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Chapter 7

Quantum Theory and Atomic Structure. Chapter 7. Goals & Objectives. See the following Learning Objectives on pages 283 and 322. Understand these Concepts: 7.1-7, 9-12; 8.1-8. Master these Skills: 7.1-2, 5; 8.1-2. The Wave Nature of Light.

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Chapter 7

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  1. Quantum Theory and Atomic Structure Chapter 7

  2. Goals & Objectives • See the following Learning Objectives on pages 283 and 322. • Understand these Concepts: • 7.1-7, 9-12; 8.1-8. • Master these Skills: • 7.1-2, 5; 8.1-2.

  3. The Wave Nature of Light Visible light is a type of electromagnetic radiation. • The wave properties of electromagnetic radiation are described by three variables: • frequency (n), cycles per second • wavelength (l), the distance a wave travels in one cycle • amplitude, the height of a wave crest or depth of a trough. The speed of light is a constant: c = nx l = 3.00 x 108 m/s in a vacuum

  4. Figure 7.1 The reciprocal relationship of frequency and wavelength.

  5. Figure 7.2 Differing amplitude (brightness, or intensity) of a wave.

  6. Figure 7.3 Regions of the electromagnetic spectrum.

  7. PROBLEM: A dental hygienist uses x-rays (l= 1.00Å) to take a series of dental radiographs while the patient listens to a radio station (l = 325 cm) and looks out the window at the blue sky (l= 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, 3.00x108 m/s.) PLAN: Use the equation c = nl to convert wavelength to frequency. Wavelengths need to be in meters because c has units of m/s. wavelength in units given wavelength in m c n = l frequency (s-1 or Hz) Sample Problem 7.1 Interconverting Wavelength and Frequency use conversion factors 1 Å = 10-10 m

  8. 325 cm x 473 nm x = 6.34 x 1014 s-1 l = 1.00 Å x = 1.00 x 10-10 m 10-10 m 1 Å = = 3.00 x 1018 s-1 10-9 m 1 cm 10-2 m 1 cm 3.00 x 108 m/s 1.00 x 10-10 m = 9.23 x 107 s-1 3.00 x 108 m/s 3.00 x 108 m/s = = c c c n = n = n = l l l Sample Problem 7.1 SOLUTION: For the x-rays: For the radio signal: For the blue sky:

  9. The Quantum Theory of Energy Any object (including atoms) can emit or absorb only certain quantities of energy. Energy is quantized; it occurs in fixed quantities, rather than being continuous. Each fixed quantity of energy is called a quantum. An atom changes its energy state by emitting or absorbing one or more quanta of energy. DE = nhn where n can only be a whole number.

  10. PROBLEM: A cook uses a microwave oven to heat a meal. The wavelength of the radiation is 1.20 cm. What is the energy of one photon of this microwave radiation? PLAN: We know l in cm, so we convert to m and find the frequency using the speed of light. We then find the energy of one photon using E = hn. hc l E = hn = = (6.626 x 10-34) J∙s)(3.00 x 108 m/s) (1.20 cm) 10-2 m ( ) 1 cm Sample Problem 7.2 Calculating the Energy of Radiation from Its Wavelength SOLUTION: = 1.66 x 10-23 J

  11. Figure 7.8A The line spectrum of hydrogen.

  12. Figure 7.8B The line spectra of Hg and Sr.

  13. 1 1 1 l n12 n22 = R - Rydberg equation R is the Rydberg constant = 1.096776x107 m-1 Figure 7.9 Three series of spectral lines of atomic hydrogen. for the visible series, n1 = 2 and n2 = 3, 4, 5, ...

  14. Bohr’s atomic model postulated the following: The H atom has only certain energy levels, which Bohr called stationary states. Each state is associated with a fixed circular orbit of the electron around the nucleus. The higher the energy level, the farther the orbit is from the nucleus. When the H electron is in the first orbit, the atom is in its lowest energy state, called the ground state. The Bohr Model of the Hydrogen Atom 1913

  15. The atom does not radiate energy while in one of its stationary states. The atom changes to another stationary state only by absorbing or emitting a photon. The energy of the photon (hn) equals the difference between the energies of the two energy states. When the E electron is in any orbit higher than n = 1, the atom is in an excited state.

  16. Figure 7.10 A quantum “staircase” as an analogy for atomic energy levels.

  17. Figure 7.11 The Bohr explanation of three series of spectral lines emitted by the H atom.

  18. strontium 38Sr copper 29Cu Fireworks display emissions similar to those seen in flame tests. The flame color is due to the emission of light of a particular wavelength by each element. Tools of the Laboratory Figure B7.1 Flame tests and fireworks. copper 29Cu

  19. Tools of the Laboratory Figure B7.2 Emission and absorption spectra of sodium atoms.

  20. Light Waves Atomic Emission Spectra Light in discrete lines; individual wavelengths, not continuum http://jersey.uoregon.edu/elements/Elements.html

  21. Tools of the Laboratory Figure B7.3 Components of a typical spectrometer.

  22. Tools of the Laboratory Figure B7.4 Measuring chlorophyll a concentration in leaf extract.

  23. m = mass u = speed in m/s h mu l = The Wave-Particle Duality of Matter and Energy Matter and Energy are alternate forms of the same entity. E = mc2 Allmatter exhibits properties of both particles and waves. Electrons have wave-like motion and therefore have only certain allowable frequencies and energies. Matter behaves as though it moves in a wave, and the de Broglie wavelength for any particle is given by:

  24. Dx∙mD u ≥ h x = position u = speed 4p Heisenberg’s Uncertainty Principle Heisenberg’s Uncertainty Principle states that it is not possible to know both the position and momentum of a moving particle at the same time. The more accurately we know the speed, the less accurately we know the position, and vice versa.

  25. The uncertainty in the speed (Du) is given as ±1% (0.01) of 6x106 m/s. We multiply u by 0.01 and substitute this value into Equation 7.6 to solve for Δx. PLAN: h h 4pmDu Dx∙mDu ≥ 4p 6.626x10-34 kg∙m2/s ≥ Dx≥ 4p (9.11x10-31 kg)(6x104 m/s) Sample Problem 7.5 Applying the Uncertainty Principle PROBLEM: An electron moving near an atomic nucleus has a speed 6x106 m/s ± 1%. What is the uncertainty in its position (Dx)? Du = (0.01)(6x106 m/s) = 6x104 m/s SOLUTION: ≥ 1x10-9 m

  26. The Quantum-Mechanical Model of the Atom The matter-wave of the electron occupies the space near the nucleus and is continuously influenced by it. The Schrödinger wave equation allows us to solve for the energy states associated with a particular atomic orbital. The square of the wave function gives the probability density, a measure of the probability of finding an electron of a particular energy in a particular region of the atom.

  27. The Shrodinger Atom (1926) • Quantum Mechanical Model • Wave equation is solved to obtain • Wave function (orbital)  • Square of wave function 2 gives probability of finding an electron in a certain space

  28. Figure 7.16 Electron probability density in the ground-state H atom.

  29. The Shrödinger Atom (1926) • Wave function for each electron in an atom contains four quantum numbers • n Principle • l Angular momentum • ml Magnetic • ms Spin

  30. Quantum Numbers and Atomic Orbitals An atomic orbital is specified by three quantum numbers. The principal quantum number (n) is a positive integer. The value of n indicates the relative size of the orbital and therefore its relative distance from the nucleus. The angular momentum quantum number (l) is an integer from 0 to (n -1). The value of l indicates the shape of the orbital. The magnetic quantum number (ml) is an integer with values from –l to +l The value of ml indicates the spatial orientation of the orbital.

  31. Quantum Numbers and The Exclusion Principle Each electron in any atom is described completely by a set of four quantum numbers. The first three quantum numbers describe the orbital, while the fourth quantum number describes electron spin. Pauli’s exclusion principle states that no two electrons in the same atom can have the same four quantum numbers. An atomic orbital can hold a maximum of two electrons and they must have opposing spins.

  32. 0 1 2 0 -1 0 +1 -1 0 +1 -2 -1 0 +1 +2 Table 7.2 The Hierarchy of Quantum Numbers for Atomic Orbitals Name, Symbol (Property) Allowed Values Quantum Numbers Principal, n (size, energy) Positive integer (1, 2, 3, ...) 1 2 3 Angular momentum, l (shape) 0 to n – 1 0 0 1 0 0 Magnetic, ml (orientation) -l,…,0,…,+l

  33. PROBLEM: PROBLEM: What values of the angular momentum (l) and magnetic (ml) quantum numbers are allowed for a principal quantum number (n) of 3? How many orbitals are allowed for n = 3? What values of the angular momentum (l) and magnetic (ml) quantum numbers are allowed for a principal quantum number (n) of 3? How many orbitals are allowed for n = 3? PLAN: Values of l are determined from the value for n, since l can take values from 0 to (n-1). The values of ml then follow from the values of l. Sample Problem 7.6 Determining Quantum Numbers for an Energy Level SOLUTION: For n = 3, allowed values of l are = 0, 1, and 2 For l = 0 ml = 0 For l = 1 ml = -1, 0, or +1 For l = 2 ml = -2, -1, 0, +1, or +2 There are 9 mlvalues and therefore 9 orbitals with n = 3.

  34. 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 PLAN: Combine the n value and l designation to name the sublevel. Knowingl, we can find ml and the number of orbitals. n l sublevel name possible ml values # of orbitals (a) 3 2 3d -2, -1, 0, 1, 2 5 (b) 2 2s 0 1 0 (c) 5 1 5p -1, 0, 1 3 (d) 4 3 4f -3, -2, -1, 0, 1, 2, 3 7 Sample Problem 7.7 Determining Sublevel Names and Orbital Quantum Numbers SOLUTION:

  35. PROBLEM: What is wrong with each of the following quantum numbers designations and/or sublevel names? n l ml Name (a) 1 0 1p 1 (b) 4 3 +1 4d (c) 3 1 -2 3p Sample Problem 7.8 Identifying Incorrect Quantum Numbers SOLUTION: (a) A sublevel with n = 1 can only have l = 0, not l = 1. The only possible sublevel name is 1s. (b) A sublevel with l = 3 is an f sublevel, to a d sublevel. The name should be 4f. (c) A sublevel with l = 1 can only have mlvalues of -1, 0, or +1, not -2.

  36. Figure 7.17 The 1s, 2s, and 3s orbitals.

  37. Figure 7.18 The 2p orbitals.

  38. Figure 7.19 The 3d orbitals.

  39. Figure 7.20 The 4fxyz orbital, one of the seven 4f orbitals.

  40. The Shrödinger Atom (1926) • Magnetic quantum number ml • Value = –l to + l • Properties of orbital: orientation in space

  41. The Shrödinger Atom (1926) • Spin quantum number ms • Value = +1/2 or – 1/2 • Two electrons can occupy an orbital one has spin +1/2, the other has spin – 1/2

  42. The Shrödinger Atom (1926) • Subshell capacities in each shell (n) • s (l=0) 1 orbital, 2 electrons • p (l=1) 3 orbitals, 6 electrons (n≥2) • d (l=2) 5 orbitals, 10 electrons (n≥3) • f (l=3) 7 orbitals, 14 electrons (n≥4)

  43. principal n positive integers (1, 2, 3, …) orbital energy (size) angular momentum l integers from 0 to n-1 orbital shape (The l values 0, 1, 2, and 3 correspond to s, p, d, and f orbitals, respectively.) magnetic ml integers from -l to 0 to +l orbital orientation spin ms +½ or -½ direction of e-spin Table 8.1 Summary of Quantum Numbers of Electrons in Atoms Name Symbol Permitted Values Property

  44. Orbitals: Order of Filling • Lowest energy orbitals are filled first • In subshells with multiple orbitals, start filling at lowest ml. • In subshells with multiple orbitals, each orbital receives one electron (ms = +1/2) until all are filled, • then second electron electron goes in • (ms = –1/2).

  45. Electron Configurations and the Periodic Table • 3Li 1s22s1 • 11Na 1s22s22p63s1 • 19K 1s22s22p63s23p64s1

  46. Electron Configurations and the Periodic Table • 9F 1s22s22p5 • 17Cl 1s22s22p63s23p5 • 35Br 1s22s22p63s23p64s23d104p5

  47. Electron Configurations and the Periodic Table • 10Ne 1s22s22p6 • 18Ar 1s22s22p63s23p6 • 36Kr 1s22s22p63s23p64s23d104p6

  48. Electron Configurations and the Periodic Table • 17Cl– 1s22s22p63s23p6 • 18Ar 1s22s22p63s23p6 • 19K+ 1s22s22p63s23p6

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