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Chapter 7 (Zumdahl) Electronic Structure of Atoms

Chemistry, The Central Science , 11th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten. Chapter 7 (Zumdahl) Electronic Structure of Atoms. John D. Bookstaver St. Charles Community College Cottleville, MO. Waves.

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Chapter 7 (Zumdahl) Electronic Structure of Atoms

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  1. Chemistry, The Central Science, 11th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Chapter 7 (Zumdahl)Electronic Structureof Atoms John D. Bookstaver St. Charles Community College Cottleville, MO © 2009, Prentice-Hall, Inc.

  2. Waves • To understand the electronic structure of atoms, one must understand the nature of electromagnetic (EM) radiation. • The distance between corresponding points on adjacent waves is the wavelength (). © 2009, Prentice-Hall, Inc.

  3. Waves • The number of waves passing a given point per unit of time is the frequency(). • For waves traveling at the same velocity, the longer the wavelength, the smaller the frequency. • Frequency is measured in the following units: s-1, = waves/s = Hz (hertz) © 2009, Prentice-Hall, Inc.

  4. Electromagnetic Radiation • All electromagnetic radiation travels at the same velocity: the speed of light (c), 3.00  108 m/s. • Therefore, c =  © 2009, Prentice-Hall, Inc.

  5. Sample Exercise 7.1 (p 292) • (Copy from text into notes.) • Follow as I review the book solution to this problem. © 2009, Prentice-Hall, Inc.

  6. Practice Problems: Text, p 337,#39-40 • #39: The laser in an audio compact disc player uses light with a wavelength of 7.80 x 102 nm. Calculate the frequency of this light. • Solution: label your given(s) & your unknown. •  = given = unknown • Identify the connection between given(s) and unknown. (c= ) • Recognize that you know the value of c for all EM radiation ( c =3.00 x 108 m/s) • Substitute in all given/known values & solve for the unknown. (NOTE: Convert nm to m first!) © 2009, Prentice-Hall, Inc.

  7. The Nature of Energy • The wave nature of light does not explain how an object can glow when its temperature increases. • Max Planck explained it by assuming that energy comes in packets called quanta. © 2009, Prentice-Hall, Inc.

  8. The Nature of Energy • Einstein used this 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. • He also concluded that wave energy has mass! © 2009, Prentice-Hall, Inc.

  9. The Nature of Energy • Therefore, if one knows the wavelength of light, one can calculate the energy in one photon, or packet, of that light (also known as a “quantum” of energy): c =  E = h © 2009, Prentice-Hall, Inc.

  10. Sample Exercise 7.2p 293-294 • Copy this problem and solution. • Follow along as we walk thru the solution together. • The blue color in fireworks is often achieved by heating copper(I) chloride (CuCl) to about 1200 C. Then the compound emits blue light having a wavelength of 450 nm. What is the increment of energy (the quantum) that is emitted by CuCl? © 2009, Prentice-Hall, Inc.

  11. Solution: label your given(s) & your unknown. •  = given E= unknown • Identify the connection between given(s) and unknown. • There is not a direct connection. No equation we have includes both E and . • However, we know that E = h. • We also know the value of h (=6.626 x 10−34 J-s). • If we know the value of , we can calculate E. Can we find ? (is there a connection between  and? • YES! c= . • Substitute given values into this second equation and solve for . (Don’t forget to convert from nm to m first!) • When  is obtained, substitute it and the value of h into the first equation and solve for E. © 2009, Prentice-Hall, Inc.

  12. Practice Problemsp 337, #41-44 © 2009, Prentice-Hall, Inc.

  13. The Nature of Energy Another mystery in the early 20th century involved the emission spectra observed from energy emitted by atoms and molecules. © 2009, Prentice-Hall, Inc.

  14. The Nature of Energy • For atoms and molecules one does not observe a continuous spectrum, as one gets from a white light source. • Only a line spectrum of discrete wavelengths is observed. © 2009, Prentice-Hall, Inc.

  15. The Nature of Energy • Niels Bohr adopted Planck’s assumption and explained these phenomena in this way: • Electrons in an atom can only occupy certain orbits (corresponding to certain energies). © 2009, Prentice-Hall, Inc.

  16. The Nature of Energy • Niels Bohr adopted Planck’s assumption and explained these phenomena in this way: • Electrons in permitted orbits have specific, “allowed” energies; these energies will not be radiated from the atom. © 2009, Prentice-Hall, Inc.

  17. The Nature of Energy • Niels Bohr adopted Planck’s assumption and explained these phenomena in this way: • Energy is only absorbed or emitted in such a way as to move an electron from one “allowed” energy state to another; the energy is defined by E = nh Where n=main energy level © 2009, Prentice-Hall, Inc.

  18. Solving for e- energy if  & are unknown • The energy of an electron in a hydrogen atom is • E= (-2.178 x 10-18 J) Z2n2 [ ] Where Z= nuclear charge (atomic #) and n = main energy level and RH = the Rydberg constant, 2.18  10−18 J © 2009, Prentice-Hall, Inc.

  19. 12 nf2 ( ) - E = −RH 12 ni2 The Nature of Energy The energy absorbed or emitted from the process of electron promotion or demotion can be calculated by the equation: ni and nf are the initial and final energy levels of the electron. And Z = 1 (atomic # of hydrogen!) © 2009, Prentice-Hall, Inc.

  20. Sample Exercise • P 338 #51a. • Calculate the energy emitted when the following transition occurs in a hydrogen atom: n=3  n= 2 © 2009, Prentice-Hall, Inc.

  21. Solution • Our givens are • nfinal (level 2) • ninitial (level 3) • Our unknown is the energy released (E) • Identify the connection between given(s) & unknown:E= -RH(1/nf2-1/ni2) • Substitute given values & solve for unknown! © 2009, Prentice-Hall, Inc.

  22. Calculating  of light from energy released • Continue w/#51 from p 338 • Calculate the wavelength of light emitted when the following transition occurs in the hydrogen atom. What type of EM radiation is emitted? • Givens: E (from prior slide) • Unknown:  • Connection: E = h to solve for  • then use c= to solve for  Substitute & Solve: © 2009, Prentice-Hall, Inc.

  23. ALTERNATIVE APPROACH • Derive formula that includes all the variables you are using: , , & E • Use both eqns E = h & c=  • Set both eqns equal to  • E =  & c=  h  • Set both eqns equal to one another • E =c h  • Isolate , since this is your unknown. •  = hc E 4. Substitute values (c, h, E) to solve for . © 2009, Prentice-Hall, Inc.

  24. Practice Problems: p 338, #51, 52, 53, 54, 55, 56 © 2009, Prentice-Hall, Inc.

  25. h mv  = The Wave Nature of Matter • Louis de Broglie posited that if light can have material properties, matter should exhibit wave properties. • He demonstrated that the relationship between mass and wavelength was This formula is on the AP exam formula sheet Please note: the base units for Joules are kg●m2/s2 . This will be important for calculations © 2009, Prentice-Hall, Inc.

  26. Wave-Particle Duality • Site with Cool Visuals & Explanation of Wave Interference, Including electron waves © 2009, Prentice-Hall, Inc.

  27. Sample Exercise (from Brown/Lemay) p 223 • What is the wavelength of an electron moving with a speed of 5.97 x 106 m/s? The mass of the electron is 9.11 x 10-31 kg. SOLUTION: Givens: melectron = 9.11 x 10-31 kg velectron = 5.97 x 106 m/s Unknown =  CONNECTION between givens & unknown:  = h mv Substitute given & constant values, solve for . © 2009, Prentice-Hall, Inc.

  28. Sample Exercise, cont. Substitute given & constant values, solve for .  = h mv • = 6.626 x 10-34 kg●m2/s (9.11 x 10-31 kg)(5.97 x 106 m/s)  = 1.22 x 10-10 Where did the exponent go? J= kg●m2/s2 © 2009, Prentice-Hall, Inc.

  29. Practice Problems 6.41) Determine the wavelength of the following objects • An 85-kg person skiing at 50 km/hr • A 10.0 g bullet fired at 250 m/s. • A lithium atom moing at 2.5 x 105 m • An ozone molecule in the upper atmosphere moving at 550 m/s. © 2009, Prentice-Hall, Inc.

  30. Practice Problems: • 6.42) Among the elementary subatomic particles of physics is the muon, which decays within a few nanoseconds after formation. The muon has a mass 206.8 times that of an electron. Calculate the deBroglie wavelength associated with a muon traveling at 8.85 x 105 cm/s. © 2009, Prentice-Hall, Inc.

  31. h 4 (x) (mv)  The Uncertainty Principle • Heisenberg showed that the more precisely the momentum of a particle is known, the less precisely is its position known: • In many cases, our uncertainty of the whereabouts of an electron is greater than the size of the atom itself! © 2009, Prentice-Hall, Inc.

  32. Sample ExerciseP 338, # 59 • Using Heisenberg uncertainty principle, calculate ∆x for each of the following: a) electron with a v = 0.100 m/s © 2009, Prentice-Hall, Inc.

  33. Practice Problems • Complete p 338 #59 & 60 © 2009, Prentice-Hall, Inc.

  34. Quantum Mechanics • Erwin Schrödinger developed a mathematical treatment into which both the wave and particle nature of matter could be incorporated. • It is known as quantum mechanics. © 2009, Prentice-Hall, Inc.

  35. Quantum Mechanics • The wave equation is designated with a lower case Greek psi (). • The square of the wave equation, 2, gives a probability density map of where an electron has a certain statistical likelihood of being at any given instant in time. © 2009, Prentice-Hall, Inc.

  36. Quantum Numbers • Solving the wave equation gives a set of wave functions, or orbitals, and their corresponding energies. • Each orbital describes a spatial distribution of electron density. • An orbital is described by a set of three quantum numbers. © 2009, Prentice-Hall, Inc.

  37. Principal Quantum Number (n) • The principal quantum number, n, describes the energy level on which the orbital resides. • The values of n are integers ≥ 1. © 2009, Prentice-Hall, Inc.

  38. Angular Momentum Quantum Number (l) • This quantum number defines the shape of the orbital. • Allowed values of l are integers ranging from 0 to n −1. • We use letter designations to communicate the different values of l and, therefore, the shapes and types of orbitals. © 2009, Prentice-Hall, Inc.

  39. Angular Momentum Quantum Number (l) © 2009, Prentice-Hall, Inc.

  40. Magnetic Quantum Number (ml) • The magnetic quantum number describes the three-dimensional orientation of the orbital. • Allowed values of ml are integers ranging from -l to l: −l ≤ ml≤ l. • Therefore, on any given energy level, there can be up to 1 s orbital, 3 p orbitals, 5 d orbitals, 7 f orbitals, etc. © 2009, Prentice-Hall, Inc.

  41. Magnetic Quantum Number (ml) • Orbitals with the same value of n form a shell. • Different orbital types within a shell are subshells. © 2009, Prentice-Hall, Inc.

  42. s Orbitals • The value of l for s orbitals is 0. • They are spherical in shape. • The radius of the sphere increases with the value of n. © 2009, Prentice-Hall, Inc.

  43. s Orbitals Observing a graph of probabilities of finding an electron versus distance from the nucleus, we see that s orbitals possess n−1 nodes, or regions where there is 0 probability of finding an electron. © 2009, Prentice-Hall, Inc.

  44. p Orbitals • The value of l for p orbitals is 1. • They have two lobes with a node between them. © 2009, Prentice-Hall, Inc.

  45. d Orbitals • The value of l for a d orbital is 2. • Four of the five d orbitals have 4 lobes; the other resembles a p orbital with a doughnut around the center. © 2009, Prentice-Hall, Inc.

  46. Energies of Orbitals • For a one-electron hydrogen atom, orbitals on the same energy level have the same energy. • That is, they are degenerate. © 2009, Prentice-Hall, Inc.

  47. Energies of Orbitals • As the number of electrons increases, though, so does the repulsion between them. • Therefore, in many-electron atoms, orbitals on the same energy level are no longer degenerate. © 2009, Prentice-Hall, Inc.

  48. Spin Quantum Number, ms • 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. © 2009, Prentice-Hall, Inc.

  49. Spin Quantum Number, ms • This led to a fourth quantum number, the spin quantum number, ms. • The spin quantum number has only 2 allowed values: +1/2 and −1/2. © 2009, Prentice-Hall, Inc.

  50. Pauli Exclusion Principle • No two electrons in the same atom can have exactly the same energy. • Therefore, no two electrons in the same atom can have identical sets of quantum numbers. © 2009, Prentice-Hall, Inc.

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