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2 2 3 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. Symmetry Elements and Operations Groups Example of a triangular planar molecule (AB3) point group D3h. Matrix Representation for symmetry elements and corresponding operators. Example of a tetrahedral molecule. Construction of character table Decomposition of a reducible representation into irreducible representations Explanation of symbols used in a character table Method of classifying molecules into point group Properties of point groups 3 6 11 13 17 19 23 27 28 29 1. Symmetry Elements and Operations What is a symmetry operation? Symmetry operation is an operation (rotation, reflection, or a combination of both) that upon its performance on any object (molecule, figure etc) brings a new configuration which is indistinguishable (but not the same) from the original configuration. Identity operation can be thought as an exception, because it is really a do-nothing operation. Its presence has to be conceived in order to meet mathematical needs of a group. This means that every molecule has at least on symmetry operation, that being identity operation. Kinds of symmetry operations 1. Rotation, Cn 3 4 2. Reflection, 3. Rotation + Reflection (also known as Improper rotation), S n Rotation operation Rotation is done along a line (known as axis and is represented as Cn, where n = 2 / ; is the angle of rotation. This is to say that a rotation by 1200 will be C3. (2 /1200 = 3). In order to identify a 2 / rotation, its axis must be defined. An example of a square planar molecule It is easy to find that rotation by 900 along an axis passing through the center of the square and perpendicular to its plane (C4), is a symmetry operation. (The configuration brought about by C4 cannot be distinguished from the configuration before C4 was applied). If the square is rotated, as above, by 900 two times i.e., net angle of rotation being 1800, then this double rotation through 900 can be represented by a single rotation through 1800 (C2) C4.C4 = C42 = C2 .This is also a symmetry operation. Moreover, the rules of group theory dictate that the combination of two symmetry operations must also be a symmetry operation. operation. Reflection This operation is done along a plane whose both faces are mirrors. To find a reflection symmetry in a molecule, the plane is passed through the body of the molecule and the configuration of the image obtained is compared with the original configuration. Let's consider the square planar molecule again. If you imagine that the plane passes through the square such that it is perpendicular to the square and bisects sides AB and DC, then the configuration after reflection will be Similarly C4 . C4 . C4 = C43 is also a symmetry 4 5 indistinguishable from the older configuration. Reflections are represented by . Kinds of reflection Operations 1. v : In this kind of reflection, the plane contains the rotational axis of the highest order. It is also called vertical reflection. h: Horizontal reflection. The plane in this case is perpendicular to the rotational axis of the highest order. d: It is a diagonal reflection. For example, the plane passing through diagonal points A & C is d. [find out all the reflection operations in a square planar molecule] Improper Rotations Sn Sn is a combination of rotation by 2 / n followed by a reflection whose plane is perpendicular to the axis of rotation OR vice versa. Some features of Sn 1. The order is not important, i.e., it is immaterial whether you apply rotation first and then reflection; or first reflection then rotation. Cn . h = h . Cn i.e., these two operations commute strictly only for improper rotations. 2.It is not necessary that either reflection or rotation are symmetry operations by themselves. That means, the separate existence of symmetry rotation operation or symmetry reflection operation is not a precondition for Sn. 5 6 3.S2 = C2. = .C2 = i ; where is an inversion on a center. An example Square planar molecule is a very simple case to use. It already has symmetry rotation operations (C4, C42, C43). It also contains symmetry reflection operation perpendicular to the axis of rotation, h. Therefore, combination of these results in: C4 . h = S4 ; C42 . h = C2 . h = S2 ( inversion) ; C43 . n = S43 Now we know that symmetry operations are applied along a point, an axis or a plane. Such a point, line or a plane is called symmetry elements of a group. For every symmetry operation there is a symmetry element. Sometimes it is difficult to distinguish between them. They are used synonymously. The following table shows their difference : Symmetry Element Identity Plane Center of Symmetry or inversion center n-fold axis n-fold rotation reflection axis Symmetry Operation Rotation through 3600 or doing nothing; E Reflection in plane; Inversion of all atoms through a center; i One or several rotations about on axis through an angle 2 /n; Cn Rotation through an angle of 2/n followed by reflection in a plane perpendicular to the rotation axis; Sn 2. Group Group is a set of elements which satisfy the following properties: 1. There is an identity element, E, so that A . E = A for any A belonging to the group. 2. The product of two elements is also member of the group, i.e., if A and B belong to 6 7 the group then A.B will also be a member of the group. It means the group is closed under the given operation. 3. Every element has its inverse as the member of the group. i.e., if A belongs to the group, then A-1 also belongs to the group. If A.B = E it means a is the inverse of B and vise versa. 4. Group members obey the associative law, that is A(B.C) = (A.B)C. The order of a group is defined as the member of elements present in the group. As for example, the order of the above group is 8 because it has 8 elements. The element of a group can be several things as you define them. It can be integers, vectors, matrices, symmetry operations (elements) etc. One has to define the operation which goes on in the group which can be several things like addition, multiplication, symmetry operations etc. Examples of Groups a) Collection of all positive and negative integers is a group under the operation of addition. It is an infinite group. b) Collection of all numbers forms a group under the operation of multiplication . c) You will soon learn that collection of all the symmetry operations (elements) belonging to any molecule from a group. (As a matter of fact this is the reason that we are in the class on group theory and this property allows us to use mathematical tools of groups for understanding the properties. of molecules). Subgroups They are smaller groups present in the group. They obey all the rules of a group. Their order must divide the order of the group. That is, if you have a group of order seven, then you cannot have any subgroup except a subgroup of order 1 that 7 8 contains E only. Similarly, if you have a group of order 12, you should not waste your time in looking for a subgroup of order 5, 7, 8,9,10 and 11. Remember there is always a subgroup of order 1. Abelian Group: It is a group in which every element commutes with every other element, i.e., A.B = B.A for every A and B. Cyclic Group: It is a group which is generated by a single element called generator. [A, A2. A3. ............. An = E] . The order of the group is n. Cn is a generator of a group of order n and its members are Cn, Cn, Cn …… Cn = E For Sn , things are a bit complicated. If n is even, then Sn generates a group of order n and its members are : Sn, Cn2, Sn3……… (Sn)n-1= Sn n-1, (Sn)n = E S6 generates the group [S6 , C62 , S63 = S2 , C64 , S65, E ] If n is odd, Sn generates a group of order 2n. [Sn , Cn 2, Sn3…….(Sn)n = h , Cn, Sn2…… Sn n-1, (Sn)2n = E ] For example, members of the group generated by S7 are : S7, C72, S73, C74, S75, C76, h , C7, S72, C73, S74, C75, S76, E Example : Now, we consider the some of the symmetry elements belonging to a square planer figure and also examine if together they from a group. The symmetry elements are as follows: Symmetry elements 8 9 1. Identity element, E 2. Clockwise rotation thru 900, C4 3.Clockwise rotation thru 1800, C2 4. Clockwise rotation thru 2700, C3 (All rotations along the axis passing through the center) 5. Reflection about the horizontal plane thru mid points of 1 & 4 and 2 & 3 , v(1) 6. Reflection about the horizontal plane thru mid points of 1 & 2 and 3 & 4 , v(2) 7.Reflection about the diagonal plane thru points 1 and 3, d1 8. Reflection about the diagonal plane thru points 2 and 4, d2 Mathematical representations of these symmetry elements. or, (Numbers in the second row are the images of those in the first row) ; ; 1 2 3 4 1 2 3 ; v (v2) 2 1 4 4 3 2 1 Let us find out the products of elements: v (1) 4 ; 3 ; v (1).C2 v (2) 4 3 2 1 3 4 1 2 2 1 4 3 v (1).C4 d2 4 3 2 1 2 3 4 1 3 2 1 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 Similarly, we can compute the product of all elements with each other and will find 9 10 that the resulting elements are already members of the group. Inverse of C4 (C4-1) is C43 and vice-versa. Moreover, C2, v(1) ,v(2),d1, d2 and of course E are their own inverses. Multiplication Table E E C4 C2 C43 v(1) v(2) d1 d2 E C4 C2 C43 v(1) v(2) d1 d2 C4 C4 C2 C43 E d2 d1 v(1) v(2) C2 C2 C43 E C4 v(2) v(1) d2 d1 C43 C43 E C4 C2 d1 d2 v(2) v(1) v(1) v(1) d1 v(2) d2 E C2 C4 C43 v(2) v(2) d2 v(1) d1 C2 E C43 C4 d1 d1 v(2) d2 v(1) C43 C4 E C2 d2 d2 v(1) d1 v(2) C4 C43 C2 E Classes A group can be divided in several classes, also called conjugacy classes. The importance of classes will be clear in our later studies. It is time consuming to find out all classes. Choose any element, perform the so called similarity transformation, i.e., compute x-1ax, where x and a belong to the group. For each 'a' perform this computation with x being all members of the group. Let us find out classes of the above group. E-1 C4 E = C4 ; C4-1 C4 C4 = C4 ; C2-1C4C2 = C4 ; (C43)-1 C4 C43 = C4 ; v(1) -1 C4 h = v(1) C4 h = v(1)d1 = C43 ; v(2)-1 C4 v = v(2)d2 = C43 ; d1-1C4 d1 = C43 and d2-1 C4d2 = C43 This way we found out the whole class induced by C4. After the similar operations with all elements we will find that there are altogether five classes and they are: 10 11 E ; C4, C43 ; C2 ; v(1), v(2) ; d1, d2 How many classes an abelian group of order n will have? 11 12 3. Example of a triangular planar molecule AB3 (Planar molecule); (Point Group D3h) Symmetry elements belonging to this group are: 1. Identity element, E 2. Rotation about the axis passing thru the mid point and perpendicular to the molecular plane, by 1200, C32 3. Rotation about the above described axis by 2400, C32 4. Rotation thru 1800 about the axis joining the corners with the mid point, C2 and there are three of them, called C2 (1), C2 (2), and C2 (3). 5. Reflection about the plane containing the line joining the corner with the mid point and perpendicular to the molecular plane and again there are three of them, v(1), v(2) and v(3). 6. Reflection about the plane, coplanar with the molecular plane, h. 7. Rotation thru 1200 and then reflection about the molecular plane or first reflection and then rotation, S3 8. Rotation thru 2400 followed by reflection about the molecular plane or the other way around, S32. 12 13 GROUP MULTIPLICATION TABLE for D3h ( AB3 planar molecule) E E C3 C32 C2(1) C2(2) C2(3) v(1) v(2) v(3) h S3 S32 C3 C3 C32 E C2(2) C2(3) C2(1) v(2) v(3) v(1) S3 S32 h C32 C32 E C3 C2(3) C2(1) C2(2) v(3) v(1) v(2) S32 h S3 C2(1) C2(1) C2(3) C2(2) E C32 C3 h S32 S3 v(1) v(3) v(2) C2(2) C2(2) C2(1) C2(3) C3 E C32 S3 h S32 v(2) v(1) v(3) C2(3) C2(3) C2(2) C2(1) C32 C3 E S32 S3 h v(3) v(2) v(1) v(1) v(1) v(3) v(2) h S32 S3 E S32 C3 C2(1) C2(3) C2(2) v(2) v(2) v(1) v(3) S3 h S32 C3 E C32 C2(2) C2(1) C2(3) v(3) v(3) v(2) v(1) S32 S3 h C32 C3 E C2(3) C2(2) v(1) h h S3 S32 v(1) v(2) v(3) C2(1) C2(2) C2(3) E C3 C32 S3 S3 S32 h v(2) v(3) v(1) C2(2) C2(3) C2(1) C3 C32 E S32 S32 h S3 v(3) v(1) v(2) C2(3) v(1) C2(2) C32 E C3 E C3 C32 C2(1) C2(2) C2(3) v(1) v(2) v(3) h S3 S32 INVERSES C3-1 = C32 ; C2-1(1) = C2(1) ; C2-1(2) = C2(2); C2-1(3) = C2(3) ; v-1(1) = v(1) ; v-1(2) = v(2); v-1(3) = v(3) ; h-1 = h ; S3-1 = S32 ; (S32) –1 = S3 ; S32 . S3 = E. How to find all the classes ? Let us pick any element, say C3, and find out all the products of type a-1C3 a where a is any element of the group: E-1C3E = C3 ; C3-1 C3C3 = C3 ; (C32)-1 C3C32 = C3 C2-1(1)C3.C2(1) = C2(1) .C3C2(1) = C2(1)C2(3) = C32 ; C2-1 (2)C3C2(2) =C2(2)C2(1) = C32 ; C-1(3)CC2(3) = C(3)C2(2) = C32 ; v-1(1)C3v(1) =v(1) v(3) = C32 ; v-1(2)C3v(2) = v(2) v(1) = C32 ; v-1 ((3)C3v(3) =v(3) v(2) = C32 ; h-1C3h = hS3 = C3 ; S3-1C3 . S3 = S32C3 . S3 =S32.S32 = C3 ; (S32)-1C3S32 = S3h = C3 These results indicate that there is a class which contains only C3 and C32, because similarity transformations on C3 yield either C3 or C32 only. Similarly it can be demonstrated that similarity transformation on h yield only h; on C2(1), C2(2) or C2(3) yield only C2(1), C2(2) or C2(3) ; on v yield v(1), v(2) or v(3) and on S3 or 13 14 S32 yield only S3 or S32 . Based on these observations the D3h group can be classified as follows: {E}; {C3,C32}; {C2(1), C2(2), C2(3)}; {v(1), v(2), 2(3)} ; {h} and {S3, S32}. Subgroups : Orders of the possible subgroups are 1, 2, 3, 4, and 6; all divisors of the group of order 12. They are: (E), (E, C3, C32); {E, C2(1)} {E, C2 (2)} etc. There will be 6 subgroups of order 2. 4. Matrix Representation for symmetry elements and corresponding operators x1 = r cos (-) = r cos cos + r. sin . sin = x. cos + y sin y1 = r sin (-) = r sin cos - r. cos sin = y cos - x sin cos 1200 = - ½ ; sin 1200 = 3/2 14 15 EFFECTS OF THE SYMMETRY OPERATIONS ON ATOMS PLACED ON THE CORNERS ( 1, 2 and 3) of AN EQUILATERAL TRIANGLE AND ALSO ON A POINT ( x,y,z )IN THE SPACE . E C3 C32 C2(1) C2(2) C2(3) v(1) v(2) v(3) h S3 S32 1 1 2 3 1 3 2 1 3 2 1 2 3 2 2 3 1 3 2 1 3 2 1 2 3 1 3 3 1 2 2 1 3 2 1 3 3 1 2 x x ½(-x+3y) ½(-x-3y) x ½(-x-3y) ½(-x+3y) x ½(-x-3y) ½(-x+3y) x ½(-x+3y) ½(-x-3y) y y ½(-3x – y) ½(3x – y) -y ½( y -3x ) ½( y +3x) -y ½( y -3x ) ½( y +3x) y ½(-3x – y) ½(3x – y) z z z z -z -z -z z z z -z -z -z 15 16 Matrix Representation of symmetry operations of D3h 16 17 17 18 5. Example of Tetrahedral Molecule, Td Let ABCD be a tetrahedron AB, BC, CD, DA, BD and AC are the edges of the tetrahedron. Z axis passes through the midpoint of AB. Y Axis passes through the midpoint of BD; X The atom at the origin remains unshifted. Axis passes through the midpoint of BD 18 19 EFFECT OF SYMMETRY OPERATIONS ( Tetrahedral molecule) Td 1. E 2. C2 ( X ) 3. C2 ( Y ) 4. C2 ( Z ) 5. C3 ( A ) 6. C32 ( A ) 7. C3 ( B ) 8. C32 ( B ) 9. C3 ( C ) 10. C32 ( C ) 11. C3 ( D ) 12. C32 ( D ) 13. S4 ( X ) 14. S43 ( X ) 15. S4 ( Y ) 16. S43 ( Y ) 17. S4 ( Z ) 18. S43 ( Z ) 19. AB = S43 ( Z ) c2 ( Z ) 20. AC = S4 ( X ) c2 ( Y ) 21. AD = S43 ( Y ) c2 ( X ) 22. BC = S4 ( Y ) c2 ( X ) 23. BD = S43 ( X ) c2 ( Y ) 24. CD = S4 ( Z ) c2 ( X ) A A C D B A A C D D B B C B D C B C D A A A D C B B B D C A D C B B A D C A C A A D D C B D C B B A C C A B D B D D A C C A B D B D A B A D C B C A C D D B A C C B A C B A D D A C B C A B C B D A D D X X X -X -X -Z Y Z Y Z -Y -Z -Y -X -X Z -Z -Y Y Y X -Z Z X -Y Y Y -Y Y -Y X -Z X Z -X -Z -X Z -Z Z -Y -Y X -X X -Z Y Y Z -X Z Z -Z -Z Z -Y -X Y X -Y X Y -X Y -Y -X X -Z -Z Z Y -X X Y Z Question: Prepare the multiplication table for Td 19 20 6. Construction of Character Table Character Table for D3h point group E 1 1 2 1 1 2 2C3 1 1 -1 1 1 -1 3C2 1 -1 0 1 -1 0 h 1 1 2 -1 -1 -2 2S3 1 1 -1 -1 -1 1 A1' A2' E' A1" A2" E" 3v 1 -1 0 -1 1 0 x2+y2 Rz (x,y) Z (Rx,Ry) z2 (x2-y2, xy) (xz,yz) We have already described the matrix representations for the different symmetry elements . Now, first we will find out the characters of those symmetry elements. Character of a symmetry element will be defined as the sum of the diagonal elements in the matrix representing the element. The characters, χ, of the symmetry elements for the point group D3h whose matrix representations have been shown before ( page 15 & 16) are given below:χ (E)=3, χ (C3)=0, χ(C32)=0, χ {C2(1)} = -1, χ{c2(2)} = -1, χ {C2(3)} = -1 χ (S32) = -2 χ { v(1)}=1, χ { v(2)}=1, χ { v(3)}=1, χ (h) = 1, χ (S3) = -2, Mathematically it turns out that representations of a group can be expressed in terms of these characters. As for example, the characters of D3h group, taken together, will represent this group. And the representation will look like this:E Γ 3 2C3 0 2C2 -1 3v 1 h 1 2S3 -2 But this representation is reducible and hence not suitable for group theoretical treatment. A character table contains only irreducible representations. Therefore, our task is to find out the characters of elements corresponding to the irreducible 20 21 representations. The following steps are taken in preparing the character table. 1. Find the number of possible irreducible representations. The number of irreducible representations belonging to a particular group is equal to the number of classes present in that group. So, the group D 3h will be represented by 6 irreducible representations because this group has 6 classes. 2. Find the dimensions of the irreducible representations It turns out the sum of squares of the dimensions of irreducible representations is equal to the number of elements present in the group (order of the group). For the D3h group, I12 + I22 + I32 + I42 + I52 + I62 = 12; where I represents the dimension of the irreducible representation. (Generally the dimension of a representation is equal to the dimension of the matrices involved in the representation and it turns out that the dimension of an irreducible representation is equal to the character of the identity element belonging to that representation.) The only possible set of values which will satisfy the above equation is I1=1, I2=1, I3=1, I4=1, I5=1 and I6=1. So the D3h group has 6 representations, four with dimensions of one and two with dimensions of two. (See the character table given above). 3. Find the characters of elements for each irreducible representation (a) Sum of squares of characters of elements in a particular irreducible representation is equal to the number of elements present in the group. For D3h χ (E2) + 2 χ (C3)2 + 3 χ (v)2 + 3 χ (C2)2 + 1 χ (h)2 + 2 χ (S3)2 = 12 There will be at least six sets of values for these characters which will satisfy the above equation: the easiest one being where characters for all elements are unity. The above mentioned condition is also useful in determining whether a certain representation is reducible. If the sum of squares of characters of elements 21 22 corresponding to a representation is greater than the number of elements in the group, the representation is reducible, that is, it can be further broken into smaller representation. As for example, the representations for D3h derived above is a reducible one because the sum of squares of characters is: 32 + 2x0 + 3 (-1)2 + 3 (1)2 + 1 (1)2 + 2 (-2)2 = 9+0+3+3+1+8 = 24 (b) The sum of the products of characters of elements corresponding to two different irreducible representations is zero, that is, two irreducible representations are orthogonal. In essence what you do is find the product of characters of an element in two representations and add the products obtained for all elements, and the sum is zero if the representations are irreducible. The above mentioned conditions will be used to prepare the character table. If the matrices representing the similarity transformations outlined before are separated in blocks as shown in page 15 & 16, then we will have two sets of matrices representing the group, that is, the previous reducible representation is divided into two representations as shown below. On considering the characters of the matrices alone we can express the representations as shown below:Γ 2C3 C3 3v h 2S3 1 1 -1 1 -1 -1 2 -1 0 0 2 -1 ------------------------------------------------3 0 -1 1 1 -2 It can be easily tested that these two smaller representations are irreducible. It can be also tested that these representations are orthogonal too. 1(2) (1) + 2(1) (-1) + 3(1) (0) + 3(-1) (0) + 1(-1) (2) + 2(-1) (-1) = 2 + (-2) + 0 + (-2) + 2 = 0 Therefore these two representations are in fact genuine representations of the group, D3h. So far out of 6 irreducible representations we already know three of 22 23 them (marked first, second and fifth representation in the character table). To find the rest of them we have to use hit and trial method, there is no easy way out. We have to set up simultaneous equations and solve them. It needs more practice and intuition than any thing else. Let us find the remaining two one dimensional representations. To find them we have to just change the signs in such a way that we generate a new representation and check if it is orthogonal to any of the already known representations (irreducible ones). Once you find two such representations, you have found all one dimensional representations, see the table. Now the remaining task is to find the irreducible representation of dimension two because we already know one two-dimensional representation. To do this we solved the following equations:1 (2) χ (E) + 2 (1) χ (C3) + 3 (0) χ (C2) + 3 (0) χ v) + 1 (2) χ (h + 2 (-1) χ (S3) = 2.2 – 2 χ (C3) + 2 χ (h) –2 χ (S3) = 0 ....................................................................(1) This equation derives from the othogonality relation between already known two dimensional representation (fifth representation in the character table) and the representation we are trying to find. The orthogonality relation between the fourth representation and the representation we are trying to find gives:1 (1) χ (E) + 2 (1) χ (C3) + 3 (-1) χ (C2) + 3 (-1) χ( v) + 1 (1) χ (h )+ 2 (1) χ (S3) = 0 or 2 + 2 χ (C3) + 3 χ (C2) –3 χ (v) + χ (h) + 2 χ (S3) = 0 .......................................(2) Similarly the orthogonality relation between the first representation and the representation we are trying to find gives:2 + 2 χ (C3) + 3χ (C2) + 3 χ (v) + χ (h) + 2 χ (S3) = 0 .......................................................(3) Add equations (2) & (3), 4 + 4 χ (C3) + 2 χ (h) + 4 χ (S3) = 0 .......................................................(4) Multiply equation (1) by 2 and add it to equation (4), we get 23 24 8 + 4 + 6 χ (h) = 0, or χ (h) = -2. Insertion of the value for the character of h in equation (1) indicates that χ (C3) = χ (S3) in the last representation. With little manipulation by using the above conditions one will arrive a set of values for characters which gives the last sixth representation. THE CHARACTER TABLE of D3h Point Group First representation Second representation Third representation Fourth representation Fifth representation Sixth representation Γ1 Γ2 Γ3 Γ4 Γ5 Γ6 E 1 1 1 1 2 2 2C3 1 -1 1 1 -1 1 3C2 1 -1 1 -1 0 0 3v 1 -1 -1 -1 0 0 h 1 -1 -1 1 2 -2 2S3 1 -1 -1 1 -1 1 7.DECOMPOSITION OF A REDUCIBLE REPRESENTATION INTO IRREDUCIBLE REPRESENTATIONS To apply the concepts of group theory to molecules it is very necessary to break a reducible representation into irreducible ones. The importance of this step will be clear in next few chapters. The nature and largeness of a reducible representation will depend upon the kind of basis sets chosen to mathematically represent the effects of symmetry transformations. The basis sets can be any of several parameters such as atoms themselves, the coordinate of a point in the space, the atomic orbitals or their combinations, modes of vibrations, etc. We have considered the first two in our previous treatment of point group D3h. As we have seen earlier for D3h, the use of the coordinates of a point (x, y, z) for determining the effects of symmetry transformations renders a reducible representation which looks as follows:- Γ1 E 3 2C3 0 3C2 -1 3v 1 h 1 2S3 -2 24 25 Let us see what kind of representation we get if we use a pi orbital on each of three corners of the triangle, points where atoms are placed. Each p z orbital will have a positive and a negative sign and therefore there will be altogether 6 half orbitals to be used for representation; they are P1+,P1-,P2+, P2-,P3+ and P3-. Effect of symmetry operations on pi orbitals P1+ E P1+ C3 P2+ C2(1) P1v P1+ h P1S3. P2P1P1P2P1+ P1P1+ P2+ P2+ P2+ P3+ P3P3+ P2P3P2P2P3P3+ P3P2+ P3+ P3+ P3+ P1+ P2P2+ P3P1P3-. P3P1-. P2+ P2P3+. P1+ There is no need to find out the effect of C32, other C2 s, other v s and S32 because their characters can be easily determined from the information given above. After writing matrix representation of the symmetry elements as described earlier and by adding the diagonal elements of the matrices we get the characters of the symmetry operations as given below (a representation for the D3h point group). E 6 2C3 0 3C2 0 3v 2 h 0 2S3 0 Γ2 And therefore we obtained a different representation by using a new basis set. This illustrates the point we made earlier that a representation of a point group depends upon the choice of the basis set. Yet we can have a third representation if we place one pi orbital at each corner just like before without further dividing them into half orbitals. P1 E C3 C2(1) v h S3. P1 P2 -P1 P1 -P1 -P2 P2 P2 P3 P3 P3 -P2 -P3 P3 P3 P1 P2 P2 -P3 -P3 Character (χ ) 3 0 -1 -1 -3 0 25 26 Γ3 E 3 2C3 0 3C2 -10 3v 1 h -3 2S3 0 A simple rule for finding the character of an symmetry operation For each orbital or atom or any parameter unshifted by the symmetry operation, 1 is added to its character; if the parameter is shifted to its negative value, -1 is added to the character. The character for C3 is zero because none of the orbitals remain unshifted due to this operation and hence they contribute zero to the character. On the other hand, in case of C2 one orbital is changed to its negative contributing –1 to the character whereas the other two orbitals contribute nothing to the character because they are changed to orbitals other than themselves. These reducible representations are made of irreducible representations belonging to the particular point group. One does not have to deduce irreducible representations for a point group because they can be easily looked up into the character tables supplied with group theory books. We have already gone thru the procedures involved in finding out the irreducible representations belonging to a group. Once you know the irreducible representations using a simple mathematical rule you can determine how many times a particular irreducible representation appears in a reducible representation derived on the basis of the method outlined above. [Remember you will get different representations for different basis sets.] 26 27 How to break a reducible representation into irreducible ones The simple mathematical formula is: Number of times an irreducible representation appears, n = (Sum of products of characters of symmetry operation in reducible representation and irreducible representation) / (number of symmetry elements present in the group) n = χ (R) χ i(R) / g; where g = Order of the group. Let us decompose the first reducible representation. Number of times the irreducible representation A1' is present in the reducible representation = 1/12(1x3x1 + 2x0x1 + 3x-1x1 + 3x1x1 + 1x1x1x + 2x-2x1) = 0, that is, A1 is not a component of the reducible representation. A2 ' : 1/12 (1x3x1 + 2x0x1 + 3x-1x-1 + 3x1x-1 + 1x1x1 + 2x-2x1) = 0; A2 ' is not present in the reducible representation. E : 1/12 (1x3x2 + 2x0x-1 + 3x-1x0 + 3x1x0 + 1x2x1 + 2x-2x1) = 1 Therefore E occurs once in the representation. This is a bi-dimensional representation. Now we have to look for just one uni-dimensional representation. A1" ; 1/12 (1x3x1 + 2x0x1 + 3x – 1x1 + 3x1x-1 + 1x1x-1 + 2x-2x-1) = 0 A2" ; 1/12 (1x3x1 + 2x0x1 + 3x – 1x1 + 3x1x-1 + 1x-1 + 2x-2x-1) = 1, That is A2" is also present in the reducible representation. Therefore, the reducible representation = A2' + E ' Similarly, it can be shown that the second representation Γ2 = A1 + A2" + E' + E" And the third representation Γ3 = A2" + E" 27 28 8. Explanation of Symbols used in the Character Table Character Table for D3h point group E 1 1 2 1 1 2 2C3 1 1 -1 1 1 -1 3C2 1 -1 0 1 -1 0 I 3v 1 -1 0 -1 1 0 h 1 1 2 -1 -1 -2 2S3 1 1 -1 -1 -1 1 A1' A2' E' A1" A2" E" II x2+y2 Rz (x,y) z (Rx,Ry) III z2 (x2-y2, xy) (xz,yz) IV Area II Symbol A Remarks One – dimensional representation which are symmetric with respect to the rotation about the principal axis of rotation, χ (Cn) = 1. One – dimensional representation which are antisymmetric with respect to the rotation about the principal axis of rotation, χ (Cn) = -1 Two – dimensional representations. Occur in molecules having an axis higher than C2. Three – dimensional representations. Occur in molecules having more than two C3 axes. Symmetric or antisymmetric with respect to a C2 axis (or a vertical plane of symmetry) perpendicular to the principal axis. Symmetric 'g' or antisymmetric 'u' with respect to a center of symmetry, (inversion) Symmetric or antisymmetric with respect to h . χ (h) = + 1, χ (h) = 1 B E F Subscripts 1 and 2 to A and B Subscripts g and u to A and B Primes and double primes with A and B Area III In this area we will always find six symbols: x, y, z, Ry, Ry, Rz. The first three present the coordinates x, y, and z, while R's stand for rotation about the axes specified in the subscripts. They form the basis for the irreducible representation specified. As for example, in Dh case, 28 29 the coordinate z forms a basis for the A2" representation. The another way of saying the same thing is z transforms as (or according to) A2". Similarly, x and y together from the basis for the E' representation, that is, x and y together transform as or according to E' representation. In the same manner Rz will form the basis for the A2' representation. The rotations around x and y axes together (Rx,Ry) from the basis for E" representation. i.e. they transform as the E" representation. Area IV In this part of the table are listed all of the squares and binary products of coordinates according to their transformation properties. In the present example, (x2 + y2) form a basis for the A1' representation. The pair (xz, yz) transforms according to the representation E". Similarly, (x2 – y2, z2 ) forms the basis for the representation of E'. 9. Method of Classifying Molecules into Point Groups 29 30 10. Properties of Point Groups Point group C2v C3v C4v C5v C6v C2h C3h C4h C5h C6h D2h D3h D4h D5h D6h D2d D3d D4d D5d D6d Symmetry elements of the point group E; C2 ; v ; v E; C3 ; C32 ; 3v E; C4 ; C43 ; C42 = C2 ; 2 v ; 2 d E; C5 ; C54 ; C52 , C53 ; 5 v E; C6 , C65 ; C62 = C3 , C64 = C32 ; 5 v ; 5 d ; C2 E; C5 ; i ; h E; 2C3 ; h; S3, S35 = (C32 . h) E; 2C4; C2; h; i ; S4, S43 E; 2C5; 2C52 ; h; 2S5, 2S53 E; 2C6; 2C3 ; C2; h; i ; 2S6, 2S3 E; C2;C2 ; C2 ; h ; i ; v; 2v E; C3 ; 3C2 ; h ; 2S3 ; 3v E; 2C4; C2 ; 2C2; 2C2; h; i ; 2S4, 2v ; 2d E; 2C5; 2C52 ; 5C2; 2S5, 2S53 ; 5v E; 2C6; 2C3 ; 3C2; 3C2; 2S6, 2S3 ; h ; 3v ; 3d E; C2; 2C2; 2S4; 2d E; 2C3 ; 3C2; 2S6; i ; 3d E; 2C4 ; C2; 4C2; 2S8; 4d E; 2C5; 2C52; 5C2; 2S10; 2S103; i ; 6d E; 2C6; 2C3 ; C2; 2C2; 2S12; 2S123; 2S125; 6d No. of elements 4 6 8 10 12 4 6 8 10 12 8 12 16 20 24 8 12 16 20 24 No. of classes 4 3 5 4 6 4 4 6 6 8 8 6 10 8 12 5 6 7 8 9 Special Groups Td E; 8C3 ; 3C2; 6S4; 6d 24 5 30 31 Some examples of Point Groups Element H2O Benzene ferrocene, Staggered configuration ethane, Staggered planar boric acid Methane Dichloromethane trans-CrBr2(H2O)4 1,3,5-tribromobenzene cyclohexane (Chair form) naphthalene p-dichlorobenzene phenanthrene acetylene anthracene o-dichlorobenzene p-dichlorobenzene Point Group (C2v) (D6h) (D5d) (D3d) (C3h) (Td) (C2V) (D4h) (D3h) (D3d) (D2h) (D2h) (C2V) (Dαh) (D2h) (C2V) D2h Element NH3 PF5 , trigonal bipyramidal ferrocene, eclipsed configuration eclipsed ethane ethylene chloromethane chloroform gauche-CH2ClCH2Cl IF5 B2H6 thiophene CCl4 planar formic acid H2 Chlorobenzene CO m-dichlorobenzene Point Group (C3v) (D3h) (D5h) (D3h) (D2h) (C3V) (C3V) (C2) (C4V) (D2h) (C2V) (Td) (Cs) (Dαh) (C2V) (Cαv) (C2v) 31 32 Analysis of Normal Modes of Molecular Vibration (ANMMV) Normal vibrations belonging to a molecule form bases of Irreducible Representations of a Character Table. A combination of group theory and experimental data of infra-red (IR) and Raman Spectroscopy can help choose a structure of a molecule from a list of possible structures that could be assigned to the molecule. For example, an AB3 type molecule can have either planar triangular (D3h ) or triangular pyramidal ( C3v ) structures. Group theory can help select one of these two possible structures, provided spectral data are available. Moreover, characters for symmetry operations can give insight as to what irreducible representations stand for symmetric and antisymmetric stretching and in-plane and out-ofplane bending vibrations. With the Character Table one can even predict that which vibrations are IR active and which are Raman active. Following steps are taken for ANVMM : (1) (2) (3) (4) (5) (6) Determination of the Point Group for the molecule. Regeneration of a Reducible Representation Decomposition of the Reducible Representation into Irreducible Representations Reduction of the Reducible Representation to the Vibrational Representation Selections Rules to distinguish IR and Raman active vibrations Determination of nature of Vibrational Modes 1. Determination of the Point Group for the molecule : Remember it will depend upon possible structures that can be assigned to the molecule. 2. Regeneration of a Reducible Representation As you know, a basis set will be required for this purpose. For application in the field of ANMMV, a set of displacement coordinate vectors ( xi,yi,zi) is placed on each and every atom. Matrices representing symmetry operations are constructed and their characters are tabulated to create a reducible representation. In short, atoms that are moved by the symmetry operation contribute zero to the character and one that remain unshifted contribute values to the character as given in the following table : Contribution made by an unshifted atom to the character of a symmetry operation Symmetry Operation E σ i C2 C3 Character (χ) 3 1 -3 -1 0 Symmetry Operation C4 S3 S4 S6 Character (χ) 1 -2 -1 0 32 33 Consider an example of an AB3 molecule having D3h point group E No. of unshifted atoms Г 4 12 2C3 1 0 3C2 2 -2 3σv 2 2 σh 4 4 2S3 1 -2 3. Decomposition of the Reducible Representation into Irreducible Representations A reducible representation can be broken down in Irreducible representations by following method described in earlier chapters ( see page 23 ). The above representation Г = A'1 + A'2 + 2A''2 + 3E' + E'' Note that sum of the dimensions ( A & B = 1, E = 2 ) is equal 12 given 3n rule. 4. Reduction of the Reducible Representation to the Vibrational Representation Irreducible representations one each for translational modes ( Tx,Ty,Tz) and for rotational modes ( Rx,Ry,Rz) are subtracted from the total reducible representation. The irreducible representations having x, y & z in the corresponding third column of the Character Table are marked for translational modes. Similarly, irreducible representations containing Rx, Ry & Rz in their third column are marked for rotational modes. In case E has (x,y) or (Rx,Ry) in the third column, removal of E will will take care of both Tx & Ty or Rx & Ry, respectively. For AB3 molecule (D3h point group) Translation : Irreducible representations (combined) for Tx, Ty is E' and for Tz is A''2 Rotation : Irreducible representation (combined ) for Rx, Ry is E'' and for Rz is A'2 Г = A'1 + A'2 + 2A''2 + 3E' + E'' Therefore, Гvib = A'1 + A''2 + 2 E' ; Note the sum of dimensions of Irreducible representations is 6 = 3n – 6 5. Selections Rules to distinguish IR and Raman active vibrations I. R. active : Remaining irreducible representations having x,y,z in their third column in the Character Table represent normal modes of vibrations that are infra-red active. 33 34 Raman active : Remaining irreducible representations having sqare and binary products of the coordinates and their combinations ( x2, y2, z2, xy, xz, yz, x2+ y2, x2- y2 etc.) in the fourth column of the Character Table represent Raman active vibrations. An irreducible representation can represent both I.R. and Raman active vibration. For AB3 molecule (D3h point group) A'1 – Raman active vibration; A''2 – I. R. active vibration and E' – both Raman and I. R. active vibrations 6. Determination of nature of Vibrational Modes The aim of this section is to get an idea about the nature of normal modes of vibration i.e. whether they involve symmetric or assymetric stretching of bonds or in-plane or out-of-plane bending of bond angles. First it is determined whether one of the irreducible representations represent an out-of plane bending vibration. While these vibrations change the bond angles , they leave the molecular plane. Out – of – plane bending ( deformation) vibration is represented by an irreducible representation having negative value of χ (σh). Since out-of-plane bending disturbs the symmetry of the molecule, it is not further considered for vibrational analysis Next, bond distances and bond angles are chosen as internal coordinates for basis set. Effect of symmetry operations on this basis set is determined ( bonds or angles that remain unshifted contribute 1 and those that move contribute zero to the character) and then compiled to generate a reducible representation which is decomposed into irreducible representations by established method. The resulting irreducible representations represent vibrations resulting from stretching or bending depending on whether bonds or angles were selected as basis sets. Let's consider again the example of AB3 (D3h) Гvib = A'1 + A''2 + 2 E' Identify the out-of-plane bending vibration and remove it. From the examination of the Character Table we can easily determine that A'2 represents a out – of – plane bending vibration E Гvib A''2 6 1 2C3 0 1 3C2 0 -1 3σv 2 1 σh 4 -1 2S3 -2 -1 34 35 Гvib - A''2 5 -1 1 1 5 -1 Bond distance : Consider the three A-B bond distances as a basis set of the representation. Compile the effects of symmetry operation to form a reducible representation ГAB. E ГAB 3 2C3 0 3C2 1 3σv 1 σh 3 2S3 0 Гvib - A''2 - ГAB 2 -1 0 0 2 -1 Decomposition will reveal that ГAB = A'1 + E' which would mean that A'1 and one of the E' represent stretching vibrations; stretching vibration for A'1 is symmetric one. Bond angles : We can consider only two bond angles for the representation because the third would depend upon the other two ( not independent ). The logical conclusion is that the remaining irreducible representation ( Гvib - A''2 - ГAB ) should represent the ГBAB. By examination of the Character Table it is easily found that this representation is simply just another E.' So , the following can be concluded regarding the nature of normal modes of vibration : A'1 --- Symmetric Stretching Vibration E' ---E' ---Assymetric Stretching Vibration In – Plane Bending Vibration A''2 ---- Out – of – plane Bending Vibration 35 36 Problems 1. Determine point groups of (a) benzene, (b) planar boric acid, (c) 1,3,5 tribromobenzene , (d) naphthalene , (f) p-dichlorobenzene, (g) PCl3 ( pyramidal) , (h) S2Cl2 (nonplanar), (i) formaldehyde, (j) CH2FCl and (k) CO32-. 2. How many symmetry operations and how many classes of symmetry operations are there in (a) C2v point group and (b) C3v point group ? 3. What is the relation between the number of symmetry species and the number of classes of symmetry operations ? . 4. For the point group C3v is it true, as it was for the C2v point group, that each symmetry operation is its own inverse ? Give examples. 5. Generate a reducible representation for eclipsed and staggered ethanes. 6. Assume a molecule of PF5 can have (a) triangular bipyramidal or (b) square planar pyramidal structure. Describe the application of group theory in the determination of PF5 , provided data of vibrational spectroscopy are available. 7. Suppose H2O2 can have either one of the cis – or trans – planar structures. Describe methods step be step for selecting one of the above structures if vibrational data are available. 36 37 8. None of the normal modes of vibration for a c3v molecule involves an out-of-plane bending , verify. 9. Compare results obtained by analysis of normal modes of vibration of chloroform and ammonia. 37

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