangement of electron decreases the overall energy of the orbital or atom we can see and hence the stability we have like this we won't say that fully filled and half filled configuration has lesser energy overall the energy decreases and stability increases see all our electrons right anura so if you have unsymmetrical distribution correct so there will be some repulsion in the electron we have to find out the state the arrangement of atoms where the repulsion is minimum then only the stability will be more so repulsion will be minimum when the arrangement is symmetrical if suppose this one is repelling this electron right so other one it is going towards the other electron right so net thing is what net arrangement the net force or net repulsive or attractive force whatever is present over there that decreases the energy of the atom right so it is basically the arrangement of electrons which affects the energy this one you see you have two possibilities okay this is 4s this is 3d this is 4s orbital and this is 3d orbital we have 3d shall we have so one arrangement is this when the 4s has two electron and 3d has four electron right so here we have some repulsion in two electrons and the other arrangement if you talk about and the other arrangement is this when this electron jumps from here into this orbit so this has certain arrangement of electrons and here none of the orbital has two electrons right so obviously the repulsion will be less here in comparison to the other configuration that's why this one is more stable this is the easier way you can understand this here we have two electron more repulsion less repulsion you can get it but the thing is here the overall energy decreases because of the symmetrical arrangement of it right that is what the thing we have now electron see electron see first of all electron fills into the orbital in order to get you know in order to get more stability of the atom so obviously the electron first go into 4s and then 3d but in the case of chromium and copper the electron won't go into 4s it goes first it fills 3d because it is getting symmetrical arrangement here more stability and then it is left with one electron that goes into 4s a little bit of different arrangement we have here the overall objective is what to get lesser energy state which is more stable in order to get more stability only the distribution of electron takes place in this way okay understood now this is a hun suit we have discussed it last part of this chapter we need to discuss that is nodal surface correct but before that one small thing we have to understand here two small thing in fact so write down just one formula we have here that you need to keep in mind write down that is spin multiplicity spin multiplicity we calculate this with the formula to summation of s plus one this is the formula of spin multiplicity how do we calculate this suppose we need to find out the spin multiplicity of carbon in excited state right for carbon in excited state so carbon has six electron we draw the electronic configuration of this that is 1s2 1s2 2s2 and 2p2 the electronic configuration is this if you draw the box diagram of carbon that would be this 1 2 this has two electron and 2p has one one electron right this is the thing we have this is in ground state this is in ground state in excited state what happens one electron jumps over here to make a bond this carbon goes into excited state and one electron jumps into the vacant p orbital and in this case it has four unpaired electron one two three four this is the excited state we have excited state okay so spin multiplicity is what you see here this is 2s this is 2p this is also 2s 2p and 1s we also have here which has two electron like so spin multiplicity is this and how do we calculate this you see we have two summation of s means the number of electrons with clockwise orientation that is 1 2 3 4 5 so 5 electron with clockwise orientation means plus half is the spin multiplicity for this 5 electron plus one electron has anti-clockwise orientation 1 into minus half bracket close plus one this is the spin multiplicity of carbon and when you solve this you will get 5 into plus half 2.5 2.5 minus 0.5 2 that would be 5 a spin multiplicity of carbon atom is 5 that is how we calculate transition of electron requires some energy so when the carbon atom is going for reaction so in the reaction we need to provide that energy then only the electron excites no we do not have orbitals here it is a it has second shell and there is no orbital after 2p no chromium was not an excited state the configuration of chromium is 4s 1 3d 5 it is not excited state 2p electron if it gets excited the electron should go into the higher orbital no do we have orbital here no we do not have do we have 3d second shell 3d orbital is not there no third energy shell we do not have over third energy shell third energy shell we do not have it there it is only for second shell we have see transition of electrons takes place in the same sub-shell first of all right transition of electron takes place in the same sub-shell if you change the shell or sub-shell right means within second energy shell transition takes place i'm not talking about chromium and copper here chromium and copper has an exceptional case due to symmetrical arrangement correct but when the electron goes into the higher energy shell it does not mean that from 2p it will go into 3s or 4s or 5s that is not possible because all these transitions requires energy right electron transits from one orbital to another depending upon the energy available there right so in 2s and 2p the electron can transfer from 2s to 2p and what is the point of this first of all 2p to 3s you're talking about there is no possibility why because 2p has only one electron it won't go into 3s because 3s has higher energy than 2p here what happens when the electrons present here so we'll have certain repulsion and when you provide energy to this then this electron transfers over here and hence it has four unpaired electron that's why carbon can form four bonds with any other element and it is tetravalent in nature right so its transition is not according to your choice or our choice right from one shell to another shell transition is difficult first of all it requires high energy number of unpaired electrons with that only we'll find out with this particular formula spin multiplicity this is the formula we have used you see five clockwise orientation five into plus half one anti clockwise orientation one into minus half plus one into two excuse yes but spin multiplicity we do not count only for unpaired electrons these electrons also we have to consider okay you said 2p has higher energy than 2s you know the difference between 2s and 2p and 3s and 2p you know this difference if the transition of electron is possible in 3s so we'll have certain energy required for this if that energy is more than the energy required for this transition then which transition takes place anura tell me 3s and 2p had some energy difference 2s and 2p had some energy difference that is what usually what happens 2p to 3s transition is very difficult and it won't happen before that only 2s 2p transition takes place and suppose hypothetically if i assume your point that the electron transfers into 3s then also it has three electron unpaired electron so carbon in that case will form only three bonds that is not possible carbon is a prevalent forms four bond and four bond it can form only when it has four unpaired electron and to get four unpaired electron this electron has to excite over here okay so this is the thing we have transition of electrons in excited state takes place in the same sub shell okay sub shell we can say or we can say in the same shell that is a better word okay 3s 4s 3d transition is kind of exceptional case we have in case of copper and chromium right symmetrical and I mean more stability i think much we have this is how we calculate the spin multiplicity with this formula okay this is one thing now one last thing here we need to discuss before nodal plane that is that is the orbital wave function we have discussed it already the orbital wave function psi write down one last thing one question they ask on this orbital wave function this psi i have already told you that it contains all the informations of electrons it contains the energy like it contains the information related to energy of an electron because in case of chromium and copper the transition of electron is possible and it gives symmetrical arrangement it depends upon two three factors one is the difference of the orbital that is four is in 3d right and transition of electrons also so in that case only that energy gap is not that high and hence electron can jump into the higher energy orbital that is 3d orbital that is why it is possible for chromium and copper only see the question is why those two elements the exception we have it is it is because of 3s and 4d electron sorry 4 3d and 4s electron the difference in the in these two orbitals in these two substance the difference in energy is not that great and hence the transition of electron from 4s to 3d is possible over here if you go to higher energy orbital like suppose we have next is 4d and 5s suppose i am taking for example then in that case also the energy difference is there and that is you know that difference does not let this 5s electron to jump into 4d so it is one kind of exception we have that we observe only in the case of 4s and 3d nothing much no in this we do not have in transition element we have a bit here and there we will discuss that in grade 12 not now okay so we know this orbital wave function it contains all the informations about energy and the distance of electron from the nucleus and this also gives us the probability of finding an electron around the nucleus right this wave function has two parts actually here this is the radial part psi r and it has the angular part angular part we represent by the angle it is made by it theta and phi right radial part and angular part only two things you need to know here nothing much the angular part that is the distance of the electron from the nucleus r this is this psi r is the is the radial part we have the radial part and theta and phi is the angular part radial part and angular part this radial part it depends upon depends on the value of n and l principal quantum number and azimuthal quantum number right so this radial part gives us an idea about the size idea about the size this is the angular part and this gives us first of all it depends upon depends on the value of l and m the quantum number size of the orbital l and m and it gives us the idea of the shape shape of orbital okay so size means what suppose n is the is the principal quantum number so if the value is we have n 1s 2s then the shape that the size is what size would be like this 1s will be like this and 2s would be like this this is the size of the two orbital right if you have 2s 2p sorry if you have suppose 2p and 3p the orbital we have 2p and 3p so both has both has double shape so 2p suppose this then 3p would be larger than this this is what it means right that's why the that's why the strudinger wave equation it gives us an idea of the three quantum number n l and m sir so using the radial part we can't uh compare the size of 1s and 2p 1s and 2p no spherical we cannot compare no dumbbell we can compare with okay only the same shape we can compare same shape we can compare means how do we compare a sphere and a dumbbell that's not possible right we can say 3p has a larger low than 2p got it this is the two things we have now you'll get this question in the exam and they'll ask questions like this they'll give you they'll give you the notation here like the question is we'll have psi and here they'll write down some number like this suppose 4 2 0 like this some number they'll give you and they'll ask you that which orbital is this means orbital related question they'll ask you which orbital is this right so you should compare this with this one when you have psi and here the number which is written it is in this order n l and m means if you compare this to what is the n value for l value is 2 and m value is 0 which orbital is this this is 4 l value is 2 with sub shell with sub shell d m is equals to 0 it is z square so it is 4 d z square the orbital is understand clear so only this question you'll get nothing more than this okay and and m value the last topic for this chapter we have write down write down nodal surface just a second i'm coming wait okay write down the definition nodal surface write down it is the surface at which it is the surface at which the probability of finding an electron is zero nodal surface the surface at which the probability of finding an electron is zero this is a nodal surface so we have a relation of psi that is wave function because we are talking about the probability of finding an electron is zero correct probability of finding an electron is zero so we must have to use the psi square function right that is a variety density function or wave function psi we can say right so what happens here different different orbital this nodal surface we have two types means basically you have two formulas which with that you can easily find out the nodal surface we have two types of nodal surface actually write down of nodal surface we have the first one is the first one is a spherical node spherical node we also call it as radial node spherical or radial node number of spherical or radial node is given by n minus l minus one these are the quantum numbers okay the second type we have that is non-spherical non-spherical or angular node or angular node and its formula is l these are the two types of nodes we have they may ask you what is the number of nodal plane for this particular you know orbital you're right these are the two nodes if I ask you what is the total number of nodes that would be the sum of these two n minus one this is the formula we have okay so first of all you should know this formula okay how to find out the nodes in any correct this is one part of it we have a graphical representation of this also right how do we represent the graph okay that is the relation of psi and r then psi square r square and then the probability with this okay so basically if I ask you find out the node for one s orbital first of all one thing you see this spherical node we can represent on graph we can show this on graph that at certain distance we have this surface spherical surface where the probability is zero angular node we don't represent on graph basically right so if it is one s orbital so number of a spherical node is what for one s orbital a spherical node for one s orbital is n minus l minus one so n value is one l value is zero minus one that is zero there's no spherical node angular node if you can count here is also zero there is no node here for one s right now based on this if you draw the graph little bit I am taking the help of experimental relation because we do not know the relation of psi and r but if you draw the graph of psi and r the graph you will get like this I'm just drawing it down three graphs I am drawing here what happened okay this side we have psi that is a wave function and it is the radial distance distance from the nucleus this is r this is the probability density function psi square and this is r and this is the probability 4 pi r square dr into psi square this is the probability we have this axis and this is the radius did you understand this term 4 pi square is a surface area of a surface of a sphere and dr is the thickness of that so 4 pi r square dr into what I'll explain you wait suppose we have a surface here a spherical surface we have like this and the radius is r so what is the surface area for this 4 pi r square 4 pi r square surface area so this is this is the radius r we have from this point this is the radius r okay and the thickness of this radius of this surface is dr this is dr this thickness this is the thickness we can assume as dr right this surface area if you multiply this with the thickness it gives you volume right so volume of this sphere is what the surface area of the sphere 4 pi r square into dr correct because the thickness I am assuming here as dr this thickness is dr I am assuming correct and we already know that this psi square is the probability density function right so the probability for the probability of finding an electron is equals to the probability density function psi square it is probability per unit volume into the volume and volume is for the surfaces 4 pi r square dr so when you integrate this you'll get the probability is it clear enough tell me yes so this graph is with probability with radius it is a probability density function with r and this is wave function with r for 1s right this is for 1s we have so the graph we here we have you see the graph for psi and r with logically we cannot find out because we do not know the relation of psi and r but this graph is observed to be like this psi square and r graph also goes deeper and it goes like this goes like this won't touch the x axis go like this okay this is the probability density a probability graph probability and this so at some point you know in it is a nucleus here in the nucleus there is no electron present so it starts from the nucleus right it starts from the nucleus goes to the increases like this and at some point the probability is maximum then it comes down and goes like this just a second right so it can go like this increases and comes down it won't touch the x axis because at this point the probability is zero but for 1s orbital there is no nodal surface here that's why it is not the it won't touch the x axis yes it won't touch the y axis it is asymptote here so this is the graph of 1s orbital this is the probability graph which is the important one sometimes what happens they'll give you this graph and they'll ask you to find out the which for which orbital the graph is right so since this graph does not touch x axis that is this one hence the probability is never zero here and that is possible with 1s orbital did you get it okay your copy this down i'll discuss the another one you will have a clear you will have more clarity into this one copy this down no this we can draw the graph with the relation of psi and r the radial relation that we have obviously the equation we do not have hence we cannot predict this okay the next orbital you see suppose the second one we are taking 2s orbital so for 2s orbital what we can write the number of spherical node could you tell me tell me the number of spherical node a spherical node is equals to n minus l minus 1 2 minus 0 minus 1 that is 1 and the number of angular node that is l value which is 0 there's no angular node here okay so when you draw the graph of this one also i'll draw it quickly here so this graph is again psi and r this one is psi square this one is the probability that is psi square 4 pi r square dr and this axis is the radius we have r okay so here the psi square and r graph goes like this from here it comes down like this and it goes up but it won't touch the x axis like this it won't touch this okay wave function can be positive can be negative and that's why we have the graph like this but if you draw this psi square so here psi square cannot be negative it is always positive this negative part will draw the mirror image of this to draw the graph of psi square that would be like this it comes down and at this point it goes up comes down and go like this okay so this point is the node we have this point is the node one node we have this is the point where this psi square is 0 means probability is 0 so when you represent this in probability graph we have one node here the graph goes like this it will go up comes down we have a node so it will touch the x axis and then again it goes up to maxima and comes down and it goes like this this point is the node so obviously it has a distance certain distance the node is present yes so won't won't the node touch the like won't the node be a nodal region won't it like how would it be a point sir because the probability on the graph see on the graph this means what from the nucleus at this distance we have a node that is what it means but when you draw this the surface I'll show you how do we draw it it just represent that at this distance one nodal surface is present and what is the nodal surface it is a speaker surface no sir I meant that shouldn't the this graph touch the x axis for not only at one point at like it should touch the x axis for some time and then go up because it should be a nodal region where the probability is 0 right wait your voice is feeble can you talk louder yes sir I meant that in the graph of the probability of finding the electron so shouldn't the the graph touch the x axis not only at one point at the node it should touch at the full at the nodal region which is extended in the x axis sir because the probability of finding electron is zero for like between first option and second option so the radius is not only at like see particular radius it's like a region from like say uh two millimeters to three millimeters yeah no see uh you are getting it wrong I'll explain this right see you have a certain distance this distance correct now this means what we have a nucleus first of all the atom suppose the atom is this let me draw the diagram here suppose the atom is this right in this atom we have a nucleus which is here and around the nucleus at this distance from the nucleus this distance we have a reason we have a spherical reason actually we have a spherical reason where on this reason the probability of finding an electron is means whatever the distance we have suppose this distance is uh uh I'm assuming this this distance is suppose r one we have this distance is r one same distance we have here also this is r one so at this distance from the nucleus this is r one you will have a spherical surface obviously we have some thickness of this also you will have a spherical surface and on that surface the probability of finding an electron is zero so it is a reason only it is not a point okay sir but then see sir if you draw the first shell that n equal to one and n equal to two so then between those two shells that there's a region right it's not only a single radius there's a region yes spherical spherical surface yeah yeah sir like the two concentric circles are there the region between that correct that that consists of more than one value of r right no for a one particular value that's what i'm telling you for this value we'll have a surface we'll have a reason spherical surface where the probability is zero it's clear no we have only we have only one node correct so we have only one spherical surface what distance from the nucleus this distance from the nucleus no sir but then it says that the probability is zero to find the electron yes so that is in the full region between the two electron clouds right no it's not the full reason why you are taking you're you're talking about this entire reason the probability is zero possible one is in two as if it is yeah i am telling you what that in between this we have a spherical surface in between this at a distance this r one on that sphere the probability is zero it's not the reason it's not the entire reason between the two so so in the only in the middle of the two there is it is exactly zero let's not let's not phrase is this way middle of the two you simply can say that at this distance from the nucleus you will have a spherical surface where the probability is zero that is it when you say spherical surface it is the reason only it is not a circle no sir but tell me sir sir like see imagine so the first circle you drew is one s and the second circle you drew was two s two s correct so so the region between that is empty right there's no electron over there that we cannot say that we cannot say if this is the sphere of this distance then it is possible if the distance suppose i'm giving you a value suppose this is five meter for example correct and this is suppose we have 12 meter the radius of this problem correct and we have a nodal surface at seven meter then the nodal surface will be seven meter from this point you can draw a circle a sphere so but isn't the probability of finding electron zero from five to twelve no that we cannot say that's what you are actually you are not even expecting accepting that definition of it whatever the see we can have a sphere of this radius and this thickness possible this thickness and this radius we can have a sphere of this radius and this thickness where the probability is zero we can have it but it is not like we have always this condition i am giving you one general and simple answer for this what i'm giving you is the graph that you drew here it must have certain distance from the nucleus so at this radial distance you will have a spherical surface within the atom where the probability of finding an electron is zero you are talking about that we can have a surface here here we have electron and here we have electron but in between these two we do not have electron which maybe may not be true depends upon the case okay so so in this in the graph of uh size square the point where it touches the x axis is seven meters not five yes yes exactly that's what i'm point that the point i'm trying to make whatever the distance it has to touch only at one point right whatever the distance we have here at the same distance we have a spherical surface that's what i have written r1 and r1 you see this r1 and r1 at the same distance we have a spherical surface where the probability of electron is zero so this is the nodal surface if you if you can write down this this is a nodal surface we have where there's no electron this circle according to the graph i am drawing here with respect to that this is it must have certain thickness right because in sphere and this is the nodal surface we have precisely you can see so anything except that seven meters the probability is not zero but it's very close to zero yes anything that is not our no concern okay it can be anything but not zero yes you understood again you see i'm explaining just i just i'm just you know concluding this you will have this graph and at this point the size square graph is cutting the x what what does it mean that this distance from the nucleus psi square is zero and when psi square is zero the probability is zero that's why we have one nodal surface here why it touches only once this x axis because we have only one spherical node and remember i told you already that we can draw only a spherical surface here angular we cannot draw okay two things you should keep in mind the probability graph always starts from origin any any confusion in this the probability graph always starts from origin any doubt in this because in origin is nucleus and nucleus electron can't present correct so this graph always starts with nucleus and the number of and the number of peak that you are getting right it is one more than the number of node present into this we have one node and we have two peak these two points write down the probability graph always starts with zero origin the probability graph always starts with zero second point the number of peak the number of peak just a second stress is one more than the number of node the number of node psi is the wave function stress okay wave function is like amplitude function for a wave so wave travels like this correct up and down like this so above the axis we consider this as positive and below the axis we consider this as negative right that's what it means amplitude can be positive and negative because it goes up and down that's the reference we are taking above the axis is positive below the axis is negative sir yes so and also for the psi square 4 pi r square dr graph so shouldn't the the peaks be opposite like the because the if the r is below the axis hello yeah tell me yes i was telling for the third graph so if the r is less then the probability then the max probability will be higher than when the r is more right i didn't get you max probability will be higher when the r is more why why is that no sir when when r is less when when the uh shell is closer to the nucleus so then the probability will be of finding electron will be more than then the shell is farther from the nucleus right so shouldn't the peak be opposite take the no it's other way as you go closer to the nucleus the probability decreases because electron is not cannot present in the nucleus and this graph this is just a representation this graph it could be like this also could be like this also yes i shouldn't the second mountain be smaller the second the first possible depends upon the atom both kind both possibility in the book if you see they have drawn the graph like this one okay but it is possible that depends upon the atom that we are going away from the nucleus the probability decreases similarly if you come closer to the nucleus we know in the nucleus also the probability is zero for finding an electron right so we cannot always say as we go away from the nucleus the probability of finding an icon is zero always so this could be anything it can be down it can be up depending upon the atom but yes we definitely have two peaks one more than the number of node present okay tell me psi square is mother psi square is the probability density function i have discussed this in schrodinger wave equation okay psi square is like it represents the electron the intensity of electron at a point if psi square is maximum probability of finding an electron is maximum because the intensity is maximum of electron you can compare this with psi you can compare with the amplitude of a wave light wave right and psi square you can compare with the intensity of a light wave no we won't do that that is not in our status no psi square cannot be negative that's why we are not drawing the graph below the axis here negative probability does not make any sense correct and that is the reason we have taken psi square here yes that is what so fine we have drawn it no you've drawn it no but this negative part in terms of probability it does not make any sense that is why we have taken psi square here that whatever the negative thing is there it becomes positive like this which graph this one and this one it represents the intensity of electron at any point right it represents the probability of electron at any point around the nucleus this is intensity this represents probability right so here the intensity is zero means probability is zero mathematically also it is correct psi square is zero this term is zero this is yes these two I have explained this one is the amplitude right of the electron wave amplitude of the electron wave which can be positive and negative so this will go below the axis as well that's why we have written this yes actually this is a bit you know a confusing if you try to understand each and every point here at this level it is a bit confusing because it is all like I said it is quantum mechanics but to simply like simply what you can understand with these two formulas just whatever if you're not getting it like you can think about it and then we can discuss this later you need probably more time on it just a second just a second what you should do here just last we are doing the last part let me finish this what you should do here mostly they'll ask you one question that I'll give you after this when we finish this I'll show you what kind of question they ask you should know what is nodal surface you should know the formula with which we'll find out the nodal surface correct with a given orbital what nodal surface is there that that also I will discuss with graph what you can conclude that you should know this represents probability when where the graph touches the axis the probability is zero and with with the help of this graph you can easily find out that this graph is for which orbital like suppose I have given you this graph and I ask you this represents which orbital 2s 2p 3s 2p 3s 2s or 1s what is your answer for this graph let it be whatever I have written here just remove it this graph we have and you have four options a b cd this graph represents which orbital what you will do you will find out the number of nodes for this orbital the number of nodes for 3s is what the number of the number of nodes for 3s is 2 the number of nodes for 2p total nodes you tell me we have only one angular nodes right only one angular nodes we have in 2p we have 2s one spherical node 1s orbital has no nodes is it clear are you comfortable with these values tell me all of you please respond yes so obviously you can eliminate these two possibilities right yes or no guys you are not involving please respond we can easily eliminate these two correct now in this one in this one because in the graph we have only one node correct but we know this is the angular node an angular node we cannot represent on the graph we cannot show this because angular node has complex shape right we have a you know if we are double double double leaf like a structure angular node has a complex shape if you are going to draw the angular node you will be drawing the angular nodes only in the exam right it's not about the drawing test right that's why we don't draw angular nodes on paper because it's complex one simple thing right so since the graph is given so we only have the spherical node possible which is there in 2s hence the answer is this so this kind of questions you will get based on the graph you can have many questions but those things they won't ask that is the best part and you will not lose marks on this I can bet you on this it's very simple kind of question there okay one more example I'll show you for 3s and then we'll move on from this you see in the graph they won't give you they won't show angular node at all if only angular node is there you can eliminate that option easily okay fine now you see if I draw the graph for for 3s orbital I want all of you to try this on your own 3s orbital try this first of all you find out the number of spherical node that is n minus l minus 1 3 minus 0 minus 1 it is 2 and the number of angular node is l and that is zero two spherical nodes now we'll draw the graph the three graphs will draw here see so we can only draw the graph for s orbital it's not for p e d n f right no for 2p also we can draw for 2p also we can draw so but then for that for 2p angular angular node is one right so you told we can't represent angular node so a spherical node we can represent angular we won't represent for 2p suppose there is no spherical node so the graph won't touch the axis right and angular node we won't represent I'll draw that also wait it's the same graph it goes like this for 2p a curve like this a mountain like this for psi psi square and probability three graphs we'll have like this I'll show you just so this value is zero okay this is psi and this is r psi square r and this is psi square 4 pi square dr integration and r see the number of node that you are getting equal time this graph will cut the x axis goes like this two nodes so two time it will touch the x axis so it is plus this is minus and this is plus further it won't touch the x axis right now the psi square of this will just get the mirrored image of this one everything will be as it is it comes like this such the axis it goes up comes down goes up comes down like this of course further it won't touch the x axis here also the probability curve it will touch the x axis twice like this we can say okay so these are the nodes we have spherical nodes these are the nodes we have spherical so whatever this distance is r1 and r2 you can draw two spherical surface at this distance r1 and r2 that would be the angular node sorry spherical node for 3s orbital okay now a few you know facts here you should know sorry suppose we have 2p orbital we won't draw the graph here if you want you can draw it is a mountain like this it comes psi like this psi square also like this and probability also like this only we get okay because it has only angular nodes that we cannot represent here we won't we won't draw that angular node so the number of a spherical node here is spherical node is n minus l minus 1 n is 2 l is 1 minus 1 is 0 angular node is equals to 1 okay so if it is px orbital if it is px orbital so it has one angular node for this so for px orbital the angular node is yz plane it is the angular node means the perpendicular plane we can say node is this yz plane is the node because the electron is in this x orbital so perpendicular for this plane is yz plane there is no electron present hence it is the yz plane similarly if you take py orbital perpendicular to this plane is what zx plane and that is the node we have pz orbital we have xy plane and that is the node okay so this you must keep in mind similarly for 3d orbital you see for 3d orbital the number of spherical node could you tell me what is the number of spherical node and angular node angular node would be 2 spherical node would be 3 minus 2 minus 1 is 0 right so it has two nodal plane nodal spherical node right so if it is if it is dxy orbital so perpendicular plane for this is yz and zx these two are nodal plane for this orbital that is dxy similarly for dyz you can have xy and zx dxz you can have yz and xy plane perpendicular plane you can take here we have only one exception write down this point note dz square there is no nodal plane for dz square this you have to memorize it's an exception there is no nodal plane for dz square but the angular nodes of this the angular nodes of this is is a cone in this shape we'll have we don't have a plane but the angular node of this is a cone and that's why we don't represent it like i said it's a complex shape and hence we do not represent the angular nodes there correct so these are the things you should keep in mind for nodal plane very direct and simple question they'll ask you it's not much difficult okay very simple question you won't lose marks over here okay now one kind of question that i have already told you that graphical questions how they ask okay another type of question okay you see this question what is this t o s e anurag yz plane for dx square y square yz plane is the node yz and zx also we can consider yes yz zx what 11th slide this is second there's nothing in 11th slide this one it's just a second this one what is the nodal plane for s orbital which s orbital you are talking about can be anything 1 s 2 s 3 s can be anything dx square y square has two nodes okay the one nodal plane is directly coming towards the observer you can imagine okay perpendicular to the plane of dx y y square the other one is horizontal to us by affecting the x y axis means horizontal and perpendicular you can imagine you want to which s orbital you are talking about 3 s here i have done this it means that we have our surface a spherical surface at this distance where there is no electron present no it's it's 3d anurag you are not getting it it's 3d actually like i said one nodal plane is bisecting the x y axis horizontal left to right you bisect it right other one is perpendicular to the plane coming towards us so when you have imagine a plane it is like this yeah that's what the perpendicular to the plane we yeah see how can i draw this uh see first of all if you have axis like this x y axis so one is bisecting this x how do we draw this there's a three-dimensional if it if you bisect this half this side above the this thing uh screen and one half below the screen that is one thing one is along this line if you bisect this perpendicular towards us observer that is the two nodal plane we have it's three-dimensional i cannot draw this in 2d solve this question you see the wave function is given and with this question i think you will have a little bit of clarity how do we get those graphs and all see the wave function is given wave function for 3s orbital is given and that is that is psi of 3s equals to 1 by 81 root under 3 pi 1 by a naught 3 by 2 open bracket 27 minus 18 r by a naught r by a naught plus 2 r by a naught q bracket close e to the power minus r divided by 3 a naught this is the this is the wave function of 3s so like this will have the wave function of each and every orbital plane okay it is given and it says the wave function of 3s is this it has a node it has a node r equals to r naught it is the radius and a naught is the first radius first board radius we can say first radius we can say the ground state find the relation between relation between r naught and a naught could you think of this question what is the possibility we have a naught and r naught r is the radius and a naught is the first radius we have r naught r naught is the information given like we have no okay okay it's like a variable yeah that at that particular distance r naught there's a nodal surface okay yeah done yes that's correct because r naught is the nodal surface so you have to put psi is equals to zero very simple question it looks very complex here right it appears to be very complex but question is the quite simple question we have you see it is simply given that at r is equals to r naught we have a nodal surface correct so when you substitute here r as r naught then psi 3s would be equals to zero we can rewrite this at r is equals to r naught 3s equals to zero and we substitute this in this equation so we get 27 minus 18 r naught by a naught plus 2 r naught by a naught cube equals to zero I think there's a mistake here it should be square not cube it should be square here okay now you can solve this quadratic for r naught by a naught that would be minus b plus minus b square minus 4ac so 2054 into 4 is 216 okay 216 root over of it divided by 2a that is 2 into 2 4 we have right now you can solve this you'll get the relation of you can solve this you'll get the relation of r naught and a naught you'll get approximately when you solve this you'll get 18 plus 10.4 28.4 divided by 4 so this would be no this divided by 2 we'll get here so okay you'll get this 56.8 you can solve this you'll get the answer and that would be 14.2 approximately so r naught value is r naught value is 14.2 a naught this is the relation we have one relation is this because since we have plus minus here so we'll get one more value of it we get this means exact value I'll write down how do we get this the exact value we get here is like this answer will be this only you'll have 18 plus minus 10.4 by 2 so when you solve this you'll get 18 plus this that is 14.2 when you take positive sign or when you take negative sign it would be 11 and 7 so 7.6 by 2 that is 3.9 approximately yes 3.8 approximately so r naught value is this or r naught value is 3.8 a naught when you solve this you'll get this expression calculation I can't do anything no it's not a cube it's a square that's a mistake in the question I mistake I've written it to cube it's a square and then you can solve this for a quadratic equation nothing I haven't done whenever you get this question right you have no other choice but to put psi is equals to zero that is it they can also ask you that what is the you know nodal radius for this particular condition so in that case you need to find out r value correct whenever it is nodal plane psi must be zero so I say psi is zero here and then it is a a square plus 2ab plus a square plus 2ab plus b square it is in that form right or simply it is a quadratic so you can use it for x equals to minus b plus minus b square minus 4ac root over by 2a this is the simplest one it is don't think about this equation it is not there in our syllabus right just you focus on it is talking about a node at this point so at node psi value would be zero so when you substitute r is equals to r0 here this becomes a node and psi would be zero when you substitute zero all these things will go on only this thing is left when you solve this simple when you solve this you'll get pranav when you solve this you'll get probably you'll get it I haven't solved it right you check the cruise cross check it the value I have written directly here okay wait I'll solve this again just a second I'll solve this for you leave it see what we get here minus 216 you'll get 108 so we'll get here 18 plus minus root over of 108 okay divided by 4 root over of 108 is what what is the value of root over of 108 10.3 18 plus minus 10.3 divided by 4 so we'll get here 28.3 divided by 4 or we'll get 7.7 divided by 4 so what is this value we are getting 7. something 28.3 divided by 4 it is 7.07 this value is and this value is 7.7 divided by 4 1.9 1.92 the value is r0 is equals to 7.07 a0 and r0 equals to 1.92 this is the possible relation of r0 and a0 here what is see this is the spherical this question is not talking about the spherical node why do you want to it's not like that you know something then you need to find out in every question the question is not about what kind of node it is it is the spherical node or angular node it is not the question is what we need to find out the relation of r0 and a0 and it is given that we have a node at this particular point right at this distance from the nucleus we'll apply the condition of it when r becomes r0 psi becomes 0 we'll use that obviously because we have 3s so 3s will have spherical node only there is no angular node for 3s so we have two spherical node right how many spherical node we have for 3s how many spherical node we have two right so we should get two values of radius where the probability is 0 so this is the two value we are getting you see this it's very simple we'll have the nodal surface at these two point these distance from the nucleus mathematically also you are getting it it is proved 3s has two spherical node hence two value of r we are getting understood all of you understood this tell me no doubt one last question for this you need to find out how many nodal planes are there how many nodal planes are there in the atomic orbitals in the atomic orbitals for for the principal quantum number principal quantum number n equals to 3 yes they'll give you the formula if they ask you this question the relation of psi and r they'll give you because we do not know see one more thing I'll tell you here you don't have to worry about this relation okay because the graph that we draw it is based upon the relation of psi and r so this kind of complex complex equation we'll get when you solve the equation correct so this solution is not there in the syllabus obviously they have to give this relation in the exam otherwise you won't be able to solve this don't worry for that tell me the number of nodal planes total nodal planes you have to tell me how many total nodal planes means with n is equals to 3 whatever orbital is associated count the nodal plane for them no it's not it's not no it's not it's not see what I said no Anurag what I said the orbitals associated with third shell we have 3 s 3 p 3 d count the nodal plane for each of this orbital no it's not that you are you are you are you know saying all the answers probably no that is not that is something else sorry n minus one is not the formula as we got the right answer very good as big product close but not correct no namada no see here simple one which all are sub shell we have in third shell we have 3 s we have 3 p we have 3 d correct so 3 s has only 1 s orbital right 1 s orbital and it has zero nodal plane right we know s has no nodal plane 3 p has 3 orbital 3 p x 3 p y and 3 p yes Anurag correct 3 p z and each of this orbital has one nodal plane each has one nodal plane it means we have total three nodal plane here it has five orbital 3 d z x x x y y z x square y square and 3 d z square now in this this one has no nodal plane it has nodal cone right not a plane no nodal plane and this each of this has two two nodal plane total 8 we have over here right each of these orbitals has two two nodal plane d z square has no nodal plane so total orbital total nodal plane is what 8 plus 3 that is 11 yeah 3 s has a spherical node it is not a nodal plane question is about nodal plane here a spherical node we have right sphere is not a plane that's why we are considering it is 0 answer would be for this question the answer is 11 so this kind of question they ask in nodal surface they can ask you what is the nodal plane for this orbital how many nodal plane present d z square one is important it has no nodal plane very important this one and that equation related question they ask that is it plus how many node how many nodal plane present into this nodal surface we have into this the radial angular node formula you know with that you can find out these three four types of question they ask nothing much clear understood okay so fine we are done with this chapter we'll take a break now we'll resume the session at 6 50 and we'll start a new chapter okay take a break periodic properties we'll start take a break guys hello can you hear me okay okay so the next chapter we are going to start with yeah the next chapter we are going to start with is periodic properties okay so this particular chapter it is basically the first chapter of the inorganic chemistry okay the first chapter of inorganic chemistry so basically here we are not going to see any