1 / 21

Types of Symmetry in Molecules

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

gizela
Download Presentation

Types of Symmetry in Molecules

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

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

  2. Symmetry Operations Cn―rotation by 2p/n radians gives an indistinguishable view of molecule 6 X C2 .. N H H1 X C3 H 1 X C6 prinicpal axis

  3. Symmetry Operations s―reflection through molecular plane gives an indistinguishable view of the molecule 1 x sh .. N H H3 X Cv H 6 X sv

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

  5. Symmetry Operations Sn― Rotation by 2p/n & reflection through a plane ┴to axis of rotation gives an indistinguishable view of the molecule S2 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/C1 Cs (C1h) Ci (S2) CnCnvCnh DnDnhDndSn Td OhIhRh

  7. no yes yes Point Group Flow chart Linear? i? D∞h no C∞v yes yes yes ≥ 2Cn, n > 2? i? C5? Ih no no Dnh no Td Oh yes yes yes Cn? Select highest Cn nC2 to Cn? sh? no no no yes yes s? Dnd nsd? Cs yes no Cnh sh? no Dn yes i? Ci no yes no no S2n? nsv? Cnv C1 yes no Cn S2n

  8. Polyatomic Molecules: BeH2 no yes yes Point Group Flow chart Linear? i? D∞h D∞h no C∞v

  9. Polyatomic Molecules: BeH2 Minimum Basis Set? Be H  (HA1s + HB1s), (HA1s - HB1s) Be1s, Be2s, Be2pz HA1s, HB1s (HA1s + HB1s), (HA1s - HB1s) Be1s, Be2s, Be2pz  Be2px, Be2py How can you keep from telling H atoms apart? Separate into symmetric and antisymmetric functions? Minimum Basis Set g = (HA1s + HB1s), Be1s, Be2s, u = Be2pz &(HA1s - HB1s) u = Be2px & Be2py

  10. AO energy levels Does LCAO with HA and HB change energy? -13.6 eV = (HA1s + HB1s) and (HA1s - HB1s)

  11. Polyatomic Molecules — BeH2 u = Be2px & Be2py u = Be2pz | (HA1s - HB1s) g = Be1s | Be2s | (HA1s + HB1s) Spartan – MNDO semi-empirical 7.2eV u* = 0.84Be2pz + 0.38(HA1s - HB1s) 3.0eV g* = 0.74Be2s - 0.48(HA1s + HB1s) 2.5eVu = (0.95Be2px - 0.30Be2py)& (0.30Be2px + 0.95Be2py) -12.3 eVu = -0.51Be2pz+ 0.59 (HA1s - HB1s) -13.8 eVg = -0.67Be2s - 0.52(HA1s + HB1s) -115 eV g = Be1s HF SCF calculation : J. Chem. Phys. 1971 u = 0.44(Be2pz) + 0.44 (HA1s - HB1s) g = -0.09(Be1s) + 0.40(Be2s) + 0.45 (HA1s + HB1s) g = 1.00(Be1s) + 0.016(Be2s) -0.002 (HA1s + HB1s)

  12. 2su* Be HA & HB 3sg* 1pu 1su 2sg 1sg 7.2 eV 3.0 eV 2.5 eV -3.7 eV -6.7 eV -12.3 eV -13.6 eV -13.8 eV -115 eV

  13. Dipole Moments & Electronegativity CH2O Geometry 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 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 without lower E AO’s  = H(1s), F(1s), F(2s), F(2pz)  = F(2px), F(2py)  = H(1s), F(2s), F(2pz)  = F(2px), F(2py)

  15. HF 13.6eV 12.9eV 0.98 H1s - 0.19 F2pz 4* 18.6eV H1s xpy 19.3eV F2pz 0.19 H1s + 0.98 F2pz 3 2 F2s

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

  17. Polyatomic Molecules: BeH2 Minimum Basis Set g = (HA1s + HB1s), Be1s, Be2s, u = Be2pz &(HA1s - HB1s) u = Be2px & Be2py

  18. Polyatomic Molecules — BeH2 What is point group? What are the basis set AOs for determining MOs ? u = Be2px & Be2py u = Be2pz | (HA1s - HB1s) g = Be1s | Be2s | (HA1s + HB1s) HF SCF calculation : J. Chem. Phys. 1971 u* = C7 Be2pz- C8(HA1s - HB1s) g* = C5Be2s + C6(HA1s + HB1s) u = Be2px & Be2py u = C3Be2pz + C4 (HA1s - HB1s) g = C1Be2s + C2 (HA1s + HB1s) g = Be1s u = 0.44(Be2pz ) + 0.44 (HA1s - HB1s) g = -0.09(Be1s) + 0.40(Be2s) + 0.45 (HA1s + HB1s) g = 1.00(Be1s) + 0.016(Be2s) -0.002 (HA1s + HB1s)

  19. 2su* Be HA & HB 3sg* 1pu - 1su + 2sg 1sg

  20. HF 1s22s23s2(1p22p2)4s0 Semi-empirical treatment of HF from Spartan (AM1) 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 H1s F2p F2s One simpler treatment of HF is given in Atkins on page 428 gives the following results.... 4s = 0.98 (H1s) - 0.19(F2pz) -13.4 eV px = py = F2px and F2py -18.6 eV 3s = 0.19(H1s) + 0.98(F2pz) -18.8 eV 2s = F2s ~ -40.2 eV 1s = F1s << -40.2 eV

  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 CH4 - localized bonding MO J. Chem. Phys. 1967 C - HA MO = .... 0.02(C1s) + 0.292(C2s) + 0.277(C2px + C2py + C2pz) + 0.57(HA1s) - 0.07(HB1s + HC1s + HD1s)

More Related