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Structural Chemistry

课程教学组 : 吕鑫 教授 Tel: 2181600 (mobile/office) email: xinlu@xmu.edu.cn 化学楼 236 室 http://pcss.xmu.edu.cn/users/xlu/ 教辅： 王雪 wangxue@stu.xmu.edu.cn 张嘉伟 bosszjw@126.com 课程主页 : http://pcss.xmu.edu.cn/users/xlu/group/courses/structurechem/index.html. 参考书 :

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Structural Chemistry

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  1. 课程教学组: 吕鑫 教授 Tel: 2181600 (mobile/office) email: xinlu@xmu.edu.cn 化学楼236室 http://pcss.xmu.edu.cn/users/xlu/ 教辅： 王雪 wangxue@stu.xmu.edu.cn 张嘉伟 bosszjw@126.com 课程主页: http://pcss.xmu.edu.cn/users/xlu/group/courses/structurechem/index.html 参考书: <<结构化学>>厦门大学化学系物构组编 <<Physical Chemistry>> P.W. Atkins <<结构化学基础>> 周公度编著, 北京大学出版社 《化学键的本质》鲍林 上海科学技术出版社 《结构化学习题解析》周公度 段连运 课程QQ群: 8675318 (加入时请注明个人学号) Structural Chemistry

  2. 两点说明： • 作业要求 （交作业时间、理科作业规范） • 考勤、考试方式与成绩 （期中/期末/随堂考试，期平成绩等）

  3. What is Chemistry The branch of natural science that deals with composition, structure, properties of substances and the changes they undergo.

  4. Types of substances Geometric Structure Atoms Molecules Clusters Congeries Size makes the difference Nano materials Bulk materials Electronic Structure

  5. Structure vs. Properties Structure determines properties Properties reflect structures

  6. Inorganic Chemistry Organic Chemistry Catalysis Electrochemistry Bio-chemistry etc. Material Science Surface Science Life Science Energy Science Environmental Science etc. Structural Chemistry

  7. Role of Structural Chemistry in Surface Science

  8. Low-index surface： NCSLIS < NCB. fcc(100) fcc(111) • High-index surface： • Abundant edge sites. • NCes < NCSLIS < NCB • Lower NC ~ higher reactivity. fcc(775) fcc(10 8 7) Surface structures of Pt single crystal

  9. (111) (775) (100) (10 8 7) Different surfaces do different chemistry. Structure-sensitive Catalysis!

  10. 418 25 1 Another example of Structure-sensitive Catalysis Surface Structure vs. Catalytic Activity N2 + 3H2 2NH3 Fe single crystal， 20atm/700K

  11. Role of Structural Chemistry in Material Science

  12. C Crystal Structures • Graphite & Diamond Structures • Diamond:Insulator or wide bandgap • semiconductor:  • Graphite:Planar structure:  • sp2 bonding 2d metal (in plane) • Other Carbon allotropes • “Buckyballs” (C60, C70 etc)  • “Buckytubes” (nanotubes), other fullerenes  Structure makes the difference

  13. Zheng LS (郑兰荪), et al. Capturing the labile fullerene[50] as C50Cl10 SCIENCE 304 (5671): 699-699 APR 30 2004 The pentagon-pentagon fusions in pristine C50-D5h are sterically strained and highly reactive. Perchlorination of these active sites stabilizes the labile C50-D5h.

  14. Nature Materials, 2008, 7, 790.

  15. Chlorofullerenes featuring triple sequentially fused pentagonsXie, S.Y., Lu, X., Zheng, L.S. et alNature Chemistry, in press. #540C54Cl8 (a), #864C56Cl12(b), #4,169C66Cl6 (c), #4,169C66Cl10 (d).

  16. Endohedral Metallofullerene:Sc4@C82 (C3v) vs. Sc4C2@C80 (Ih) QM-predicted Proposed • (Sc2+)4@C828- DE = 0.0 kcal/mol C26-@(Sc3+)4@C806--Ih A Russian-Doll endofullerene Sc4@C82 DE = 28.8 kcal/mol X.Lu, J. Phys. Chem. B. 2006, 110, 11098; Lu & Wang, J. Am. Chem. Soc. 2009, 131, 16646. Highlighted by C&EN and Nat. Chem. Shinohara, H. Rep. Prog. Phys. 2000, 63, 843.

  17. Role of Structural Chemistry in Life Science

  18. What do proteins do ? • Proteins are the basis of how biology gets things done. • As enzymes, they are the driving force behind all of the biochemical reactions which makes biology work. • As structural elements, they are the main constituents of our bones, muscles, hair, skin and blood vessels. • As antibodies, they recognize invading elements and allow the immune system to get rid of the unwanted invaders.

  19. What are proteins made of ? • Proteins are necklaces of amino acids, i.e. long chain molecules.

  20. Definition of Structural Chemistry • It is a subject to study the microscopic structures of matters at the atomic/molecular level using Chemical Bond Theory. • Chemical bonds  structures  properties.

  21. Objective of Structural Chemistry • Determining the structure of a known substance • Understanding the structure-property relationship • Predicting a substance with specific structure and property

  22. Outline Chapter 1 basics of quantum mechanics 4 Chapter 2 Atomic structure 4 Chapter 3 Symmetry 3 Chapter 4 Diatomic molecules 3 Chapter 5/6 Polyatomic structures 5 Chapter 7 Basics of Crystallography 4 Chapter 8 Metals and Alloys 1 Chapter 9 Ionic compounds 3 Special talk

  23. Chapter 1 The basic knowledge of quantum mechanics • 1.1 The origin of quantum mechanics--- The failures of classical physics • Black-body radiation, Photoelectric effect, Atomic and molecular spectra • Classical physics: (prior to 1900) • Newtonian classical mechanics • Maxell’s theory of electromagnetic waves • Thermodynamics and statistical physics

  24. 1.1.1 Black-body radiation Device for the experimentation of black-body radiation.

  25. Black-Body Radiation A number of experiments revealed the temperature-dependence of lmax and independence on the substance made of the black-body device. Classical Solution: Rayleigh-Jeans Law (high energy, Low T) Wien Approximation (long wave length) Wave length / m It can not be explained by classical thermodynamics and statistical mechanics.

  26. The dawn of quantum mechanics

  27. 1.1.2 The photoelectric effect

  28. The photoelectric effect

  29. The Photoelectric Effect 1. The kinetic energy of the ejected electrons depends exclusivelyand linearly on the frequency of the light. 2. There is a particular threshold frequency for each metal. 3. The increase of the intensity of the light results in the increase of the number of photoelectrons.

  30. Classical physics: The energy of light wave should be directly proportional to intensity and not be affected by frequency.

  31. Explaining the Photoelectric Effect • Albert Einstein • Proposed a corpuscular theory of light (photons) in 1905. • won the Nobel prize in 1921 1. Light is consisted of a stream of photons. The energy of a photon is proportional to its frequency.  = hh = Planck’s constant 2. A photon has energy as well as mass. Mass-energy relationship:  = mc2 m= h/c2 3. A photon has a definite momentum. p=mc= h /c=h/ 4. The intensity of light depends on the photon density.

  32. Explaining the Photoelectric Effect • Therefore, the photon’s energy is the sum of thephotoelectron’s kinetic energy and the binding energy of the electron in metal. • Ephoton = E binding + E Kinetic energy • h=W + Ek

  33. Solution: Ephoton =hn = (6.626 x 10 -34Js)(9.73 x 109/s) = 6.447098 x 10 -24J Ephoton = 6.45 x 10 - 24 J Egamma ray =hn = ( 6.626 x 10-34Js )( 1.60 x 1020/s ) = 1.06 x 10 -13J Egamma ray = 1.06 x 10 - 13J Example I: Calculation Energy from Frequency Problem: What is the energy of a photon of electromagnetic radiation emitted by an FM radio station at 97.3 x 108 cycles/sec? What is the energy of a gamma ray emitted by Cs137 if it has a frequency of 1.60 x 1020/s? Plan: Use the relationship between energy and frequency to obtain the energy of the electromagnetic radiation (E = hn).

  34. c wavelength 3.00 x108 m/s 1.22 x 10 -1m frequency = = = 2.46 x 1010 /s Example II: Calculation of Energy from Wavelength Problem: What is the photon energy of electromagnetic radiation that is used in microwave ovens for cooking, if the wavelength of the radiation is 122 mm ? Plan: Convert the wavelength into meters, then the frequency can be calculated using the relationship;wavelength x frequency = c (where c is the speed of light), then using E=hn to calculate the energy. Solution: wavelength = 122 mm = 1.22 x 10 -1m Energy = E = hn = (6.626 x 10 -34Js)(2.46 x 1010/s) = 1.63 x 10 - 23 J

  35. Example III: Photoelectric Effect • The energy to remove an electron from potassium metal is 3.7 x 10 -19J. Will photons of frequencies of 4.3 x 1014/s (red light) and 7.5 x 1014 /s (blue light) trigger the photoelectric effect? • E red = hn = (6.626 x10 - 34Js)(4.3 x1014 /s) E red = 2.8 x 10 - 19 J • E blue = hn = (6.626 x10 - 34Js)(7.5x1014 /s) E blue = 5.0 x 10 - 19 J

  36. The binding energy of potassium is = 3.7 x 10 - 19 J • The red light will not have enough energy to knock an electron out of the potassium, but the blue light will eject an electron ! • E Total = E Binding Energy + EKinetic Energy of Electron • E Electron = ETotal - E Binding Energy • E Electron = 5.0 x 10 - 19J - 3.7 x 10 - 19 J = 1.3 x 10 - 19Joules

  37. 1.1.3 Atomic and molecular spectra

  38. The Line Spectra of Several Elements An atom can emit lights of discrete/specific frequencies upon electric/photo-stimulation.

  39. Planetary model: • First proposed by Rutherford in 1911. • The electrons are like planets of the solar system --- orbit the nucleus (the Sun). • Light of energy E given off when electrons change orbits of different energies. Based on classical physics, the electrons would be attracted by the nucleus and eventually fall into the nucleus by continuously emitting energy/light!! Why do the electrons not fall into the nucleus? Why are they in discrete energies?

  40. Bohr’s atomic model • Niels Bohr, a Danish physicist, combined the Plank’s quanta idea, Einstein’s photon theory and Rutherford’s Planetary model, and first introduced the idea of electronic energy level into atomic model. • Quantum Theory of Energy. • The energy levels in atoms can be pictured as orbits in which electrons travel at definite distances from the nucleus. • These he called “quantized energy levels”, also known as principal energy levels. n :principal quantum number

  41. The electron in H atom can be promoted to higher energy levels by photons or electricity.

  42. The Energy States of the Hydrogen Atom Bohr derived the energy for a system consisting of a nucleus plus a single electron eg. He predicted a set of quantized energy levels given by : - R is called the Rydberg constant (2.18 x 10-18 J) - n is a quantum number - Z is the nuclear charge

  43. 1 n12 1 n22 Ephoton = -2.18 x10 -18J - 1 4 2 1 2 2 Ephoton = -2.18 x 10 -18J - = 4.09 x 10 -19J h c E (6.626 x 10 -34Js)( 3.00 x 108 m/s)  = = 4.09 x 10 -19J = 4.87 x 10 -7 m = 487 nm Problem: Find the energy change when an electron changes from the n=4 level to the n=2 level in the hydrogen atom? What is the wavelength of this photon? Plan: Use the Rydberg equation to calculate the energy change, then calculate the wavelength using the relationship of the speed of light. Solution:

  44. 1.1.2 The photoelectric effect A corpuscular theory of light (photons)  = hh = Planck’s constant p=h/ Lesson 1 Summary 1.1 The failures of classical physics 1.1.1 Black-Body Radiation Planck’s quanta idea E= nhn 1.1.3 Atomic and molecular spectra Planetary model: orbits of electrons around the nucleus Bohr’s atomic model: quantized energy levels of orbits

  45. 1.2 The characteristic of the motion of microscopic particles

  46. 1.2.1 The wave-particle duality of microscopic particles In classical physics, waves and particles behave differently and can be described by rather different theories. As mentioned above, Einstein’s Corpuscular Theory of Lights (photons) to explain the photoelectric effect for the first time introduced the wave-particle duality of photon. In 1924 de Broglie suggested that microscopic particles such as electron and proton might also have wave properties in addition to their partical properties.

  47. h = Planck’s constant, p = particle momentum, = de Broglie wavelength E=h p=h/ De Broglie considered that the wave-particle relationship in light is also applicable to particles of matter, i.e. • The latter equation clearly reflects the wave-particle duality by involving both a particle property (momentum) and a wave property (wavelength)! • The wavelength of a particle could be determined by • =h/p = h/mv (v: velocity, m: mass)

  48. h mv 6.626 x 10 - 34Js ( 9.11 x 10 - 31kg )( 1.00 x 106 m/s )  = = kg m2 s2 J = hence de Broglie Wavelength Example: Calculate the de Broglie wavelength of an electron with speed 3.00 x 106 m/s. electron mass = 9.11 x 10 -31 kg velocity = 3.00 x 106 m/s Wavelength  = 2.42 x 10 -10 m = 0.242 nm

  49. The moving speed of an electron is determined by the potential difference of the electric field (V) 1eV=1.602x10-19 J If the unit of V is volt, then the wavelength is:

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