1. Copyright © Cengage Learning. All rights reserved.7 | 1 Contents and Concepts Light Waves, Photons, and the Bohr Theory To understand the formation of chemical bonds, you need to know something about the electronic structure of atoms. Because light gives us information about this structure, we begin by discussing the nature of light. Then we look at the Bohr theory of the simplest atom, hydrogen. 1. The Wave Nature of Light 2. Quantum Effects and Photons 3. The Bohr Theory of the Hydrogen Atom

2. Copyright © Cengage Learning. All rights reserved.7 | 2 Quantum Mechanics and Quantum Numbers The Bohr theory firmly establishes the concept of energy levels but fails to account for the details of atomic structure. Here we discuss some basic notions of quantum mechanics, which is the theory currently applied to extremely small particles, such as electrons in atoms. 4. Quantum Mechanics 5. Quantum Numbers and Atomic Orbitals

3. 3 Maxwell (1873), proposed that visible light consists of electromagnetic waves. Electromagnetic radiation is the emission and transmission of energy in the form of electromagnetic waves. Speed of light (c) in vacuum = 3.00 x 10 8 m/s All electromagnetic radiation x  c

4. 4 Properties of Waves Wavelength ( ) is the distance between identical points on successive waves. Amplitude is the vertical distance from the midline of a wave to the peak or trough. Frequency ( ) is the number of waves that pass through a particular point in 1 second (Hz = 1 cycle/s). The speed (u) of the wave = x

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6. Copyright © Cengage Learning. All rights reserved.7 | 6 The range of frequencies and wavelengths of electromagnetic radiation is called the electromagnetic spectrum.

7. Copyright © Cengage Learning. All rights reserved.7 | 7 When the wavelength is reduced by a factor of two, the frequency increases by a factor of two.

8. Copyright © Cengage Learning. All rights reserved. 7 | 8 What is the wavelength of blue light with a frequency of 6.4  10 14 /s? = 6.4  10 14 /s c = 3.00  10 8 m/s c = so = c/ = 4.7  10 −7 m

9. ? Copyright © Cengage Learning. All rights reserved.7 | 9 What is the frequency of light having a wavelength of 681 nm? = 681 nm = 6.81  10 −7 m c = 3.00  10 8 m/s c = so = c/ = 4.41  10 14 /s

10. ? 10 x = c = c/ = 3.00 x 10 8 m/s / 6.0 x 10 4 Hz = 5.0 x 10 3 m A photon has a frequency of 6.0 x 10 4 Hz. Convert this frequency into wavelength (nm). Does this frequency fall in the visible region? = 5.0 x 10 12 nm

11. Copyright © Cengage Learning. All rights reserved.7 | 11 The photoelectric effect is the ejection of an electron from the surface of a metal or other material when light shines on it.

12. Copyright © Cengage Learning. All rights reserved.7 | 12 Einstein proposed that light consists of quanta or particles of electromagnetic energy, called photons. The energy of each photon is proportional to its frequency: E = h h = 6.626  10 −34 J  s (Planck’s constant)

13. Copyright © Cengage Learning. All rights reserved.7 | 13 Light, therefore, has properties of both waves and matter. Neither understanding is sufficient alone. This is called the particle–wave duality of light.

14. 14 E = h x E = 6.63 x 10 -34 (J s) x 3.00 x 10 8 (m/s) / 0.154 x 10 -9 (m) E = 1.29 x 10 -15 J E = h x c /  When copper is bombarded with high-energy electrons, X rays are emitted. Calculate the energy (in joules) associated with the photons if the wavelength of the X rays is 0.154 nm.

15. ? Copyright © Cengage Learning. All rights reserved.7 | 15 The blue–green line of the hydrogen atom spectrum has a wavelength of 486 nm. What is the energy of a photon of this light? = 486 nm = 4.86  10 −7 m c = 3.00  10 8 m/s h = 6.63  10 −34 J  s E = h and c = so E = hc/ = 4.09  10 −19 J

16. Copyright © Cengage Learning. All rights reserved.7 | 16 Recall Rutherford’s model of the atom. This explanation left a theoretical dilemma: -an electrically charged particle circling a center would continually lose energy as electromagnetic radiation. -But this is not the case—atoms are stable.

17. Copyright © Cengage Learning. All rights reserved.7 | 17 In addition, this understanding could not explain the observation of line spectra of atoms. A continuous spectrum contains all wavelengths of light. A line spectrum shows only certain colors or specific wavelengths of light. When atoms are heated, they emit light. This process produces a line spectrum that is specific to that atom. The emission spectra of six elements are shown on the next slide.

18. Copyright © Cengage Learning. All rights reserved.7 | 18

19. Copyright © Cengage Learning. All rights reserved.7 | 19 In 1913, Neils Bohr, a Danish scientist, set down postulates to account for 1. The stability of the hydrogen atom 2. The line spectrum of the atom

20. 20 1.e - can only have specific (quantized) energy values 2.light is emitted as e - moves from one energy level to a lower energy level Bohr’s Model of the Atom (1913) E n = -R H ( ) 1 n2n2 n (principal quantum number) = 1,2,3,… R H (Rydberg constant) = 2.18 x 10 -18 J

21. Copyright © Cengage Learning. All rights reserved.7 | 21 Energy-Level Postulate An electron can have only certain energy values, called energy levels. Energy levels are quantized. For an electron in a hydrogen atom, the energy is given by the following equation: R H = 2.179  10 −18 J n = principal quantum number

22. 22 E = h

23. Copyright © Cengage Learning. All rights reserved.7 | 23 Transitions Between Energy Levels An electron can change energy levels by absorbing energy to move to a higher energy level or by emitting energy to move to a lower energy level.

24. Copyright © Cengage Learning. All rights reserved.7 | 24 For a hydrogen electron, the energy change is given by R H = 2.179  10 −8 J, Rydberg constant

25. Copyright © Cengage Learning. All rights reserved.7 | 25 The energy of the emitted or absorbed photon is related to  E: We can now combine these two equations:

26. Copyright © Cengage Learning. All rights reserved.7 | 26 Light is absorbed by an atom when the electron transition is from lower n to higher n (n f > n i ). In this case,  E will be positive. Light is emitted from an atom when the electron transition is from higher n to lower n (n f < n i ). In this case,  E will be negative. An electron is ejected when n f = ∞.

27. Copyright © Cengage Learning. All rights reserved.7 | 27 Energy-level diagram for the hydrogen atom.

28. Copyright © Cengage Learning. All rights reserved.7 | 28 Electron transitions for an electron in the hydrogen atom.

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30. ? Copyright © Cengage Learning. All rights reserved.7 | 30 What is the wavelength of the light emitted when the electron in a hydrogen atom undergoes a transition from n = 6 to n = 3? n i = 6 n f = 3 R H = 2.179  10 −18 J = −1.816  10 −19 J 1.094  10 −6 m

31. Copyright © Cengage Learning. All rights reserved.7 | 31 Planck Vibrating atoms have only certain energies: E = h or 2h or 3h Einstein Energy is quantized in particles called photons: E = h Bohr Electrons in atoms can have only certain values of energy. For hydrogen:

32. Copyright © Cengage Learning. All rights reserved.7 | 32 Light has properties of both waves and particles (matter). What about matter?

33. Copyright © Cengage Learning. All rights reserved.7 | 33 In 1923, Louis de Broglie, a French physicist, reasoned that particles (matter) might also have wave properties. The wavelength of a particle of mass, m (kg), and velocity, v (m/s), is given by the de Broglie relation:

34. 34 = h/mu = 6.63 x 10 -34 / (2.5 x 10 -3 x 15.6) = 1.7 x 10 -32 m = 1.7 x 10 -23 nm What is the de Broglie wavelength (in nm) associated with a 2.5 g Ping-Pong ball traveling at 15.6 m/s? m in kgh in J su in (m/s)

35. ? Copyright © Cengage Learning. All rights reserved.7 | 35 Compare the wavelengths of (a) an electron traveling at a speed that is one-hundredth the speed of light and (b) a baseball of mass 0.145 kg having a speed of 26.8 m/s (60 mph). Electron m e = 9.11  10 −31 kg v = 3.00  10 6 m/s Baseball m = 0.145 kg v = 26.8 m/s

36. Copyright © Cengage Learning. All rights reserved.7 | 36 Electron m e = 9.11  10 −31 kg v = 3.00  10 6 m/s Baseball m = 0.145 kg v = 26.8 m/s 2.43  10 −10 m 1.71  10 −34 m

37. 37 Chemistry in Action: Electron Microscopy STM image of iron atoms on copper surface e = 0.004 nm Electron micrograph of a normal red blood cell and a sickled red blood cell from the same person

38. Copyright © Cengage Learning. All rights reserved.7 | 38 Building on de Broglie’s work, in 1926, Erwin Schrödinger devised a theory that could be used to explain the wave properties of electrons in atoms and molecules. The branch of physics that mathematically describes the wave properties of submicroscopic particles is called quantum mechanics or wave mechanics.

39. Copyright © Cengage Learning. All rights reserved.7 | 39 Quantum mechanics alters how we think about the motion of particles. In 1927, Werner Heisenberg showed how it is impossible to know with absolute precision both the position, x, and the momentum, p, of a particle such as electron. Because p = mv this uncertainty becomes more significant as the mass of the particle becomes smaller.

40. 40 Schrodinger Wave Equation In 1926 Schrodinger wrote an equation that described both the particle and wave nature of the e - Wave function (  ) describes: 1. energy of e - with a given  2. probability of finding e - in a volume of space Schrodinger’s equation can only be solved exactly for the hydrogen atom. Must approximate its solution for multi-electron systems.

41. Copyright © Cengage Learning. All rights reserved.7 | 41 The wave function for the lowest level of the hydrogen atom is shown to the left. Note that its value is greatest nearest the nucleus, but rapidly decreases thereafter. Note also that it never goes to zero, only to a very small value.

42. Copyright © Cengage Learning. All rights reserved.7 | 42

43. Copyright © Cengage Learning. All rights reserved.7 | 43 According to quantum mechanics, each electron is described by four quantum numbers: 1. Principal quantum number (n) 2. Angular momentum quantum number ( l ) 3. Magnetic quantum number (m l ) 4. Spin quantum number (m s ) The first three define the wave function for a particular electron. The fourth quantum number refers to the magnetic property of electrons.

44. 44 Schrodinger Wave Equation  is a function of four numbers called quantum numbers (n, l, m l, m s ) principal quantum number n n = 1, 2, 3, 4, …. n=1 n=2 n=3 distance of e - from the nucleus

45. 45 Where 90% of the e - density is found for the 1s orbital

46. Copyright © Cengage Learning. All rights reserved.7 | 46 Shells are sometimes designated by uppercase letters: Letter n K1K1 L2L2 M3M3 N4N4...

47. Copyright © Cengage Learning. All rights reserved.7 | 47 Angular Momentum Quantum Number, l This quantum number distinguishes orbitals of a given n (shell) having different shapes. It can have values from 0, 1, 2, 3,... to a maximum of (n – 1). For a given n, there will be n different values of l, or n types of subshells. Orbitals with the same values for n and l are said to be in the same shell and subshell.

48. 48 quantum numbers: (n, l, m l, m s ) angular momentum quantum number l for a given value of n, l = 0, 1, 2, 3, … n-1 n = 1, l = 0 n = 2, l = 0 or 1 n = 3, l = 0, 1, or 2 Shape of the “volume” of space that the e - occupies l = 0 s orbital l = 1 p orbital l = 2 d orbital l = 3 f orbital Schrodinger Wave Equation

49. 49 l = 0 (s orbitals) l = 1 (p orbitals)

50. 50 l = 2 (d orbitals)

51. Copyright © Cengage Learning. All rights reserved.7 | 51 Subshells are sometimes designated by lowercase letters: l Letter 0s0s 1p1p 2d2d 3f3f... n ≥n ≥1234 Not every subshell type exists in every shell. The minimum value of n for each type of subshell is shown above.

52. Copyright © Cengage Learning. All rights reserved.7 | 52 Magnetic Quantum Number, m l This quantum number distinguishes orbitals of a given n and l —that is, of a given energy and shape but having different orientations. The magnetic quantum number depends on the value of l and can have any integer value from – l to 0 to + l. Each different value represents a different orbital. For a given subshell, there will be (2 l + 1) values and therefore (2 l + 1) orbitals.

53. 53 quantum numbers: (n, l, m l, m s ) magnetic quantum number m l for a given value of l m l = -l, …., 0, …. +l orientation of the orbital in space if l = 1 (p orbital), m l = -1, 0, or 1 if l = 2 (d orbital), m l = -2, -1, 0, 1, or 2 Schrodinger Wave Equation

54. 54 m l = -1, 0, or 1 3 orientations is space

55. 55 m l = -2, -1, 0, 1, or 25 orientations is space

56. 56 (n, l, m l, m s ) spin quantum number m s m s = +½ or -½ Schrodinger Wave Equation m s = -½m s = +½

57. 57 Existence (and energy) of electron in atom is described by its unique wave function . Pauli exclusion principle - no two electrons in an atom can have the same four quantum numbers. Schrodinger Wave Equation quantum numbers: (n, l, m l, m s ) Each seat is uniquely identified (E, R12, S8) Each seat can hold only one individual at a time

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59. 59 Energy of orbitals in a single electron atom Energy only depends on principal quantum number n E n = -R H ( ) 1 n2n2 n=1 n=2 n=3

60. 60 Energy of orbitals in a multi-electron atom Energy depends on n and l n=1 l = 0 n=2 l = 0 n=2 l = 1 n=3 l = 0 n=3 l = 1 n=3 l = 2

61. 61 “Fill up” electrons in lowest energy orbitals (Aufbau principle) H 1 electron H 1s 1 He 2 electrons He 1s 2 Li 3 electrons Li 1s 2 2s 1 Be 4 electrons Be 1s 2 2s 2 B 5 electrons B 1s 2 2s 2 2p 1 C 6 electrons ??

62. 62 C 6 electrons The most stable arrangement of electrons in subshells is the one with the greatest number of parallel spins (Hund’s rule). C 1s 2 2s 2 2p 2 N 7 electrons N 1s 2 2s 2 2p 3 O 8 electrons O 1s 2 2s 2 2p 4 F 9 electrons F 1s 2 2s 2 2p 5 Ne 10 electrons Ne 1s 2 2s 2 2p 6

63. 63 Order of orbitals (filling) in multi-electron atom 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s

64. 64 Electron configuration is how the electrons are distributed among the various atomic orbitals in an atom. 1s 1 principal quantum number n angular momentum quantum number l number of electrons in the orbital or subshell Orbital diagram H 1s 1

65. 65 What is the electron configuration of Mg? Mg 12 electrons 1s < 2s < 2p < 3s < 3p < 4s 1s 2 2s 2 2p 6 3s 2 2 + 2 + 6 + 2 = 12 electrons Abbreviated as [Ne]3s 2 [Ne] 1s 2 2s 2 2p 6 What are the possible quantum numbers for the last (outermost) electron in Cl? Cl 17 electrons1s < 2s < 2p < 3s < 3p < 4s 1s 2 2s 2 2p 6 3s 2 3p 5 2 + 2 + 6 + 2 + 5 = 17 electrons Last electron added to 3p orbital n = 3l = 1m l = -1, 0, or +1m s = ½ or -½

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67. 67 Paramagnetic unpaired electrons 2p Diamagnetic all electrons paired 2p