1 Group Theory and Molecular Symmetry Molecular Symmetry Symmetry Elements and Operations Identity element – E - Apply E to object and nothing happens. Object is unmoved. Rotation axis – Cn - Rotation of object through 360/n yields an indistinguishable object. C 3 axis C4 axis C a x i s 2 - C2 C2 E , C3 C3 C3 E , etc … - The Cn axis with the highest n is an object’s principal axis. Reflection plane - - vertical - v - plane is parallel to principal axis - 1,1,1 trichloroethane (eclipsed or staggered) is molecule with only v Cl ClCl H ClCl H HH Cl HH - horizontal - h - plane is perpendicular to principal axis - ethane (eclipsed) is molecule with h - dihedral - d - plane is parallel to principal axis and bisects angle between two C2 axes. - staggered ethane is molecule with d - Examples of composite operations - E , d C2 E , C 4 V C 2 Center of inversion – i - turns molecule “inside out” - make following replacements x x , y y and z z - staggered ethane has i, eclipsed ethane does not. 2 Improper rotation axes – Sn - reflection followed by rotation through 360/n - object may have Sn without having h or Cn separately. - Allene is an example. H C C H C H H Symmetry Groups - all symmetry operations on a molecule form a closed group called a symmetry group. Example: C2v C2v E C2 v v' E E C2 v v' C2 C2 E v' v v v v' E C2 v v' v' v C2 E v’ H H O C2 Special Groups C1 – has no symmetry H Cl H C O N O C HH Ci – has only inversion symmetry Br Cl Cl Br Cs – has only mirror plane Br Br Cl Cl 3 Cn Groups Cn – has E and (n-1) Cn symmetry elements H C H H O O H H C O H C3 Cnv – has E, (n-1) Cn and n v symmetry elements N H H H C3v Cnh – has E, (n-1) Cn, (n-1) Sn, h, et. al. symmetry elements H Cl Cl H C2h Dn Groups – similar to Cn, but has n C2 axes perpendicular to Cn axis Dn – has E and (n-1) Cn, n C2 axes principal axis, et. al. symmetry elements H H H H neither staggered nor eclipsed H H D3 Dnh – has E, (n-1) Cn, n v, h et. al. symmetry elements Ru D5h Dnd – has E, (n-1) Cn, v, d, et. al. symmetry elements HH H H HH D3d 4 Sn groups Sn – has E, (n-1) Sn symmetry elements Cubic groups T, Td, Th – tetrahedral groups F F Si F F Td O, Oh – octahedral groups Cl Cl Cl Co Cl Cl Cl Oh Icosahedral group Ih – icosahedral group Buckminsterfullerene Linear groups Cv - Heteronuclear diatomic molecules are most important example. Dh - Homonuclear diatomic molecules are most important example. Immediate Consequences of Symmetry 1. Polarity - Molecule cannot have dipole moment perpendicular to Cn axis. - Molecule may have dipole moment parallel to principal axis if no C2 axes are perpendicular to principal axis - Molecule with Sn axis may not have dipole moment. - Thus only C1, Cs, Cn and Cnv molecules can have dipole moment. 2. Chirality - Molecule without Sn axis may be chiral - A chiral molecule cannot have center of inversion. 5 Representations of Groups Recall that the symmetry elements for a C2v molecule form a group that, by definition, is closed under multiplication. C2v E E E C2 C2 v v v' v' C2 C2 E v' v v v v' E C2 v' v' v C2 E v v’ H H O C2 Rather than relying on our ability to visualize the spatial changes that occur during a pair of symmetry operator, we would like to use numerical representations for the group. Consider the following vector to represent the nine coordinates that describe the position of a water molecule. Hx1 Hy1 Hz1 Ox 2 v Oy 2 Oz 2 Hx 3 Hy3 Hz 3 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 vector v by the matrix E. Indeed, E is an identity element. 1 0 0 0 E 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 1 0 0 0 Ev 0 0 0 0 0 0 0 0 0 0 0 0 0 Hx1 Hx1 1 0 0 0 0 0 0 0 Hy1 Hy1 0 1 0 0 0 0 0 0 Hz1 Hz1 0 0 1 0 0 0 0 0 Ox 2 Ox 2 0 0 0 1 0 0 0 0 Oy 2 Oy 2 0 0 0 0 1 0 0 0 Oz 2 Oz 2 0 0 0 0 0 1 0 0 Hx 3 Hx 3 0 0 0 0 0 0 1 0 Hy3 Hy3 0 0 0 0 0 0 0 1 Hz3 Hz3 6 With the C2 rotation, H1 and H3 switch places and the x and y directions change sign. The following matrix represents the C2 rotation. 0 0 0 0 0 0 0 0 C2 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 C2 v 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 Hx1 Hx 3 0 0 1 0 Hy1 Hy3 0 0 0 0 0 1 Hz1 Hz3 1 0 0 0 0 0 Ox 2 Ox 2 0 1 0 0 0 0 Oy 2 Oy 2 0 0 1 0 0 0 Oz 2 Oz 2 0 0 0 0 0 0 Hx 3 Hx1 0 0 0 0 0 0 Hy3 Hy1 0 0 0 0 0 0 Hz3 Hz1 0 0 0 0 Thus v and v also have 9 9 matrix representations. 0 0 0 0 0 0 0 0 v 0 0 0 0 1 0 0 1 0 0 1 0 0 0 v 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 v v 0 0 0 0 1 0 0 1 0 0 1 0 0 0 v v 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 1 0 0 Hx1 Hx 3 0 0 1 0 Hy1 Hy3 0 0 0 0 1 Hz1 Hz3 0 0 0 0 0 Ox 2 Ox 2 1 0 0 0 0 Oy 2 Oy 2 0 1 0 0 0 Oz 2 Oz 2 0 0 0 0 0 Hx 3 Hx1 0 0 0 0 0 Hy3 Hy1 0 0 0 0 0 Hz3 Hz1 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Hx1 Hx1 0 0 Hy1 Hy1 0 0 0 Hz1 Hz1 0 0 0 Ox 2 Ox 2 0 0 0 Oy 2 Oy 2 0 0 0 Oz 2 Oz 2 1 0 0 Hx 3 Hx 3 0 1 0 Hy3 Hy3 0 0 1 Hz3 Hz3 0 0 7 These four matrices faithfully reproduce the spatial manipulations of each symmetry element. ***But the big question is; when we multiply the symmetry elements together, will we reproduce the multiplication table for the C2v group?*** C2v E E E C2 C2 v v v' v' v v v' E C2 C2 C2 E v' v v v' v' v C2 E v’ H H O C2 Let’s look at an example. 1 0 0 0 v v 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 C2 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 We could go through the other pairs of operations to verify that these four matrices do faithfully reproduce the group multiplication table above. 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 (whatever 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 3 3 matrix of the oxygen in rows 4, 5 and 6 and columns 4, 5 and 6 of the 9 9 matrices we just used. 1 0 0 E 0 1 0 0 0 1 1 0 0 C2 0 1 0 0 0 1 1 0 0 v 0 1 0 0 0 1 1 0 0 v 0 1 0 0 0 1 1 0 0 1 0 0 1 0 0 v v 0 1 0 0 1 0 0 1 0 C2 0 0 1 0 0 1 0 0 1 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 C2v group. O C2v E,C2 , v , v 1 0 0 1 0 0 1 0 0 1 0 0 O C2v 0 1 0 , 0 1 0 , 0 1 0 , 0 1 0 0 0 1 0 0 1 0 0 1 0 0 1 What if we separated the z component from the rest of the components? 1 0 0 1 0 E 0 1 0 , 0 1 0 0 1 1 0 0 1 0 C2 0 1 0 , 0 1 0 0 1 1 1 0 0 1 0 v 0 1 0 , 0 1 0 0 1 1 1 0 0 1 0 v 0 1 0 , 0 1 0 0 1 1 1 Would the two groups of matrices reproduce the group multiplication table? 1 0 1 0 1 0 1 0 A C2v , , , 0 1 0 1 0 1 0 1 1 0 1 0 1 0 v v C2 0 1 0 1 0 1 You can check the other pairs of symmetry operations to see that A is a faithful representation. 9 z C2v 1 , 1 , 1 , 1 v v 1 1 1 C2 While the example using z seems trivial, it is important because it works! 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 z is an irreducible representation of the C2v symmetry group. It is also the totally symmetric representation of the group. More about that later. Let’s keep going with A. Let’s break apart its x and y components. 1 0 E 0 1 1 0 C2 0 1 1 , 1 1 0 v 0 1 1 , 1 1 0 v 0 1 1 , 1 1 , 1 x C2v 1 , 1 , 1 , 1 y C2v 1 , 1 , 1 , 1 Are these faithful representations for C2v? x C2v 1 , 1 , 1 , 1 v v 1 1 1 C2 y C2v 1 , 1 , 1 , 1 v v 1 1 1 C2 After checking with the other combinations of symmetry elements, we can see that both x and y are faithful representations of the algebra of the C2v symmetry group. They are also two more of the irreducible representations of the group. There is one more irreducible representation for the C2v symmetry group that we haven’t found yet and that is, A2 C2v 1 , 1 , 1 , 1 Thus the C2v symmetry group has four irreducible representations. 10 Character Tables and Symmetry Labels Definitions Order – number of total symmetry operations in a symmetry group Class – subset of symmetry operations that transform into each other. - Every Cn axis will have n symmetry operations that belong to the same class. Character – sum of diagonal elements in a matrix (also known as trace) 1 0 1 0 A B A 2 B 0 0 1 0 1 Symmetry Label – label given to an object (such as an MO or a vibration) 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. - One can loosely think of the symmetry labels as a basis set to describe to shape of an object in a particular symmetry group. Character Table Example: C2v z C2v A1 A2 B1 B2 E 1 1 1 1 C2 1 1 -1 -1 v 1 -1 1 -1 v' 1 -1 -1 1 v z x y v’ z2, y2, x2 xy xz yz H H y O C2 A1, A2, B1 and B2 are symmetry labels Each row in the character table represents an irreducible representation. Interpretation of the Character Table 1.) If a molecule at equilibrium in C2v has an object (such as an MO or a vibration) associated with that have the same C2v symmetry; the object is given the A1 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. 2.) If an object changes parity for each reflection, the object is given the A2 symmetry label. - The –1 value 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 px orbitals on the H atoms of water. O H H Orbitals plane. 11 3.) If an object changes its parity for the C2 rotation and the v reflection, then the object is given the B2 symmetry label. - Example: Consider a molecular orbital made from a set of pz orbitals on the H atoms of water H H O 4.) If an object changes its parity for the C2 rotation and the v reflection, then the object is given the B1 symmetry label. - Example: Consider a molecular orbital made from a set of px orbitals on the H atoms of water. O H H Orbitals plane. 5.) Functions on the right of character table indicate to which symmetry they belong. For C2v molecule, z is along C2 axis and x is perpendicular to plane. Another Character Table Example: C3v C3v A1 A2 E E 1 1 2 2C3 1 1 -1 3v 1 z -1 0 (x,y) z2, x2+y2 (xz,yz), (xy, x2-y2) 6.) Coefficients in header row indicate number of operations within each class - C3 , C32 are members of a single class. - v , v and v are members of a single class. - Adding the numbers of elements of each class yields the order of the group. - For C3v, order = 1 + 2 + 3 = 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 have two (or more) parts. - The representation entangles two (or more) objects that cannot be separated. - Note for C3v, the pair of functions (x, y) cannot be separated. - Symmetry operations must operate on both together. - x x cos y sin - y x sin y cos - Objects that have E symmetry (such as orbitals or vibrations) must have degenerate energies. - Td, Oh and Ih objects can have T symmetries that are triply degenerate. 12 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 have 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 3 2C3 0 3v 1 Reducing a Reducible Representation The number of times of particular irreducible representation can be found within a reducible representation, ai, is found by using the very intimidating formula: 1 R i R h R a i – the number of irreducible representations h – the order of the group R – a symmetry operation, thus is sum of all symmetry operations ai R R – characters of the reducible representation (3 0 1) from example above i R – characters of the ith irreducible representation from the character table A1 = (1 21 31) = (1 2 3) A2 = (1 21 3-1) = (1 2 -3) E = (2 2-1 30) = (2 -2 0) - note the number of symmetry elements of each class are included. 13 One can think of taking a “dot product” of the reducible representation with the irreducible representation then dividing by the order of the group to find the number of times the irreducible representation is found. C3v A1 A2 E E 1 1 2 2C3 1 1 -1 E 3 3v 1 z -1 0 (x,y) 2C3 0 z2, x2+y2 (xz,yz), (xy, x2-y2) 3v 1 1 1 3 11 0 2 1 1 31 6 1 1 2 3 6 1 1 a A2 3 11 0 2 1 1 3 1 0 0 6 6 1 1 a E 3 1 2 0 2 1 1 3 0 6 1 6 6 a A1 Thus the reducible representation, 3 0 1 is reduced to A1 E - Of the orbitals made from the perpendicular p orbitals of the cyclopropenyl anion, one will have A1 symmetry and two will have 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 a 0 f x dx 2 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.) 14 For an integral (such as overlap integral or a matrix element), any physically meaningful value the character of the integral must be independent of any symmetry operation upon it. (No matter how integral is oriented, it must have the same value.) Thus the representation of any nonzero integral must contain the totally symmetric representation (A or A1 or Ag etc… depends on specific point group). For an overlap integral, , Find the symmetry of each orbital, , Multiply characters of the orbitals together. Find the irreducible representation. 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 value. 1. 2. 3. 4. Example: Consider the overlap integral, 1s 2pz in a C2v molecule. 1s 1 1 1 1 A1 2pz 1 1 1 1 A1 nonzero overlap possible 1 1 1 1 A1 Example: Consider the overlap integral, 1s 2px in a C2v molecule. 1s 1 1 1 1 A1 2px 1 1 1 1 B1 overlap equals zero 1 1 1 1 B1 Example: Consider the overlap integral, 2p x 2p y in a C4v molecule. C4v A1 A2 B1 B2 E E 1 1 1 1 2 C2 1 1 1 1 -2 2 C4 1 1 -1 -1 0 2 v 1 -1 1 -1 0 2 d 1 -1 -1 1 0 z x2+y2, z2 x2-y2 xy (x,y) (xz, yz) 2px 2 2 0 0 0 E 2p y 2 2 0 0 0 E 4 4 0 0 0 15 At this point, we must reduce the representation into its irreducible parts. 1 1 a A1 4 11 4 11 0 2 1 0 2 1 0 2 1 8 1 11 2 2 2 8 1 1 a A2 4 11 4 11 0 2 1 0 2 1 0 2 1 8 1 11 2 2 2 8 1 1 a B1 4 11 4 11 0 2 1 0 2 1 0 2 1 8 1 11 2 2 2 8 1 1 a B2 4 11 4 11 0 2 1 0 2 1 0 2 1 8 1 11 2 2 2 8 1 1 aE 4 1 2 4 1 2 0 2 0 0 2 0 0 2 0 0 0 11 2 2 2 8 Thus the representation, 4 4 0 0 0 , can be reduced to A1 A2 B1 B2 . The representation contains A1; therefore, the overlap integral, 2p x 2p y , could be nonzero. For a transition dipole matrix element, ˆ , 1. Find the symmetry of each orbital, , 2. Find the characters of the components of the dipole operator, er e xiˆ yˆj zkˆ 3. 4. 5. 6. - 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. Find the irreducible representation. If the irreducible representation does not contain the totally symmetric representation, the integral is zero. Thus no transition can occur. 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 , are allowed. 16 Example: Determine the electronic transitions that are allowed in a C2v molecule based on the symmetry of the electronic states. C2v A1 A2 B1 B2 E 1 1 1 1 v 1 -1 1 -1 C2 1 1 -1 -1 v' 1 -1 -1 1 z x y z2, y2, x2 xy xz yz A1 x A1 A1 1 1 1 1 B1 1 -1 1 -1 A1 1 1 1 1 B1 1 -1 1 -1 transition not allowed A1 y A1 A1 1 1 1 1 B2 1 -1 -1 1 A1 1 1 1 1 B2 1 -1 -1 1 transition not allowed A1 z A1 1 1 1 1 1 1 1 1 1 1 1 1 transition allowed A1 x A2 A1 1 1 1 1 B1 1 -1 1 -1 A2 1 1 -1 -1 B2 1 -1 -1 1 transition not allowed A1 A1 A1 A1 1 1 1 1 While the above process will eventually yield the correct answers, we could choose a more direct route. Let us think in terms of solving for an unknown. For example, we choose two electronic states and ask what is the only operator that will make the integral have A1 symmetry. A1 ? B1 A1 x B1 A1 1 1 1 1 A1 1 1 1 1 ? B 1 -1 1 -1 1 B1 1 -1 1 -1 B1 1 -1 1 -1 A1 1 1 1 1 A1 1 1 1 1 The only operator with B1 symmetry is x; therefore, the only allowed transition is A1 x B1 . A1 ? B2 1 1 1 A1 y B2 A1 1 A1 1 1 1 1 ? 1 -1 -1 1 B2 B2 1 -1 -1 1 B2 1 -1 -1 1 A1 1 1 1 1 A1 1 1 1 1 The only operator with B2 symmetry is y; therefore, the only allowed transition is A1 y B2 . 17 A2 ? A2 1 1 -1 A2 z A2 A2 -1 A2 1 1 -1 -1 ? 1 1 1 1 A1 A2 1 1 -1 -1 A2 1 1 -1 -1 A1 1 1 1 1 A1 1 1 1 1 The only operator with A1 symmetry is z; therefore, the only allowed transition is A2 z A2 . A2 ? B1 1 1 -1 A2 y B1 A2 -1 A2 1 1 -1 -1 ? 1 -1 -1 1 B2 B1 1 -1 1 -1 B1 1 -1 1 -1 A1 1 1 1 1 A1 1 1 1 1 The only operator with B2 symmetry is y; therefore, the only allowed transition is A2 y B1 . A2 ? B2 1 1 -1 A2 x B2 A2 -1 A2 1 1 -1 -1 ? 1 -1 1 -1 B1 B2 1 -1 -1 1 B2 1 -1 -1 1 A1 1 1 1 1 A1 1 1 1 1 The only operator with B1 symmetry is x; therefore, the only allowed transition is A2 x B2 . B1 ? B1 1 -1 1 B1 z B1 B1 -1 B1 1 -1 1 -1 ? A 1 1 1 1 1 B1 1 -1 1 -1 B1 1 -1 1 -1 A1 1 1 1 1 A1 1 1 1 1 The only operator with A1 symmetry is z; therefore, the only allowed transition is B1 z B1 . B1 ? B2 1 -1 1 B1 y B2 B1 -1 B1 1 -1 1 -1 ? 1 1 -1 -1 A2 B2 1 -1 -1 1 B2 1 -1 -1 1 A1 1 1 1 1 A1 1 1 1 1 There is no dipole operator with A2 symmetry; therefore, an electronic transition between a B1 and B2 state is not allowed. In other words, B1 ˆ B2 0