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 :  (x,y,z,t)  2 (x,y,z,t)  x  y  z : H  = E 

2. Quantum Mechanics can be extended to explains and predicts Molecular Orbital (MO) – Quantal Picture of chemical bond Lewis model and VSEPR model rationalized the structure and bonding qualitatively.. H2H2 CH 4

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 E el (R nu ) to be the potential energy function fo nuclei. H2H2 CH 4

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: H 2 + ion Energy level.

6. 2. LCAO Approximation : The result of H 2 + indicates that, for  g1s MO, when the electron is near one nucleus A the ground state wave function resembles a 1s atomic wave function  A 1s when the electron is near one nucleus A the ground state wave function resembles a 1s atomic wave function  B 1s For all the other Mo’s we can apply the same approximation and LCAO can predict Homonuclear Diatomic Molecules : H 2, He 2 LCAO:  mol = C A  A 1s  C B  B 1s For homodinuclear molecule, then, wave functions are electron probability becomes cf. electron probability of none interacting system

7. Correlation diagramEnergy level diagram For H 2 +

8. Correlation diagramEnergy level diagram For H 2

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

10. Homonuclear Diatomic Molecules : Li 2, Be 2, B 2, C 2, N 2, O 2, F 2, Ne 2 LCAO:  mol = C A1  A 1s  C B1  B 1s + C A2  A 2s + C B2  B 2s + C A3  A 2px + ………….. Another approximation 1.Two atomic orbitals contribute significantly to bond formation only if

11. Homonuclear Diatomic Molecules : Li 2, Be 2, B 2, C 2, N 2, O 2, F 2, Ne 2 LCAO:  mol = C A1  A 1s  C B1  B 1s + C A2  A 2s + C B2  B 2s + C A3  A 2px + ………….. Another approximation. Two atomic orbitals contribute significantly to bond formation only if

12. Bonding orbital Antibonding orbital Exact solution for Molecular Orbitals: H 2 + ion Energy level.

13. Homonuclear Diatomic Molecules : Li 2, Be 2, B 2, C 2, N 2, O 2, F 2, Ne 2 LCAO:  mol = C A1  A 1s  C B1  B 1s + C A2  A 2s + C B2  B 2s + C A3  A 2px + ………….. Another approximation 1.Two atomic orbitals contribute significantly to bond formation only if their atomic energy levels are very close. 2.Two atomic orbitals contribute significantly to bond formation only if they have substantial overlap. 3.Molecular orbitals form only with atomic orbitals of the most outer shell.

14. Molecular Orbitals from p-orbitals

17. Another approximation 1.Two atomic orbitals contribute significantly to bond formation only if their atomic energy levels are very close. 2.Two atomic orbitals contribute significantly to bond formation only if they have substantial overlap. 3.Molecular orbitals form only with atomic orbitals of the most outer shell. 1.the average energy level of a bonding-antibonding pair of molecular orbital is close to the original atomic orbitals 2.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 Li, Be, B, C, N For O, F, Ne

21. Heteronuclear Diatomic Molecules : CO, CN, NO, NF, etc… 1.Two atomic orbitals contribute significantly to bond formation only if their atomic energy levels are very close. 2.Two atomic orbitals contribute significantly to bond formation only if they have substantial overlap. HF

22. Molecular Orbitals in Polyatomic Molecules : Valence Bond Model Make BeH 2 From VSEPR, we know BeH 2 is a linear molecule The electronic configuration of BeH 2 is from

25. Rules for Hybridization Example : BeH 2 1. Assign Geometry using VSEPR Theory BeH 2 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. 1. Assign Geometry using VSEPR Theory BH 3 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 sp 2 Hybrid Make BH 3

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. 1. Assign Geometry using VSEPR Theory CH 4 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 sp 3 Hybrid Make CH 4

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 CO 2 1. Assign Geometry using VSEPR Theory CH 4 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  bonding orbital.

31. Remaining p orbitals of CO 2

32. Triatomic nonhydride : follow the same hybridization rules for the central atom + p orbitals from others Make NO 2 - 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 sp 2 Hybrid 4. Combine hybrid orbitals with other atom’s orbitals using diatomic MO theory. One p orbital from oxygen atom makes  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  bonding orbital. That makes  bond.

33. Triatomic nonhydride : follow the same hybridization rules for the central atom + p orbitals from others Make NO 2 - 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 sp 2 Hybrid 4. Combine hybrid orbitals with other atom’s orbitals using diatomic MO theory. One p orbital from oxygen atom makes  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  bonding orbital. That makes  bond. Overall, NO 2 - makes two  bonds and one  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 C 2 H 6 Analyze one carbon at a time. C has SN=4 i.e. it’s the same as CH 4 sp 3 hybridize the Cadd hydrogens Do the same to the other C Bring these two together All bonds are  bonds, and freely rotates along the axis.

35. Make C 2 H 4 Analyze one carbon at a time. C has SN=3 i.e. it’s the same as BH 3 From VSEPR Do the same to the other C Bring these two together sp 2 orbitals form  bonds. sp 2 hybridize with extra p orbitaladd hydrogens

36. Make C 2 H 4 Remaining p orbitals form  bond. If the bond rotates….. Therefore all atoms in the molecule are constrained to lie in a plain.

37. Structural isomerizm For CH 3 CHCHCH 3 two structures are possible and they are not interconvertible! geometrical isomer: optical isomer (chirality) :

38. Make C 2 H 4 Analyze one carbon at a time. C has SN=2 i.e. it’s the same as BeH 2 From VSEPR Do the same to the other C Bring these two together sp orbitals form  bonds, and rotationally symmetric. sp hybridize with two extra p orbitaladd hydrogens

39. Make C 2 H 2 Remaining p orbitals form  bond.

40. Conjugated molecules When two or more double bonds or triples bonds occur close to each other in a molecule, For CH 2 CHCHCH 2

41. C 6 H 6 Benzene ultimate conjugated system. From VSEPR model, benzene has resonance structures Benzene has 6 sp 2 centers P orbitals form all delocalized  bonding orbitals ( total 6 molecular orbitals)

42.  molecular orbitals of benzene ( total 6)  molecular orbitals of CH 2 CHCHCHCHCH 2 LUMO HOMO HOMO-LUMO gap of benzene is much larger than hexatriene. i.e. extra stable.  H= -29.2 Kcal/mol No reaction !  H= 2 Kcal/mol

43. Fullerenes : represented by buckminsterfullerene, C 60 An allotrope of C Found in universe 20 hexagons + 12 pentagons all sp 2 carbons 30  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 16 장 : 8, 14, 16, 28, 34, 36, 42 제출기한 : 9 월 26 일