1. Quantum Theory and the Electronic Structure of Atoms Chapter 7 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2. What we're going to learn Basic calculations involving waves Planck’s Quantum Theory: Radiation energy is dependent on wavelength Energy is emitted and absorbed in discrete units called quanta Einstein's explanation of the photoelectric effect Bohr’s description of the hydrogen atom Particle/wave duality of electromagnetic radiation and electrons Heisenberg’s uncertainty principle: a tool or the limits of knowledge? Schrodinger’s wave equations explaining quantized energy levels for an atom’s electrons known as orbitals Use this information to begin building elements

3. Properties of Waves Wavelength ( ) is the distance between identical points on successive waves. Amplitude (A) is the vertical distance from the midline of a wave to the peak or trough. 7.1 http://www.ies.co.jp/math/products/trig/applets/graphSinX/graphSinX.html

4. Properties of Waves Frequency ( ) is the number of waves that pass through a particular point in 1 second (Hz = 1 cycle/second (units: sec -1 ) The speed (u) of the wave = x Speed = wavelength X frequency u = 7.1 Examples of waves: sound, water ripples and surf, earthquakes, weather, jump ropes, springs and slinkies, gravitational waves, inertial waves, you? ….

5. Speed = wavelength X frequency u = Example 1; sound: frequency (20 to 20,000 Hz); speed ~ 340 m/s or 770 mph; wavelength ( 17 m to 1.7 cm) (In dry air, at a temperature of 21 °C). Example 2; visible light: frequency (Hz); ~ 10 14 Hz speed ~ 300,000 m/s; ~ 3 X 10 8 m/s) wavelength (400 to 700 nm; nm = 10 -9 m) (in a vacuum). #1

6. Visible light

7. 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 7.1

8. Electromagnetic Radiation

9. x = c = c/ = 3.00 x 10 8 m/s / 6.0 x 10 4 Hz = 5.0 x 10 3 m Radio wave An EM wave 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 7.1 #2-3

10. Classical Physics: Planetary Model of the Atom 7.3 “Electrical force is like gravity.” The planets orbit the much more massive Sun in (mostly) circular orbits. The electrons should orbit the much more massive nucleus in (mostly) circular orbits. The total energy of a planet only depends on its orbital radius (higher energy, bigger radius); the same should be true of the electrons in an atom.

11. “Classical” physics applied to atoms 2. Photoelectric Effect. has some serious problems 1.Thermal radiation. –The energy given off is called thermal radiation because it depends on temperature. –Accelerating electric charges give off light and lose energy. –Kinetic molecular theory says that atoms in a gas are constantly being accelerated (changing direction, changing speed in collisions). –Classical physics says that thermal radiation should get more powerful at higher frequency, and the total energy emitted should be infinite! –Shining light on certain metals causes them to emit electrons –The speed of the emitted electrons depends on color of the light, not its brightness. –Brightness only affects number of electrons coming off. –Called “black body problem” or “UV catastrophe”.

12. 4.Line spectra. 3.Atoms should be unstable. –Electrons should quickly (within milliseconds) spiral to their doom in the nucleus. –Moving in circles is a constant acceleration (change of direction). –Electrons in atoms should constantly give off light and lose energy. –Electrically energized gas atoms give off light that is composed only of a set of particular colors/frequencies. –Classical physics says they should give off all frequencies (continuous spectra).

13. 1.Thermal radiation. –Classical physics says that thermal radiation should get more powerful at higher frequency, and the total energy emitted should be infinite! 2. Photoelectric Effect. –Shining light on certain metals causes them to emit electrons –The speed of the emitted electrons depends on color of the light, not its brightness. –Brightness only affects number of electrons coming off. 3.Atoms should be unstable. –Electrons in atoms should constantly give off light and lose energy. 4.Line spectra. –Electrically energized gas atoms give off light that is composed only of a set of particular colors/frequencies. –Classical physics says they should give off all frequencies (continuous spectra).

14. Solution? Then (perhaps)… energy is emitted or absorbed in discrete indivisible portions called “quanta”. E = h x Planck’s constant (h) h = 6.63 x 10 -34 J s 7.1 The “Quantum” Theory of Physics If… matter comes in discrete indivisible portions called “atoms”…, Light energy quanta are called “photons” Light of one color/frequency is composed of photons of one fixed energy (E). #4

15. u = c = Equation review Speed = (wavelength)X(frequency) m/s = (m) X (sec -1 ) E = h  = h c/ C = 3 X 10 8 m/sec; h = 6.63 x 10 -34 (J s) General form For electromagnetic waves Energy as a function of wavelength or frequency Units?

16. 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 /  7.2 When copper is bombarded with high-energy electrons, X-rays are emitted with a wavelength of 0.154 nm. Calculate the energy (in joules) associated with the photons. Light has both: 1.wave nature 2.particle nature #4

17. Mystery #1, “Ultraviolet Catastrophe” Solved by Planck in 1900 7.1 Classical physics says that atoms can emit or absorb an arbitrary amount of radiant energy. Planck said that atoms emit and absorb on in discrete units within a limited ranges characteristic of a particular atom or molecule, making the total emitted radiation finite. Does the “quantum” theory solve the four problems of “classical” physics?

18. h = KE + BE Mystery #2, “Photoelectric Effect” Solved by Einstein in 1905 Photon is a “particle” of light KE = h - BE h KE e - 7.2 Where BE is the binding energy of the atom, i.e. a measure of how strongly the electron is held in the atom. Number of ejected electrons proportional to the intensity, BUT the energy of the emitted electrons is a function of the frequency and only occurred above a certain ‘threshold’ frequency

19. What about Mysteries #3 & #4? 7.1 Atomic stability? Line spectra?

20. What Newton demonstrated: The Sun’s emission spectra

21. 7.3 Line Emission Spectrum of Hydrogen Atoms Hydrogen emission spectra

22. 7.3

23. 1.e - can only have specific (quantized) energy values 2.Light, as a single photon, 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,… integer values only R H (Rydberg constant) = 2.18 x 10 -18 J (for Hydrogen) 7.3

24. E = h 7.3

25. E photon = -  E = E i - E f E f = -R H ( ) 1 n2n2 f E i = -R H ( ) 1 n2n2 i i f  E = R H ( ) 1 n2n2 1 n2n2 n f = 1 n i = 2 n f = 1 n i = 3 n f = 2 n i = 3 7.3 #5

26. E photon = 2.18 x 10 -18 J x (1/16 - 1/36) E photon = -  E = 7.57 x 10 -20 J = 6.63 x 10 -34 (Js) x 3.00 x 10 8 (m/s)/7.57 x 10 -20 J = 2630 nm Calculate the wavelength (in nm) of a photon emitted by a hydrogen atom when its electron drops from the n = 6 state to the n = 4 state. E photon = h x c /  = h x c / E photon ( ) 1 n2n2 1 n2n2 if  E = R H E photon = 7.3 What color is emitted or absorbed for n=2 and n=5?

27. De Broglie (1924) reasoned that if light waves can have properties of particles then perhaps electrons can have properties of waves. e - is both particle and wave, specifically a ‘standing wave’ 2  r = n = h/mu r = radius of the orbit u = velocity of e - m = mass of e - Why is e - energy quantized? 7.4 nodes Relationship between the circumference of an allowed obit and an electron’s wavelength Relationship between wavelength and mass of a particle

28. = 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) 7.4

29. Chemistry in Action: Laser – The Splendid Light Laser light is (1) intense, (2) monoenergetic, and (3) coherent

30. 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. 7.5

31. Schrodinger Wave Equation  f(n, l, m l, m s ) principal quantum number n n = 1, 2, 3, 4, …. n=1 n=2 n=3 7.6 distance of e - from the nucleus

32. e - density (1s orbital) falls off rapidly as distance from nucleus increases Where 90% of the e - density is found for the 1s orbital 7.6

33.  = f(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 l describes the 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 7.6

34.  = f(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 7.6

35. l = 0 (s orbitals) l = 1 (p orbitals) 7.6

36. l = 2 (d orbitals) 7.6

38.  = f(n, l, m l, m s ) spin quantum number m s m s = +½ or -½ Schrodinger Wave Equation m s = -½m s = +½ 7.6

39. 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  = f(n, l, m l, m s ) Each seat is uniquely identified (E, R12, S8) Each seat can hold only one individual at a time 7.6

40. Schrodinger Wave Equation  = f(n, l, m l, m s ) Shell – electrons with the same value of n Subshell – electrons with the same values of n and l Orbital – electrons with the same values of n, l, and m l How many electrons can an orbital hold? If n, l, and m l are fixed, then m s = ½ or - ½  = (n, l, m l, ½ ) or  = (n, l, m l, - ½ ) An orbital can hold 2 electrons 7.6

41. Summary An electron’s wavefunction  (x,y,z) gives the probability of finding the electron in a small volume of space  V centered at point (x,y,z). Probability =  (x,y,z)| 2  V probability density The more intense  is in a region of space, the more likely it is that the electron will be found there if you look for it. Wavefunctions are solutions to the Schrodinger Wave Equation, the basic equation of quantum mechanics. They mathematically describe the behavior of subatomic particles. Quantum mechanics is the theory of physics that best describes the behavior of atomic and sub-atomic particles, like electrons.

42. Summary Electrons obey the Pauli Exclusion Principle. Every electron in an atom must be in a different state, i.e. each electron has a unique wavefunction (from the allowed set), labeled by a unique set of quantum numbers. For electrons bound to nuclei, there is a set of possible wavefunctions an electron may have. Each  is labeled by a set of four “quantum” numbers (n,l,m l,m s ) representing the different allowed behaviors of an atom-bound electron (called states).

43. Summary A wavefunction can have non-zero values at every point in space. There is some chance of finding an electron anywhere. In QM, one cannot pin down the location of an electron to any region of space with certainty. This reflects the Heisenberg Uncertainty Principle of quantum mechanics. However, an atom-bound electron is by far most likely to be found somewhere near (within a few nanometers of) the nucleus of an atom.  drops in intensity rapidly with distance from the nucleus.

44. Summary To visualize where an electron in a given state is most likely to be found, we define a boundary surface diagram. The BSD encloses the region of space where an electron in that state is 90% likely to be found. The BSD is always centered on the nucleus. There is a different BSD for each possible state an electron can have.

45. Summary The principal quantum number n indicates the size of the region the electron is most likely in, i.e. the size of the BSD. It takes positive integer values: n = 1, 2, 3, etc. The angular momentum quantum number l indicates the shape of the region the electron is most likely in. It takes non-negative integer values less than n: l = 0, 1, …, n-1 n=3 n=1 n=2 l = 0 or s l = 1 or p l = 2 or d

46. Summary The magnetic quantum number m l relates to the orientation of the region the electron is most likely in. It takes integer values between –l and +l: m l = -l, -l+1,…, 0,…, +l-1, +l. The spin quantum number m s does not relate to where an electron is likely to be found in space. It refers to the orientation of the electron’s magnetic field. It takes values m s = +1/2 and -1/2, sometimes called “up” and “down”. pxpx pzpz pypy l = 1 or p m l = -1, 0, 1

47. All electrons with the same n are said to be in the same shell. All electrons with the same n and l are said to be in the same subshell. Every shell has n subshells (possible values of l). How many subshell are there in the n = 4 shell of an atom? If n = 4, then l = 0, 1, 2, or 3 The 4s, 4p, 4d, and 4f subshells

48. How many 2p orbitals are there in an atom? 2p n=2 l = 1 If l = 1, then m l = -1, 0, or +1 3 orbitals 7.6 All electrons with the same n, l, and m l are said to be in the same orbital. Every subshell has 2l+1 orbitals (possible values of m l ). (2 x 1+1 = 3)

49. How many electrons can be placed in the 3d subshell? 3d n=3 l = 2 If l = 2, then m l = -2, -1, 0, +1, or +2 5 orbitals which can hold a total of 10 e - Each orbital can only hold at most two electrons, one in the “up” state (m s = +1/2) and one in the “down” state (m s = -1/2).

50. 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 7.7

51. 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 7.7

52. Paramagnetic unpaired electrons 2p Diamagnetic all electrons paired 2p 7.8

53. “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 ?? 7.7

54. 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 7.7

55. Order of orbitals (filling) in multi-electron atom 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s 7.7

56. What is the electron configuration of Ti? Ti 22 electrons 1s < 2s < 2p < 3s < 3p < 4s < 3d [Ti] = 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 2 2 + 2 + 6 + 2 + 6 + 2 + 2 = 22 electrons 7.7 Abbreviated as [Ar]4s 2 3d 2 [Ar] = 1s 2 2s 2 2p 6 3s 2 3p 6 What are the possible quantum numbers for the last (outermost) electron in Cl? Cl 17 electrons1s < 2s < 2p < 3s < 3p < 4s [Cl] = 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

57. Outermost subshell being filled with electrons 7.8

58. Tab. 7.3 Slightly more stability in half- filled and completely filled d orbitals 1 st Transition metal series: Incompletely filled d-subshells Lanthanides (rare earths) : Incompletely filled 4f-subshells

60. Eqn. 7.1 KEY EQUATIONS c =

61. Eqn. 7.2 Calculating energy in joules from frequency (sec -1 ) h (Planck’s constant = 6.63 X 10 -34 Joule secs)

62. Eqn. 7.3 Calculating energy in joules from wavelength (m) and the speed of light (c = 3 X 10 8 m/sec) E = h  = h c/

63. Eqn. 7.4 Calculating energy in joules of a particular shell of the hydrogen atom using the principal quantum number (n) and Rydberg’s constant (R H = 2.18 x 10 -18 J)

64. Eqn. 7.5 Calculating energy in joules and the frequency of a transition of an electron from one shell to another in a hydrogen atom using the principal quantum number (n) and Rydberg’s constant (R H = 2.18 x 10 -18 J)

65. Eqn. 7.6 De Broglie’s equation for calculating the wavelength of a moving mass

66. "The more precisely the POSITION is determined, the less precisely the MOMENTUM is known" Heisenberg’s Uncertainty Principle  x  p  h / 4 

67. I knew of [Heisenberg's] theory, of course, but I felt discouraged, not to say repelled, by the methods of transcendental algebra, which appeared difficult to me, and by the lack of visualizability. -Schrödinger in 1926 The more I think about the physical portion of Schrödinger's theory, the more repulsive I find it...What Schrödinger writes about the visualizability of his theory 'is probably not quite right,' in other words it's crap. --Heisenberg, writing to Pauli, 1926 Also a battle of personalities, jobs and money…

68. I believe that the existence of the classical "path" can be pregnantly formulated as follows: The "path" comes into existence only when we observe it. --Heisenberg, in uncertainty principle paper, 1927

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70. The colors in order of discovery were blue (Mercury), white (CO2), gold (Helium), red (Neon), and then different colors from phosphor-coated tubes.

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