Quantum Mechanics & Molecular Structure

1. Quantum Mechanics & Molecular Structure Quantum Mechanics : Quantum mechanics is the foundation of all chemistry and biology. Statistical mechanics rests on the foundation of quantum mechanics and provides the basis of thermodynamics. “The ultimate aim of the modern movement in biology is in fact to explain all biology in terms of Physics and Chemistry….. Quantum mechanics, together with our empirical knowledge of chemistry, appears to provide us with a ‘foundation of certainty’ on which to build biology.” Francis Crick Schrodinger equation : Y(x,y,z,t) Y2(x,y,z,t)DxDyDz : HY = EY

2. Quantum Mechanics can be extended to explains and predicts Lewis model and VSEPR model rationalized the structure and bonding qualitatively. . Molecular Orbital (MO) – Quantal Picture of chemical bond CH4 H2

3. Quantum Mechanical solution of molecules : 1. Born-Oppenheimer Approximation : Frozen Nuclei, Fleeing Electrons Then, 1) consider the nuclei “frozen” and solve the Schrodinger Eqn. for electrons. Then, 2) consider the function Eel(Rnu) to be the potential energy function fo nuclei. CH4 H2

4. Exact solution for Molecular Orbitals: This is the case that the exact solution can be obtained with Born-Oppenheimer approximation.

5. Exact solution for Molecular Orbitals: H2+ ion Energy level.

6. 2. LCAO Approximation : The result of H2+ indicates that, for sg1s MO, when the electron is near one nucleus A the ground state wave function resembles a 1s atomic wave function YA1s when the electron is near one nucleus A the ground state wave function resembles a 1s atomic wave function YB1s For all the other Mo’s we can apply the same approximation and LCAO can predict Homonuclear Diatomic Molecules : H2, He2 LCAO: Ymol = CAYA1s + CBYB1s For homodinuclear molecule, then , wave functions are electron probability becomes cf. electron probability of none interacting system

7. Correlation diagram Energy level diagram For H2+

8. Correlation diagram Energy level diagram For H2

9. For He+2 Stabilization through the formatioin of molecule is DE Bond order = Summary : in forming LCAO

10. Homonuclear Diatomic Molecules : Li2, Be2, B2, C2, N2, O2, F2, Ne2 LCAO: Ymol = CA1YA1s + CB1YB1s + CA2YA2s + CB2YB2s + CA3YA2px + ………….. Another approximation • Two atomic orbitals contribute significantly to bond formation only if

11. Homonuclear Diatomic Molecules : Li2, Be2, B2, C2, N2, O2, F2, Ne2 LCAO: Ymol = CA1YA1s + CB1YB1s + CA2YA2s + CB2YB2s + CA3YA2px + ………….. Another approximation • . • Two atomic orbitals contribute significantly to bond formation only if

12. Exact solution for Molecular Orbitals: H2+ ion Energy level. Antibonding orbital Bonding orbital

13. Homonuclear Diatomic Molecules : Li2, Be2, B2, C2, N2, O2, F2, Ne2 LCAO: Ymol = CA1YA1s + CB1YB1s + CA2YA2s + CB2YB2s + CA3YA2px + ………….. Another approximation • Two atomic orbitals contribute significantly to bond formation only if their atomic energy levels are very close. • Two atomic orbitals contribute significantly to bond formation only if they have substantial overlap. • Molecular orbitals form only with atomic orbitals of the most outer shell.

14. Molecular Orbitals from p-orbitals

17. Another approximation • Two atomic orbitals contribute significantly to bond formation only if their atomic energy levels are very close. • Two atomic orbitals contribute significantly to bond formation only if they have substantial overlap. • Molecular orbitals form only with atomic orbitals of the most outer shell. • the average energy level of a bonding-antibonding pair of molecular orbital is close to the original atomic orbitals • the energy difference between a bonding-antibonding pair becomes greater as the overlap of the atomic orbital increases.

18. Magnetic properties Paramagnetic – Diamagnetic –

19. For O, F, Ne For Li, Be, B, C, N

21. Heteronuclear Diatomic Molecules : CO, CN, NO, NF, etc… HF • Two atomic orbitals contribute significantly to bond formation only if their atomic energy levels are very close. • Two atomic orbitals contribute significantly to bond formation only if they have substantial overlap.

22. Molecular Orbitals in Polyatomic Molecules : Valence Bond Model Make BeH2 From VSEPR, we know BeH2 is a linear molecule The electronic configuration of BeH2 is from

25. Rules for Hybridization Example : BeH2 1. Assign Geometry using VSEPR Theory BeH2 SN=2 geometry = linear 2. Write electronic configuration of the atom to be hybridized Be: (1s)2(2s)2(2p)0 3. Draw energy diagram from said atom and “decouple” paired electrons 4. Take linear combination of the atomic orbitals participating in the bond to make hybrid orbitals. 5. Combine hybrid orbitals with other atom’s orbitals using diatomic MO theory.

26. Make BH3 1. Assign Geometry using VSEPR Theory BH3 SN=3 geometry = trigonal planar 2. Write electronic configuration of the atom to be hybridized B: (1s)2(2s)2(2p)1 3. Draw energy diagram from said atom and “decouple” paired electrons sp2 Hybrid

27. 4. Take linear combination of the atomic orbitals participating in the bond to make hybrid orbitals. 5. Combine hybrid orbitals with other atom’s orbitals using diatomic MO theory.

28. Make CH4 1. Assign Geometry using VSEPR Theory CH4 SN=4 geometry = tetrahedral 2. Write electronic configuration of the atom to be hybridized C: (1s)2(2s)2(2p)2 3. Draw energy diagram from said atom and “decouple” paired electrons sp3 Hybrid

29. 4. Take linear combination of the atomic orbitals participating in the bond to make hybrid orbitals. 5. Combine hybrid orbitals with other atom’s orbitals using diatomic MO theory.

30. Triatomic nonhydride : follow the same hybridization rules for the central atom + p orbitals from others Make CO2 1. Assign Geometry using VSEPR Theory CH4 SN=2 geometry = linear 2. Write electronic configuration of the atom to be hybridized C: (1s)2(2s)2(2p)2 3. Draw energy diagram from said atom and “decouple” paired electrons sp Hybrid 4. Combine hybrid orbitals with other atom’s orbitals using diatomic MO theory. 5. The remaining p orbitals are all perpendicular to s bonding orbital.

31. Remaining p orbitals of CO2

32. Triatomic nonhydride : follow the same hybridization rules for the central atom + p orbitals from others Make NO2- 1. Assign Geometry using VSEPR Theory SN=3 geometry = trigonal planar 2. Write electronic configuration of the atom to be hybridized N: (1s)2(2s)2(2p)3 3. Draw energy diagram from said atom and “decouple” paired electrons sp2 Hybrid 4. Combine hybrid orbitals with other atom’s orbitals using diatomic MO theory. One p orbital from oxygen atom makes s bond. That leaves one p orbital from N and one p orbital from O as unpaired electrons 5. The remaining p orbitals are all perpendicular to s bonding orbital. That makes p bond.

33. Triatomic nonhydride : follow the same hybridization rules for the central atom + p orbitals from others Make NO2- 1. Assign Geometry using VSEPR Theory SN=3 geometry = bent 2. Write electronic configuration of the atom to be hybridized N: (1s)2(2s)2(2p)3 3. Draw energy diagram from said atom and “decouple” paired electrons sp2 Hybrid 4. Combine hybrid orbitals with other atom’s orbitals using diatomic MO theory. One p orbital from oxygen atom makes s bond. That leaves one p orbital from N and one p orbital from O as unpaired electrons 5. The remaining p orbitals are all perpendicular to s bonding orbital. That makes p bond. Overall, NO2- makes two s bonds and one p bonds. Total bond order is 3, bond order for each bond is 1.5. clearly bent structure.

34. Structure and bonding in Organic Molecules Make C2H6 Analyze one carbon at a time. C has SN=4 i.e. it’s the same as CH4 sp3 hybridize the C add hydrogens Do the same to the other C Bring these two together All bonds are s bonds, and freely rotates along the axis.

35. Make C2H4 From VSEPR Analyze one carbon at a time. C has SN=3 i.e. it’s the same as BH3 Do the same to the other C sp2 hybridize with extra p orbital add hydrogens Bring these two together sp2 orbitals form s bonds.

36. Make C2H4 Remaining p orbitals form p bond. If the bond rotates….. Therefore all atoms in the molecule are constrained to lie in a plain.

37. Structural isomerizm For CH3CHCHCH3 two structures are possible and they are not interconvertible! geometrical isomer: optical isomer (chirality) :

38. Make C2H4 From VSEPR Analyze one carbon at a time. C has SN=2 i.e. it’s the same as BeH2 sp hybridize with two extra p orbital add hydrogens Do the same to the other C Bring these two together sp orbitals form s bonds, and rotationally symmetric.

39. Make C2H2 Remaining p orbitals form p bond.

40. Conjugated molecules When two or more double bonds or triples bonds occur close to each other in a molecule, For CH2CHCHCH2

41. C6H6 Benzene ultimate conjugated system. From VSEPR model, benzene has resonance structures Benzene has 6 sp2 centers P orbitals form all delocalized p bonding orbitals ( total 6 molecular orbitals)

42. p molecular orbitals of benzene ( total 6) p molecular orbitals of CH2CHCHCHCHCH2 LUMO HOMO HOMO-LUMO gap of benzene is much larger than hexatriene. i.e. extra stable. DH= -29.2 Kcal/mol No reaction ! DH= 2 Kcal/mol

43. Fullerenes : represented by buckminsterfullerene, C60 An allotrope of C Found in universe 20 hexagons + 12 pentagons all sp2 carbons 30 p orbitals Fullerenes can have more than 400 carbons. electronically useful – superconductivity, wire in nano-science can trap other molecules or ion inside. can form nano-tubes

44. 숙제 16장 : 8, 14, 16, 28, 34, 36, 42 제출기한 : 9월 26일