Group Theory: Matrix Representation & Consequences of Symmetry Matrix Representation of Group Theory Reducible and Irreducible Representations The Great Orthogonality Theorem The ive Rules The Standard Reduction ormula (Derivation) Applications of Group Theory Dipole Moments and Chirality (15.3) Molecular orbitals and symmetry (15.4) Selection rules and symmetry (15.5)
Symmetry & Dipole Moments The electric dipole moment is a measure of charge asymmetry in a molecule, the magnitude of which is expressed by µ. Since the dipole moment can also be a vector quantity, boldfaced µ is used to denote this. *- (a) (b) (c) O *+ *- N *+ *+ *- Cl (d) (e) (f) *- N C C *+ C 3 (g) (h) (i) *+ C C *- C C B
Examples of Electric Dipole Moments (a) and (b): 2 O and N 3 Direction of the dipole moment is along the C 2 and C 3 axes, respectively, with the negative end of the dipole moment in each case directed towards the unpaired electrons. (c): 2-chlorofluorobenzene Charge asymmetry can be associated with a particular bond in a molecule, giving rise to the bond dipole moment or bond moment. In (c), the dipole moment can be estimated from vector sum of C- and C-Cl bond moments found in C 3 and C 3 Cl, respectively (i.e., similar molecules). (d): 1,2-cis-difluoroethene Bond moments involving atoms are relatively small, so µ in (d) is dominated by the large bond moments of the C- bonds (*- points towards electronegative atoms). (e) and (f): N 3 and toluene Both have small dipole moments along symmetry axes, and it is not evident which way they point since effect of negative atoms, for instance, are balanced by lone pairs. (g): 1-fluoroethene The large C- bond moment dominates µ, which lies in the symmetry plane, but with *- slightly towards the atom (h) and (i): 1,2-trans-difluoroethene and B 3 Neither have permanent dipole moments
Quantitation of Dipole Moments Electric dipole moments are very simply products of their charges and their separations, so that for charges -q and +q separated by distance r: µ = qr and for a molecule with many nuclei and electrons with i charges q i, we can write the simple formula: µ ' j i q i r i where r i can be calculated from Cartesian coordinates of any atom from an origin as r i = (x i 2 + y i 2 + z i 2 ) 1/2 Magnitude of dipole moments is normally given in Debye units (D), which are still much more commonly used than the cumbersome SI units, since: 1 D 3.335 64 10-30 C m or instance, some dipole moments are µ(n 3 ) = 1.47 D, µ(n 3 ) = 0.2 D and µ(c 6 5 C 3 ) = 0.36D. There are many different experiments for determining the magnitude of the dipole moment, but there is no method for determining the orientation of the dipole moment in the molecular frame. Such orientations are normally obtained from valence theory calculations or other high level ab initio calculations.
Existence of Dipole Moments ortunately, group theory provides us with a simple means of determining whether or not a molecule has a permanent electric dipole moment (sometimes for trickier molecules determination by inspection can be difficult). The dipole moment vector is a totally symmetric entity, and therefore must be symmetric to all of the operations of the point group of the molecule (otherwise it could change directions and disappear! this can t happen!!) Since µ = µ x + µ y + µ z, consider the following examples: (b) and (e): N 3 and N 3 The C 3 axis is the z-axis, and µ z 0, but µ x and µ y = 0 (a) and (d): 2 O and 1,2-cis-difluoroethene The C 2 axis is the z-axis, and µ z 0, but µ x and µ y = 0 (g): 1-fluoroethene µ is in the xy-plane, so µ x and µ y 0 but µ z = 0, CClBr T µ not constrained by symmetry at all, so µ x, µ y, µ z 0 The dipole moment vector has the same symmetry species as translation vectors, that is: '(µ x ) = '(T x ) '(µ y ) = '(T y ) '(µ z ) = '(T z ) Since µ must be totally symmetric, a molecule has a permanent electric dipole moment if any of µ x, µ y, µ z is totally symmetric! T T T
Permanent Dipoles & Symmetry The rule therefore is: A molecule has a permanent electric dipole moment if any of the translational symmetry species of the point group to which the molecule belongs is totally symmetric. Looking at the character tables, it is easy to see that only the molecules belonging to the following point groups have permanent electric dipole moments: C 1 C s C n '(T x ) = '(T y ) = '(T z ) = A '(T x ) = '(T y ) = AN '(T z ) = A C nv '(T z ) = A 1 (or E + in C 4v ) or example, CClBr (C 1 ) has all three translations as being totally symmetric, therefore µ 0. But the direction of µ is not obvious N 3 (C 3v ) has '(T z ) = A 1, so µ 0, and along z-axis. B 3 (D 3h ) has no translational symmetry species which are totally symmetric, so µ = 0. Also, bond moments show that µ = µ B - 2µ B cos(60 o ) = 0 (since cos(60 o ) = -0.5) Symmetry properties tell us only of the existence of an electric dipole moment, but nothing about its magnitude. Experimentally, they are most accurately determined by µ- wave and mm-wave (rotational) spectroscopy.
Chirality & Optical Activity A chrial molecule is one which exists in two forms known as enantionmers. Each enantiomer is optically active, meaning it can rotate a plane of plane-polarized light to the right (clockwise) or to the left (counter-clockwise). These enantiomers are referred to as the dextro- and levo- (d- and l-) forms, or the (+)- and (-)-forms. Plane-polarized light consists of two circularly polarized components of equal intensity but opposite handedness. Circularly polarized light is defined as right-handed when the electric and magnetic vectors rotate clockwise as viewed by an observer facing the direction of light propagation (i.e., the source), and the frequency of the rotation is related to the frequency of the light d-circularly polarized light plane polarized light source If plane-polarized light passes through an optically isotropic medium, it remains plane polarized. If it passes through an optically anisotropic medium, i.e., one for which the refractive indices of d- and l-polarized light differ, the plane of the polarized light will be rotated by " = (n l - n d )/8 for light of wavelength 8.
More on Chirality & Enantiomers A sample of chiral molecules may exist as an equimolar mixture of (+)- and (-)-enantiomers, which will not rotate the plane of polarized light. This is mixture is referred to as racemic, and is denoted (±). Enantiomers are sometimes found separately and despite their chemical similarity have distinct ways of interacting or reacting with different enantiomeric reagents. or instance, (+)-glucose, which is metabolized by animals and fermented by yeast, but (-)-glucose has neither of these properties. (+)-carvone smells like caraway seeds, but (-)- carvone smells like spearmint! A rule that is applicable to any molecule, not just carbon CWXYZ type molecules is: A molecule that is not superimposable on its mirror image is a chiral molecule. Br Cl Cl Br mirror plane The rules of group theory and symmetry give us a much more definitive way of determining whether or not a molecule is chiral: A molecule is chiral if it does not have any S n symmetry element with any value of n (includes S 1 /, S 2 / i)
Examples of Chiral Molecules 2,3-difluorobutane is a good place to consider some chiral and achiral forms of an organic molecule. The most stable structure is considered to be the one where the C 3 's are trans w.r.t. one another, as are the s: C 3 i C 3 (a) all trans, not chiral C 3 C 3 (b) all gauche, chiral C 2 C 2 C 3 C 3 C 3 C 3 C 3 C 2 C 3 (c) C 3 's trans, chiral (d) 's trans, chiral (e) 's trans, chiral Since this stable structure (a) has a centre of inversion, it is achiral, and is called a meso structure. (a) and (b) are interconvertible by rotation about the central C-C bond, as are (c), (d) and (e). The following molecules may also be regarded as chiral, though for the first two, enantiomers have never been separated: Cl N O O Co
The LCAO Approximation Molecular orbital (MO) theory says that MO s can be written as a linear combination of atomic orbitals (LCAO). Denoting the ith AO, N i, and the kth MO as R k : R k ' j i c ik N i The AO s, N i, are the basis set, and satisfy the usual condition of orthonormality: * m N i N j ' * ij, * ij ' 0 if i j, * ij 0 if i'j, Using the LCAO-MO s, a particular form of the wave equation known as the secular equation (i.e., equation not dependent upon time) is developed as follows. The wave equation is written as:,r & ER ' (, &E)R ' 0 and the LCAO expression for MO R is j i c i (, &E)N i ' 0 or instance, the MO s of 2 O, labelled with their symmetry species, can be constructed from linear combinations of 1s, O2s and O2p AO s (note there are 6 MO s, since 6 AO s are used to construct them). Also, MO symmetry labels are lowercase!
Secular Equation Consider a 2 atomic orbital MO to make our discussion simpler. The LCAO expression is written for 2 centres as: c 1 (,&E)N 1 % c 2 (,&E)N 2 ' 0 The above equation is multiplied by N 1, and the LS is integrated over all space: c 1 m N * (,&E)N % c 1 1 m N * 2 (,&E)N 1 2 ' 0 To simplify notation, the following definitions are used: ii ' N* m i,n i dj energy of AO N i * ij ' m N i,n j dj * S ij ' m N i N j dj energy of interaction btw. N i & N j overlap integral Since E is a scalar value, we can write: * * m N i EN j dj ' E m N i N j dj ' ES ij c 1 ( 11 & E) % c 2 ( 12 & ES 12 ) ' 0 c 1 ( 21 & ES 21 ) % c 2 ( 22 & E) ' 0 which have trivial solutions c 1 = c 2 = 0. Non-trivial solutions exist if the c i s form a determinant matrix equal to zero (Cramer s theorem). So the secular equation is: 11 & E 12 & ES 12 /0 21 & ES 21 22 & E /0 ' 0 which is solved to have two roots: E 1 & E 2, which are upper limits to the ground and 1st excited states (variation theorem) *
Symmetry & Molecular Orbitals Consider the N 3 molecule, which has C 3v symmetry. C 3v E 2C 3 3σ v A 1 1 1 1 z A 2 1 1-1 R z E 2-1 0 (x,y), (R x, R y ) The order of the group is h = 6. Some typical symmetry adapted linear combinations (SALC s) of orbitals are shown for a C 3v molecule (we will talk about SALC s later) AO s like s and p z transform as the totally symmetric species, A 1. p x and p y jointly span an irreducible representation of symmetry species E. The same transformations may be applied to MO s. or example, the LCAO R 1 = N A + N B + N C for the 3 1s orbitals on s in N 3 remains unchanged under all of the symmetry operations of the group, so P(E) = 1, P(C 3 ) = 1 and P( v ) = 1. This MO is of symmetry species A 1 and contributes to a 1 MO s in N 3.
Symmetry and Overlap Integrals of the forms shown below are very common in quantum mechanics: I ' m f 1 f 2 dj I ' m f 1 f 2 f 3 dj The integral on the LS is often used as a means of determining overlap between two atomic orbitals. If it is known that the integral is equal to zero, then we can immediately say that a molecular orbital in the molecule does not result from overlap of atomic orbitals 1 and 2. The key is dealing with an integral like this is that the overlap of any orbitals, and therefore the overlap integral, should be independent of the orientation of the molecule. Since the volume element dj must be invariant to any symmetry operation, it follows that f 1 f 2 must be unchanged by any symmetry operation in the group. If the integrand changes signs under a symmetry operation, the integral would be the sum of equal and opposite contributions and be equal to zero. Thus, in order for I not to be zero for any integral of the above types, f 1 f 2 or f 1 f 2 f 3 must span the symmetry species A 1 (or other totally symmetric symmetry species, depending on the point group). This means that under any symmetry operation of the point group of the molecule, f 1 f 2 6 f 1 f 2 or f 1 f 2 f 3 6 f 1 f 2 f 3 (all characters equal to +1)
Symmetry Species & Direct Product 1. Decide on the symmetry species of the functions f 1 and f 2, write their characters in two rows. 2. Multiply the numbers in each column 3. Inspect the row produced to see if it can be expressed as sum of characters (may use standard reduction formula) Say f 1 is the 2s N AO and f 2 is the LC s 3 = s B - s C. 2s N spans A 1 and s 3 is a member of the basis spanning E: C 3v E 2C 3 3σ v f 1 1 1 1 A 1 f 2 2-1 0 E f 1 f 2 2-1 0 E Since the characters of the direct product span E alone (irreducible), the integrand does not span A 1, and the integral must be zero Say f 1 is the 2s N AO and f 2 is the LC s 1 = s A + s B + s C. 2s N spans A 1 and s 3 is a member of the basis spanning A 1 : C 3v E 2C 3 3σ v f 1 1 1 1 A 1 f 2 1 1 1 A 1 f 1 f 2 1 1 1 A 1 Thus, s 1 and 2s N may have non-zero overlap; however, it may be zero for other reasons (distance, etc.)
Orbitals with Non-Zero Overlap We just saw AO s with non-zero overlap: i.e., 2s N overlaps with s 1 = s A + s B + s C. So, bonding and anti-bonding MO s can form from (2s N, s 1 ) overlap: i.e., they form a 1 (bonding) and a 1 * (anti-bonding) MO s. Only orbitals of the same symmetry species may have non-zero overlap, so that only orbitals of the same symmetry species form bonding and anti-bonding combinations. The s 2 and s 3 linear combinations have symmetry species E, and do not overlap with N2s. Symmetry and intuition suggest that the N2p x and N2p y AO s, which belong jointly to symmetry species E, may have overlap with s 2 and s 3. This can be verified by the direct product of E E = A 1 + A 2 + E, and the two bonding e MO s are shown to the left, (b) and (c). Note there are also two e* anti-bonding MO s. So three bonding orbitals may be constructed from (N2s, 1s), (N2p x, 1s) and (N2p y, 1s) overlap. Orbitals not involved in any kind of overlap are called non-bonding orbitals.
Symmetry Adapted Linear Combinations ortunately, we do not have to guess at or predetermine combinations of AO s with certain symmetry properties (like s 1, s 2 and s 3 ), as group theory provides the mechanism to provide us with an arbitrary basis set of atomic orbitals. These are still LCAO s, but they are adapted to the symmetry of the molecule, and are called symmetry adapted linear combinations (SALC). 1. Construct a table showing the effect of each symmetry operation on each AO of the original basis. 2. To generate the combination of a specified symmetry species, take each column in turn and: i) Multiply each member of the column by the character of the corresponding operation ii) Add the AO s in each column with factors as determined in step i) iii) Divide the sum by the order of the group. Use the basis set of AO s: N2s, s A, s B, s C (abbr:s N, s A, s B, s C ), C 3v s N s A s B s C E s N s A s B s C C 3 N s N s B s C s A C 3 O s N s C s A s B σ v s N s A s C s B σ v N s N s B s A s C σ v O s N s C s B s A
SALC s, continued Take the characters for A 1 (1,1,1,1,1,1) and perform steps i) & ii): R % s N % s N %... ' 6s N since h = 6, then R = s N. or the column under s A R = 1 6 (s A % s B % s C % s A % s B % s C ) ' 1 3 (s A % s B % s C ) which is also obtained from the other two columns. So we have formed the s 1 combination we used before. So the overall MO with symmetry a 1 is R ' c N s N % c 1 s 1 or degenerate symmetry species like E and T, representations are in two dimensions or more, so the rules for generating SALC s actually generate sums of SALC s. In C 3v, the E characters are (2, -1, -1, 0, 0, 0), and the s N column gives R ' 1 6 (2s N & s N & s N % 0 % 0 % 0) ' 0 and the other columns give: 1 6 (2s & s & s ) 1 A B C 6 (2s & s & s ) 1 B A C 6 (2s & s & s ) C B A Any one of these three expressions can be written as a sum of the other two (not linearly independent). The difference of the second and third and the first give 1 2 (s & s ) 1 B C 6 (2s & s & s ) A B C which are linearly independent SALC s we have used in the previous discussion of e orbitals.
Symmetry & Selection Rules As seen, the integrals below are important in MO theory. or the integral on the RS to be non-zero, f 1 f 2 f 3 must span A 1. If f 1 f 2 f 3 does not span A 1, the integral is zero. I ' m f 1 f 2 dj I ' m f 1 f 2 f 3 dj Recall that the intensity of a spectral line arising from a molecular transition from an initial state R i to a final state R f depends on the (electric) transition dipole moment, µ. The z-component of µ is defined as: µ z ' &e m R ( f zr i dj where e is the charge on an electron. Once the symmetry species of the states involved in the proposed transition have been identified, group theory can be used to determine whether the transitions is allowed (non-zero integral) or forbidden (integral equal to zero). Can an electron in an a 1 orbital in 2 O (C 2v ) make a electric dipole allowed transition to a b 1 orbital? In fact, what we will see is that certain polarizations of EM radiation produce certain transitions (allowed), and then certain transitions are forbidden.
Selection Rules Example, C 2v Can an electron in an a 1 orbital in 2 O (C 2v ) make a electric dipole allowed transition to a b 1 orbital? We must examine the x, y and z components of the transition dipole moment, taking f 1 as the b 1 orbital, f 3 as the a 1 orbital, and f 2 as x, y and z in turn: x-component y-component z-component E C 2 σ v σ v N E C 2 σ v σ v N E C 2 σ v σ v N f 3 1-1 1-1 1-1 1-1 1-1 1-1 B 1 f 2 1-1 1-1 1-1 -1 1 1 1 1 1 f 1 1 1 1 1 1 1 1 1 1 1 1 1 A 1 f 1 f 2 f 3 1 1 1 1 1 1-1 -1 1-1 1-1 Only the first product (with f 2 = x) spans A 1, so only the x- component of the transition dipole moment may be nonzero. Thus, transitions between a 1 and b 1 are allowed. The radiation emitted (or absorbed) is x-polarized and has its electric field vector in the x-direction, since that form of the radiation couples with the x-component of the transition dipole. Try examples 15.6 and 15.7 on pp. 446-447 of Atkins 6th edition to better understand this procedure.
Key Concepts 1. The electric dipole moment is a measure of charge asymmetry in a molecule, the magnitude of which is expressed by µ. A molecule has a permanent electric dipole moment if any of the translational symmetry species of the point group to which the molecule belongs is totally symmetric. 2. A molecule that is not superimposable on its mirror image is a chiral molecule. A molecule is chiral if it does not have any S n symmetry element with any value of n (includes S 1 /, S 2 / i) 3. Molecular orbital (MO) theory says that MO s can be written as a linear combination of atomic orbitals (LCAO). The method of symmetry adapted linear combinations (SALCs) can be used to determine an arbitrary basis set of atomic orbitals which are adapted to the symmetry of the molecule. 4. Only orbitals of the same symmetry species may have non-zero overlap, so that only orbitals of the same symmetry species form bonding and anti-bonding combinations. Orbitals not involved in any kind of overlap are called non-bonding orbitals. 5. If the (electric) transition dipole moment integral is non-zero, the transition is allowed; if the integral is zero, the transition is forbidden.