Lecture Notes on Molecular Symmetry and Group Theory | CLASS 40, Study notes of Classical Literature

Download Lecture Notes on Molecular Symmetry and Group Theory | CLASS 40 and more Study notes Classical Literature in PDF only on Docsity! 1 Molecular Symmetry and Group Theory Robert L. Carter Wiley Paperback (Optional text for Chem 176B (Ford’s class)) ~$40 UCEN bookstore--also can get online http://www-theory.mpip-mainz.mpg.de/~gelessus/group.html http://www.cryst.ehu.es/rep/point http://electron6.phys.utk.edu/qm2/modules/m4/wigner.htm 2 Representations of Groups Character Tables and Their Applications References Carter…… Chapters 2 and 3; Chapter 6; Appendices A and C Shriver….. Chapter 7, Student Aid http://www.hull.ac.uk/php/chsajb/symmetry&spectroscopy/index.html http://www.hull.ac.uk/php/chsajb/symmetry/revision.html http://bcs.whfreeman.com/ichem4e/default.asp?uid=0&rau=0 3 C1(1) E h=1 A 1 Cs=Ch(m) E h A 1 1 x, y, Rz x 2, y2, z2, xy A 1 –1 z, Rx, Ry yz, xz Ci = S2 (1) E i Ag 1 1 Rx, Ry, Rz x 2, y2, z2, xy, xz, yz Au 1 –1 x, y, z Group Character tables -- Shriver Resource Section 4 Page 786 Where do they come from? How do you use them? 4 Consists of elements and a combining operation with the following properties: 1. Closure Ai ! Aj = some element of the group " AiAj 2. Identity element E EAi = AiE = Ai (so that EE = E 2 = E)) 3. Associative operation (AiAj)Ak = Ai(AjAk) 4. Reciprocal elements for all Ai exist, Ak = Ai –1 , such that AiAi –1 = Ai –1Ai = E WHAT IS A GROUP? 9 For the point group Cs with mirror plane % to c axis (z coordinate) 1 0 0 0 1 0 0 0 1 ! " # # # $ % & & & x y z # $ # $ = x y z # $ # $ &h representation for &h Trace of a matrix is the sum of the diagonal elements i.e. for the &hmatrix above = 1 + 1 + 1 = 1 = trace of matrix 10 1 0 0 0 1 0 0 0 1 ! " # # # $ % & & & E 1 0 0 0 1 0 0 0 1 ! " # # # $ % & & & &h x,y,z are independent of each other in Cs and C2v since no symmetry operation of either group interchanges x and y Therefore can break down the above representations into independent irreducible (can’t be broken down further) representations that are sub-block diagonal matrices, e.g. for Cs , 1-d irreducible matrices. Not the simplest representations... 11 x y z (1) (1) (1) &h E (1) (1) (1) Trace of the matrices (in this case 1-D irreducible) representations gives characters x y z +1 +1 -1 &h E +1 +1 +1 character table which gives 12 Characters describe the effect of the symmetry operations on x,y,z and other specific functions, Characters form representation of the group Cs E &h A' +1 +1 A( +1 -1 x,y z x y z +1 +1 -1 &h E +1 +1 +1 character table for Cs can better summarize character table Prime (‘) and (“) double prime in the symmetry representation label indicates “symmetric” or “anti-symmetric” with respect to the &h Mulliken symbols x y z Notation 13 x and y have the same symmetry properties for all operations (E and &h) z has different symmetry properties than x and y and a different irreducible representation and the same irreducible representation Cs E &h A' +1 +1 x,y, Rz x 2, y2, z2, xy xz2, yz2, x2y, xy2, x3, y3 A( +1 -1 z, Rx, Ry yz, xz linear functions rotations quadratic functions cubic functions Character table for point group Cs z3, xyz, y2z, x2z Summarize as follows: 14 C2 E C2 A +1 +1 z, Rz x 2, y2, z2, xy z3, xyz, y2z, x2z B +1 -1 x, y, Rx, Ry yz, xz xz 2, yz2, x2y, xy2, x3, y3 linear functions rotations quadratic functions cubic functions Character tables for point group Cs and C2 Cs E &h A' +1 +1 x,y, Rz x 2, y2, z2, xy xz2, yz2, x2y, xy2, x3, y3 A( +1 -1 z, Rx, Ry yz, xz linear functions rotations quadratic functions cubic functions z3, xyz, y2z, x2z Notation “A” means symmetric with regard to rotation about the principle axis “B” means anti-symmetric with regard to rotation about the principle axis The infrared (IR) active molecular motions are linear The Raman active molecular rotations are quadratic 19 ! ni = 1 h gc "i "r c # ni = number of times irreducible representation i occurs in the reducible representation h = order of the group = number of elements in the group c = class of operations gc = number of operations in the class *i = character of the irreducible representation for the operations of the class *r = character for the reducible representation for the operations of the class Now need to find the irreducible reps that make up reducible rep 20 Class structure of a group In the group tables in your text the symmetry operations are collected together into classes C3v E 2C3 (z) 3&v linear functions quadratic cubic rotations functions functions A1 +1 +1 +1 z x 2+y2, z2 z3, x(x2-3y2), z(x2+y2) A2 +1 +1 -1 Rz y(3x 2-y2) E +2 -1 0 (x, y) (Rx, Ry) (x 2-y2, xy) (xz2, yz2) [xyz, z(x2-y2)] (xz, yz) [x(x2+y2), y(x2+y2)] di = dimension of a representation = value of the character of the identity operation class #1 class #2 class #3 + di 2 = h = the order of the group = # elements in the group # irreducible rep’s If the symmetry operations are in the same class they have the same character!! The number of classes = number of irreducible representations !! In an Abelian Group, every element is in its own class!!! 21 11-13)r &v'&vC2EC2v Reducible representation of C2v “Tabular” method for reducing )r 1111A1 1-1-11B2 -11-11B1 -1-111A2 &v'&vC2EC2v 22 +/4+ B2 B1 A2 A1 11-13)r &v'&vC2EC2v Worksheet for reducing )r ! ni = 1 h gc "i "r c # enter gc*i*r values in table sum and divide by h = order = 4 23 1 1 0 1 +/4 4 4 0 4 + 1-113B2 -1113B1 -1-1-13A2 11-13A1 11-13)r &v'&vC2EC2v Completed worksheet )r = A1 + B1 + B2 + must be divisible by group order = 4 dimension of )r = dr=+ndi i i.e. 3 = 1x1 + 1x1 +1x1 If characters are used, the dimension of a representation = character of the identity operation Note: 24 C3v E 2C3 (z) 3&v linear functions quadratic cubic rotations functions functions A1 +1 +1 +1 z x 2+y2, z2 z3, x(x2-3y2), z(x2+y2) A2 +1 +1 -1 Rz y(3x 2-y2) E +2 -1 0 (x, y) (Rx, Ry) (x 2-y2, xy) (xz2, yz2) [xyz, z(x2-y2)] (xz, yz) [x(x2+y2), y(x2+y2)] Frequently run into degenerate representations, such as the E irreducible rep for C3v 29 )m &vC3EC3v ! "1/2 " 3 2 0 3 2 "1/2 0 0 0 1 ! 1 0 0 0 1 0 0 0 1 ! 1 0 0 0 "1 0 0 0 1 Vector (x, y, z) transformation with C3v Block diagonalize by inspection to give irreducible representation character table C3v E 2C3 3&v 2 -1 0)x,y )z 1 1 1 30 C3v E 2C3 3&v 2 -1 0)x,y )z 1 1 1 In C3v there is no difference in symmetry between x and y directions, and they may be treated as equivalent and indistinguishable. Every property associated with the x direction will be equivalent and indistinguishable from the corresponding property along the y direction, which means that the properties are degenerate. A1 +1 +1 +1 z x 2+y2, z2 z3, x(x2-3y2), z(x2+y2) A2 +1 +1 -1 Rz y(3x 2-y2) E +2 -1 0 (x, y) (Rx, Ry) (x 2-y2, xy) (xz2, yz2) [xyz, z(x2-y2)] (xz, yz) [x(x2+y2), y(x2+y2)] C3v E 2C3 (z) 3&v Need to label the character table, and include one additional irreducible representation, A2 31 More complex groups (OPTIONAL) . = exp(i/) = cos / + isin / is a rotation operator .p = exp(2!pi/n) = cos(2!p/n) + isin(2!p/n) where p = 1,2, … n Common procedure is to modify the group table by adding the degenerate characters to make a reducible E representation that is real. C3 E C3 C3 2 {E} 2 2cos2!/3 2cos2!/3 .* = exp(-i/) = cos / – isin / is a rotation through -/ 32 More complex groups Thus, for C3, use the modified character table to determine the symmetries contained in a reducible representation C3 E C3 C3 2 {E} 2 2cos2!/3 2cos2!/3 A 1 1 1 Only catch is that n ({E}) will be twice as large as it should be, and must divide by 2 to get the number of time the true E irreducible representation is contained in the reducible representation ! ni = 1 h gc "i "r c # 33 Example: C4h . = exp(i/) = cos / + isin / / = 2!/4 = !/2 so that .1 = i and e1* = -i ; .3 = -i and e3* = i 34 Now suppose have a property that transforms as the reducible representation )r Use Carter’s tabular method using the modified C4h character table to reduce )r This gives )r = 2Bg + 2{Eg} + Au But as noted above when we add true Eg irreducible representation characters to give the modified character table, the contributions of {Eg} and {Eu} are two times too large. Therefore the correct result is )r = 2Bg + Eg + Au