Lecture 12. Symmetry Operations. NC State University

Transcription

1 Chemistry 431 Lecture 12 Group Theory Symmetry Operations NC State University

2 Wave functions as the basis for irreducible representations The energy of the system will not change when symmetry Operations are applied. If we consider atomic orbitals on Each atom as the basis vectors we state generally that RΨ = +Ψ or Ψ S-orbitals are always + since they are spherical P-orbitals can be either + or -. RΨ = +Ψ RΨ = Ψ

3 Mixing of s- and p-orbitals in diatomic molecules Using the concepts developed in the hydrogen molecule ion we can consider combinations of all of the atomic orbitals that are important for diatomic molecules. Only orbitals that have the same irreducible representation can mix. Mathematically this statement is the same as saying the the overlap integrals will be non-zero only if they contain the totally symmetric representation.

4 Homonuclear diatomic molecules Homonuclear diatomic molecules belong to the D h. We will determine the reducible representations of each of the combinations shown below.

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6 Homonuclear diatomic molecules The D h symmetry operations are E, σ v, C, i, 2S, C 2. The reducible representations are: E` σ v C i 2S C 2. s p z p x, p y 4 4cosφ The s irrep reduces to Σ g+ to Σ u+. p g u The p z irrep reduces to Σ g+ to Σ u+. The p x, p y irreps reduce to Π g to Π u.

7 Homonuclear diatomic molecules The D h symmetry operations are E, σ v, C, i, 2S, C 2.

8 Polyatomic molecules For polyatomic molecules appropriate atomic orbitals must be combined into symmetry adapted linear combinations (SALCs). H 2 O belongs to the C 2v point group. Valence orbitals 2 H(1s) + O(2s) + O(2p x ) + O(2p y ) + O(2p z ) Since the oxygen is unmoved by every symmetry operation the character can be determined directly from the character table. However, the hydrogens need to be formed as SALCs first.

9 Polyatomic molecules The reducible representations for hydrogen SALCs are a 1 and b 1. The oxygen orbitals (read from the character Table are: 2s = a 1 2p z = a 1 2p x =b 1 2p y = b 2 z y x Note that the energies are obtained from a calculation. However, the forms of the orbitals can be predicted based on symmetry considerations.

10 Ethylene Molecules with p-systems are particularly important in Organic chemistry. The simplest of these is ethylene. H H C Ethylene belongs to the D 2h point group. The carbons are symmetrically equivalent. The orbital set on each consists of C(2s) + C(2p x ) + C(2p y ) + C(2p z ) C H H

11 Ethylene The atomic orbitals are shown below: The reducible representations of the symmetrically equivalent sets formed for the valence ao s of ethylene can be determined and decomposed in the following groups.

12 Ethylene Since the members of all of the sets can be exchanged completely with only the rotational operations, only the C 2 characters of the above character table need be considered.

13 Ethylene The normalized SALCs can now be constructed:

14 Ethylene The normalized SALCs can now be constructed:

15 Huckel MO Theory 1. Determine the molecular point group. 2. Determine the normalized symmetry adapted linear combinations (SALCs) using the projection operator method. If there is only one set of SALCs then these are also the MOs. 3. Use the SALCs to determine the elements of the Huckel secular determinant det H ij S ij E o =0 j 4. Simplify the mathematical formalism by letting H ii = α and H ij = β. Divide each element by β and let x = (α -E)/β 5. For each eigenvalue solve for the eigenvectors (coefficients). These are the contributions of each SALC to the MO. N Σ H ij S ij E o C ij j = 1 N Σ C ij 2 j =1 =1 j =0

16 2 3 Butadiene and 3 are symmetry related 1 and 4 are symmetry related Both transform as E C 2v σ(xz) σ(yz) Γ

17 Butadiene E C 2v σ(xz) σ(yz) Γ A Total = 0 E C 2v σ(xz) σ(yz) Γ A Total = 4 E C 2v σ(xz) σ(yz) Γ B Total = 0 E C 2v σ(xz) σ(yz) Γ B Total = 4 The order is 4 so the character sums are divided by 4. The irreducible representations are: Γ = a 2 + b 2

18 Butadiene SALC 14 (b 2 ) = p 1 +p 4 +p 4 + p 1 SALC 14 (b 2 )= 1 2 p 1 + p 4 SALC 14 (a 2 ) = p 1 p 4 p 4 + p 1 SALC = 1 14 (a 2 ) 2 p 1 p 4

19 Butadiene SALC 23 (b 2 ) = p 2 +p 3 +p 3 + p 2 SALC 23 (b 2 )= 1 2 p 2 + p 3 SALC 23 (a 2 ) = p 2 p 3 p 3 + p 2 SALC 23 (a 2 ) = 1 2 p 2 p 3

20 Construct symmetrized Huckel Determinant The unsymmetrized Huckel determinant is a 4x4 matrix. The solution is a fourth order polynomial. By factoring according to symmetry the determinant can be simplified into 2 2x2 matrices. H 11 = 1 p p 4 H p 1 + p 4 dτ = 2 1 p 1Hp 1 dτ p 4Hp 4 dτ p 1Hp 4 dτ p 4Hp 1 dτ = α H 22 = 1 p p 3 H p 2 + p 3 dτ = 1 2 p 2Hp 2 dτ p 3Hp 3 dτ p 2Hp 3 dτ p 3Hp 2 dτ = α + β

21 Construct symmetrized Huckel Determinant H 12 = 1 2 p 1 + p 4 Hp 2 + p 3 dτ = 1 H H H H 2 p 1Hp 2 dτ p 4Hp 2 dτ p 1Hp 3 dτ p 4Hp 3 dτ = β H 21 = H 12 H 33 = 1 2 p 1 p 4 Hp 1 p 4 dτ = 1 p Hp 1 dτ p 2 4 Hp 1 dτ p 2 4 Hp 1 dτ + p 2 4 Hp 4 dτ = α

22 Construct symmetrized Huckel Determinant H 34 = 1 2 p 1 p 4 Hp 2 p 3 dτ = 1 2 p 1Hp 2 dτ 1 2 p 4Hp 2 dτ 1 2 p 1Hp 3 dτ p 4Hp 3 dτ = β H 43 = H 34 H 44 = 1 2 p 2 p 3 Hp 2 p 3 dτ = 1 p Hp 2 dτ + p 2 3 Hp 3 dτ p 2 2 Hp 3 dτ p 2 3 Hp 2 dτ = α β H 13 = H 31 = H 24 = H 42 = H 14 = H 41 = H 23 = H 32 =0

23 Construct symmetrized Huckel Determinant H 11 E H 12 H 13 H 14 H 21 H 22 E H 23 H 24 H 31 H 32 H 33 E H 34 H 41 H 42 H 43 H 44 E α E β 0 0 β α+β E α E β 0 0 β α β E Divide by β and let x = α E β

24 Solve each symmetry block separately det x x x x 1 = 0 b 2 block x 1 1 x +1 = xx+1 1=x2 + x 1,x = 1± = 5 ±1 2 a 2 block x 1 = xx 1 1=x2 x 1, x = 1 x 1 1± 1+4 = 5 ±1 2 2

25 Determine the orbital energies β =1.62β 5 1 β = 0.62β β = 0.62β 5 1 β = 1.62β 2

26 Determine the eigenvectors Only the SALCs from the same irreducible ibl representation ti Can contribute to the same MO. For the b 2 block and x = we have: c 11 c 21, c 11 c 21 =0 1.62c 11 +c 21 =0 c c 21 =0 c 21 = 1.62c 11 c 11 = 0.62c 21 and c c 2 21 =1 so c c 2 11 = 3.62c 2 11 =1 2 = 1/3.62 = and finally c 11 = c 11 this implies thatt c 21 = 1.62(0.525) = 0.85 The MO is p 1 + p p 2 + p 3 =0.37p 1 + p p 2 + p 3 =0.37p p p p 4

27 Determine the eigenvectors Only the SALCs from the same irreducible ibl representation ti Can contribute to the same MO. For the b 2 block and x = 0.62 we have: c c, c c 21 =0 0.62c 11 +c 21 =0 c c 21 =0 c 21 = 0.62c 11 c 11 = 1.62c 21 and c c 2 21 =1 so c c 2 11 =2.38c 2 11 =1 2 = 1/2.38 = and finally c 11 =0.85 c 11 this implies thatt c 21 = 0.62(0.85) = 0.53 The MO is p 1 + p p 2 + p 3 =0.60p 1 + p p 2 + p 3 =0.60p p p p 4

28 Bond order The bond order can be calculated from the MO coefficients. The p-bond order is the product of the coefficients of two adjacent atoms times the number of electrons in the respective orbitals. Bond order = occup Σ k =1 n k c ik c jk Where nk is the number of electrons in the MO and cik and cjk Are the coefficients. For butadiene the bond order for p1-p2 p2 and p3-p4 p4 is: Bond order =2(0.37)(0.60)+2(0.60)(0.37)=0.888 The bond order for p2-p3 p3 is: Bond order = 2(0.60)(0.60) + 2(0.37)( 0.37) = 0.446

29 Bond order The bond order of the butadiene double bond is 88.8% as large as that of ethylene. However, the total π-bonding energy is 2(0.888) = 2.22, 22 which is greater than two ethylene molecules. This is an example of stabilization by delocalization. The total energy is the energy of each MO times its occupancy. Total energy = occup Σ k =1 n k E k For butadiene this is: E =2( 1.62) + 2( 0.62) = 4.48 (units of β)

30 Benzene Most of the spectroscopy and reactivity of benzene is attributable to its π system. To a good approximation the σ and π orbitals can be separated. We consider how the 6 p-orbitals transform under D 6h symmetry. In this case we do not need to construct SALCs since there is only one type of orbital. Γ

31 Projection operator approach The operation required to carry the reference p1 orbital into any of the others. b 2g C 6

32 Projection operator approach The operation required to carry the reference p1 orbital into any of the others. b 2g C 3

33 Projection operator approach The operation required to carry the reference p1 orbital into any of the others. b 2g C 2

34 D 6h character table

35 Molecular orbitals constructed using the projection operator approach

36 H 2 O H 2 O H 2 O Mn OH 2 OH 2 In the crystal field picture the repulsion of the oxygen lone pairs causes the d orbitals to split. We consider each water as a point in the Oh point group. OH 2 E 6 C 3 0 C 2 0 C 4 2 C 2d 2 i 0 S 4 0 S 6 0 σ h 4 σ d 2