1. Chapter 6 Electronic Structure of Atoms: Quantum Mechanics and Atomic Orbitals

2. Quantum Theory and Atomic Structure

3. Figure 1 Frequency and Wavelength c =  The Wave Nature ofLight

4. Figure 2 Amplitude (intensity) of a wave. Amplitude  Light Intensity

5. Figure 3 Regions of the electromagnetic spectrum.

6. Sample Problem 1 SOLUTION:PLAN: Interconverting Wavelength and Frequency wavelength in units given wavelength in m frequency (s -1 or Hz) = c/ Use c = 10 -2 m 1 cm 10 -9 m 1 nm = 1.00x10 -10 m = 325x10 -2 m = 473x10 -9 m == 3x10 8 m/s 1.00x10 -10 m = 3x10 18 s -1 == == 3x10 8 m/s 325x10 -2 m = 9.23x10 7 s -1 3x10 8 m/s 473x10 -9 m = 6.34x10 14 s -1 PROBLEM: A dental hygienist uses x-rays ( = 1.00 Å ) to take a series of dental radiographs while the patient listens to a radio station ( = 325 cm) and looks out the window at the blue sky ( = 473 nm). What is the frequency (in s -1 ) of the electromagnetic radiation from each source? (Assume that the radiation travels at the speed of light, 3.00x10 8 m/s.) 1 A = 10 -10 m 1 cm = 10 -2 m 1 nm = 10 -9 m o 325 cm 473nm 1.00A o 10 -10 m 1A o

7. Figure 4 Different behaviors of waves and particles. Refracted-bent at an anglecurved path diffracted through opening forms circular waves moves through opening

8. Figure 5 The diffraction pattern caused by light passing through two adjacent slits.

9. Three phenomena involving matter and light were especially confounding to physicists of the early 20th century: (1) Black body radiation (2) The photoelectric effect (3) Atomic Spectra

10. Figure 6 Blackbody radiation E = n h n=integer q.n.  E =  n h  E = h when n = 1 Max Planck - proposed that a hot glowing Object could emit (or absorb) only certain quantities of energy. Energy is quantized! ~1000K smoldering coal red glow ~1500K, electric heating element more orange. > 2000K light bulb filament

11. The Photoelectric Effect and the Photon Theory of Light Einstein used Planck’s assumption to explain the photoelectric effect. He concluded that energy is proportional to frequency: E = h  where h is Planck’s constant, 6.626  10 -34 J-s. 1/2mv 2 = h - h o

12. Figure 7 Demonstration of the photoelectric effect.

13. Sample Problem 2 SOLUTION: PLAN: Calculating the Energy of Radiation from Its Wavelength PROBLEM:A cook uses a microwave oven to heat a meal. The wavelength of the radiation is 1.20cm. What is the energy of one photon of this microwave radiation? After converting cm to m, we can use the energy equation, E = h combined with = c/ to find the energy. E = hc/ E = 6.626X10 -34 J*s3x10 8 m/s 1.20cm 10 -2 m cm x = 1.66x10 -23 J

14. The Nature of Energy Another mystery involved the emission spectra; observed from energy emitted by atoms and molecules.

15. Figure 8 The line spectra of several elements.

16. = RRydberg equation - 1 1 n22n22 1 n12n12 R is the Rydberg constant = 1.096776x10 7 m -1 Figure 9 Three series of spectral lines of atomic hydrogen. for the visible series, n 1 = 2 and n 2 = 3, 4, 5,...

17. Figure 10 Quantum staircase. Neils Bohr: Energy levels, or stationary states are quantized. These energy levels are present at specific instances from the hydrogen nucleus.

18. Figure 11 The Bohr explanation of the three series of spectral lines.

19. Bohr Model

20. Figure B6.2 Figure B6.1 Emission and absorption spectra of sodium atoms. Flame tests. strontium 38 Srcopper 29 Cu

22. The wave nature of Electrons and the Particle Nature of Photons

23. Figure 13 Wave motion in restricted systems. Quantized electronic motion is one such restricted system. In the Bohr Theory each permitted orbit was equal to a whole number of wavelengths.

24. The Wave Nature of Matter Louis de Broglie proposed that if light can have particle properties, matter should exhibit wave properties. He demonstrated that the relationship between mass and wavelength was… = h mv E=mc 2 E=h =hc/

25. Table 1 The de Broglie Wavelengths of Several Objects SubstanceMass (g)Speed (m/s) (m) slow electron fast electron alpha particle one-gram mass baseball Earth 9x10 -28 6.6x10 -24 1.0 142 6.0x10 27 1.0 5.9x10 6 1.5x10 7 0.01 25.0 3.0x10 4 7x10 -4 1x10 -10 7x10 -15 7x10 -29 2x10 -34 4x10 -63  h/mv

26. Sample Problem 3 SOLUTION: PLAN: Calculating the de Broglie Wavelength of an Electron PROBLEM:Find the deBroglie wavelength of an electron with a speed of 1.00x10 6 m/s (electron mass = 9.11x10 -31 kg; h = 6.626x10 -34 kg·m 2 /s). Knowing the mass and the speed of the electron allows to use the equation = h/m v to find the wavelength. = 6.626x10 -34 kg·m 2 /s 9.11x10 -31 kgx1.00x10 6 m/s = 7.27x10 -10 m

27. Figure 14 Comparing the diffraction patterns of x-rays and electrons. x-ray diffraction of aluminum foilelectron diffraction of aluminum foil

28. CLASSICAL THEORY Matter particulate, massive Energy continuous, wavelike Since matter is discontinuous and particulate perhaps energy is discontinuous and particulate. ObservationTheory Planck: Energy is quantized; only certain values allowed (quanta) blackbody radiation Einstein: Light has particulate behavior (photons)photoelectric effect Bohr: Energy of atoms is quantized; photon emitted when electron changes orbit. atomic line spectra Figure 15 Summary of the major observations and theories leading from classical theory to quantum theory.

29. Since energy is wavelike perhaps matter is wavelike ObservationTheory deBroglie: All matter travels in waves; energy of atom is quantized due to wave motion of electrons Davisson/Germer: electron diffraction by metal crystal Since matter has mass perhaps energy has mass ObservationTheory Einstein/deBroglie: Mass and energy are equivalent; particles have wavelength and photons have momentum. Compton: photon wavelength increases (momentum decreases) after colliding with electron Figure 15 continued QUANTUM THEORY Energy same as Matter particulate, massive, wavelike

30. The Heisenberg Uncertainty Principle  x · m  v ≥ h 44 Werner Heisenberg postulated the uncertainty principle, which states that it is impossible to know simultaneously the exact position and the momentum of a particle, like an electron.

31. Sample Problem 4 SOLUTION: PLAN: Applying the Uncertainty Principle PROBLEM:An electron moving near an atomic nucleus has a speed 6x10 6 ± 1% m/s. What is the uncertainty in its position (  x )? The uncertainty (  v ) is given as ±1%(0.01) of 6x10 6 m/s. Once we calculate this, plug it into the uncertainty equation, we can find the uncertainty in the electrons position.  v = (0.01)(6x10 6 m/s) = 6x10 4 m/s  x · m  v ≥ h 44 x ≥x ≥ 4  (9.11x10 -31 kg)(6x10 4 m/s) 6.626x10 -34 kg·m 2 /s = 9.52x10 -9 m The uncertainty in the electrons position is about 10 times greater than the diameter of the entire atom (10 -10 m)!

32. The Schrödinger Equation H  = E  d2d2 dy 2 d2d2 dx 2 d2d2 dz 2 ++ 82m82m h2h2 (E-V(x,y,z)  (x,y,z) = 0 + how  changes in space mass of electron total quantized energy of the atomic system potential energy at x,y,zwave function Each energy state (solution) of the atom is associated with a given wave function, an atomic orbital. While the wave function,  has no physical meaning, the square of the wave function,  , does.   expresses the probability that the electron is at a particular volume in the atom.

33. Figure 16 Electron probability in the ground-state H atom.

34. Quantum Numbers and Atomic Orbitals An atomic orbital is specified by three quantum numbers. n the principal quantum number is a positive integer, n= 1,2,….  l the angular momentum quantum number - an integer from 0 to n -1 m l the magnetic moment quantum number - an integer from - l, 0 + l For a total of 2l+1 possible values m s the spin quantum number ± 1/2

35. s Orbitals Spherical in shape. Radius of sphere increases with increasing value of n.

36. Figure 17 1s2s3s S-Orbitals

37. p Orbitals Have two lobes with a node between them.

38. Figure 18 The 2p orbitals. P-Orbitals

39. Figure 19 The 3d orbitals.

40. Figure 19 continued D-Orbitals

41. Figure 20 One of the seven possible 4f orbitals.

42. Table 2 The Hierarchy of Quantum Numbers for Atomic Orbitals Name, Symbol (Property) Allowed ValuesQuantum Numbers Principal, n (size, energy) Angular momentum, l (shape) Magnetic, m l (orientation) Positive integer (1, 2, 3,...) 0 to n-1 - l,…,0,…,+ l 1 0 0 2 01 0 3 0 12 0 0 +1 0+1 0 +2-2

43. Sample Problem 5 SOLUTION: PLAN: Determining Quantum Numbers for an Energy Level PROBLEM:What values of the angular momentum ( l ) and magnetic (m l ) quantum numbers are allowed for a principal quantum number (n) of 3? How many orbitals are allowed for n = 3? Follow the rules for allowable quantum numbers found in the text. l values can be integers from 0 to n-1; m l can be integers from -l through 0 to + l. For n = 3, l = 0, 1, 2 For l = 0 m l = 0 For l = 1 m l = -1, 0, or +1 For l = 2 m l = -2, -1, 0, +1, or +2 There are 9 m l values and therefore 9 orbitals with n = 3.

44. Sample Problem 6 SOLUTION: PLAN: Determining Sublevel Names and Orbital Quantum Numbers PROBLEM:Give the name, magnetic quantum numbers, and number of orbitals for each sublevel with the following quantum numbers: (a) n = 3, l = 2(b) n = 2, l = 0(c) n = 5, l = 1(d) n = 4, l = 3 Combine the n value and l designation to name the sublevel. Knowing l, we can find m l and the number of orbitals. n l sublevel namepossible m l values# of orbitals (a) (b) (c) (d) 3 2 5 4 2 0 1 3 3d 2s 5p 4f -2, -1, 0, 1, 2 0 -1, 0, 1 -3, -2, -1, 0, 1, 2, 3 5 1 3 7

45. 7-45 The Aufbau (build-up) Principle Electrons enter each orbital of the same energy singly, beginning with the lowest energy orbital before they pair.

46. 7-46 Energies of Atomic Orbitals For a one-electron hydrogen atom, orbitals within the same principal energy level, n, have the same energy. That is, they are degenerate.

47. 7-47 Energies of Orbitals in multi-electron Atoms As the number of electrons increases, though, so does the repulsion between them. Therefore, in many- electron atoms, different types of orbitals in the same principal energy level are no longer degenerate. E ns < E np < E nd < E nf

48. 7-48 1s < 2s < 2p < 3s < 3p <4s < 3d <4p < 5s < 4d < 5p < 6s < 4f <5d < 6p < 7s < 5f < 6d < 7p < 8s...

49. 7-49 The Electron Spin Quantum Number, m s In the 1920s, it was discovered that two electrons in the same orbital do not have exactly the same energy. The “spin” of an electron describes its magnetic field, which affects its energy. + ½ more stable - 1/2m s =

50. 7-50 Pauli Exclusion Principle No two electrons in the same atom can have identical sets of quantum numbers! This restricts the number of electrons allowed per orbital to two, with anti-parallel spins. Since the two spin states have different energies, no two electrons in the same atom will have exactly the same energy.

51. 7-51 Electron Configurations Distribution of all electrons in an atom Consist of  Number denoting the energy level = n

52. 7-52 Electron Configurations Distribution of all electrons in an atom Consist of  Number denoting the energy level = n.  Letter denoting the type of orbital = l.

53. 7-53 Electron Configurations Distribution of all electrons in an atom. Consist of  Number denoting the energy level = n.  Letter denoting the type of orbital = l.  Superscript denoting the number of electrons in those orbitals = m l.

54. 7-54 Orbital Diagrams Each box represents one orbital. Half-arrows represent the electrons. The direction of the arrow represents the spin of the electron.

55. 7-55

56. 7-56 Hund’s Rule “For degenerate orbitals, the lowest energy is attained when the number of electrons with the same spin is maximized.” This means that when electrons enter a degenerate set of orbitals, they do so one at a time, with their spins up, before they pair!

57. 7-57 Periodic Table We fill orbitals in increasing order of energy. Different blocks on the periodic table, then correspond to different types of orbitals.

58. 7-58 Some Anomalies Some irregularities occur when there are enough electrons to half-fill s and d orbitals on a given row.

59. 7-59 Some Anomalies For instance, the electron configuration for chromium is [Ar] 4s 1 3d 5 rather than the expected [Ar] 4s 2 3d 4.

60. Figure B.3 The main components of a typical spectrophotometer. Monochromator (wavelength selector) disperses incoming radiation into continuum of component wavelengths that are scanned or individually selected. Sample in compartment absorbs characteristic amount of each incoming wavelength. Computer converts signal into displayed data. Source produces radiation in region of interest. Must be stable and reproducible. In most cases, the source emits many wavelengths. Lenses/slits/collimaters narrow and align beam. Detector converts transmitted radiation into amplified electrical signal.

61. 7-61

62. 7-62 1s < 2s < 2p < 3s < 3p <4s < 3d <4p < 5s < 4d < 5p < 6s < 4f <5d < 6p < 7s < 5f < 6d < 7p < 8s...