Chapter 3 - Molecular Symmetry Symmetry helps us understand molecular structure, some chemical properties, and characteristics of physical properties (spectroscopy) used with group theory to predict vibrational spectra for the identification of molecular shape, and as a tool for understanding electronic structure and bonding. Symmetrical: implies the species possesses a number of indistinguishable configurations. 1
Group Theory: mathematical treatment of symmetry. symmetry operation an operation performed on an object which leaves it in a configuration that is indistinguishable from, and superimposable on, the original configuration. symmetry elements the points, lines, or planes to which a symmetry operation is carried out. Element peration Symbol Identity Identity E Symmetry plane Reflection in the plane σ Inversion center Inversion of a point,y,z to -,-y,-z i Proper ais Rotation by (360/n) C n Improper ais 1. Rotation by (360/n) 2. Reflection in plane perpendicular to rotation ais S n Proper aes of rotation (C n ) Rotation with respect to a line (ais of rotation). C n is a rotation of (360/n). C 2 = 180 rotation, C 3 = 120 rotation, C 4 = 90 rotation, C 5 = 72 rotation, C 6 = 60 rotation Each rotation brings you to an indistinguishable state from the original. owever, rotation by 90 about the same ais does not give back the identical molecule. Therefore 2 does NT possess a C 4 symmetry ais. XeF 4 is square planar. It has four different C 2 aes. A C 4 ais out of the page is called the principle ais because it has the largest n. By convention, the principle ais is in the z-direction 2
BF 3 3
Reflection through a planes of symmetry (mirror plane) If reflection of all parts of a molecule through a plane produced an indistinguishable configuration, the symmetry element is called a mirror plane or plane of symmetry. σ v (vertical): plane colinear with principal ais Reflection through a planes of symmetry (mirror plane) σ d (dihedral) Vertical, parallel to principal ais, σ parallel to C n and bisecting two C 2 ' aes 4
Reflection through a planes of symmetry (mirror plane) σ h (horizontal): plane perpendicular to principal ais Inversion, Center of Inversion (i) A center of symmetry: A point at the center of the molecule. (,y,z) (-,-y,-z). It is not necessary to have an atom in the center (e.g. benzene). Tetrahedrons, triangles, and pentagons don't have a center of inversion symmetry. C 2 6 Ru(C) 6 C 4 4 Cl 2 F 2 5
Rotation-reflection, Improper rotation(s n ) This is a compound operation combining a rotation 360 /n (C n ) with a reflection through a plane perpendicular to the C n ais σ h.(c n followed by σ h ) σc n =S n An improper rotation (or rotation reflection), S n, involves rotation about 360 /n followed by reflection through a plane that is perpendicular to the rotation ais. 6
Identity (E) Simplest symmetry operation. All molecules have this element. If the molecule does have no other elements, it is asymmetric. The identity operation amounts to doing nothing to a molecule and so leaves any molecule completely unchanged. CFClBr SFCl Successive perations 7
Symmetry Point Groups Symmetry of a molecule located on symmetry aes, cut by planes of symmetry, or centered at an inversion center is known as point symmetry. Collections of symmetry operations constitute mathematical groups. Each symmetry point group has a particular designation. C n, C nh, C nv D n, D nh, D nd S 2n C v, D h I h, I T d, T h,t h, C 1,C i, C s 8
Cl F 2 C v D h Linear molecular species can be classified according to whether they possess a centre of symmetry (inversion centre) or not. 9
Tetrahedral Geometry P 4 B 4 Cl 4 ctahedral Geometry Icosahedral Geometry [W(C) 6 ] [B 12 12 ] 2 10
Identify the symmetry elements that are present in benzene. Scheme for assigning point groups of molecules and molecular ions 11
C n Point Groups C n n E C 1 PBrClF 2 2 C 2 As(C 6 5 ) 3 C 3 M(N 2 C 2 C 2 ) 4 C 4 C nh Point Groups The direction of the C n ais is take as vertical, so a symmetry plane perpendicular to it is a horizontal plane, h. Cl N 2 F The point group is called C s (C 1h ) BBrClF N 2 F 2 B() 3 C 2h C 3h 12
C nv Point Groups If a mirror plane contains the rotational ais, the group is called a C nv group. 2 SF 4 C 2v C 2v NF 3 C 3v CCl 3 C 3v SF 5 Cl C 4v D n and D nh Point Groups Adding a C 2 ais perpendicular to a C n ais generates one of the dihedral groups. Angle between rings not 0 or 90 D 2 D 3 There must be n C 2 aes perpendicular to C n Adding a h to a D n group generates a D nh group. C 2 4 D 2h [PtCl 4 ]2- D 4h 13
BF 3 D 3h D nd Point Groups Adding a vertical mirror plane to a D n group in such a fashion to bisect adjacent C 2 aes generates a D nd group. D 2d Ferrocene Fe(C 5 5 ) 2 D 5d 14
S n groups For odd n, (S n ) n = h For even n, (S n ) n = E The S n for odd n is the same as the C nh. 1,3,5,7-tetrafluorocyclooctatetraene Absence of mirror planes distinguish S n groups from D nd groups. Linear Groups -Cl = C v D h as a h igh Symmetry Molecules CCl 4 SF 6 C 60 T d h I h 15
The letter A. tetrachloroplatinate(ii) Ge 3 F SF 6 Cl Se 3 F B() 3 B AsBr 5 (trigonal bipyramid) B 3 trans rotamer of Si 2 6 B Crown-shaped S 8 Ni(en) 3 16
Characteristic symmetry elements of some important classes of point groups. 17
Character Tables Character tables contain, in a highly symbolic form, information about how something of interest (an bond, an orbital, etc.) is affected by the operations of a given point group. Each point group has a unique character table, which is organized into a matri. Column headings are the symmetry operations, which are grouped into classes. orizontal rows are called irreducible representations of the point group. The main body consists of characters (numbers), and a section on the right side of the table provides information about vectors and atomic orbitals. C 2v E C 2 σ v (z) σ v (yz) A 1 1 1 1 1 z 2, y 2, z 2 A 2 1 1-1 1 R z y B 1 1-1 1 1, R y z B 2 1 1 1 1 y, R yz C 3v E 2C 3 3σ v A 1 1 1 1 z 2 + y 2, z 2 A 2 1 1-1 R z E 2-1 0 (,y), (R, R y ) ( 2 -y 2, y)(z, yz) Symmetry elements possessed by the point group are in the top row Left hand column gives a list of symmetry labels Gives information about degeneracies (A and B indicate nondegenerate, E refers to doubly degenerate, T means triply degenerate) Main part of table contains characters (numbers) to label the symmetry properties (of M s or modes of molecular vibrations) 18
To obtain from this total set the representations for vibration only, it is necessary to subtract the representations for the other two forms of motion: rotation and translation. z z z y z y y y Cartesian displacement vectors for a water molecule Translational Modes Rotational Modes A mode in which all atoms are moving in the same direction, equivalent to moving the molecule. A mode in which atoms move to rotate (change the orientation of) the molecule. There are 3 rotational modes for nonlinear molecules, and 2 rotational modes for linear molecules. 19
Selection Rules: Infrared and Raman Spectroscopy Infrared energy is absorbed for certain changes in vibrational energy levels of a molecule. -for a vibration to be infrared active, there must be a change in the molecular dipole moment vector associated with the vibration. For a vibration mode to be Raman active, there must be a change in the net polarizability tensor -polarizability is the ease in which the electron cloud associated with the molecule is distorted For centrosymmetric molecules, the rule of mutual eclusion states that vibrations that are IR active are Raman inactive, and vice versa The transition from the vibrational ground state to the first ecited state is the fundamental transition. The two bending modes require the same amount of energy and are therefore degenerate. 20
C 2v E C 2 σ v (z) σ v (yz) A 1 1 1 1 1 z 2, y 2, z 2 A 2 1 1-1 1 R z y B 1 1-1 1 1, R y z B 2 1 1 1 1 y, R yz If the symmetry label (e.g. A 1, B 1, E) of a normal mode of vibration is associated with, y, or z in the character table, then the mode is IR active. If the symmetry label (e.g. A 1, B 1, E) of a normal mode of vibration is associated with a product term ( 2, y) in the character table, then the mode is Raman active. C 2v E C 2 σ v (z) σ v (yz) A 1 1 1 1 1 z 2, y 2, z 2 A 2 1 1-1 1 R z y B 1 1-1 1 1, R y z B 2 1 1 1 1 y, R yz 21
D 3h E 2C 3 3C 2 h 2S 3 3 3 v v A1 1 1 1 1 1 1 2 + y 2, z 2 A 2 E A 1 A 2 E 1 1-1 1 1-1 R z 2-1 0 2-1 0 (, y ) ( 2 y 2, 2y ) 1 1 1-1 -1-1 1 1-1 -1-1 1 z 2-1 0-2 1 0 (R, R y ) (y, yz ) 22
The vibrational modes of C 4 (T d ), only two of which are IR active. CCl 4 http://fy.chalmers.se/ldusers/brodin/molecularmotions/ccl4spectra.html 23
vibrational modes of [PtCl 4 ] 2 (D 4h ) 24
Point groups of octahedral metal carbonyl complees 25
Enantiomers A pair of enantiomers consists of two molecular species which are mirror images of each other and are nonsuperposable. 26
Group Theory Supplemental Material Up to this point, we have considered symmetry operations only insofar as they affect atoms occupying points in molecules. Consider a water molecule ( 2 ). E Character Table Development C 2 v (z) v (yz) Coordinates are assigned according to the convention that the highest fold ais of rotation is aligned with the z-ais, and the ais is perpendicular to the plane of the molecule. Let the translation of the molecule in the +y direction be represented by unit vectors on the atoms, and consider how they change when undergoing the C 2v symmetry operations. z y B 2 = +1-1 -1 +1 C 2 v (z) z v (yz) y 27
The set of four labels (+1, -1, -1, +1) generated in the analysis constitutes one irreducible representation within the C 2v point group. It is irreducible in the sense that it cannot be decomposed into a simpler or more fundamental form. Not only does it describe the effects on the y translation but also on other y-vector functions such as a p y orbital. Therefore, y is understood to serve as a basis function for this irreducible representation within the C 2v point group. Effect of a symmetry rotation about the z-ais. z y E C 2 v (z) v (yz) A 2 = +1 +1-1 -1 Translation of the molecule in the + direction z B 1 = y E C 2 v (z) +1-1 +1-1 v (yz) Translation of the molecule in the +z direction z y E C 2 v (z) v (yz) A 1 = +1 +1 +1 +1 28
C 2v E C 2 σ v (z) σ v (yz) A 1 1 1 1 1 z 2, y 2, z 2 A 2 1 1-1 1 R z y B 1 1-1 1 1, R y z B 2 1 1 1 1 y, R yz The resulting character table for C 2v is shown above. The column heading are classes of symmetry operations for the group, and each row depicts one irreducible representation. The +1 and -1 numbers, which correspond to symmetric and antisymmetric behavior, are called characters. Columns on the right are some of the basis functions which have the symmetry properties of a given irreducible representation. R, R y, R z stand for rotations about the specified aes. Symbols in the column on the far left are Mulliken Labels. Reducible Representations When applying the methods of group theory to problems related to molecular structure or dynamics, the procedure that is followed usually involves deriving a reducible representation for the phenomenon of interest, such as a molecular vibration, and then decomposing it into its irreducible components. A reducible representation will always be a sum of irreducible representations. In some cases (simple molecules with few bonds) we can perform the decomposition by inspection, for the more general case (complicated molecule with many bonds), we can use the reduction formula. 29
Reducible Representations The reduction can be achieved using the reduction formula. It is a mathematical way of reducing that will always work when the answer cannot be spotted by eye. It is particularly useful when there are large numbers of bonds involved. The vibrational modes of the molecule are reduced to produce a reducible representation into the irreducible representations. This method uses the following formula reduction formula: N 1 h n N is the number of times a symmetry species occurs in the reducible representation, h is the order of the group : simply the total number of symmetry operations in the group. The summation is over all of the symmetry operations. For each symmetry operation, three numbers are multiplied together. These are: Χ r character of the reducible representation for the operations of the class Χ i character of the irreducible representation for the operations of the class n is the number of symmetry operations in the class r i The characters of the reducible representation can be determined by considering the combined effect of each symmetry operation on the atomic vectors. z z Atomic contributions, by symmetry operations, to the reducible representation for the 3N degrees of freedom for a molecule. peration Contribution per atom* E 3 C 2-1 C 3 0 C 4 1 C 6 2 1 i -3 S 3-2 S 4-1 S 6 0 *C n = 1 + 2cos(360/n); S n = -1 + 2cos(360/n) z z y y (z) i y y E y z y z 30
z Derivation of reducible representation for degrees of freedom in 2 y Unshifted atoms E 3 a b a b C 2 1 a b b a (z) 1 a b b a (yz) 3 a b a b btain the reducible representation (for 2 ) by multiplying the number of unshifted atoms times the contribution per atom. E C 2 v (z) v (yz) Unshifted Atoms 3 1 1 3 Contribution per atom 3-1 1 1 tot 9-1 1 3 31
Reducible Representations The reduction can be achieved using the reduction formula. It is a mathematical way of reducing that will always work when the answer cannot be spotted by inspection. It is particularly useful when there are large numbers of atoms and bonds involved. The vibrational modes of the molecule are reduced to produce a reducible representation into the irreducible representations. This method uses the following formula reduction formula: N 1 h n N is the number of times a symmetry species occurs in the reducible representation, h is the order of the group : simply the total number of symmetry operations in the group. The summation is over all of the symmetry operations. For each symmetry operation, three numbers are multiplied together. These are: Χ r is the character for a particular class of operation in the reducible representation Χ i is the character of the irreducible representation. n is the number of symmetry operations in the class r i Tabulate our known information. Reducible Representation (for 2 ) irrep 1) 2) E C 2 v (z) v (yz) r 9-1 1 3 N 1 h n A 1 : (1/h)[( r E )( i E )(n E ) + ( r C2 )( i C2 )(n C2 ) + ( r v(z) )( i v(z) )(n v(z) ) + ( r v(yz) )( i v(yz) )(n v(yz) )] Character Table C 2v E C 2 σ v (z) σ v (yz) A 1 1 1 1 1 z 2, y 2, z 2 A 2 1 1 1 1 R z y B 1 1-1 1 1, R y z B 2 1 1 1 1 y, R yz h = 4 A 1 : (1/4)[(9)( i E )(n E ) + (-1)( i C2 )(n C2 ) + (1)( i v(z) )(n v(z) ) + (3)( i v(yz) )(n v(yz) )] A 1 : (1/4)[(9)(1)(n E ) + (-1)(1)(n C2 ) + (1)(1)(n v(z) ) + (3)(1)(n v(yz) )] r i A 1 : (1/4)[(9)(1)(1) + (-1)(1)(1) + (1)(1)(1) + (3)(1)(1)] = 3 32
Calculate irreducible representation A 2 A 2 : (1/h)[( r E )( i E )(n E ) + ( r C2 )( i C2 )(n C2 ) + ( r v(z) )( i v(z) )(n v(z) ) + ( r v(yz) )( i v(yz) )(n v(yz) ) A 2 : (1/4)[(9)(1)(1) + (-1)(1)(1) + (1)(-1)(1) + (3)(-1)(1) = 1 Calculate irreducible representation B 1 B 1 : (1/4)[(9)(1)(1) + (-1)(-1)(1) + (1)(1)(1) + (3)(-1)(1) = 2 Calculate irreducible representation B 2 B 2 : (1/4)[(9)(1)(1) + (-1)(-1)(1) + (1)(-1)(1) + (3)(1)(1) = 3 The reducible representation E C 2 v (z) v (yz) r 9-1 1 3 is resolved into three A 1, one A 2, two B 1, and three B 2 species. C 2v E C 2 σ v (z) σ v (yz) A 1 1 1 1 1 z 2, y 2, z 2 A 2 1 1 1 1 R z y B 1 1-1 1 1, R y z B 2 1 1 1 1 y, R yz tot = 3A 1 + A 2 + 2B 1 + 3B 2 -[ trans = A 1 + B 1 + B 2 ] -[ rot = A 2 + B 1 + B 2 ] vib = 2A 1 + B 2 Notice this is the same result we obtained by analyzing the symmetries of the vibrational modes. -each mode is IR active 33
Consider only the stretches in 2 -consider the number of unchanged - bonds under the symmetry operations of the point group C 2v E C 2 σ v (z) σ v (yz) Unchanged bonds 2 0 0 2 Reducible Representation C 2v E C 2 σ v (z) σ v (yz) Character Table A 1 1 1 1 1 z 2, y 2, z 2 A 2 1 1 1 1 R z y B 1 1-1 1 1, R y z B 2 1 1 1 1 y, R yz By inspection, the reducible representation is composed of the A 1 and B 2 representation. E C 2 σ v (z) σ v (yz) A 1 1 1 1 1 B 2 1-1 -1 1 Sum of rows 2 0 0 2 General Method: Determine stretching in 2 using reducible representations and reduction formula C 2v E C 2 σ v (z) σ v (yz) Coefficient 1 1 1 1 rder of group 4 Unchanged bonds () 2 0 0 2 C 2v E C 2 σ v (z) σ v (yz) A 1 1 1 1 1 z 2, y 2, z 2 A 2 1 1 1 1 R z y B 1 1-1 1 1, R y z B 2 1 1 1 1 y, R yz Using reducible representations and the reduction formula, one obtains A 1 + B 2 modes. 34
Derive tot for BCl 3 given the character table for D 3h Derive the number of vibrational modes and assign modes for BCl 3. E 2C 3 3C 2 h 2S 3 3 v Unshifted atoms 4 1 2 4 1 2 Contribution per atom 3 0-1 1-2 1 tot 12 0-2 4-2 2 D 3h E 2C 3 3C 2 h 2S 3 3 3 v v A1 1 1 1 1 1 1 2 + y 2, z 2 A 2 E A 1 A 2 E h 1 1-1 1 1-1 R z 2-1 0 2-1 0 (, y ) ( 2 y 2, 2y ) 1 1 1-1 -1-1 1 1-1 -1-1 1 z 2-1 0-2 1 0 (R, R y ) (y, yz ) Results of using the reduction formula. A 1 A E 2 A 1 A 2 E Χ r *Χ i *n 1/h Sum Total 12 0-6 4-4 6 0.083333 12 1 12 0 6 4-4 -6 0.083333 12 1 24 0 0 8 4 0 0.083333 36 3 12 0-6 -4 4-6 0.083333 0 0 12 0 6-4 4 6 0.083333 24 2 24 0 0-8 -4 0 0.083333 12 1 Therefore, we have determined tot = A 1 + A 2 +3 E +2 A 2 + E but, subtract off the translational representations. -[ trans = E + A 2] and subtract off the rotational representations. -[ rot = A 2 + E ] vib = A 1 +2 E + A 2 ] Each E representation describes two vibrational modes of equal energy. 35
Symmetrical stretching. ut-of-plane bending mode. Unsymmetrical stretching. In-plane bending mode. Raman active. IR active. Raman and IR active. We can use isotopic substitution to interpret spectra, since the characteristic frequency of the mode will depend on the masses of the atoms moving in that mode. Review: What do I do when I need to? Assign symmetry labels to vibrational modes? If the vibrational mode is known and illustrated, sketch the resulting vibrational mode before and after each symmetry operation of the point group. Using the character table, assign the symmetry label and identify if the mode is IR and/or Raman active. Determine the symmetries of all vibrational modes and if the modes are IR and/or Raman active? Determine how many atoms are left unchanged by each symmetry operation. Find the reducible representation and reduce into the irreps. Subtract translational and rotational modes Identify which modes are IR and/or Raman active. Determine the symmetries of only the stretching modes and if the modes are IR and/or Raman active? Determine how many bonds are left unchanged by each symmetry operation. Find the reducible representation and reduce into the irreps. Identify which are IR and/or Raman active. Develop a character table? Determine the effect of each symmetry operation on the, y, z translation and the rotation R, R y, and R z. The resulting set of characters correspond to an irrep in the character table. 36
Determine the number of and assign the vibrational modes of the following: ow many peaks in the (1) IR spectra and (2) Raman spectra 1. N 3 2. C 4 3. [PtCl 4 ] 2-4. SF 6 5. SF 5 Cl Determine the symmetries and number of vibrational modes, and number IR and Raman peaks for a) C stretching modes b) all vibrational modes: 1. Mn(C) 6 2. Mn(C) 5 Cl 3. trans-mn(c) 4 Cl 2 4. cis-mn(c) 4 Cl 2 5. fac-mn(c) 3 Cl 3 6. mer-mn(c) 3 Cl 3 37