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

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2 Properties of Waves Wavelength ( ) is the distance between identical points on successive waves. Amplitude is the vertical distance from the midline of a wave to the peak or trough. Frequency ( ) is the number of waves that pass through a particular point in 1 second (Hz = 1 cycle/s). The speed (u) of the wave = x

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3 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

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4 Energy (light) is emitted or absorbed in discrete units (quantum). E = h x Planck’s constant (h) h = 6.63 x 10 -34 J s When solids are heated, they emit electromagnetic radiation over a wide range of wavelengths. Radiant energy emitted by an object at a certain temperature depends on its wavelength.

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5 Light has both: 1.wave nature 2.particle nature h = KE + W Photon is a “particle” of light KE = h - W where W (the binding energy for the emitted photoelectron) is the work function and depends how strongly electrons are held in the metal KE: Kinetic Energy of the emitted photoelectron

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6 E = h x E = 6.63 x 10 -34 (J s) x 3.00 x 10 8 (m/s) / 0.200 x 10 -9 (m) E = 9.95 x 10 -16 J E = h x c / When copper is bombarded with high-energy electrons, X rays are emitted. Calculate the energy (in joules) associated with the photons if the wavelength of the X rays is 0.200 nm.

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7 Line Emission Spectrum of Hydrogen Atoms

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8 1.e - can only have specific (quantized) energy values 2.light 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,… R H (Rydberg constant) = 2.18 x 10 -18 J

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9 E photon = E = E f - E i E f = -R H ( ) 1 n2n2 f E i = -R H ( ) 1 n2n2 i i f E = R H ( ) 1 n2n2 1 n2n2

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10 E photon = 2.18 x 10 -18 J x (1/16 - 1/4) E photon = E = -4.09 x 10 -19 J = 6.63 x 10 -34 (Js) x 3.00 x 10 8 (m/s)/4.09 x 10 -19 J = 486.6 nm Calculate the wavelength (in nm) of a photon emitted by a hydrogen atom when its electron drops from the n = 4 state to the n = 2 state. E photon = h x c / = h x c / E photon i f E = R H ( ) 1 n2n2 1 n2n2 E photon =

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11 De Broglie (1924) reasoned that e - is both particle and wave. Why is e - energy quantized? u = velocity of e- m = mass of e- 2 r = n = h mu

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13 = h/mu = 6.63 x 10 -34 / (3.7 x 10 -3 x 22.5) = 7.96 x 10 -33 m = 7.96 x 10 -24 nm What is the de Broglie wavelength (in nm) associated with a 3.7 g Ping-Pong ball traveling at 22.5 m/s? m in kgh in J su in (m/s)

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14 What are the quantum numbers? They are a set of solutions to equations that give the most likely location of electrons in an atom They do not give exact answers as that is not possible

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15 Schrodinger Wave Equation is a function of four numbers called quantum numbers (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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16 quantum numbers: (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 Schrodinger Wave Equation

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17 l = 0 (s orbitals) l = 1 (p orbitals)

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18 l = 2 (d orbitals)

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19 l = 3 (f orbitals)

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21 quantum numbers: (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

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22 (n, l, m l, m s ) spin quantum number m s m s = +½ or -½ Schrodinger Wave Equation m s = -½m s = +½

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23 Pauli exclusion principle :- no two electrons in an atom can have the same four quantum numbers. Schrodinger Wave Equation quantum numbers: (n, l, m l, m s )

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25 Schrodinger Wave Equation quantum numbers: (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 An orbital can hold 2 electrons

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26 How many 2p orbitals are there in an atom? 2p2p n=2 l = 1 If l = 1, then m l = -1, 0, or +1 3 orbitals How many electrons can be placed in the 3d subshell? 3d3d 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 -

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

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28 Electron configuration is how the electrons are distributed among the various atomic orbitals in an atom. 1s 1 principal quantum number n angular momentum quantum number l number of electrons in the orbital or subshell Orbital diagram H 1s 1

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29 What is the electron configuration of Na? Na 11 electrons 1s < 2s < 2p < 3s < 3p < 4s 1s 2 2s 2 2p 6 3s 1 2 + 2 + 6 + 2 = 12 electrons Abbreviated as [Ne]3s 1 [Ne] 1s 2 2s 2 2p 6 What are the possible quantum numbers for the last (outermost) electron in Cl? Cl 17 electrons1s < 2s < 2p < 3s < 3p < 4s 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

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