Symmetry: Translation and Rotation

Transcription

1 Symmetry: Translation and Rotation The sixth column of the C 2v character table indicates the symmetry species for translation along (T) and rotation about (R) the Cartesian axes. y y y C 2 F v (x) T x -1 1 C 2 F v (x) T y -1-1 C 2 F v (x) T 1 1 y y y C 2 F v (x) R x -1-1 C 2 F v (x) R y -1 1 C 2 F v (x) R 1-1 The symbol stands for the representation of... and is used below as an irreducible representation. (T x ) = B 1 (T y ) = B 2 (T ) = A 1 (R x ) = B 2 (R y ) = B 1 (R ) = A 2 Note the fast way which these are assigned with just the C 2 and F v (x) elements (since they generate F v (y)).

2 Symmetry: Vibrations The seventh column of the table contains the symmetry species of the components of the symmetric hyperpolariability tensor, ", which are important in vibrational Raman spectroscopy (this is covered later in the course). Out of the total 3N degrees of freedom for an N-atom nonlinear molecule, there are 3N-6 normal modes of vibration (each atom requires 3 coordinates to specify vibrational motion), 3 for translation and 3 for rotation. For a linear molecule, there are only 3N-5 normal modes of vibration (no degree of freedom for vibration perpendicular to the cylindrical internuclear axis). H 2 O, N = 3, normal modes of vibration: 3 y y y C 2 F v (x) v C 2 F v (x) v C 2 F v (x) v (R v(1) ) = A 1 (R v(2) ) = A 1 (R v(3) ) = B 2 Sometimes choice of axes is difficult: some conventions include along the principal axis, x to molecular plane (note if we switched x and y, the B 1 s would become B 2 s and vice versa - it is important to label these axes!!)

3 Symmetry of Normal Modes In order to consider the symmetry of the normal modes of vibration, we must consider the following: 1. The number of normal vibrational modes (3N - 6 if non-linear, 3N - 5 if linear), where N = # of atoms. 2. The point group of the molecule 3. The order (total # of operations) of the point group, h 4. The number of classes in the group, N c 5. The number of elements or members in each class, g n 6. The reducible representation for all possible motions of atoms in the molecule, tot or cart (i.e., Cartesian coordinates), which has characters P n (R) 7. The irreducible representations of the symmetry species of the point groups that produce the reducible representation, which has the characters P "n (R) 8. The vibrational modes which will be visible in IR or Raman spectra. This looks pretty complicated, but it is not that bad. This is referred to as normal mode analysis. The easiest basis set for this type of analysis is a set of N Cartesian coordinate systems, with one centred on each atom in the molecule. The dimension of this representation will be a 3N 3N matrix. 3 degrees of freedom will be assigned to translation, and 3 to rotation

4 Normal Mode Analysis of H 2 O Here we have N = 3 atoms, so 3N - 6 = 3 vibrational modes. The order of the group is h = 4, the number of classes in the group is N c = 4 (there are no degenerate symmetry elements, so g n = 1 in each case) x C C 2 1 x C y x C y 2 3 y The trace of the 9 9 reducible representation matrix will give us the characters of the reducible representation. Below, results of the C 2 and F v (x) operations are shown on the left and right, respectively. C 2 3 F v (x) 3 B C 2 B B 1 2 C C 1

5 Matrix Representations These transformations can be represented as: x ) 1 x ) 1 y ) 1 ) 1 x ) 2 y ) 2 ) 2 x ) 3 y ) x 1 y 1 1 x 2 y 2 2 x 3 y 3 3 y ) 1 ) 1 x ) 2 y ) 2 ) 2 x ) 3 y ) & & & & & & x 1 y 1 1 x 2 y 2 2 x 3 y 3 3 ) 3 P(E) = Tr(E) = 9 P(C 2 ) = Tr(C 2 ) = -1 ) 3 x ) 1 x ) 1 y ) 1 ) & x 1 y 1 1 y ) 1 ) 1 & x 1 y 1 1 x ) x 2 x ) & x 2 y ) 2 0 & y 2 y ) y 2 ) 2 x ) 3 y ) & x 3 y 3 3 ) 2 x ) 3 y ) & x 3 y 3 3 ) 3 P(F v (x)) = Tr(F v ) = 1 P(F v N(y)) = Tr(F v N) = 3 ) 3 C 2v E C 2 σ v (x) σn v (y) Γ tot

6 Irreducible Representations C 2v E C 2 σ v (x) σn v (y) Γ tot C 2v E C 2 F v (x) FN v (y) A A R B x, R y B y, R x To determine the number of irreducible representations it takes to make the reducible representations, we use the standard reduction formula: m " 1 h j N c n=1 P n (R)g n P "n (R) m A1 ¼[9@1@1 % (&1)@1@1 % 1@1@1 % 3@1@1] 3 m A2 ¼[9@1@1 % (&1)@1@1 % 1@1@(&1) % 3@1@(&1)] 1 m B1 ¼[9@1@1 % (&1)@1@(&1) % 1@1@1 % 3@1@(&1)] 2 m B2 ¼[9@1@1 % (&1)@1@(&1) % 1@1@(&1) % 3@1@1] 3 tot 3A 1 % A 2 % 2B 1 % 3B 2 From the character table, we can determine the symmetries of translational and rotational degrees of freedom: x B 1 R x B 2 y B 2 R y B 1 A 1 R A 2

7 Vibrational Modes tot 3A 1 % A 2 % 2B 1 % 3B 2 vib tot & trans, rot 3A 1 % A 2 % 2B 1 % 3B 2 & (A 1 % A 2 % 2B 1 % 2B 2 ) 2A 1 % B 2 To identify the symmetries of the normal modes, we use a basis set of internal displacement coordinates: r 1 and r 2 are along the O-H bonds, and 2 is the HOH angle (these are the three vibrational degrees of freedom in this molecule) No operation interchanges the r vectors with the angle 2, so they form independent basis sets 3 r 1 r (E) (C 2 ) r ) 1 r ) 2 2 ) r ) 1 r ) 2 2 ) r 1 r 2 2 r 1 r 2 2 C 2v E C 2 σ v (x) σn v (y) Γ int (F v ) r ) 1 r ) 2 2 ) r 1 r 2 2 Γ r Γ θ (F ) v ) r ) 1 r ) 2 2 ) r 1 r 2 2 Either by the standard reduction formula or by comparison with the character table, it is easy to see that r A 1 % B 2 2 A 1 One vibrational mode is purely stretching of bonds: v 3 is B 2 Other vibrational modes are mixes of bending and stretching: v 1 and v 2 are A 1

8 Direct Products In many cases, it will be necessary to obtain the direct product of symmetry species. For example, say H 2 O is vibrationally excited with two quanta simutaneously: one that influences the v 1 and one the v 3 wavefunctions. The total symmetry species for the vibrational wavefunction of this combination state is (R v ) = A 1 B 2 = B 2 To obtain this direct product of the two species, we multiply the characters under each element using the rules: (+1) (+1) = 1; (+1) (-1) = -1; (-1) (-1) = 1; C 2v E C 2 σ v (x) σn v (x) A B A 1 B 2 = B If the two quanta excite vibrations of species v 3, then (R v ) = B 2 B 2 = A 1 (a) The direct product of any species with the totally symmetric species leaves it unchanged (b) The direct product of any species with itself gives the totally symmetric species Also, for C 2v : A 2 B 1 = B 2 and A 2 B 2 = B 1

9 C 3v E 2C 3 3σ v A T α xx +α yy, α A R The C 3v Character Table E (T x,t y ), (R x, R y ) (α xx -α yy, α xy ), (α x, α y ) Some differences from C 2v : 1. Grouping together of all elements in the same class. 2. E, the doubly degenerate symmetry species whose characters are not always +1 or -1 (also 2 and 0)! Two elements P and Q belong to the same class if: P = R -1 Q R For example, for the C 3v point group: C 3 = F v -1 C 3 2 F v so C 3 and C 3 2 belong to the same class. H 1 F v N H 3 H 2 F v -1 C 3 2 F v 3 1 C Elements belonging to the same class have the same characters. The number of symmetry species is equal to the number of classes, N c = 3, for both degenerate and nondegenerate point groups. The order, h, of the point group is the total number of operations: h(c 2v ) = 4, h(c 3v ) = 6

10 x C y Normal Mode Analysis: C 3v side view C y x top view C C C Let s generate the tot for this C 3v molecule. E leaves all of the Cartesian vectors alone, so P(E) = 12. C 3 rotation moves the three hydrogen atoms, so none of their coordinates contribute to the trace, and we can forget about them. C 3 leaves the vector on N alone (+1) and rotates the x and y vectors by 2B/3. x y C 3 2 = 2B/3 Projection of yn on the x and y axes is xn = x cos(2b/3) + y sin(2b/3) = -0.5x y and yn = -x sin(2b/3) + y cos(2b/3) = x - 0.5y We are concerned only with what happens with the x and y portions of this rotational transform (on the diagonal), so x and y are -0.5 and Thus, P(C 3 ) = = 0. The same holds true for the C 3 2 operator. xn yn x y

11 Normal Mode Analysis: C 3v, Part 2 Reflection, F v, exchanges positions of two of the hydrogen atoms, so we forget about these completely. y y x F v x The reflection changes the sign of the two y vectors (-2), but the two x and two vectors are unaffected (2+2 = 4). Thus, P(F v ) = 2, and C 3v E 2C 3 3σ v Γ tot Using the standard reduction formula: m " 1 h j N c n=1 P n (R)g n P n" (R) m A1 (1/6)[12@1@1 % 0@2@1 % 2@3@1] 3 m A2 (1/6)[12@1@1 % 0@2@1 % 2@3@(&1)] 1 m E (1/6)[12@1@2 % 0@2@(&1) % 2@3@0] 4 tot 3A 1 % A 2 % 4E

12 Normal Mode Analysis: C 3v, Part 3 C 3v E 2C 3 3σ v A A R E (x,y), (R x, R y ) From the character table, we can determine the symmetries of translational and rotational degrees of freedom: x R E transform x E y together R y A 1 R A 2 vib tot & trans, rot 3A 1 % A 2 % 4E & (A 1 % A 2 % 2E) 2A 1 % 2E C 3v E 2C 3 3σ v Γ r Γ θ transform together The obvious coordinates for describing internal motions are to have three bond vectors r 1, r 2 and r 3, and three inter-bond angles 2 1, 2 2 and 2 3. Each set of three forms an independent basis set with representations: which can be reduced as above to yield: r = A 1 + E and 2 = A 1 + E both A 1 and E are IR and Raman active (from character table), but observed bands in the spectrum are a mix of bending & stretching

13 Degenerate Symmetry Species The characters 0 and 2 of the E species can be understood by considering the normal vibrations of NH 3 The observed bands are neither from purely N-H stretching or H-N-H angle deformation (bending), but rather from a combination of both, since r = A 1 + E and 2 = A 1 + E. v 1 (A 1 ) v 2 (A 1 ) v 3a v 3 (E) v 3b v 4a v 4 (E) v 4b Both v 1 and v 2 are A 1, v 3a and v 3b are degenerate, as are v 4a and v 4b. The degenerate vibrations are excited by the same quantum of energy.

14 Direct Products and C 3v With the exception of E E, direct products are: A 1 A 2 = A 2 ; A 2 A 2 = A 1 ; A 1 E= E; A 2 E = E For E E, the result depends on whether we require (R v ) when: (a) one quantum each of two different E vibrations is excited (a combination level); or, (b) two quanta of the same E vibration level are excited (an overtone level) For (a), such as for the combination v 3 + v 4, the product is E E and we obtain this by squaring the characters: C 3v E 2C 3 3σ v E E The characters 4, 1, 0 form a reducible representation in the C 3v point group, which can be reduced to: E E = A 1 + A 2 + E For (b), such as for 2v 3, the product is written as (E) 2 where (E) 2 = A 1 + E which is the symmetric part of E E (symmetric to particle exchange). One of the species of the E E product is forbidden: In this case it is A 2, which is non-totally symmetric, and therefore is referred to as the antisymmetric part of E E Tables are available with all of the symmetry species of vibrational combination states resulting from (a) and (b).

15 The C 4v Character Table C 4v E 2C 4 φ... 4σ v A 1 /Σ x 2 +y 2, 2 A 2 /Σ R E 1 /Π 2 2cosφ... 0 (x, y), (R x, R y ) (x, y) E 2 / 2 2cos2φ... 0 (x 2 +y 2, 2 ) E 3 /Φ 2 2cos3φ... 0!!!!! There is an infinite number of classes for this point group since rotation about the C 4 axis may be by any angle N, and each C 4 N element belongs to a different class. Thus, there are an infinite number of symmetry species. Symmetry species are: A 1, A 2, E 1, E 2,..., E 4 according to general symmetry rules. However, there is an older convention which is still used, notably in electronic spectroscopy of diatomic molecules: Electronic states are labelled according to the electronic orbital angular momentum number 7 = 0, 1, 2, 3,...for diatomic molecules: E, A, ), M,... The symbols + and - are equivalent to 1 and 2 in the usual notation.

16 The C 4v Character Table, 2 Multiplication of the symmetry species follow the usual rules: E % E & E & ; E & A A ; E % ) ) For the product A A, the reducible representation is: C 4v E 2C 4 φ 4σ v Π Π 4 4cos 2 φ (= 2+2cos2φ) 0 This can be reduced by inspection: A A E % % E & % ) Recall that (A) 2 will be the symmetric part of A A (the part that is symmetric w.r.t. particle exchange). One species of the A A product is forbidden: in this case it is the E - species, which is non-totally symmetric. So: (A) 2 E % % )

17 Mulliken Symbols for Symmetry The Mulliken symbols are used to denote all of the representations that occur in the character tables: Symbol Meaning A (Σ) B E (Π,, Φ,...) T Non-degenerate representation, symmetric w.r.t. C n Non-degenerate representation of a finite order point group, antisymmetric with respect to C n Doubly degenerate representation Triply degenerate representation 1 (+) Symmetric w.r.t. σ v 2 (-) Antisymmetric w.r.t. σ v N O g u Symmetric w.r.t. σ h Antisymmetric w.r.t. σ h Symmetric w.r.t. i Antisymmetric w.r.t. i First representations: total symmetric irreducible representations: A, A 1, AN, A 1 N, A g, A 1g, E +, E g +

18 Key Concepts: Group Theory 1. Group theory is a mathematical treatment that provides us with a formal means of describing the symmetry of objects such as molecules. 2. A symmetry operation is an operation which moves the molecule into a new orientation equivalent to its original orientation. A symmetry element is a point, line or plane with respect to which a symmetry operation is performed. 3. The symmetry elements include E (identity), F (plane of symmetry), C n (proper axis of rotation), i (centre of inversion) and S n (improper rotation). 4. If a molecule has an n-fold rotational axis, C n, rotation of the molecule by 2B/n radians (n = 1, 2, 3,..., 4) generates an orientation of the molecule which is indistinguishable from the original orientation. 5. Reflections may be labelled as: F v (reflection plane vertical w.r.t. principal axis), F h (reflection plane horiontal w.r.t. principal axis) and F d (reflection planes which bisect angles between dihedral axes). 6. All of the elements of symmetry which any molecule may have constitute a point group, which is used to accurately determine the symmetry of a molecule.

19 Key Concepts: Character Tables 1. The symmetry elements of a molecule can be classified using character tables. 2. Each row in a character table is the irreducible representation for a particular symmetry element. The irreducible representation is composed of different characters, which describe the effects of the symmetry operations on different objects: e.g., wavefunctions, vectors, orbitals, etc. 3. The character table also contains information on symmetry species for translational and rotational motion, as well as polariability. 4. Out of the total 3N degrees of freedom for an N-atom non-linear molecule, there are 3N-6 modes of vibration (each atom requires three coordinates which can vary to specify the position of each), 3 for translation and 3 for rotation. Linear molecules have 3N-5 modes of vibration. 5. Normal mode analysis can be used to create a total reducible representation for the molecule, which can then be reduced to identify the various modes of vibration and classify them as symmetry species. 6. Direct products are used to determine the total representation of combination states.