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AP Chemistry Quantum Theory and Atomic Structure.

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Presentation on theme: "AP Chemistry Quantum Theory and Atomic Structure."— Presentation transcript:

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2 AP Chemistry Quantum Theory and Atomic Structure

3 Late 1880s…physicists thought they knew everything…discouraged students from studying physics in college CLASSICAL THEORY OF PHYSICS MATTERENERGY -particulate-continuous -massive-wavelike

4 Background: WAVES

5 A disturbance traveling through a medium by which energy is transferred from one particle of the medium to another without causing any permanent displacement of the medium itself. Waves

6 Parts of A Wave wavelength distance between two consecutive peaks or troughs in a wave symbol: lambda, units: meters (m) frequency the number of waves passing a point in a given amount of time symbol: nu, units: cycles/sec, 1/sec, sec -1, Hertz (Hz) peak/crest trough node

7 Figure 7.1 Frequency and Wavelength c = 

8 Figure 7.2 Amplitude (Intensity) of a Wave

9 Speed of radiation = wavelength · frequency c =  where c = 3 X 10 8 m/s (speed of light, all EM radiation in a vacuum) Electromagnetic Radiation

10 The Electromagnetic Spectrum

11 Sample Problem 7.1Interconverting Wavelength and Frequency PROBLEM: A dental hygienist uses x-rays ( = 1.00A) to take a series of dental radiographs while the patient listens to a radio station ( = 325cm) and looks out the window at the blue sky ( = 473nm). 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.)

12 Energy as a Wave Diffraction of Light Change in the directions and intensities of a group of waves after passing by an obstacle or through an aperture (slit) whose size is approximately the same as the wavelength of the waves

13 Figure 7.4 Different behaviors of waves and particles.

14 Prism used to separate light into wavelengths

15 A single point source and the resulting wave fronts

16 Figure 7.5 The diffraction pattern caused by light passing through two adjacent slits.

17 Wave interference pattern produced by two point sources

18 Constructive Interference

19 Destructive Interference http://www.youtube.com/watch?v=P_rK66GFeI4 Stadium waves

20 Young’s Double Slit Exp. (1802) Confirms that light is an electromagnetic wave

21 Energy as a PARTICLE Heat a solid and it starts to glow Why does the wavelength change with Temp? Hot coals ~1000 KElectric heating coil ~1500 KLightbulb filament~2000 K

22 Max Planck (1903) - black body radiation Observations: Energy is gained and lost in whole numbers Intensity and frequency (color) of radiation depend on temperature Black body: a perfect absorber of radiation; object appears Black when cold and emits a temperature dependent spectrum of light Snow is a black body for IR radiation, but not for visible

23 Energy as a PARTICLE Heat a solid and it starts to glow Why does the wavelength change with Temp? If an atom can only absorb certain wavelengths of energy, then it can only emit certain wavelengths Hot coals ~1000 KElectric heating coil ~1500 KLightbulb filament~2000 K

24 Planck’s Quantum theory of EM waves Light energy is transmitted in discrete “packets” (photons) called quanta (singular is quantum) The energy of one quantum: E = h  h = Planck’s constant 6.63 x 10 -34 J·s Niels Bohr and Max Planck at MIT

25 When EM radiation above a certain frequency is shined on the device, an electric current registers on the meter As frequency increases, the current increases BUT – below the cutoff frequency, no current is obtained, even at very high intensities!! “Photoelectric Effect” Einstein (1905) Conclusion: Photon is a “particle” of light with energy E= h

26 Sample Problem 7.2Calculating 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?

27 Planck and Einstein at Nobel Conference

28 Dual Nature of Light Dual Nature of Light Light has both: wave nature particle nature

29 ATOMIC THEORY

30 Determination of the Structure of the Atom Rutherford and Geiger. 1897- J.J. Thompson- Cathode Ray Tube Determined the charge/mass ratio of an electron. Suggested plum pudding model for atom 1911- Ernest Rutherford- Gold Foil Experiment Suggested presence of a positively charged nucleus

31 Figure 7.8 The line spectra of several elements

32 Colors of “neon” lights Red = Ne Others = Ar, Hg, phosphor

33 Continuous Spectrum vs. Line Spectrum

34 e - can only have specific (quantized) energy values The e-’s energy correspond to orbits around the nucleus. Outer orbits have higher energy light is emitted as e - moves from higher energy level to lower energy level Photon E= h Niels Bohr’s Model of the Atom (1913) Ground State: n = Excited State: n Ionized: n = 1 >1 ∞

35 Figure 7.10 Quantum staircase

36 Problem – model does not work for atoms with more than 1 e- !!! E = h

37 = RRydberg equation - 1 1 n22n22 1 n12n12 R is the Rydberg constant = 1.09 × 10 7 m -1 Figure 7.9 Three series of spectral lines of atomic hydrogen for the visible series, n 1 = 2 and n 2 = 3, 4, 5,...

38 Figure 7.11 The Bohr explanation of the three series of spectral lines.

39 Figure B7.2 Figure B7.1 Emission and absorption spectra of sodium atoms. Flame tests strontium 38 Srcopper 29 Cu

40 7.3 Bright Line Spectra of Several Elements

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42 E n = -R H ( ) Z2Z2 n2n2 R H = 2.18 x 10 -18 J n (principal quantum number) = 1,2,3,… Z = Atomic Number (Hydrogen, Z = 1) E n = -R H ( ) 1 n2n2

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44 Sample Problem 7.3 A hydrogen atom absorbs a photon of visible light and its electron transitions from the n = 1 to n = 4 energy level. Calculate (a)The change of energy of the atom, and (b)The wavelength (in nm) of the photon

45 Figure B7.3 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.

46 - Maybe e - is also a wave!!! Q: Why is e - energy quantized? Louis De Broglie (1924) 1802 - T. Young - Light is a wave 1905 - A. Einstein - Light is also a particle 1899 – J.J. Thomson - e - is a particle

47 = h /mu u = velocity of e - m = mass of e - Wave-likeParticle-likeProperties only certain frequencies can work in a circle with a particular radius

48 Sample Problem 7.4 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).

49 Table 7.1 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/m u

50 G.P. Thomson (1925) Thomson’s diffraction apparatus The e - is also a wave! Exhibits properties similar to those of x-rays

51 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 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 7.15 Summary of the major observations and theories leading from classical theory to quantum theory.

52 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 (momentumdecreases) after colliding with electron Figure 7.15 continued QUANTUM THEORY Energy same as Matter particulate, massive, wavelike

53 AP Chemistry Heisenberg Uncertainty Principle  x  p > h /4  "The more precisely the _____________ of the e- is determined, the less precisely the _________________ is known" POSITION MOMENTUM Bohr, Heisenberg, and Pauli Investigates limitations in pinpointing the position of an e- Werner Heisenberg, 1926  xm  u > h /4 

54 A photon of “light” strikes an electron and is reflected (left). In the collision the photon transfers momentum to the electron. The reflected photon is seen through the microscope, but the electron is out of focus (right). Its exact position cannot be determined. Uncertainty Principle

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56 Erwin Schroedinger (1926) Once at the end of a colloquium I heard Debye saying something like: “Schroedinger, you are not working right now on very important problems… why don’t you tell us some time about that thesis of de Broglie’s… In one of the next colloquia, Schroedinger gave a beautifully clear account of how de Broglie associated a wave with a particle, and how he could obtain the quantization rules… When he had finished, Debye casually remarked that he thought this way of talking was rather childish… To deal properly with waves, one had to have a wave equation. Felix Bloch, Address to the American Physical Society, 1976 Attempts to incorporate statistics with DeBroglie’s and Heisenberg’s work

57 In the span of about two weeks, Schroedinger develops his wave equation Schroedinger visits the University of Wisconsin, Madison to deliver a series of lectures in January and February 1927, introducing his his work on wave mechanics to American physicists and chemists “ epoch-making work” M. Planck “true genius” A. Einstein  “Wave Function”  2  -A region in space in which there is a 90% chance of finding an electron -Also called an _______________ ORBITAL

58 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

59 Figure 7.16 Electron probability in the ground-state H atom

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

61 “Anyone who is not shocked by quantum theory does not understand it” --Niels Bohr, 1897 “No one understands quantum theory” --Richard Feynman, 1967

62 Existence (and energy) of electron in atom is described by its unique wave function . No two electrons in an atom can have the same four quantum numbers. Schrodinger Wave Equation  = fn(n, l, m l, m s ) Each seat is uniquely identified (Camp Randall, E, R12, S8) Each seat can hold only one individual at a time

63 Quantum Numbers and Atomic Orbitals An atomic orbital is specified by three quantum numbers. n the principal quantum number - a positive integer l the angular momentum quantum number - an integer from 0 to n-1 m l the magnetic moment quantum number - an integer from - l to + l

64 Principal Quantum Number (Shell)  fn(n, l, m l, m s ) principal quantum number = n n = 1, 2, 3, 4, …. n=1 n=2 n=3 distance of e - from the nucleus

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66 Figure 7.17 1s2s3s

67  = fn(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 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 Angular Momentum Quantum Number (Subshell)

68 l = 0 s orbital l = 1 p orbital l = 2 d orbital l = 3 f orbital

69 Figure 7.18 The 2p orbitals

70 Figure 7.15 A Cross Section of the Electron Probability Distribution for a 3p Orbital

71 Figure 7.19 The 3d orbitals

72 Figure 7.19 continued

73 Figure 7.20 One of the seven possible 4f orbitals

74  = fn(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 Magnetic Quantum Number

75 Quantum Numbers and Atomic Orbitals An atomic orbital is specified by three quantum numbers. n the principal quantum number - a positive integer l the angular momentum quantum number - an integer from 0 to n-1 m l the magnetic moment quantum number - an integer from - l to + l

76 Table 7.2 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

77 Sample Problem 7.5 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.

78 Sample Problem 7.6 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

79 Sample Problem 7.6 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 3 1 3 7

80 Sample Problem 7.8: Identifying Incorrect Quantum Numbers What is wrong with each of the following quantum number designations and/or sublevel names? nlmlml Name a)1101p b)43+14d c)31-23p

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82 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

83 Writing Orbital Diagrams and Electron Configurations Aufbau Principle: Electrons fill from lowest to highest energy levels Hund’s Rule: Each orbital must half fill before it completely fills Pauli Exclusion Principle: No 2 e - s can have the same quantum numbers Each orbital can hold 2 electrons with opposite spin NOT

84 Practice: Write Electron Configurations for: Al Al 3+ Br Br - Cr Cu Cu 1+ Zn 2+ 1s 2 2s 2 2p 6 3s 2 3p 1 1s 2 2s 2 2p 6 OR [Ne]3s 2 3p 1 OR [He]2s 2 2p 6 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 5 OR [Ar]4s 2 4p 5 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 [Ar]4s 2 4p 6 [Ar]4s 1 3d 10 OR [Ar]4s 1 3d 5 [Ar]3d 10 Why does 4s fill before 3d?

85 Chapter 7: Atomic Structure and Periodicity AP Chemistry

86 Outermost subshell being filled with electrons

87 Terms –Isoelectronic: Atoms/Ions that have the same electron configurations –Core electrons: Electrons in the lower energy levels –Valence electrons: Electrons in the outer energy level

88 Quantum Numbers A series of numbers that describe the properties of an orbital The “address of an electron”

89  = fn(n, l, m l, m s ) spin quantum number m s m s = +½ or -½ Spin Quantum Number m s = -½m s = +½ 7.6

90 Paramagnetism attracted to a magnetic field unpaired electrons Diamagnetism repelled by a magnetic field paired electrons Para/Diamagnetism

91 Paramagnetic unpaired electrons 2p Diamagnetic all electrons paired 2p 7.8

92 Substance is placed between electromagnets. If the substance appears heavier, it is attracted in the magnetic field, and is paramagnetic. If the substance appears lighter, it is diamagnetic. Determining Para/Diamagnetism

93 Magnet Video Nitrogen Video Oxygen Video These links need to be fixed

94 NaCl Na + Para or Dia? Cl - CaSO 4 Ca 2+ ZnSO 4 Zn 2+ CuSO 4 Cu 2+ MnSO 4 Mn 2+ MnO 2 Mn ? KMnO 4 Mn ? Write electron configuration for each and predict para- or dia-magnetic 1s 2 2s 2 2p 6 Dia 1s 2 2s 2 2p 6 3s 2 3p 6 Dia 1s 2 2s 2 2p 6 3s 2 3p 6 Dia 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 Dia 1s 2 2s 2 2p 6 3s 2 3p 6 3d 9 Para 1s 2 2s 2 2p 6 3s 2 3p 6 3d 5 Para 1s 2 2s 2 2p 6 3s 2 3p 6 3d 3 1s 2 2s 2 2p 6 3s 2 3p 6 +4 +7 Para Dia


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