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Chapter 7 The Quantum– Mechanical Model of the Atom Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA Chemistry: A Molecular Approach,

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Presentation on theme: "Chapter 7 The Quantum– Mechanical Model of the Atom Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA Chemistry: A Molecular Approach,"— Presentation transcript:

1 Chapter 7 The Quantum– Mechanical Model of the Atom Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA Chemistry: A Molecular Approach, 2nd Ed. Nivaldo Tro

2 Electromagnetic Radiation or “Light” is composed of two orthogonal vectors: electric wave magnetic wave An electric wave and a magnetic wave. Electromagnetic Radiation

3 Light is characterized by its wavelength and frequency. Electromagnetic Radiation

4 wavelength Ultraviolet radiation Amplitude Node The intensity of light is a function of the wave’s amplitude. A point of zero amplitude is called a “node”. Electromagnetic Radiation

5 The frequency of light is represented by the Greek letter “nu”, has units of “cycles per sec” or Hertz (s -1 ) All radiation obeys the relationship:   = c Long wavelength, low frequency Short wavelength, high frequency Wavelength and frequency are inversely proportional. Electromagnetic Radiation

6 Wavelength has units of length: m…  m… nm… pm Frequency has units of inverse time: s -1 or Hz (hertz) (m)  (s –1 ) = c (m s –1 ) c c, the speed of electromagnetic radiation (light) moving through a vacuum is: 2.99792458  10 8 m/s or Electromagnetic Radiation

7 Long wavelength = low frequency Short wavelength = high frequency increasing frequency increasing wavelength Electromagnetic Radiation

8 The visible region of the electromagnetic spectrum is only a small portion of the entire spectrum. Electromagnetic Radiation

9 Problem: Visible red light has a wavelength ( of 685 nm, calculate the frequency. Electromagnetic Radiation

10 Max Planck (1858-1947) proposed that light waves existed as discrete packets of energy, “quanta” in order to account for the “ultraviolet catastrophe” predicted by classical physics. The “ultraviolet catastrophe” arises from the classical theory for the energy emitted by an ideal black- body governed by the Rayleigh-Jeans law. According to classical physics, the intensity of emitted light approaches infinity as the wavelength of the light approaches zero, hence the term catastrophe. Quantization of Energy

11 E = h · E = h · Energy of radiation is proportional to frequency h = Planck’s constant = 6.6262 x 10 -34 J·s QUANTA An object can gain or lose energy by absorbing or emitting radiant energy in QUANTA. A quanta of energy is the smallest unit of energy that may be exchanged between oscillators or emitted as radiation. It is too small to be observed in the classical world in which we live. Quantization of Energy

12 Light with a short (high ) has a high Energy Light with long (low ) has a low Energy E = h · Quantization of Energy

13 Practice – Calculate the wavelength of a radio signal with a frequency of 100.7 MHz the unit is correct, the wavelength is appropriate for radiowaves Check: Solve: ∙ = c, 1 MHz = 10 6 s −1 Conceptual Plan: Relationships: = 100.7 MHz, (m) Given: Find:  (MHz) (s −1 ) (m) 13 Tro: Chemistry: A Molecular Approach, 2/e

14 Blue Light, Blue Light, (higher frequency) has more energy than Red Light, Red Light, with a lower frequency. and yields: frequencyAs the frequency of light increases, the energy of the photon increases wavelengthAs the wavelength of light increases, the energy of the photon decreases. Planck’s Law

15 Problem: Calculate the energy of 1.00 mol of photons of red light at 685 nm units of kJ/mol. Energy of Radiation The energy of a photon in units of kJ/mol can be determined by converting units of wavelength to frequency to energy using Planck’s law and Avogadro's number via dimensional analysis.

16 Energy of Radiation Problem: Calculate the energy of 1.00 mol of photons of red light at 685 nm units of kJ/mol.

17 Problem: Energy of Radiation

18 Problem: Calculate the energy of 1.00 mol of photons of red light at 685 nm units of kJ/mol. Energy of Radiation

19 Problem: Calculate the energy of 1.00 mol of photons of red light at 685 nm units of kJ/mol. Energy of Radiation

20 Problem: Calculate the energy of 1.00 mol of photons of red light at 685 nm units of kJ/mol. This is in the range of energies that can break chemical bonds! Energy of Radiation The units cancel leaving J/mol of photon.

21 21 Practice – What is the frequency of radiation required to supply 1.0 x 10 2 J of energy from 8.5 x 10 27 photons? Solve: E=h, E total = E photon ∙# photons Conceptual Plan: Relationships: E total = 1.0 x 10 2 J, number of photons = 8.5 x 10 27 Given: Find: (s −1 ) E photon number photons Tro: Chemistry: A Molecular Approach, 2/e

22 Certain metals will release (eject) electrons when light strikes the metal surface. The energy of the light must exceed a minimum or “threshold energy” for this to occur. Any excess energy beyond this minimum goes into the kinetic energy of the ejected electron. (They fly away with greater velocity) A. Einstein (1879-1955) Photoelectric Effects

23 Classical theory suggests that energy of an ejected electron should increase with an increase in light intensity. This however is not experimentally observed! No ejected electrons were observed until light of a certain minimum energy is applied. Number of electrons ejected depends on light intensity so long as the light is above a minimum energy. (This “minimum energy” is also the ionization energy of the metal.) A. Einstein (1879-1955) Photoelectric Effect

24 Experiment demonstrates the particle like nature of light. Photoelectric Effect

25 Conclusion: There is a one-to-one correspondence between ejected electrons and light waves. “PHOTONS”This can only occur if light consists of individual units called “PHOTONS”. A Photon is a packet of light of discrete energy. Photoelectric Effect

26 Problem: You are an engineer that is designing a switch that works via the photoelectric effect. The metal used requires 6.7  10 -19 J/atom to eject electrons. Will the switch work with light of 540 nm or greater? Why or why not? Photoelectric Effect

27 Problem: You are an engineer that is designing a switch that works via the photoelectric effect. The metal used requires 6.7  10 -19 J/atom to eject electrons. Will the switch work with light of 540 nm or greater? Why or why not? Solution: Determine if the energy of the light is greater than the minimum threshold energy. If so, then electrons will be ejected, if not, then the switch will not work. Photoelectric Effect

28 Problem: You are an engineer that is designing a switch that works via the photoelectric effect. The metal used requires 6.7  10 -19 J/atom to eject electrons. Will the switch work with light of 540 nm or greater? Why or why not? Photoelectric Effect

29 Problem: You are an engineer that is designing a switch that works via the photoelectric effect. The metal used requires 6.7  10 -19 J/atom to eject electrons. Will the switch work with light of 540 nm or greater? Why or why not? 540 nm = 3.7  10  19 J Photoelectric Effect

30 Conclusion: The energy of the light is below the minimum threshold. No ejection of electrons will occur. The incident light must have a  297 nm to eject electrons. (Confirm this on your own.) 3.7  10 -19 J < 6.7  10 -19 J Photoelectric Effect Problem: You are an engineer that is designing a switch that works via the photoelectric effect. The metal used requires 6.7  10 -19 J/atom to eject electrons. Will the switch work with light of 540 nm or greater? Why or why not?

31 line spectra Bohr is credited with the first modern model of the hydrogen atom based on the “line spectra” of atomic emission sources. planetary He proposed a “planetary” structure for the atom where the electrons circled the nucleus in defined orbits. In this model, the attractive electrostatic forces of the electron and nucleus were balanced by the centripetal forces of the orbiting electron. Niels Bohr (1885-1962) Atomic Line Emission Spectra and Niels Bohr

32 Spectrum of White Light When white light passes through a prism, all the colors of the rainbow are observed.

33 When the light from a discharge tube containing a pure element (hydrogen in this case) is passed through the same prism, only certain colors (lines) are observed. Recall that color (wavelength) is related to energy via Planck’s law. Spectrum of Excited Hydrogen Gas

34 Excited atoms emit light of only certain wavelengths The wavelengths of emitted light are unique to each individual element. Line Emission Spectra of Excited Atoms

35 + orbits Bohr asserted that line spectra of elements indicated that the electrons were confined to specific energy states called orbits. The orbits or energy levels are “quantized” such that only certain levels are allowed. n = 1, 2, 3...  The Bohr Model: r n = n 2 a o a o = Bohr radius (53 pm) Atomic Spectra & Bohr Model

36 + orbits Bohr asserted that line spectra of elements indicated that the electrons were confined to specific energy states called orbits. The lines (colors) corresponded to “jumps” or transitions between the levels. Atomic Spectra & Bohr Model

37 Quantum Mechanical Explanation of Atomic Spectra Each wavelength in the spectrum of an atom corresponds to an electron transition between orbitals When an electron is excited, it transitions from an orbital in a lower energy level to an orbital in a higher energy level When an electron relaxes, it transitions from an orbital in a higher energy level to an orbital in a lower energy level When an electron relaxes, a photon of light is released whose energy equals the energy difference between the orbitals 37 Tro: Chemistry: A Molecular Approach, 2/e

38 Balmer The “Balmer” series for the hydrogen atom is in the visible region of the spectrum. A “series” of transitions end with a common lower level. Origin of Line Spectra

39 Each element has a unique line spectrum. The lines indicate that the electrons can only make “jumps” between allowed energy levels. Knowing the color (wavelength) on can determine the magnitude of the energy gaps using Planck's Law. Line Spectra of Other Elements

40 BALMER Visible lines in H atom spectrum are called the BALMER series. High E Short High Low E Long Low Line Emission Spectra of Excited Atoms

41 The energy of each level is given by: R = Rydberg constant (1.097  10 7 m -1 ) h = Planck’s constant (6.626  10 -34 Js) c = speed of light (2.997  10 8 ms -1) n = the quantum level of the electron (1, 2, 3…  ) The sign of E n is negative because the potential energy between the electron and the nucleus is attractive. The Bohr Model of the Atom

42 The location of electrons in an energy level is indicated by assigning a number n. The value of n can be 1, 2, 3, 4, etc. “shell”The higher the n value, the higher is the energy of the “shell” that that the electrons occupy. shell Each shell can be thought of as a step on a ladder n = 1 n = 2 n = 3 n =  Energy Levels

43 The spacing between adjacent levels is given by: (as n increases, the levels get closer together) virtual continuum of levels n = 1 n = 2 n = 3 n = 4 Energy n =  between n = 1 and 2: between n = 2 and 3: Energy Levels

44 Key terms and Vocabulary: Ground State Ground State: The lowest energy level (n = 1) Excited State Excited State: A subsequently higher energy level. n = 2 is the “first excited state” and so on. Absorption Absorption: An electron moving from a lower energy level to a higher energy level via excitation. Emission Emission: An electron moving from a higher to a lower energy level accompanied by the release of a photon. Energy Levels

45 Since the gaps between states get closer and closer together with increasing n, the frequency of the light emitted changes. Energy Absorption/Emission

46 Problem: Determination the photon wavelength of a transition between two energy levels:

47 Problem:

48 Problem:

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50 Problem:

51 Problem:

52 Problem:

53 Problem:

54 Problem:

55 Problem: The sign of  E tells the direction: (+) indicated absorption (-) indicates emission take the absolute value to assure a positive wavelength.

56 Problem: Determination the photon wavelength of a transition between two energy levels: n final = 5n initial = 2 The value of  E is positive because this is an absorption. n =2 n =5 energy in

57 Problem: Determination the photon wavelength of a transition between two energy levels: Watch your math… =

58 Problem: Determination the photon wavelength of a transition between two energy levels: = 434.0 nm

59 Louis de Broglie in response to Planck & Einstein’s assertion that light was “particle-like” (photon) stated that small particles should exhibit a characteristic wavelength. L. de Broglie (1892-1987) Particle-Wave Duality: A Prelude to Quantum Mechanics

60 L. de Broglie (1892-1987) Conclusion: Light waves have mass, particles have a wavelength. Particle-Wave Duality: A Prelude to Quantum Mechanics

61 Problem: What is the wavelength associated with an 80g tennis ball (d = 8 cm) moving at 115 miles per hour? (tennis ball) = 2 x 10 –34 m The ball is a bajillion times the size of the wavelength !!! Compare that to an electron moving at the same velocity: (electron) = 1.4 x 10 –5 m By golly the electron looks a little like a wave !!! No wonder we can’t see it.

62 Determinacy vs. Indeterminacy According to classical physics, particles move in a path determined by the particle’s velocity, position, and forces acting on it –determinacy = definite, predictable future Because we cannot know both the position and velocity of an electron, we cannot predict the path it will follow –indeterminacy = indefinite future, can only predict probability The best we can do is to describe the probability an electron will be found in a particular region using statistical functions 62 Tro: Chemistry: A Molecular Approach, 2/e

63 Trajectory vs. Probability 63 Tro: Chemistry: A Molecular Approach, 2/e

64 According to Heisenberg the limits of quantum mechanics state that it is impossible to determine simultaneously both the position and velocity of an electron or any other particle with any great degree of accuracy or certainty. Therefore an electron is both a particle and a wave simultaneously. W. Heisenberg 1901-1976 Uncertainty Principle

65 Taking on the ideas of Bohr, de Broglie and Heisenberg, Irwin Schrödinger proposed that matter can be described as a wave. In this theory, the electron is treated as both a wave and a particle. Wave FunctionAn electron is described by a Wave Function “  ” that completely defines a system of matter. E. Schrodinger 1887-1961 Wave or Quantum Mechanics

66 Schrödinger’s Equation Schödinger’s Equation allows us to calculate the probability of finding an electron with a particular amount of energy at a particular location in the atom Solutions to Schödinger’s Equation produce many wave functions,  A plot of distance vs.  2 represents an orbital, a probability distribution map of a region where the electron is likely to be found 66 Tro: Chemistry: A Molecular Approach, 2/e

67 Solutions to the Wave Function,  Calculations show that the size, shape, and orientation in space of an orbital are determined to be three integer terms in the wave function –added to quantize the energy of the electron These integers are called quantum numbers –principal quantum number, n –angular momentum quantum number, l –magnetic quantum number, m l 67 Tro: Chemistry: A Molecular Approach, 2/e

68 Wave motion: wave length and nodes “Quantization” in a standing wave

69 The mental picture of an electron corresponds to a wave superimposed upon the radial trajectory of a particle orbiting the nucleus. The position of an electron is found from the solutions to the Schrödinger Wave Equation which predict the “probability” in a region space where the electron is likely to be found. The orbitals are really Probability Distributions Wave Functions, Ψ

70 Types of Orbitals The solutions to the Schrödinger equation yields the probability in 3-dimensons for the likelihood of finding and electron about the nucleus. It is these probability functions that give rise to the familiar hydrogen- like orbitals that electrons occupy.

71 Principal Quantum Number, n Characterizes the energy of the electron in a particular orbital –corresponds to Bohr’s energy level n can be any integer  1 The larger the value of n, the more energy the orbital has Energies are defined as being negative –an electron would have E = 0 when it just escapes the atom The larger the value of n, the larger the orbital As n gets larger, the amount of energy between orbitals gets smaller 71 Tro: Chemistry: A Molecular Approach, 2/e

72 Principal Energy Levels in Hydrogen 72 Tro: Chemistry: A Molecular Approach, 2/e

73 Angular Momentum Quantum Number, l The angular momentum quantum number determines the shape of the orbital l can have integer values from 0 to (n – 1) Each value of l is called by a particular letter that designates the shape of the orbital –s orbitals are spherical –p orbitals are like two balloons tied at the knots –d orbitals are mainly like four balloons tied at the knot –f orbitals are mainly like eight balloons tied at the knot 73 Tro: Chemistry: A Molecular Approach, 2/e

74 Magnetic Quantum Number, m l The magnetic quantum number is an integer that specifies the orientation of the orbital –the direction in space the orbital is aligned relative to the other orbitals Values are integers from −l to +l –including zero –gives the number of orbitals of a particular shape »when l = 2, the values of m l are −2, −1, 0, +1, +2; which means there are five orbitals with l = 2 74 Tro: Chemistry: A Molecular Approach, 2/e

75 n defines the Principal energy level “shell” There are n “sub–shells” for each n – level corresponding to l n = 1l = 0 n = 2l = 0 & 1 n = 3l = 0, 1 & 2 Each l is divided into (2l + 1) m l “orbitals” separated by orientation. l = 0m l = 0 l = 1 m l = 0, ±1 l = 2 m l = 0, ±1, ±2 if “n” equals: “l” can have values of: if “l” equals: “m l ” can have values of: Quantum Numbers & Electron Orbitals

76 Each “l” within an “n-level” represents a sub-shell. Each “l” sub-shell is divided into m l degenerate orbitals, where m l designates the spatial orientation of each orbital. Type of orbital# of orbitals l = 0“s” sub-shell (sharp)1 l = 1“p” sub-shell (Principal)3 l = 2“d” sub-shell (diffuse)5 l = 3“f” sub-shell (fine)7 each subshell contains 2l+1 orbitals Quantum Numbers & Electron Orbitals

77 Energy Shells and Subshells 77 Tro: Chemistry: A Molecular Approach, 2/e

78 s orbital p orbital d orbital Types of Orbitals

79 l = 0, m l = 0 2l+1 = 1 one s-orbital that extends in a radial manner from the nucleus forming a spherical shape. s-Orbitals

80 All s-orbitals have “n-1” spherical nodes. A 1s-orbital has none, a 2s has 1 and so on. These nodes represent a sphere of zero probability for finding an electron 2 s orbital Spherical Nodes

81 1s, 2s, 3s-Orbitals

82 When n = 2, then l = 0 and 1 Therefore, in 2 nd shell there are 2 types of orbitals, i.e. 2 sub- shells For l = 0m l = 0 (s-orbital) For l = 1 m l = -1, 0, +1 2(1) + 1 or 3 p-orbitals When n = 2, then l = 0 and 1 Therefore, in 2 nd shell there are 2 types of orbitals, i.e. 2 sub- shells For l = 0m l = 0 (s-orbital) For l = 1 m l = -1, 0, +1 2(1) + 1 or 3 p-orbitals Each p-orbital is characterized by a “nodal plan” p-Orbitals

83 The three degenerate p-orbitals spread out on the x, y & z axis, 90° apart in space. p-Orbitals

84 2p x - & 3p x -Orbital

85 When n = 3, l = 0, 1, 2 resulting in 3 sub-shells within the shell. For l = 0, m l = 0 s-sub-shell with a single s-orbital For l = 1, m l = -1, 0, +1 p-sub-shell with 3 p-orbitals For l = 2, m l = -2, -1, 0, +1, +2 d-sub-shell with 5 d-orbitals 2(2)+1 = 5 d-Orbitals

86 s-orbitals have no nodal planes (l = 0) p-orbitals have one nodal plane (l = 1) d-orbitals therefore have two nodal planes (l = 2) d-Orbitals

87 3d xy, 3d xz, 3d yz -Orbital

88 3d x 2 - y 2 & 3d z 2 -Orbital

89 When n = 4, l = 0, 1, 2 & 3 resulting in 4 sub-shells within the shell. For l = 0, ml = 0 s-sub-shell with a single s-orbital s-sub-shell with a single s-orbital For l = 1, ms = -1, 0, +1 p-sub-shell with 3 p-orbitals p-sub-shell with 3 p-orbitals For l = 2, ms = -2, -1, 0, +1, +2 d-sub-shell with 5 d-orbitals d-sub-shell with 5 d-orbitals For l = 3, ms = -3, -2, -1, 0, +1, +2, +3 f-sub-shell with 7 f-orbitals f-sub-shell with 7 f-orbitals 2(3)+1 = 7 2(3)+1 = 7 f-Orbitals

90 All have 3 nodal planes. One of the 7 possible f-orbitals. f-Orbitals

91 Summary of Quantum Numbers

92 Each orbital can accommodate no more than 2 electrons Since each electron is unique, we need a way to distinguish the individual electrons in an orbital from one another. “m s ”This is done via the 4 th quantum number, “m s ”. Arrangement of Electrons in Atoms

93 N S Detector Source of electrons When an atom with one unpaired electron is passed through a magnetic field, the path is altered into two directions. This means that the electrons have a magnetic moment! Electron Spin

94 Since there were 2 pathways in the experiment, there must be 2 spins affected by the magnetic field. One spinning to the right, one spinning to the left. Each “spin state” is assigned a quantum number m s = ± ½ + ½ for “spin up”  ½ for “spin down” Electron Spin

95 Electron Spin Quantum Number, m s The experiment results indicate that electron has an intrinsic property referred to as “spin.” Two spin directions are given by m s where m s = +1/2 and -1/2. Electron Spin

96 It is not that the electrons are actually spinning on axis... Rather it is that the mathematics that describe the electrons “looks” like they are spinning on axis. Electron Spin

97 Diamagnetic SubstancesDiamagnetic Substances: Are NOT attracted to a magnetic field Paramagnetic SubstancesParamagnetic Substances: ARE attracted to a magnetic field. unpaired electronsSubstances with unpaired electrons are paramagnetic. Electron Spin & Magnetism

98 When a substance containing unpaired electrons is place into a magnetic field, it is attracted to that magnetic field. The effect is proportional to the number of unpaired electrons. Measuring Paramagnetism

99 Principal (n) shellsn = 1, 2, 3, 4,... Angular (l) sub-shellsl = 0, 1, 2,... n - 1 Magnetic (m l ) orbitals m l = - l... 0... + l Spin (m s ) individual electrons m s =  ½ Electron Quantum Numbers


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