1. Chapter 7 Quantum Theory of the Atom

2. Copyright © Cengage Learning. All rights reserved.7 | 2 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

3. Copyright © Cengage Learning. All rights reserved.7 | 3 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

4. Copyright © Cengage Learning. All rights reserved.7 | 4 Learning Objectives Light Waves, Photons, and the Bohr Theory 1.The Wave Nature of Light a.Define the wavelength and frequency of a wave. b.Relate the wavelength, frequency, and speed of light. c.Describe the different regions of the electromagnetic spectrum.

5. Copyright © Cengage Learning. All rights reserved.7 | 5 2.Quantum Effects and Photons a.State Planck’s quantization of vibrational energy. b.Define Planck’s constant and photon. c.Describe the photoelectric effect. d.Calculate the energy of a photon from its frequency or wavelength.

6. Copyright © Cengage Learning. All rights reserved.7 | 6 3.The Bohr Theory of the Hydrogen Atom a.State the postulates of Bohr’s theory of the hydrogen atom. b.Relate the energy of a photon to the associated energy levels of an atom. c.Determine the wavelength or frequency of a hydrogen atom transition. d.Describe the difference between emission and absorption of light by an atom.

7. Copyright © Cengage Learning. All rights reserved.7 | 7 Quantum Mechanics and Quantum Numbers 4.Quantum Mechanics a.State the de Broglie relation. b.Calculate the wavelength of a moving particle. c.Define quantum mechanics. d.State Heisenberg’s uncertainty principle. e.Relate the wave function for an electron to the probability of finding the electron at a location in space.

8. Copyright © Cengage Learning. All rights reserved.7 | 8 5.Quantum Numbers and Atomic Orbitals a.Define atomic orbital. b.Define each of the quantum numbers for an atomic orbital. c.State the rules for the allowed values for each quantum number. d.Apply the rules for quantum numbers. e.Describe the shapes of s, p, and d orbitals.

9. Copyright © Cengage Learning. All rights reserved.7 | 9 A wave is a continuously repeating change or oscillation in matter or in a physical field. Light is an electromagnetic wave, consisting of oscillations in electric and magnetic fields traveling through space.

10. Copyright © Cengage Learning. All rights reserved.7 | 10 A wave can be characterized by its wavelength and frequency. Wavelength, symbolized by the Greek letter lambda,, is the distance between any two identical points on adjacent waves.

11. Copyright © Cengage Learning. All rights reserved.7 | 11 Frequency, symbolized by the Greek letter nu,, is the number of wavelengths that pass a fixed point in one unit of time (usually a second). The unit is 1 / S or s -1, which is also called the Hertz (Hz).

12. Copyright © Cengage Learning. All rights reserved.7 | 12 Wavelength and frequency are related by the wave speed. The speed of light, c, is 3.00 x 10 8 m/s. c = The relationship between wavelength and frequency due to the constant velocity of light is illustrated on the next slide.

13. Copyright © Cengage Learning. All rights reserved.7 | 13 When the wavelength is reduced by a factor of two, the frequency increases by a factor of two.

14. ? Copyright © Cengage Learning. All rights reserved.7 | 14 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

15. ? Copyright © Cengage Learning. All rights reserved.7 | 15 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

16. Copyright © Cengage Learning. All rights reserved.7 | 16 The range of frequencies and wavelengths of electromagnetic radiation is called the electromagnetic spectrum.

17. Copyright © Cengage Learning. All rights reserved.7 | 17 When frequency is doubled, wavelength is halved. The light would be in the blue-violet region. Concept Check 7.1 Laser light of a specific frequency falls on a crystal that converts a portion of this light into light with double the original frequency. How is the wavelength of this frequency- doubled light related to the wavelength of the original laser light? Suppose the original laser light was red. In which region of the spectrum would the frequency-doubled light be? (If this is in the visible region, what color is the light?)

18. Copyright © Cengage Learning. All rights reserved.7 | 18 One property of waves is that they can be diffracted—that is, they spread out when they encounter an obstacle about the size of the wavelength. In 1801, Thomas Young, a British physicist, showed that light could be diffracted. By the early 1900s, the wave theory of light was well established.

19. Copyright © Cengage Learning. All rights reserved.7 | 19 The wave theory could not explain the photoelectric effect, however.

20. Copyright © Cengage Learning. All rights reserved.7 | 20 The photoelectric effect is the ejection of an electron from the surface of a metal or other material when light shines on it.

21. Copyright © Cengage Learning. All rights reserved.7 | 21 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)

22. Copyright © Cengage Learning. All rights reserved.7 | 22 Einstein used this understanding of light to explain the photoelectric effect in 1905. Each electron is struck by a single photon. Only when that photon has enough energy will the electron be ejected from the atom; that photon is said to be absorbed.

23. Copyright © Cengage Learning. All rights reserved.7 | 23 Light, therefore, has properties of both waves and matter. Neither understanding is sufficient alone. This is called the particle–wave duality of light.

24. ? Copyright © Cengage Learning. All rights reserved.7 | 24 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

25. Copyright © Cengage Learning. All rights reserved.7 | 25 In the early 1900s, the atom was understood to consist of a positive nucleus around which electrons move (Rutherford’s model). This explanation left a theoretical dilemma: According to the physics of the time, an electrically charged particle circling a center would continually lose energy as electromagnetic radiation. But this is not the case—atoms are stable.

26. Copyright © Cengage Learning. All rights reserved.7 | 26 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.

27. Copyright © Cengage Learning. All rights reserved.7 | 27

28. Copyright © Cengage Learning. All rights reserved.7 | 28 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

29. Copyright © Cengage Learning. All rights reserved.7 | 29 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

30. Copyright © Cengage Learning. All rights reserved.7 | 30 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.

31. Copyright © Cengage Learning. All rights reserved.7 | 31 For a hydrogen electron, the energy change is given by R H = 2.179  10 −8 J, Rydberg constant

32. Copyright © Cengage Learning. All rights reserved.7 | 32 The energy of the emitted or absorbed photon is related to  E: We can now combine these two equations:

33. Copyright © Cengage Learning. All rights reserved.7 | 33 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 = ∞.

34. Copyright © Cengage Learning. All rights reserved.7 | 34 Energy-level diagram for the hydrogen atom.

35. Copyright © Cengage Learning. All rights reserved.7 | 35 Electron transitions for an electron in the hydrogen atom.

36. ? Copyright © Cengage Learning. All rights reserved.7 | 36 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

37. Copyright © Cengage Learning. All rights reserved.7 | 37 Concept Check 7.2 An atom has a line spectrum consisting of a red line and a blue line. Assume that each line corresponds to a transition between two adjacent energy levels. Sketch an energy-level diagram with three energy levels that might explain this line spectrum, indicating the transitions on this diagram. Consider the transition from the highest energy level on this diagram to the lowest energy level. How would you describe the color or region of the spectrum corresponding to this transition?

38. Copyright © Cengage Learning. All rights reserved.7 | 38 The red line corresponds to the smaller energy difference in going from n = 3 to n = 2. The blue line corresponds to the larger energy difference in going from n = 2 to n = 1. n = 1 n = 2 n = 3 A minimum of three energy levels are required.

39. Copyright © Cengage Learning. All rights reserved.7 | 39 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:

40. Copyright © Cengage Learning. All rights reserved.7 | 40 Light has properties of both waves and particles (matter). What about matter?

41. Copyright © Cengage Learning. All rights reserved.7 | 41 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:

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

43. Copyright © Cengage Learning. All rights reserved.7 | 43 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

44. Copyright © Cengage Learning. All rights reserved.7 | 44 Concept Check 7.3 A proton is approximately 2000 times heavier than an electron. How would the speeds of these particles compare if their corresponding wavelengths were about equal?

45. Copyright © Cengage Learning. All rights reserved.7 | 45

46. Copyright © Cengage Learning. All rights reserved.7 | 46 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.

47. Copyright © Cengage Learning. All rights reserved.7 | 47 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.

48. Copyright © Cengage Learning. All rights reserved.7 | 48 Quantum mechanics allows us to make statistical statements about the regions in which we are most likely to find the electron. Solving Schrödinger’s equation gives us a wave function, represented by the Greek letter psi, , which gives information about a particle in a given energy level. Psi-squared,  2, gives us the probability of finding the particle within a region of space.

49. Copyright © Cengage Learning. All rights reserved.7 | 49 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.

50. Copyright © Cengage Learning. All rights reserved.7 | 50 Two additional views are shown on the next slide. Figure A illustrates the probability density for an electron in hydrogen. The concentric circles represent successive shells. Figure B shows the probability of finding the electron at various distances from the nucleus. The highest probability (most likely) distance is at 50 pm.

51. Copyright © Cengage Learning. All rights reserved.7 | 51

52. Copyright © Cengage Learning. All rights reserved.7 | 52 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.

53. Copyright © Cengage Learning. All rights reserved.7 | 53 A wave function for an electron in an atom is called an atomic orbital (described by three quantum numbers—n, l, m l ). It describes a region of space with a definite shape where there is a high probability of finding the electron. We will study the quantum numbers first, and then look at atomic orbitals.

54. Copyright © Cengage Learning. All rights reserved.7 | 54 Principal Quantum Number, n This quantum number is the one on which the energy of an electron in an atom primarily depends. The smaller the value of n, the lower the energy and the smaller the orbital. The principal quantum number can have any positive value: 1, 2, 3,... Orbitals with the same value for n are said to be in the same shell.

55. Copyright © Cengage Learning. All rights reserved.7 | 55 Shells are sometimes designated by uppercase letters: Letter n K1K1 L2L2 M3M3 N4N4...

56. Copyright © Cengage Learning. All rights reserved.7 | 56 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.

57. Copyright © Cengage Learning. All rights reserved.7 | 57 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.

58. Copyright © Cengage Learning. All rights reserved.7 | 58 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.

59. Copyright © Cengage Learning. All rights reserved.7 | 59 Let’s summarize: When n = 1, l has only one value, 0. When l = 0, m l has only one value, 0. So the first shell (n = 1) has one subshell, an s- subshell, 1s. That subshell, in turn, has one orbital.

60. Copyright © Cengage Learning. All rights reserved.7 | 60 When n = 2, l has two values, 0 and 1. When l = 0, m l has only one value, 0. So there is a 2s subshell with one orbital. When l = 1, m l has only three values, -1, 0, 1. So there is a 2p subshell with three orbitals.

61. Copyright © Cengage Learning. All rights reserved.7 | 61 When n = 3, l has three values, 0, 1, and 2. When l = 0, m l has only one value, 0. So there is a 3s subshell with one orbital. When l = 1, m l has only three values, -1, 0, 1. So there is a 3p subshell with three orbitals. When l = 2, m l has only five values, -2, -1, 0, 1, 2. So there is a 3d subshell with five orbitals.

62. Copyright © Cengage Learning. All rights reserved.7 | 62 We could continue with n = 4 and 5. Each would gain an additional subshell (f and g, respectively). In an f subshell, there are seven orbitals; in a g subshell, there are nine orbitals. Table 7.1 gives the complete list of permitted values for n, l, and m l up to the fourth shell. It is on the next slide.

63. Copyright © Cengage Learning. All rights reserved.7 | 63

64. Copyright © Cengage Learning. All rights reserved.7 | 64 The figure shows relative energies for the hydrogen atom shells and subshells; each orbital is indicated by a dashed- line.

65. Copyright © Cengage Learning. All rights reserved.7 | 65 Spin Quantum Number, m s This quantum number refers to the two possible orientations of the spin axis of an electron. It may have a value of either + 1 / 2 or - 1 / 2.

66. ? Copyright © Cengage Learning. All rights reserved.7 | 66 Which of the following are permissible sets of quantum numbers? n = 4, l = 4, m l = 0, m s = ½ n = 3, l = 2, m l = 1, m s = -½ n = 2, l = 0, m l = 0, m s = ³/ ² n = 5, l = 3, m l = -3, m s = ½ (a) Not permitted. When n = 4, the maximum value of l is 3. (b) Permitted. (c) Not permitted; m s can only be +½ or –½. (b) Permitted.

67. Copyright © Cengage Learning. All rights reserved.7 | 67 Atomic Orbital Shapes An s orbital is spherical. A p orbital has two lobes along a straight line through the nucleus, with one lobe on either side. A d orbital has a more complicated shape.

68. Copyright © Cengage Learning. All rights reserved.7 | 68 The cross-sectional view of a 1s orbital and a 2s orbital highlights the difference in the two orbitals’ sizes.

69. Copyright © Cengage Learning. All rights reserved.7 | 69 The cutaway diagrams of the 1s and 2s orbitals give a better sense of them in three dimensions.

70. Copyright © Cengage Learning. All rights reserved.7 | 70 The next slide illustrates p orbitals. Figure A shows the general shape of a p orbital. Figure B shows the orientations of each of the three p orbitals.

71. Copyright © Cengage Learning. All rights reserved.7 | 71

72. Copyright © Cengage Learning. All rights reserved.7 | 72 The complexity of the d orbitals can be seen below: