molecules X square minus y square and XY are degenerate orbitals in the case of C3v orbital. Okay, and here this A square plus Y square and Z square. Okay, and this is all about the character table for Z3 viewpoint group. I think you have understood these things and now you work out this and practice C for V point group yourself. I think I need not discuss a policy for a point group. You can do it by yourself. Okay, you do it and practice it. Okay, thank you.

of this 2 diameter representation both x and y together because here X&Y weeks to integrate of c3 and Sigma V dash exedra X&Y mix together that is why the off-diagonal elements are nonzero okay x and y mix together therefore basis for this two-dimensional representation is XY together that is why we are writing like this that is X comma Y it is a two-dimension representation if x and y together forms the basis for the representation we are they putting it in a bracket separated by a comma X comma Y in bracket similarly the basis for this a two representations are said our set is a rotation about the z-axis thus we can see be simpler to be already discussed about these things in c2v point group consider a XYZ axis and consider a rotation SN are represented as an arrow by transforming e no change C 3 C 3 set down there for passing through this axis no change in this rotation vector and Sigma V the sine Sigma V means passing through this a plane passing through the

minus 2 3 D is equal to 0. D equal to 0, so this D is equal to 0. Okay, so we get a 1. B is equal to minus 1. C is equal to 1. D is equal to 0. We get our character trouble. This is gamma 1 is equal to 1 into 1 into 1. Gamma 2 is equal to 1 do 1 into minus 1 and a gamma 3 is equal to 2 into minus 1 into 0. These are two one-dimensional representations. This is a two-dimensional representation. Okay, and this is beta notation gamma 1 gamma 2 gamma 3 and substituting Mulliken symbol for this we can try Mulliken symbol and this is one time. This gamma 1 is a one-dimensional representation, so it will be if Mulliken symbol will be a or b. And how can you decide whether it is a or b? Look at the character of the principal axis. It is symmetric one means symmetry. If it

z-axis. Therefore, this rotation or the arrow changes its direction. Therefore, this minus 1. So, our that is the basis for this representation. In a similar manner, we can understand that in the area for that is actually this X X pi set the represent transformation of Cartesian coordinate XY sets or the vectors X Phi set translational vectors or the exit X PRIZE at the dipole moment. The vectors in the X Y Z direction rx ry are said to represent the rotation about the x-rays at axis okay. And this area I have already told you is important for vibration spectroscopy and rotational spectroscopy. And this is the area important for the Raman spectroscopy. And also here we can see that these are the actually transformation of D orbital ok. I am NOT going into details actually x square minus y square and XY together forms the basis or together transforms says a II and exit and vice it transforms s II. That means exit and why I said Rd generate orbital in the c3v point guru c3v.

therefore here only one element 1e. Therefore, 1 square plus 2 a square plus 3 B square is equal to 6. And to satisfy this, satisfy this equation, a and B should be either 1 or minus 1. Ok, 1 and minus 4, minus 1 both are possible here because square of 1 and minus 1 is equal to 1 in cells. Therefore, 1 plus 2 plus 3 is equal to 6. But we can't say which one is 1, which one is minus 1, that we don't know. And then we will get the orthogonality rule, that is, the other rule, that is, the Sigma GI kar in duke IJ or is equal to J when I know T equal to 0. That is, different appreciation representations are different representations, are different representations, are orthogonal. That is, we are, if we are considering two representations, they are orthogonal. That is, by multiplying their characters and adding from all the elements, we will get zero. That is, GI kar in Duke Isaiah R is

character that is common for C3 in 2C3 and C3 and C32. There is common for both this operation. Okay, so we want to multiply it with GI, that is, the number of elements in D, either representation or indict the class. Okay, okay, then this rule is already satisfied by our gamma 1. This is gamma 1, that is, 1 square plus 2 into 1 square. Here, 1, 2 into G is equal to 2, where 2 elements are 2C3CR2 into 1 square plus 3 into 1 square. That is, 1 plus 2 plus 3 is equal to 6. That is already satisfied there in the case of gamma 1. And for gamma 2, we can write 1 square plus 2 into a square. Okay, 1 square plus 2a square plus 3b square because your three elements are here. Look here, there are three elements here. Here, two elements in this class.

Okay, welcome to your students. Our discussion on group theory. This is the fourth video. In the last video, we had discussed the construction of the character table of the C2V point group, and today we are discussing the character table for the C3V point group. The character table for the C3V point group. We know that the elements in the C3V point group are E, C3, C3^2, Sigma V dash, Sigma V double dash, and Sigma V triple dash. There is a total of 6 elements in the C3V point group, and the number of classes in the C3V point group is 3. That is, E is a separate class, C3 and C3^2 together form a class, we are denoting it as 2C3, Sigma V dash, Sigma V double dash, and Sigma V triple dash forms another class, and it is denoted as 3Sigma V. Then we solved together three classes are present in the C3V point group, and we know that according

symmetric. That means all the characters are equal to 1. Okay, then which in every point group, there will be one here that is totally symmetric, that is common for all point groups. Common for all point groups. Therefore, the posterior gamma 1 is stored ray symmetric. There is a general case in all point groups. The irreducible person is a totally symmetric, that is, all the characters are equal to one. Therefore, the first irreducible representation is one, one, one. Okay, and we know that the character of E is equal to the dimension of the representation. So, this is so first one damaged the representation character of each one. Second one is also one-dimensional. We already learned that two irreducible representations are one-dimensional, and third one is two-dimensional. Therefore, here character is equal to one, and this is a two-dimensional representation. So, character of is equal to two. Okay, now the other characters we are arbitrarily denoted as ABCD. We want to determine what are the values of a, b, c, and d. Okay, now we are

L3 square is equal to 6, and there are 3, i.e., their dimensions are L1, L2, and L3. Therefore, L1 square plus L2 square plus L3 square will be equal to 6. The upper equation would be satisfied only if the values of Li are 1, 1, and 2. This equation would be satisfied; the values would be 1, 1, and 2. That is, 1 square plus 1 square plus 2 square is equal to 6. 1 plus 1 plus 4 is equal to 6. There is the only possibility; that means there would be two irreducible representations that are one-dimensional, and the other one is two-dimensional. That is, out of the three irreducible representations, two of them are one-dimensional, and the third one will be a two-dimensional representation. Okay, and we have studied that in any group, there would be a one-dimensional representation, which is totally

equal to zero when I know T equal to Z and for gamma 1 and gamma 2 we are here we are applying this rule for gamma 1 and gamma 2 so 1 into 1 here 1 into 1 and here 1 in 2 in do because here 2 C 3 is here that is G is equal to 2 2 into 1 into a ok 2 into 1 into a plus 3 into 1 into B 3 into 1 into B that is equal to 0 that is 1 plus 2 a plus 3 B is equal to 0 ok then what would be the values of a and B very clear here should be 1 B should be minus 1 when a is equal to 1 we get 1 plus a is 1/2 when B is equal to minus 1 we get minus 3 that is 1 plus 2 3 3 minus 3 is equal to 0 there is only possibility other

equation is equation 1 to C plus 3 D is equal to minus 2 and one. Now for gamma 2 and gamma 3, that is this 1 gamma 2 gamma 3. So 1 into 2 again 1 into 2 and 2 into a into C 2 into a into C 2 in do sorry is 1 is 1 B is equal to minus 1 that we got you already got this a is equal to 1 and B is equal to minus 1 over 2 into 1 into C 2 into 1 into C and 3 into minus 1 into D 3 into minus 1 into D because this B is equal to minus 1 we already code from here B is equal to minus 1 is equal to 1 okay. So we get 2 plus 2 C minus 3 D equal to 0 or 2 C minus 3 d this to the other side we get minus 2 to C minus 3 0.

possibilities are there, but their fractions we already are known. Only it is a one-dimensional representation; therefore, the characters would be 1 or -1. Okay, so A is equal to 1 and B is equal to -1. That is, this is 1, this is -1. Okay, that we get. Now we want to guess, we want to get C and D, and applying the orthogonality rule for gamma 1 and gamma 3. Gamma 1 and gamma 3. Okay, gamma 1 and gamma 3. Look here, gamma 1, a plane orthogonal. Do so, 1 into 2, 1 into 2, plus 2 into 1 into C, 2 into 1 into C, plus 3 into 1 into D, 3 into 1 into D. That is equal to 0 according to orthogonal 0. That is, 2 plus 2C plus 3D is equal to 0, or 2C plus 3D is equal to this to become a other side. We get -2 and put this.

going to the we are here. We are deriving this ABCD, finding out these values using the rules from the orthogonality theorem. We know that according to the rule Sigma GI ki R square is equal to H, that is equal to 6. There is sum of the squares of the characters in a particular irreducible representation is equal to order of the group and order of C 3 viewpoint group e 6. That is, we see this is our rule. We have studied Sigma R Chi I R square is equal to H. That is a very important rule. That is, sum of the squares of the characters in a particular representation is equal to order of the group. Okay, Here I am adding this term GI for GI is the number of elements in a particular class, number of elements. We want to multiply the number of elements. Okay, we are right. We know that in a elements in a particular group, we have a common character. That is why we are writing the character of only one of the elements. Actually, here to see three steps, but we are writing only one. One is the

is symmetric with the principal axis character will be a yes it is a and there is no subsidiary axis so we are considering the vertical plane. If it is symmetrical with the vertical plane, then we are putting the subscript 1, so it is a 1. This gamma 1 is a 1 actually, okay, in a gamma. This is also one-dimensional because the character of E is 1, so either A or B. And here C 3 is again symmetric, therefore again A, not B, it is A itself, but it is anti-symmetric to this vertical plane, therefore A 2, subscript is 2, and this one is a 2-dimensional representation, so we are using the symbol Mulliken symbol notation E, 2-dimensional representation. We are using the symbol E, 3-dimensional negative T. It is two-dimensional. In the case of E, we are not using the subscript 1 and 2 like that. That is only for the one-dimensional personation. That is no, not needed here because only two, one, two-dimensional personation is here, so no subscript is needed. We simply

write e ok then this is the area 3, and this is area of 4, and the square we can understood already. We have discussed about the in the C 2 V character table. I explained only something, some details that is already. We c3v, we had discussed earlier that is a simple form of character table. We have discussed in the first video, I think. Okay, these are the three-dimensional representations of c3v elements. That is e, C 3, C 3 2, Sigma V, Sigma V dash, Sigma V double - I think you have remember that. This can be block factored into a 2-dimensional representation and a one-dimensional representation like this. Okay, two-dimensional, one-dimensional representation. Okay, so the one-dimension representation, this one, one chiral cross, C 3 and C through the same 1 character of Sigma V, Sigma V dash, is my double dash is also same 1, and that is 1, 1, 1, and the base is 4. We know that this is the transformation of x coordinate. This

minus 2. That is equation 2. So adding 1 + 2, 1 + 2, a DM. Then we get this 2C plus 2C. That is equal to 4C. 4C. 3D minus 3D cancelled. Ok, so minus 2 here, minus 2 here, minus 2. Therefore, minus 4. So C is equal to minus 4 by 4 is equal to 1. So we get C is equal to 1. C is equal to 1. Okay, and putting the value of C in equation 1. This is equation 1. Here we are substituting the value of C. C is equal to 1. So the minus 1. So 2 into minus 1. That is minus 2. 2 into 2 into minus 1 plus 3D is equal to here - - ok. How about 3D is equal to this minus 2. Become other side. Become plus 2. Therefore, minus 2 plus 2 equal to 0. So D equal to 0. Minus 2 plus 3D is equal to

is a transformation y coordinate. This is a transformation of said Cora. These are all set transformations of said coordinates. Okay, so 1, 1, 1 is actually the transformation of said coordinate, the hero's edge. Okay, and look at the two-dimensional representation, and that is this 1, 1, 0, 0, 1, and this character YZ 1 into 1. So the one plus one character is a sum of the demo diagonal elements, that is 2 here, minus 2, minus or plus, minus off is equal to 1, sorry, minus 1 here. Also, it is minus 1 because they are in the same class, minus 1. And this is 1 minus 1, 0, minus 1/2 plus off 0, and the world will be 0 because they are the same class. They should have the same character. So we get to these are the same character. This is the same class, 2 minus 1, 0. This is this one, 2 minus 1, 0. So what is the basis? What is the basis?

to the rule derived from the orthogonality theorem. The number of your reducible representation is equal to the number of classes. There is a rule we have studied: the number of IR is equal to the number of classes, and here there are three classes. Therefore, the number of IRS in the C 3 V point group is 3. The number of IR since III v per group will be 3. There are 3 AR, 3 irreducible representation in the C 3 V point group. And suppose let the l1, l2, and l3 be the dimensions of the three, say gamma 1, gamma 2, gamma 3. Suppose these irreducible additional represented as gamma 1, gamma 2, gamma 3. Their dimensions will be B l1, l2, and l3. Okay, so according to the rule from the orthogonality theorem, we know that Sigma and I square is equal days. There is sum of the squares of the dimensions of all the irreducible representation will be equal to the order of the group, and here water is equal to 6. There is L 1 square plus L 2 square plus

Group Theory-4 ( Construction of Character table of C3v Point Group)

Group Theory-4 ( Construction of Character table of C3v Point Group)

Scripts