Group Theory and Molecular Symmetry

Group Theory and Molecular Symmetry Molecular Symmetry Symmetry Elements and perations Identity element E - Apply E to object and nothing happens. bject is unmoed. Rotation axis C n - Rotation of object through 60 /n yields an indistinguishable object. C axis C axis C axis 4 - C C E, C C C E, etc - The C n axis with the highest n is an object s principal axis. Reflection plane - σ - ertical - σ - plane is parallel to principal axis -,, trichloroethane (eclipsed or staggered) is molecule with only σ - horizontal - σ h - plane is perpendicular to principal axis - ethane (eclipsed) is molecule with σ h - dihedral - σ d - plane is parallel to principal axis and bisects angle between two C axes. - staggered ethane is molecule with σ d - Examples of composite operations - σ σ E, σd C E, C4 σ V Center of inersion i - turns molecule inside out - make following replacements x x, y y and z z - staggered ethane has i, eclipsed ethane does not. C

Improper rotation axes S n - reflection followed by rotation through 60 /n - object may hae S n without haing σ h or C n separately. - Allene is an example. C C C Symmetry Groups - all symmetry operations on a molecule form a closed group called a symmetry group. Example: C C E C σ σ ' E E C σ σ ' C C E σ ' σ σ σ σ ' E C σ ' σ ' σ C E σ σ Special Groups C has no symmetry C C N C C i has only inersion symmetry Br Br C s has only mirror plane Br Br

C n Groups C n has E and (n-) C n symmetry elements C C C C n has E, (n-) C n and n σ symmetry elements N C C nh has E, (n-) C n, (n-) S n, σ h, et. al. symmetry elements C h D n Groups similar to C n, but has n C axes perpendicular to C n axis D n has E and (n-) C n, n C axes principal axis, et. al. symmetry elements neither staggered nor eclipsed D D nh has E, (n-) C n, n σ, σ h et. al. symmetry elements Ru D 5h D nd has E, (n-) C n, σ, σ d, et. al. symmetry elements D d

S n groups S n has E, (n-) S n symmetry elements 4 Cubic groups T, T d, T h tetrahedral groups F F Si F F, h octahedral groups T d Co h Icosahedral group I h icosahedral group Buckminsterfullerene Linear groups C - eteronuclear diatomic molecules are most important example. D h - omonuclear diatomic molecules are most important example. Immediate Consequences of Symmetry. Polarity - Molecule cannot hae dipole moment perpendicular to C n axis. - Molecule may hae dipole moment parallel to principal axis if no C axes are perpendicular to principal axis - Molecule with S n axis may not hae dipole moment. - Thus only C, C s, C n and C n molecules can hae dipole moment.. Chirality - Molecule without S n axis may be chiral - A chiral molecule cannot hae center of inersion.

Representations of Groups Recall that the symmetry elements for a C molecule form a group that, by definition, is closed under multiplication. 5 C E C σ σ ' E E C σ σ ' C C E σ ' σ σ σ σ ' E C σ ' σ ' σ C E σ σ C Rather than relying on our ability to isualize the spatial changes that occur during a pair of symmetry operator, we would like to use numerical representations for the group. Consider the following ector to represent the nine coordinates that describe the position of a water molecule. x y z x y z x y z If we wanted to represent the identity element with a matrix, the following matrix E would be a sensible choice. Note the effect of multiplying our coordinate ector by the matrix E. Indeed, E is an identity element. 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 E 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 x x 0 0 0 0 0 0 0 0 y y 0 0 0 0 0 0 0 0 z z 0 0 0 0 0 0 0 0 x x E 0 0 0 0 0 0 0 0 y y 0 0 0 0 0 0 0 0 z z 0 0 0 0 0 0 0 0 x x 0 0 0 0 0 0 0 0 y y 0 0 0 0 0 0 0 0 z z

With the C rotation, and switch places and the x and y directions change sign. The following matrix represents the C rotation. 6 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 C 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 x x 0 0 0 0 0 0 0 0 y y 0 0 0 0 0 0 0 0 z z 0 0 0 0 0 0 0 0x x C 0 0 0 0 0 0 0 0 y y 0 0 0 0 0 0 0 0 z z 0 0 0 0 0 0 0 0 x x 0 0 0 0 0 0 0 0y y 0 0 0 0 0 0 0 0 z z Thus σ and σ also hae 9 9 matrix representations. 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 σ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 x x 0 0 0 0 0 0 0 0 y y 0 0 0 0 0 0 0 0 z z 0 0 0 0 0 0 0 0 x x σ 0 0 0 0 0 0 0 0 y y 0 0 0 0 0 0 0 0 z z 0 0 0 0 0 0 0 0 x x 0 0 0 0 0 0 0 0 y y 0 0 0 0 0 0 0 0 z z σ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 x x 0 0 0 0 0 0 0 0 y y 0 0 0 0 0 0 0 0 z z 0 0 0 0 0 0 0 0 x x σ 0 0 0 0 0 0 0 0 y y 0 0 0 0 0 0 0 0 z z 0 0 0 0 0 0 0 0 x x 0 0 0 0 0 0 0 0 y y 0 0 0 0 0 0 0 0 z z

These four matrices faithfully reproduce the spatial manipulations of each symmetry element. 7 ***But the big question is; when we multiply the symmetry elements together, will we reproduce the multiplication table for the C group?*** C E C σ σ ' E E C σ σ ' C C E σ ' σ σ σ σ ' E C σ ' σ ' σ C E σ σ C Let s look at an example. 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 σ σ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 C 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 We could go through the other pairs of operations to erify that these four matrices do faithfully reproduce the group multiplication table aboe. Therefore the four 9 9 matrices that we constructed are a representation of the group that is called a reducible representation.

8 That these matrices are a reducible representation implies that they can be reduced (whateer that means at the moment). To reduce a representation we need to ask the question, can we reproduce the group multiplication table with simpler objects. Sure we can! Let us focus on the matrix of the oxygen in rows 4, 5 and 6 and columns 4, 5 and 6 of the 9 9 matrices we just used. 0 0 E 0 0 0 0 0 0 C 0 0 0 0 0 0 σ 0 0 0 0 σ 0 0 0 0 0 0 0 0 0 0 0 0 σ σ 0 0 0 0 0 0 C 0 0 0 0 0 0 Do these four matrices faithfully reproduce the group multiplication table? After doing the other pairwise combinations we can say yes. Thus these four matrices are a representation of the C group. ( C ) { E,C,, } Γ σ σ 0 0 0 0 0 0 0 0 Γ ( C) 0 0, 0 0, 0 0, 0 0 0 0 0 0 0 0 0 0 What if we separated the z component from the rest of the components? 0 0 0 E 0 0, 0 0 0 σ () 0 0 0 0 0, 0 0 0 () 0 0 0 C 0 0, 0 0 0 0 0 0 0 0, 0 0 0 σ () () Would the two groups of matrices reproduce the group multiplication table? 0 0 0 0 Γ A( C ),,, 0 0 0 0 0 0 0 σ σ C 0 0 0 You can check the other pairs of symmetry operations to see that Γ A is a faithful representation.

z ( C) {(,,, ) ( ) ( ) ( )} Γ 9 ( ) ( ) ( ) σ σ C While the example using Γ z seems triial, it is important because it works! Γ z faithfully reproduces the group multiplication table. What is special about this representation is that it cannot be broken further into pieces that reproduce the group multiplication table. We say that Γ z is an irreducible representation of the C symmetry group. It is also the totally symmetric representation of the group. More about that later. Let s keep going with Γ A. Let s break apart its x and y components. 0 E, 0 () () 0 σ 0 (), ( ) 0 C, 0 0 σ 0 ( ) ( ) (, ) ( ) x ( C) {( ),( ),( ),( ) } Γ y ( C) {( ),( ),( ),( ) } Γ Are these faithful representations for C? x ( C) {( ),( ),( ),( ) } Γ ( ) ( ) ( ) σ σ C y ( C) {( ),( ),( ),( ) } Γ ( ) ( ) ( ) σ σ C After checking with the other combinations of symmetry elements, we can see that both Γ x and Γ y are faithful representations of the algebra of the C symmetry group. They are also two more of the irreducible representations of the group. There is one more irreducible representation for the C symmetry group that we haen t found yet and that is, A ( C) {( ),( ),( ),( ) } Γ Thus the C symmetry group has four irreducible representations.

Character Tables and Symmetry Labels Definitions rder number of total symmetry operations in a symmetry group ass subset of symmetry operations that transform into each other. - Eery C n axis will hae n symmetry operations that belong to the same class. Character sum of diagonal elements in a matrix (also known as trace) 0 0 A χ A B χ B 0 0 0 Symmetry Label label gien to an object (such as an M or a ibration) that corresponds to symmetry found in symmetry group. - Each symmetry group has fundamental shapes (symmetries) that can be added and subtracted to represent an arbitrary shape of an object. - Each fundamental shape is also known as an irreducible representation. - ne can loosely think of the symmetry labels as a basis set to describe to shape of an object in a particular symmetry group. 0 Character Table Example: C C E C σ σ ' A z z, y, x A - - xy B - - x xz B - - y yz z σ σ y A, A, B and B are symmetry labels Each row in the character table represents an irreducible representation. C Interpretation of the Character Table.) If a molecule at equilibrium in C has an object (such as an M or a ibration) associated with that hae the same C symmetry; the object is gien the A symmetry label. - The first row of the character table indicates that the object has the same shape or parity after each symmetry operation is performed. - Example: Consider a molecular orbital made from an s orbital on each atom..) If an object changes parity for each reflection, the object is gien the A symmetry label. - The alue in the second row indicate that the parity of the object has switched when the object is reflected in each plane. - Example: Consider a molecular orbital made from a set of p x orbitals on the atoms of water. rbitals plane.

.) If an object changes its parity for the C rotation and the σ reflection, then the object is gien the B symmetry label. - Example: Consider a molecular orbital made from a set of p z orbitals on the atoms of water 4.) If an object changes its parity for the C rotation and the σ reflection, then the object is gien the B symmetry label. - Example: Consider a molecular orbital made from a set of p x orbitals on the atoms of water. rbitals plane. 5.) Functions on the right of character table indicate to which symmetry they belong. For C molecule, z is along C axis and x is perpendicular to plane. Another Character Table Example: C C E C σ A z z, x +y A - E - 0 (x,y) (xz,yz), (xy, x -y ) 6.) Coefficients in header row indicate number of operations within each class C, C are members of a single class. - - σ, σ and σ are members of a single class. - Adding the numbers of elements of each class yields the order of the group. - For C, order + + 6 7. The third row of the character table is the E symmetry representation that is an example of a degenerate irreducible representation. - fundamental symmetry must hae two (or more) parts. - The representation entangles two (or more) objects that cannot be separated. - Note for C, the pair of functions (x, y) cannot be separated. - Symmetry operations must operate on both together. - x xcosθ ysinθ - y xsinθ+ ycosθ - bjects that hae E symmetry (such as orbitals or ibrations) must hae degenerate energies. - T d, h and I h objects can hae T symmetries that are triply degenerate.

Constructing a Reducible Representation from a Basis Set. By performing all of the symmetry operations upon the character of a particular basis set for an object, a reducible representation is created. Consider a basis set of p orbitals for a cyclopropenyl anion. **What are symmetries of the orbitals that can be formed from these orbitals?** Consider the effect of symmetry operations on the basis. For each symmetry operation, a) Count the number of basis objects that remain exactly the same. b) Count the number of basis objects that hae changed parity This representation is the reducible representation of the symmetries contained within this basis set of functions. For the cyclopropenyl anion and its perpendicular p orbitals, the representation is E C σ Γ 0 Reducing a Reducible Representation The number of times of particular irreducible representation can be found within a reducible representation, a i, is found by using the ery intimidating formula: ai χγ( R) χγ ( R) i h R a i the number of irreducible representations h the order of the group R a symmetry operation, thus Σ is sum of all symmetry operations R χ R characters of the reducible representation ( 0 ) from example aboe Γ ( ) χ Γ i ( R ) characters of the i th irreducible representation from the character table A ( ) ( ) A ( -) ( -) E ( - 0) ( - 0) - note the number of symmetry elements of each class are included.

ne can think of taking a dot product of the reducible representation with the irreducible representation then diiding by the order of the group to find the number of times the irreducible representation is found. C E C σ A z z, x +y A - E - 0 (x,y) (xz,yz), (xy, x -y ) E C σ Γ 0 a A { ()() 0 ( )() ( )() } ( 6) + + + + 6 a A { ()() 0 ( )() ( )( ) } ( 0) 0 6 + + 6 a E { ( )( ) 0 ( )( ) ( )( 0) } ( 6) 6 + + 6 Thus the reducible representation, ( 0 ) Γ is reduced to Γ A + E - f the orbitals made from the perpendicular p orbitals of the cyclopropenyl anion, one will hae A symmetry and two will hae E symmetry and be degenerate. Using Group Theory to Determine Nonzero Integrals The symmetry of orbitals and operators can be exploited to find zero integrals without doing the integral explicitly. (Always a good idea!) f(x) f(x) x x a f ( x) dx 0 ( ) ( ) a a a 0 a f x dx f x dx 0 Note: There is no need to explicitly do the first integral. The symmetry of the function tells us that the integral must be zero. (If the function is not perfectly symmetric about the origin then the integral is not zero.)

4 For an integral (such as oerlap integral or a matrix element), any physically meaningful alue the character of the integral must be independent of any symmetry operation upon it. (No matter how integral is oriented, it must hae the same alue.) Thus the representation of any nonzero integral must contain the totally symmetric representation (A or A or A g etc depends on specific point group). For an oerlap integral, φψ,. Find the symmetry of each orbital, φ, ψ. Multiply characters of the orbitals together.. Find the irreducible representation. 4. If the irreducible representation does not contain the totally symmetric representation, the integral is zero. 5. If the irreducible representation does contain the totally symmetric representation we must do the integral explicitly to find its alue. Example: Consider the oerlap integral, s p z in a C molecule. Γ ( s) A ( p ) A Γ z Γ A nonzero oerlap possible Example: Consider the oerlap integral, s p x in a C molecule. Γ ( s) A ( ) Γ p B x Γ B oerlap equals zero Example: Consider the oerlap integral, px p y in a C 4 molecule. C 4 E C C 4 σ σ d A z x +y, z A - - B - - x -y B - - xy E - 0 0 0 (x,y) (xz, yz) ( ) Γ p 0 0 0 E x ( y ) Γ p 0 0 0 E Γ 4 4 0 0 0

5 At this point, we must reduce the representation into its irreducible parts. a A { 4 ()() 4 ()() 0 ( )() 0 ( )() 0 ( )() } ( 8) + + + + + + + + 8 a A { 4 ()() 4 ()() 0 ( )() 0 ( )( ) 0 ( )( ) } ( 8) + + + + + + + + 8 a B { 4 ()() 4 ()() 0 ( )( ) 0 ( )() 0 ( )( ) } ( 8) + + + + + + + + 8 a B { 4 ()() 4 ()() 0 ( )( ) 0 ( )( ) 0 ( )() } ( 8) + + + + + + + + 8 a E { 4 ( )( ) 4 ( )( ) 0 ( )( 0) 0 ( )( 0) 0 ( )( 0) } ( 0) 0 + + + + + + + + 8 Thus the representation, Γ 4 4 0 0 0, can be reduced to Γ A+ A + B+ B. The representation contains A ; therefore, the oerlap integral, px p y, could be nonzero. For a transition dipole matrix element, φ µψ ˆ,. Find the symmetry of each orbital, φ, ψ. Find the characters of the components of the dipole operator, µ er e( xi ˆ + yj ˆ + zk ˆ ) - I. e., find the characters of x, y and z.. For each component of the dipole operator, multiply the characters of the component and the characters of the two orbitals. 4. Find the irreducible representation. 5. If the irreducible representation does not contain the totally symmetric representation, the integral is zero. Thus no transition can occur. 6. If the irreducible representation does contain the totally symmetric representation we must do the integral explicitly to find whether a transition between the two states φ, ψ are allowed.

6 Example: Determine the electronic transitions that are allowed in a C molecule based on the symmetry of the electronic states. C E C σ σ ' A z z, y, x A - - xy B - - x xz B - - y yz A x A A y A A A B - - B - - A A B - - B - - transition not allowed transition not allowed A z A A x A A A A B - - A A - - A B - - transition allowed transition not allowed While the aboe process will eentually yield the correct answers, we could choose a more direct route. Let us think in terms of soling for an unknown. For example, we choose two electronic states and ask what is the only operator that will make the integral hae A symmetry. A? B A x B A A? B - - B - - B - - A A The only operator with B symmetry is x; therefore, the only allowed transition is A x B. A? B A y B A A? B - - B - - B - - A A The only operator with B symmetry is y; therefore, the only allowed transition is A y B.

A? A A z A A - - A - -? A A - - A - - A A The only operator with A symmetry is z; therefore, the only allowed transition is A z A. 7 A? B A y B A - - A - -? B - - B - - B - - A A The only operator with B symmetry is y; therefore, the only allowed transition is A y B. A? B A x B A - - A - -? B - - B - - B - - A A The only operator with B symmetry is x; therefore, the only allowed transition is A x B. B? B B z B B - - B - -? A B - - B - - A A The only operator with A symmetry is z; therefore, the only allowed transition is B z B. B? B B y B B - - B - -? A - - B - - B - - A A There is no dipole operator with A symmetry; therefore, an electronic transition between a B and B state is not allowed. In other words, B ˆ µ B 0