 By now we have learnt how to generate expressions for hybrid orbitals by using linear combinations of atomic orbitals that we get by direct solution of Schrodinger equation. So, as we said earlier why is it that hybridization is taking place so that the electron density is optimized to form the bonds. So, see electron cloud is what we have we do not have rigid particles and that is the beauty of the system the electron cloud can deform and adapt to the requirement that it faces. Right now the requirement we are discussing is formation of bond. So, if say for example boron it forms a trigonal complex trigonal compound with fluorine BF3. So, it has to react with 3 fluorine atoms it has to form bonds with 3 fluorine atoms in order to satisfy its valency. You can actually go back to this very fundamental concept of valency. So, 3 bonds will be there and these 3 bonds according to Gillespie and Nyholm VSEPR have to be at 120 degrees to each other. That is why the electron cloud sort of deforms itself because the situation demands and forms these new orbitals you can think like that. Of course, this orbital is used to the hybrid orbital is used to explain the situation. Okay, it is not as if they are the cause they are the effect of the cause really is minimization of the repulsion between among electron clouds. Let us not forget that. So, you can think of these electron clouds as shape shifters. Remember, I can only remember movies from long ago. So, Terminator where this bad element from future could sort of take the form of anything or remember species or many such movies are there. So, electrons are like shape shifters and they can modify themselves according to the demand of the situation. And the demand we have discussed so far is formation of 2 bonds or 3 bonds. We had said that we will work back calculate the angle between the bonds but I had a second thought about that. We are going to do a calculation but we will do it with SP3 hybrid orbitals. Let us get ahead. So, now I want to talk about SP3 hybrid orbitals. So, SP3 hybrid orbitals means 4 orbitals are there 4 hybrid orbitals that means they are required in formation of AB4 kind of molecules where A is a central atom and they should form tetrahedra. Before doing anything else I can say that the angle has to be 109 degrees because this is how this is a one good way of drawing a tetrahedron right. Put a dot well first of all draw a box and unfortunately this has got distorted this is a all the sides are square actually. This is a cube perfect cube not cuboid. So, put dots on alternate vertices and then join those dots to the center you have your tetrahedron. So, this is what we want since we want 4 hybrid orbitals all the 3 p orbitals have to participate. And let us say to start with we have a situation in which all the 3 p orbitals make equal contribution to each of the hybrid orbitals. So, what should the coefficients be? Before showing you the results let me just work it out. So, our expression will be something like this psi hybrid is equal to C1 psi 2 s and I want equal contribution from each of the p orbitals. So, I can write plus C2 psi 2 p x plus C2 psi 2 p y plus C2 psi 2 p z. If you cannot read my handwriting please do not worry I will show you the final result in in a nice form anyway. So, how many such hybrids are there I can write 1. So, here everything will be C1 is not it and total contribution of 2 s is 1 anyway. So, what we get is 4 C1 square is equal to 1. So, C1 is equal to half simple plus half or minus half whatever does not matter for we can take either plus half or minus half all that will change is your p orbital will look either like this or like this nothing else will change. In fact, you will see a situation for coming very soon in which we are going to use a negative coefficient for psi 2 s knowingly C1 equal to half that is done. What about this C2 s? Let me say something else what will the C2 s be for one thing I can write right away this hybrid orbitals have to be normalized right. So, I can write C1 square plus 3 C2 square is equal to 1 and I know very well that C1 square is equal to 1 4th because C1 equal to half. So, I have 1 4th plus 3 C2 square is equal to 1. So, C2 square is equal to 3 4th 3 C2 square equal to 3 4th C2 is equal to plus minus half. So, one solution is plus half right I can write plus half for everything it is fine. So, let us start with that the totally symmetric answer in which every coefficient for the p orbital is plus half. So, the first wave function that we can write is and now before I say what the first wave function we can write let me just clear all this and let us go back to the presentation mode that is where things are written much more neatly. So, hopefully it will be easier for you to understand just give me a sec please all right all set. This is the first one that we have already got what will the second one be see your hybrid orbitals have to be orthogonal to each other and again I will start writing of course I will erase also later on. Suppose I write something like this phi H2 sp3 equal to this will be half not much doubt about that psi s. Now I will just write the orbitals for psi 2 p x psi 2 p y psi 2 p I have not written 2 p there but anyway psi 2 p z sorry for being a little inconsistent. Now if I just multiply them and integrate over space I am going to get 0 in any case I know that mode of this coefficient first of all I can write like this plus C3 plus C3 plus no C3 plus not a good idea to write C3 again if I write plus C4 plus C5 and we know that mod C3 equal to mod C4 equal to mod C5 is equal to half because every p orbital that is what we are considering every p orbital makes an equal contribution to the wave function. So the squares must be the squares must add up to 1 and they are equal to each other so that is how okay fine. So now the modes are same and what the other thing that I can do is I can take this integral phi h 1 phi h 2 is equal to 0 which means 1 4th plus C3 by 2 plus C4 by 2 plus C5 by 2 equal to 0. How did I get this because when I do the multiplication remember integral of say psi 2 s psi 2 p x overall space is equal to 0 right they are orthogonal to each other so that is how we have got it and then they are normalized also and then remembering that actually these are all plus half well plus or minus half what it means is that I have to put in 2 minus signs for this quantity to be equal to 0 that is what I mean the C3 by 2 C4 by 2 C5 by 2 all mode of these mode of C3 by 2 I can write like this mode of C3 by 2 is equal to mod C4 by 2 equal to mod C5 by 2 is equal to half okay. So what are the absolute values sorry is equal to 1 by 2 into 2 4 that 2 is equal to 1 by 4 so what I have here is that this is 1 4th this can be plus or minus 1 4th this can be plus or minus 1 4th this can be plus or minus 1 4th so the only way in which this will be 0 is suppose I put in a minus sign here and a minus sign here then I will get 0 right 1 4th from here plus 1 4th from here 1 4th from here 1 4th from here minus minus that will be 0 and what I can do is in the next combination I can move this plus sign here take the minus sign there so I can move the plus sign around what it means is that for the coefficients of the pure orbitals 2 of the coefficients must be equal to minus half and one of the coefficients has to be equal to plus half that is the only way in which the condition of orthogonality of the hybrid orbitals can be satisfied there is no other way so what we have done is remember we have actually constructed these wave functions under a lot of constraints we have made sure that contribution from each pure orbital is exactly the same and then we have required that they are normalized and now we are required that they are mutually orthogonal also so that gives us that leads us to the conclusion that 2 of the pure orbitals must have coefficients of minus half the 3rd will have coefficient of plus half so I first of all I can write like this and then all I do is I just move this plus sign first could I have written plus sign here to start with yes of course how does it matter right which one is h1 which one is h2 is in our hand we will name accordingly so this is the complete set orthonormal set of hybrid wave functions now let us see whether the picture we have drawn is valid equal contributions from x y and z can you see where x y and z are here maybe I will draw again this is a center this is a face center we join the face center with this this is a face center join with this x y and let us say this is the face center this is the easiest the most difficult to draw the draw the join the face center like this so this is your x axis this is y axis this is z axis let us say now you can think of these hybrid orbitals as vectors and you do not have to worry about psi s because it is non-directional anyway what we are saying is that every each of the axes x y and z contributes equally to the length of the vector right we are taking squares so plus or minus will not matter we do not need to go over length non-direction so where will this resultant vector lie actually you can think of this smaller cube within cube this is a cube not a cuboid okay smaller cube within cube the body diagonal of that smaller cube that is your hybrid orbital okay and equal contribution of x y z ensures that it is there so similarly you can see that the all the hybrid orbitals are actually along the body diagonals of the cube and it is not very difficult from your solid state geometry to work out that the angle between body diagonals of a cube is indeed 109 point whatever degrees okay so we have 25% contribution from s 75 contribution from p more importantly we have 25% contribution from each of the participating orbitals may it be s may it be p okay we are going to discuss another situation shortly all right so remember there is no unique combinational solution to change the orientation the coefficients will change what will not change is a relative contribution what will not change is the fact that they have to form an orthonormal set complete orthonormal set now that being said we have worked out the angle anyway now let us work out the angle in a little more formal manner and to do that we will work with h1 and h2 orbitals okay if you remember the formula the working formula for dot product of two vectors this is what it is you take a dot product r1 dot r2 that gives you mod r1 mod r2 magnitude of r1 magnitude of r2 multiplied by cos theta okay so what we can try to do is what is the unknown here the unknown is theta what is the angle between h1 and h2 that we do not know but knowing the vectors we can find out the dot product and we can also work out the magnitudes the length mod r1 mod r2 and we can find that product so we can try to find cos theta by from the by dividing the dot product of r1 and r2 by the product of their magnitudes to do that first of all we royally ignore the s orbital okay we have washed it out with a different color we do not care about s anymore well for some time because s is not directional it does not count as a vector it does not contribute in this discussion so the vectors that we want to work with are x y and z p x p y p z all equal magnitudes all right so the dot product here would be what now remember this cos theta business so cos theta of x and y is 0 this is well known in vector algebra but I am saying this in case some of us forgotten dot product of x and y y and z z and x all that is equal to 0 dot product of x and x is 1 okay so what will I get half into half is 1 4th I can take 1 4th common so I will be left with 1 minus 1 all that so the first term comes from p x dot p x 1 into minus 1 second term comes from p y dot p y plus 1 into minus 1 third term comes from p z dot p z that is plus 1 into plus 1 so what do I get the dot product turns out to be remember dot product is a scalar quantity not a vector quantity 1 4th multiplied by minus 1 minus 1 plus 1 that is minus 1 4th we have worked out the dot product what about the magnitudes? Magnitudes will be well these are at 90 degrees to each other right p x p y p z so magnitude will simply be well square root of half square plus half square plus half square yeah so that will be root over 3 by 2 that is the length of the vector okay magnitude so if I take product of magnitudes then I get 3 by 4 so I have got the dot product to be minus 1 4th I have got the product of the magnitudes to be 3 by 4 what is cos theta cos theta is just the dot product divided by product of magnitudes turns out to be minus 1 3rd take cos inverse of that theta turns out to be 109.5 degrees okay so we are convinced that this is an actual tetrahedron okay and everything falls together beautifully it is a tetrahedron in order to minimize depulsion between bond pairs but when we do that and we construct these hybrid orbitals it turns out that we get the correct value of the angle and this is going to be very useful this kind of calculation when in the next class we talk about non equivalent hybrid orbitals okay we will talk about water wait for it okay but let us finish this discussion what happens if I hold it in a different orientation I will not work out all of it I will work out a part of it and I will ask you to work it out work out the rest of it what happens if h1 is oriented along z axis okay and let us say I hold the other one in the zx plane so I do not know whether this figure is very clear to you this is what I mean why can I not draw a straight line I am drawing on a screen so unfortunately I cannot or maybe I can use a ruler I do not have one that is all whether straight or not this is z axis this is x axis y axis is pointing towards us let us say now I will since I can change color I will change color what I am saying is I will put h1 along your z axis this is h1 and I will keep h2 in the zx plane well this is x this is z sorry so it will be something like this h2 you know this is 109 degrees 109.5 degrees 90 so 109.5 minus 90 now so where will h3 and h4 be one will be above the plane towards us let us say there is h3 the other will be below the plane away from us h4 right this is how we have held the molecule now now we want to see what kind of coefficients we get and we want to check whether it is compatible whether it gives similar results compared to the earlier case where the coefficients were all either plus half or minus half great so how do I get the first one first one is very similar let me write a little bit so first one psi h1 let us say is equal to c1 into psi s no I will not write c1 can you tell me what I will write see does not matter how I hold it contribution of s has to remain the same right we still want this sp3 orbitals so contribution has to be same everywhere and the total contribution has to be 1 so without repeating the calculation that we have done already I will write it is going to be 1 fourth is going to be 1 fourth c1 c1 square plus c1 square plus c1 square plus c1 square equal to 1 so c1 square will be 1 fourth c1 will be equal to half so the same coefficient that we got earlier will be there for the s orbital but what about the p orbitals now they will not be the same see for h1 I have held it along the z axis so only pz will make a contribution px and py will have 0 contribution so I can write like this 0 into px psi px sorry plus 0 into psi py plus what will I write let me write what would it have been c1 c2 c3 c4 c4 into psi pz let me rewrite looks like I have written in indelibuling cannot help it so psi h1 hydribid 1 is equal to half into psi 2 s plus again I wrote to anyway does not matter plus 0 into psi 2 px plus 0 into psi 2 py plus something c4 into psi 2 pz rest is simple this is normalized so you will get 1 fourth plus c4 square is equal to 1 so c4 is equal to root over 3 divided by 2 so already worked out 1 orbital and what I see is that first of all the entire p contribution is from 1 orbital that is pz however if you take the coefficient square of coefficient of psi 2 s and square of coefficient of psi 2 pz what do you get 1 fourth is to 3 fourth that is 1 is to 3 so this is still an sp3 hybrid orbital and that is what we are trying to emphasize here if you hold the molecule in a different way the coefficients will be different but the overall picture does not change total s character total p character will not change what will change is contribution of individual p orbitals does not matter as long as the total s character and the total p character are not compromised we are absolutely fine because we do not even know which is x and which is y and which is z right we are I thought this is indelible now it is gone anyway so we are just writing things in a way to simplify problems so we are justified in simplifying the problem that is what I am trying to say it is okay if you hold the molecule in an orientation that makes our problem little simple as long as the overall picture does not get distorted okay now let us go ahead so this is what we had written and in fact we have gone further than what is written here we already know the values of c1 and c4 half and root 3 by 2 if you go to phi h2 then again c5 is going to be half this quite mundane now what will c8 be a very easy way of proceeding is by using the orthogonality of psi h1 mutual orthogonality of phi h1 and phi h2 so what you get is one fourth plus root 3 by 2 c8 is equal to 0 so c8 right we will divide it at one fourth sorry what am I doing I am multiplying phi h1 by phi h2 so this is one fourth this is also one fourth so one fourth is actually correct but it is root 3 by 2 into cth everything is correct unnecessarily I had a panic attack so c8 is equal to minus one fourth into 2 by root 3 is equal to minus 1 by actually root 6 okay it is a minus fine actually it is because it is on the other side so I mean it is towards minus z so it is fine okay what is the contribution of this pz orbital now 1 6th okay whereas contribution is one fourth so 1 6th one fourth that is not sp3 that is because this c6 is non-zero how do you find c6 now simple one fourth plus c6 square plus 1 6th is equal to 1 from normalization condition and now since I have done so many and I have made a couple of mistakes also I will not do all this arithmetic anymore you do it yourself I will erase and now just show you the final result and then I will go home okay so you know how to get it right how to get the coefficients this is how you get it and the coefficient for px here turns out to be root over 2 by 3 now the thing is simple the other two once again the easiest way of starting the problem is to start with the mutual orthogonality with the orbital that has a maximum number of 0 coefficients and then proceed then you have fewer terms to handle to start with at least but before going there just have a look here s contribution is half one fourth p contribution what is it square of this 2 3rd plus 1 4th of 3 so just work it out you will see that it comes to the same thing you still have sp3 here it is just that contribution from px and pz are no longer the same because orientation is not symmetric anymore so symmetry actually is a very important role to play in handling quantum mechanical problems we have this NPTEL course that we floated on symmetry sometime ago lectures are all available whoever is interested you are more than welcome to have a look at the lectures if there are questions I will be happy to answer them. So now the reason why tutorial problem is written here is that it is for you to do please work out the expressions for the 3rd and the 4th hybrid orbitals and I am showing you the result here what I also want you to do is I want you to work out the bond angles using the formula that we have discussed this is something that I have done earlier I do not know why you are written in bigger and smaller fonts but the point is work out the bond angles and you will see that the bond angle once again comes out to be 109.5 degrees. So once again we have a scenario where you have a regular tetrahedron we have worked with hybrid orbitals that are all equivalent to each other same as character same p character and now I hope we have a little more quantitative idea of hybridization there is only one thing left on the agenda as far as hybridization is concerned and that is to discuss non-equivalent hybrid orbitals that is what we will do in the next class.