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Chapter Seven 1 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Atomic Structure Chapter Seven

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2 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry History: The Classic View of Atomic Structure

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Chapter Seven 3 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Properties of Cathode Rays 1. Cathode rays are emitted from the cathode when electricity is passed through an evacuated tube. 2. The rays are emitted in a straight line, perpendicular to the cathode surface. 3. The rays cause glass and other materials to fluoresce. 4. The rays are deflected by a magnet in the direction expected for negatively charged particles. 5. The properties of cathode rays do not depend on the composition of the cathode. For example, the cathode rays from an aluminum cathode are the same as those from a silver cathode.

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Chapter Seven 4 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Cathode Ray Tube

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Chapter Seven 5 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry J. J. Thomson used the deflection of cathode rays and the magnetic field strength together, to find the cathode ray particle’s mass-to-charge ratio: m e /e = –5.686 × 10 –12 kg/C Investigating Cathode Rays

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Chapter Seven 6 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Mass-to-Charge Ratio of Cathode Rays The ratio m e /e for cathode rays is about 2000 times smaller than the smallest previously known m e /e (for hydrogen ions). 1. If the charge on a cathode ray particle is comparable to that on a H + ion, the mass of a cathode ray particle is much smaller than the mass of H + ; or 2. If the mass of a cathode ray particle is comparable to that of a H + ion, the charge of a cathode ray particle is much larger than the charge on H + ; or 3. The situation is somewhere between the extremes described in the first two statements. To resolve the situation we must know either the mass or the charge of the cathode ray particle.

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Chapter Seven 7 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry George Stoney: names the cathode-ray particle the electron. Robert Millikan: determines a value for the electron’s charge: e = –1.602 × 10 –19 C Millikan’s Oil Drop Experiment Charged droplet can move either up or down, depending on the charge on the plates. Magnitude of charge on the plates lets us calculate the charge on the droplet. Radiation ionizes a droplet of oil.

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Chapter Seven 8 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Thomson determined the mass-to-charge ratio; Millikan found the charge; we can now find the mass of an electron: m e = 9.109 × 10 –31 kg/electron This is almost 2000 times less than the mass of a hydrogen atom (1.79 × 10 –27 kg) Some investigators thought that cathode rays (electrons) were negatively charged ions. But the mass of an electron is shown to be much smaller than even a hydrogen atom, so an electron cannot be an ion. Since electrons are the same regardless of the cathode material, these tiny particles must be a negative part of all matter. Properties of the Electron

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Chapter Seven 9 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Thomson proposed an atom with a positively charged sphere containing equally spaced electrons inside. He applied this model to atoms with up to 100 electrons. J. J. Thomson’s Model of the Atom

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Chapter Seven 10 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry A very few “bounced back” to the source! Most of the alpha particles passed through the foil. A few particles were deflected slightly by the foil. Alpha Scattering Experiment: Rutherford’s observations Alpha particles were “shot” into thin metal foil.

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Chapter Seven 11 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Alpha Scattering Experiment: Rutherford’s conclusions If Thomson’s model of the atom was correct, most of the alpha particles should have been deflected a little, like bullets passing through a cardboard target. Most of the alpha particles passed through the foil => An atom must be mostly empty space. A very few alpha particles bounced back => The nucleus must be very small and massive. The nucleus is far smaller than is suggested here.

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Chapter Seven 12 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Rutherford’s experiments also told him the amount of positive nuclear charge. The positive charge was carried by particles that were named protons. The proton charge was the fundamental unit of positive charge. The nucleus of a hydrogen atom consists of a single proton. Scientists introduced the concept of atomic number, which represents the number of protons in the nucleus of an atom. James Chadwick discovered neutrons in the nucleus, which have nearly the same mass as protons but are uncharged. Protons and Neutrons

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Chapter Seven 13 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Research into cathode rays showed that a cathode- ray tube also produced positive particles. Positive particles Cathode rays Unlike cathode rays, these positive particles were ions. The metal of the cathode: M e – + M + Mass Spectrometry

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Chapter Seven 14 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry In mass spectrometry a stream of positive ions having equal velocities is brought into a magnetic field. All the ions are deflected from their straight line paths. The lightest ions are deflected the most; the heaviest ions are deflected the least. The ions are thus separated by mass. –Actually, separation is by mass-to-charge ratio (m/e), but the mass spectrometer is designed so that most particles attain a 1+ charge. Mass Spectrometry (cont’d)

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Chapter Seven 15 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Stream of positive ions with equal velocities Light ions are deflected greatly. Heavy ions are deflected a little bit. Ions are separated according to mass. A Mass Spectrometer

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Chapter Seven 16 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry A Mass Spectrum for Mercury Mass spectrum of an element shows the abundance of its isotopes. What are the three most abundant isotopes of mercury? Mass spectrum of a compound can give information about the structure of the compound.

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Chapter Seven 17 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Light and the Quantum Theory

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Chapter Seven 18 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Electromagnetic waves originate from the movement of electric charges. The movement produces fluctuations in electric and magnetic fields. Electromagnetic waves require no medium. Electromagnetic radiation is characterized by its wavelength, frequency, and amplitude. The Wave Nature of Light

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Chapter Seven 19 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Simple Wave Motion Notice that the rope moves only up-and- down, not from left- to-right.

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Chapter Seven 20 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry An Electromagnetic Wave The waves don’t “wiggle” as they propagate … … the amplitude of the “wiggle” simply indicates field strength.

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Chapter Seven 21 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Wavelength ( ) is the distance between any two identical points in consecutive cycles. Frequency ( v ) of a wave is the number of cycles of the wave that pass through a point in a unit of time. Unit = waves/s or s –1 (hertz). Wavelength and Frequency

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Chapter Seven 22 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry The relationship between wavelength and frequency: c = v where c is the speed of light (3.00 × 10 8 m/s) Wavelength and Frequency

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Chapter Seven 23 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Example 7.1 Calculate the frequency of an X ray that has a wavelength of 8.21 nm.

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Chapter Seven 24 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Visible light is only a tiny portion of the spectrum. UV, X rays are shorter wavelength, higher frequency radiation. Communications involve longer wavelength, lower frequency radiation. The Electromagnetic Spectrum

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Chapter Seven 25 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Example 7.2 A Conceptual Example Which light has the higher frequency: the bright red brake light of an automobile or the faint green light of a distant traffic signal?

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Chapter Seven 26 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry A Continuous Spectrum White light from a lamp contains all wavelengths of visible light. When that light is passed through a prism, the different wavelengths are separated. We see a spectrum of all rainbow colors from red to violet – a continuous spectrum.

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Chapter Seven 27 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry A Line Spectrum Light from an electrical discharge through a gaseous element (e.g., neon light, hydrogen lamp) does not contain all wavelengths. The spectrum is discontinuous; there are big gaps. We see a pattern of lines, multiple images of the slit. This pattern is called a line spectrum. (duh!)

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Chapter Seven 28 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry The line emission spectrum of an element is a “fingerprint” for that element, and can be used to identify the element! Line Spectra of Some Elements How might you tell if an ore sample contained mercury? Cadmium? Line spectra are a problem; they can’t be explained using classical physics …

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Chapter Seven 29 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry … proposed that atoms could absorb or emit electromagnetic energy only in discrete amounts. The smallest amount of energy, a quantum, is given by: E = hv where Planck’s constant, h, has a value of 6.626 × 10 –34 J·s. Planck’s quantum hypothesis states that energy can be absorbed or emitted only as a quantum or as whole multiples of a quantum, thereby making variations of energy discontinuous. Changes in energy can occur only in discrete amounts. Quantum is to energy as _______ is to matter. Planck …

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Chapter Seven 30 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry The Photoelectric Effect Light striking a photoemissive cathode causes ejection of electrons. Ejected electrons reach the anode, and the result is … … current flow through an external circuit. But not “any old” light will cause ejection of electrons …

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Chapter Seven 31 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry The Photoelectric Effect (cont’d) Light of low frequency does not cause current flow … at all. Each photoemissive material has a characteristic threshold frequency of light. When light that is above the threshold frequency strikes the photoemissive material, electrons are ejected and current flows. As with line spectra, the photoelectric effect cannot be explained by classical physics.

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Chapter Seven 32 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Albert Einstein won the 1921 Nobel Prize in Physics for explaining the photoelectric effect. He applied Planck’s quantum theory: electromagnetic energy occurs in little “packets” he called photons. Energy of a photon (E) = hv The photoelectric effect arises when photons of light striking a surface transfer their energy to surface electrons. The energized electrons can overcome their attraction for the nucleus and escape from the surface … … but an electron can escape only if the photon provides enough energy. The Photoelectric Effect

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Chapter Seven 33 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry The Photoelectric Effect Explained A long wavelength—low frequency—photon doesn’t have enough energy to eject an electron. Short wavelength (high frequency, high energy) photons have enough energy per photon to eject an electron. The electrons in a photoemissive material need a certain minimum energy to be ejected.

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Chapter Seven 34 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Imagine a car stuck in a ditch; it takes a certain amount of “push” to “eject” the car from the ditch. Suppose you push ten times, with a small amount of force each time. Will that get the car out of the ditch? Likewise, ten photons, or a thousand, each with too-little energy, will not eject an electron. Suppose you push with more than the required energy; the car will leave, with that excess energy as kinetic energy. What happens when a photon of greater than the required energy strikes a photoemissive material? An electron is ejected—but with _____ _____ as ______ _____. Analogy to the Photoelectric Effect

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Chapter Seven 35 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Example 7.3 Calculate the energy, in joules, of a photon of violet light that has a frequency of 6.15 × 10 14 s –1. Example 7.4 A laser produces red light of wavelength 632.8 nm. Calculate the energy, in kilojoules, of 1 mol of photons of this red light.

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Chapter Seven 36 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Quantum View of Atomic Structure

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Chapter Seven 37 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Niels Bohr followed Planck’s and Einstein’s lead by proposing that electron energy (E n ) was quantized. The electron in an atom could have only certain allowed values of energy (just as energy itself is quantized). Each specified energy value is called an energy level of the atom: E n = –B/n 2 –n is an integer, and B is a constant (2.179 × 10 –18 J) –The negative sign represents force of attraction. The energy is zero when the electron is located infinitely far from the nucleus. Bohr’s Hydrogen Atom

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Chapter Seven 38 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Example 7.5 Calculate the energy of an electron in the second energy level of a hydrogen atom.

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Chapter Seven 39 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Each circle represents an allowed energy level for the electron. The electron may be thought of as orbiting at a fixed distance from the nucleus. Excitation: The atom absorbs energy that is exactly equal to the difference between two energy levels. The Bohr Model of Hydrogen When excited, the electron is in a higher energy level. Emission: The atom gives off energy—as a photon. Upon emission, the electron drops to a lower energy level.

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Chapter Seven 40 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry … each electronic energy level in an atom is quantized. Since the levels are quantized, changes between levels must also be quantized. A specific change thus represents one specific energy, one specific frequency, and therefore one specific wavelength. Line Spectra Arise Because … Transition from n = 3 to n = 2. Transition from n = 4 to n = 2.

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Chapter Seven 41 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry … allows us to find the energy change ( E level ) that accompanies the transition of an electron from one energy level to another. Initial energy level: Final energy level: Bohr’s Equation … To find the energy difference, just subtract: –B E i = —— n i 2 –B E f = —— n f 2 Together, all the photons having this energy ( E level ) produce one spectral line. –B –B 1 1 E level = —— – —— = B — – — n f 2 n i 2 n i 2 n f 2

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Chapter Seven 42 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Example 7.6 Calculate the energy change, in joules, that occurs when an electron falls from the n i = 5 to the n f = 3 energy level in a hydrogen atom. Example 7.7 Calculate the frequency of the radiation released by the transition of an electron in a hydrogen atom from the n = 5 level to the n = 3 level, the transition we looked at in Example 7.6.

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Chapter Seven 43 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Energy Levels and Spectral Lines for Hydrogen What is the (transition that produces the) longest-wavelength line in the Balmer series? In the Lyman series? In the Paschen series?

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Chapter Seven 44 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry When an atom has its electrons in their lowest possible energy levels, the atom is in its ground state. When an electron has been promoted to a higher level, the electron (and the atom) is in an excited state. Electrons are promoted to higher levels through an electric discharge, heat, or some other source of energy. An atom in an excited state eventually emits a photon (or several) as the electron drops back down to the ground state. Ground States and Excited States

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Chapter Seven 45 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Example 7.8 A Conceptual Example Without doing detailed calculations, determine which of the four electron transitions shown in Figure 7.19 produces the shortest-wavelength line in the hydrogen emission spectrum.

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Chapter Seven 46 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Louis de Broglie’s hypothesis stated that an object in motion behaves as both particles and waves, just as light does. A particle with mass m moving at a speed v will have a wave nature consistent with a wavelength given by the equation: = h/mv This wave nature is of importance only at the microscopic level (tiny, tiny m). De Broglie’s prediction of matter waves led to the development of the electron microscope. De Broglie’s Equation

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Chapter Seven 47 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Example 7.9 Calculate the wavelength, in meters and nanometers, of an electron moving at a speed of 2.74 × 10 6 m/s. The mass of an electron is 9.11 × 10 –31 kg, and 1 J = 1 kg m 2 s –2.

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Chapter Seven 48 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry de Broglie just messed up the Bohr model of the atom. Bad: An electron can’t orbit at a “fixed distance” if the electron is a wave. –An ocean wave doesn’t have an exact location—neither can an electron wave. Worse: We can’t even talk about “where the electron is” if the electron is a wave. Worst: The wavelength of a moving electron is roughly the size of an atom! How do we describe an electron that’s too big to be “in” the atom?? Uh oh …

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Chapter Seven 49 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Erwin Schrödinger: We can describe the electron mathematically, using quantum mechanics (wave mechanics). Schrödinger developed a wave equation to describe the hydrogen atom. An acceptable solution to Schrödinger’s wave equation is called a wave function. A wave function represents an energy state of the atom. Wave Functions

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Chapter Seven 50 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry De Broglie ’ s Equation Louis de Broglie speculated that matter can behave as both particles and waves, just like light He proposed that a particle with a mass m moving at a speed v will have a wave nature consistent with a wavelength EOS De Broglie’s prediction of matter waves led to the development of the electron microscope

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Chapter Seven 51 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Wave Functions ( ) Quantum mechanics, or wave mechanics, is the treatment of atomic structure through the wavelike properties of the electron Erwin Schrödinger developed an equation to describe the hydrogen atom EOS A wave function is a solution to the Schrödinger equation and represents an energy state of the atom

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Chapter Seven 52 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Interpretation of a Wave Function Wave mechanics provides a probability of where an electron will be in certain regions of an atom The Born interpretation: The square of a wave function ( 2 ) gives the probability of finding an electron in a small volume of space around the atom The interpretation leads to the idea of a cloud of electron density rather than a discrete location EOS e – = ?

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Chapter Seven 53 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry The Uncertainty Principle Werner Heisenberg’s uncertainty principle states that we can’t simultaneously know exactly where a tiny particle like an electron is and exactly how it is moving EOS The act of measuring the particle actually interferes with the particle

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Chapter Seven 54 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry The Uncertainty Principle In light of the uncertainty principle, Bohr’s model of the hydrogen atom fails, in part, because it tells more than we can know with certainty EOS When the photon is detected, the electron is in a different state because of the interaction

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Chapter Seven 55 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Quantum Numbers and Atomic Orbitals The wave functions for the hydrogen atom contain three parameters that must have specific integral values called quantum numbers A wave function with a given set of these three quantum numbers is called an atomic orbital EOS These orbitals allow us to visualize the region in which there is a probability of finding an electron

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Chapter Seven 56 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry A wave function doesn’t tell us where the electron is. The uncertainty principle tells us that we can’t know where the electron is. However, the square of a wave function gives the probability of finding an electron at a given location in an atom. Analogy: We can’t tell where a single leaf from a tree will fall. But (by viewing all the leaves under the tree) we can describe where a leaf is most likely to fall. The Uncertainty Principle

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Chapter Seven 57 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry The wave functions for the hydrogen atom contain three parameters called quantum numbers that must have specific integral values. A wave function with a given set of these three quantum numbers is called an atomic orbital. These orbitals allow us to visualize the region in which the electron “spends its time.” Quantum Numbers and Atomic Orbitals

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Chapter Seven 58 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry When values are assigned to the three quantum numbers, a specific atomic orbital has been defined. Quantum Numbers: n The principal quantum number (n): Is independent of the other two quantum numbers. Can only be a positive integer (n = 1, 2, 3, 4, …) The size of an orbital and its electron energy depend on the value of n. Orbitals with the same value of n are said to be in the same principal shell.

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Chapter Seven 59 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry The orbital angular momentum quantum number (l): Determines the shape of the orbital. Can have positive integral values from 0, 1, 2, … (n – 1) Orbitals having the same values of n and of l are said to be in the same subshell. Quantum Numbers: l Each orbital designation represents a different region of space and a different shape. fdpsSubshell 3210Value of l

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Chapter Seven 60 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry The magnetic quantum number (m l ): Determines the orientation in space of the orbitals of any given type in a subshell. Can be any integer from –l to +l The number of possible values for m l is (2l + 1), and this determines the number of orbitals in a subshell. Quantum Numbers: m l

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Chapter Seven 61 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Notice: one s orbital in each principal shell three p orbitals in the second shell (and in higher ones) five d orbitals in the third shell (and in higher ones)

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Chapter Seven 62 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Example 7.10 Considering the limitations on values for the various quantum numbers, state whether an electron can be described by each of the following sets. If a set is not possible, state why not. (a) n = 2, l = 1, m l = –1 (c) n = 7, l = 3, m l = +3 (b) n = 1, l = 1, m l = +1 (d) n = 3, l = 1, m l = –3 Example 7.11 Consider the relationship among quantum numbers and orbitals, subshells, and principal shells to answer the following. (a) How many orbitals are there in the 4d subshell? (b) What is the first principal shell in which f orbitals can be found? (c) Can an atom have a 2d subshell? (d) Can a hydrogen atom have a 3p subshell?

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Chapter Seven 63 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry The 1s Orbital Spherical symmetry; probability of finding the electron is the same in each direction. The 1s orbital (n = 1, l = 0, m l = 0) has spherical symmetry. An electron in this orbital spends most of its time near the nucleus. The electron cloud doesn’t “end” here … … the electron just spends very little time farther out.

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Chapter Seven 64 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Analogy to the 1s Orbital Highest “electron density” near the center … … but the electron density never drops to zero; it just decreases with distance.

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Chapter Seven 65 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry The 2s orbital has two concentric, spherical regions of high electron probability. The region near the nucleus is separated from the outer region by a node—a region (a spherical shell in this case) in which the electron probability is zero. The 2s Orbital

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Chapter Seven 66 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry The Three p Orbitals Three values of m l gives three p orbitals in the p subshell.

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Chapter Seven 67 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry The Five d Orbitals Five values of m l (– 2, –1, 0, 1, 2) gives five d orbitals in the d subshell.

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Chapter Seven 68 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry The spin refers to a magnetic field induced by the moving electric charge of the electron as it spins. The magnetic fields of two electrons with opposite spins cancel one another; there is no net magnetic field for the pair. Electron Spin: m s The electron spin quantum number (m s ) explains some of the finer features of atomic emission spectra. The number can have two values: +½ and –½.

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Chapter Seven 69 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry The Stern-Gerlach Experiment Demonstrates Electron Spin Silver has 47 electrons (odd number). On average, 23 electrons will have one spin, 24 will have the opposite spin. The magnet splits the beam. These silver atoms each have 24 +½-spin electrons and 23 –½-spin electrons. These silver atoms each have 23 +½-spin electrons and 24 –½-spin electrons.

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Chapter Seven 70 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry CUMULATIVE EXAMPLE Which will produce more energy per gram of hydrogen: H atoms undergoing an electronic transition from the level n = 4 to the level n = 1, or hydrogen gas burned in the reaction: 2 H 2 (g) + O 2 (g) 2 H 2 O(l)?

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