1. Types of Symmetry in Molecules 1. axis of symmetry (C n ) 2. plane of symmetry (  ) 3. center of symmetry (i) 4. improper axis of symmetry (S n ) “Wavefunctions have symmetry and their symmetry can be used to understand their properties and to define and describe molecular wavefunctions more easily.”

2. Symmetry Operations C n ― rotation by 2  /n radians gives an indistinguishable view of molecule.. N H H 1 X C 3 H 6 X C 2 1 X C 6 prinicpal axis

3. Symmetry Operations  ― reflection through molecular plane gives an indistinguishable view of the molecule 1 x  h.. N H H 3 X C v H 6 X  v

4. Symmetry Operations i ― inversion through center of mass gives an indistinguishable view of the molecule

5. Symmetry Operations S n ― Rotation by 2  /n & reflection through a plane ┴ to axis of rotation gives an indistinguishable view of the molecule S2S2 Ball – table 13.1 – p423 Each symmetry element can be defined by a 3x3 matrix.

6. Ball – p421 Molecules do not have random sets of symmetry elements – only certain specific sets of symmetry elements are possible. Such sets of symmetry always intersect at a single point. Therefore the groups of symmetry elements are referred to as point groups. Character Tables are lists for a specific point group that indicates all of the symmetry elements necessary for that point group. These can be found in Ball - appendix 3, p797. The number of individual symmetry operations in the point group is the order (h) of the group. The character tables are in the form of an hxh matrix. Point Groups E/C 1 Cs (C 1h ) C i (S 2 ) C n C nv C nh D n D nh D nd S n T d O h I h R h

7. Linear? no yes no i? yes D ∞h no C ∞v ≥ 2C n, n > 2? yes i?C5?C5? no yes IhIh OhOh Cn?Cn?Select highest C n yes ?? i? no CsCs C1C1 yes CiCi nC 2  to C n ? yes h?h? D nh nd?nd? h?h? yes C nh no nv?nv? yes C nv S 2n ? yes S 2n no CnCn TdTd DnDn yes D nd Point Group Flow chart

8. Polyatomic Molecules: BeH 2 Linear? yesno i? yes D ∞h no C ∞v Point Group Flow chart D ∞h

9. Be1s, Be2s, Be2p z H A 1s, H B 1s (H A 1s + H B 1s), (H A 1s - H B 1s) Be1s, Be2s, Be2p z Polyatomic Molecules: BeH 2 Minimum Basis Set?   Be H Be2p x, Be2p y How can you keep from telling H atoms apart? Separate into symmetric and antisymmetric functions? Minimum Basis Set  g = (H A 1s + H B 1s), Be1s, Be2s,  u = Be2p z & (H A 1s - H B 1s)  u = Be2p x & Be2p y

10. BeH 2 – Minimum Basis Set BeHAHA HBHB 1s -115 eV-13.6 eV 2s -6.7 eV 2p -3.7 eV AO energy levels Does LCAO with H A and H B change energy? -13.6 eV = (H A 1s + H B 1s) and (H A 1s - H B 1s)

11. Polyatomic Molecules — BeH 2 Spartan – MNDO semi-empirical 7.2eV  u * = 0.84 Be2p z + 0.38 (H A 1s - H B 1s) 3.0eV  g * = 0.74Be2s - 0.48(H A 1s + H B 1s) 2.5eV  u = (0.95Be2p x - 0.30Be2p y )& (0.30Be2p x + 0.95Be2p y ) -12.3 eV  u = -0.51Be2p z + 0.59 (H A 1s - H B 1s) -13.8 eV  g = -0.67Be2s - 0.52(H A 1s + H B 1s) -115 eV  g = Be1s  u = Be2p x & Be2p y  u = Be2p z | (H A 1s - H B 1s)  g = Be1s | Be2s | (H A 1s + H B 1s)  u = 0.44(Be2p z ) + 0.44 (H A 1s - H B 1s)  g = -0.09(Be1s) + 0.40(Be2s) + 0.45 (H A 1s + H B 1s)  g = 1.00(Be1s) + 0.016(Be2s) -0.002 (H A 1s + H B 1s) HF SCF calculation : J. Chem. Phys. 1971

12. Be H A & H B 1g1g 2g2g 1u1u 1u1u 3g*3g* 2u*2u* -13.6 eV -115 eV -6.7 eV -3.7 eV -13.8 eV -12.3 eV 3.0 eV 2.5 eV 7.2 eV

13. Average Bond Dipole Moments in Debyes (1 D = 3.335641 Cm) H - O1.5C - Cl 1.5C = O2.5 H - N1.3C - Br 1.4C - N0.5 H - C0.4C - O 0.8 C  N 3.5 e = 1.6022 x 10 -19 Dipole Moments & Electronegativity In MO theory the charge on each atom is related to the probability of finding the electron near that nucleus, which is related to the coefficient of the AO in the MO CH 2 O Geometry Use VSEPR and SOHCAHTOA to find dipole moment in debyes.

14. Heteronuclear Diatomic Molecules MO = LCAO same type (  ) — similar energy All same type AO’s = basis set minimum basis set (no empty AO’s) Resulting MO’s are delocalized Coefficients = weighting contribution HF minimum basis set  = H(1s), F(1s), F(2s), F(2p z )  = F(2p x ), F(2p y ) without lower E AO’s  = H(1s), F(2s), F(2p z )  = F(2p x ), F(2p y )

15. HF 12.9eV 19.3eV H1s F2s F2p z 13.6eV 18.6eV **    x  y 0.19 H 1s + 0.98 F 2pz 0.98 H 1s - 0.19 F 2pz

16. Delocalized HF Molecule 1  x & 1  y = F(2p x ) & F(2p y ) 3  = -0.023F(1s) - 0.411F(2s) + 0.711F(2p z ) + 0.516H(1s) 2  = -0.018F(1s) + 0.914F(2s) +.090F(2p z ) +.154H(1s) 1  = 1.000F(1s) + 0.012F(2s) + 0.002F(2p z ) - 0.003H(1s)

17. Polyatomic Molecules: BeH 2 Minimum Basis Set  g = (H A 1s + H B 1s), Be1s, Be2s,  u = Be2p z & (H A 1s - H B 1s)  u = Be2p x & Be2p y

18. Polyatomic Molecules — BeH 2  u * = C 7 Be2p z - C 8 (H A 1s - H B 1s)  g * = C 5 Be2s + C 6 (H A 1s + H B 1s)  u = Be2p x & Be2p y  u = C 3 Be2p z + C 4 (H A 1s - H B 1s)  g = C 1 Be2s + C 2 (H A 1s + H B 1s)  g = Be1s What is point group? What are the basis set AOs for determining MOs ?  u = Be2p x & Be2p y  u = Be2p z | (H A 1s - H B 1s)  g = Be1s | Be2s | (H A 1s + H B 1s)  u = 0.44(Be2p z ) + 0.44 (H A 1s - H B 1s)  g = -0.09(Be1s) + 0.40(Be2s) + 0.45 (H A 1s + H B 1s)  g = 1.00(Be1s) + 0.016(Be2s) -0.002 (H A 1s + H B 1s) HF SCF calculation : J. Chem. Phys. 1971

19. Be H A & H B 1g1g 2g2g 1u1u 1u1u 3g*3g* 2u*2u* + -

20. MO: 1 2 3 4 5 Eigenvalues:-1.82822 -0.63290 -0.51768 -0.51768 0.24632 (ev): -49.74849 -17.22198 -14.08688 -14.08688 6.70261 A1 A1 ??? ??? A1 1 H2 S 0.37583 -0.46288 0.00000 0.00000 0.80281 2 F1 S 0.91940 0.29466 0.00000 0.00000 -0.26052 3 F1 PX 0.00000 0.00000 -0.78600 0.61823 0.00000 4 F1 PY 0.00000 0.00000 0.61823 0.78600 0.00000 5 F1 PZ -0.11597 0.83601 0.00000 0.00000 0.53631 HF 1  2 2  2 3  2 (1  2 2  2 ) 4  0 Semi-empirical treatment of HF from Spartan (AM1) One simpler treatment of HF is given in Atkins on page 428 gives the following results.... 4  0.98 (H1s) - 0.19(F2p z )-13.4 eV  x =  y = F2p x and F2p y -18.6 eV 3  = 0.19(H1s) + 0.98(F2p z )-18.8 eV 2  = F2s ~ -40.2 eV 1  = F1s << -40.2 eV H 1s F 2s F 2p

21. Localized MO’s 6e - = 6 x 6 determinant adding cst column to another column leaves determinant value unchanged adjust so resultant determinant represents localized MO’s CH 4 - localized bonding MO J. Chem. Phys. 1967 C - H A MO =.... 0.02(C1s ) + 0.292(C2s) + 0.277(C2p x + C2p y + C2p z ) + 0.57(H A 1s) - 0.07(H B 1s + H C 1s + H D 1s)

22. Calculating Dipole Moments p ´ =   *  i Q i ´ r i  d  Q i ´ = charge on particle i Semi-empirical Methods CNDO = Complete Neglect of Differential Overlap INDO = Intermediate Neglect of Differential Overlap MINDO = Modified INDO

23. CH 2 O Geometry page 659 HF real Hyper Emp Chem C-H 1.10 1.12 1.11 1.07 C=O 1.22 1.21 1.21 1.23 HCH 114.8 o 116.5 o 108.4 o 120 o dipole moment = 2.33 debyes (CRC)