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

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Quantum Theory

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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. 7.1 Electromagnetic radiation Frequency ( ) is the number of waves that pass through a particular point in 1 second (Hz = 1 cycle/s). Speed of light (c) in vacuum = 3.00 x 10 8 m/s All electromagnetic radiation x c

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x = c = c/ = 3.00 x 10 8 m/s / 500 x 10 -9 m What is the frequency of light with wavelength of 500nm? = 6.00 x 10 14 s -1 7.1 Practice problem of chap7:Q1

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7.1

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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. Calculate the energy (in joules) associated with the photons if the wavelength of the X rays is 0.154 nm. Energy of light: E = h h x c / Planck’s constant (h) h = 6.63 x 10 -34 Js

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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 7.3

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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 n f = 1 n i = 2 n f = 1 n i = 3 n f = 2 n i = 3 7.3 Transition between energy states if n i > n f, then emission occurs = h x

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E photon = 2.18 x 10 -18 J x (1/25 - 1/9) E photon = E = -1.55 x 10 -19 J = 6.63 x 10 -34 (Js) x 3.00 x 10 8 (m/s)/1.55 x 10 -19 J = 1280 nm Calculate the wavelength (in nm) of a photon emitted by a hydrogen atom when its electron drops from the n = 5 state to the n = 3 state. E photon = h x c / = h x c / E photon i f E = R H ( ) 1 n2n2 1 n2n2 E photon = 7.3

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Quantum Mechanical Description of the distribution of electrons in the atom: 4 quantum numbers: n (principal quantum number) l (angular momentum quantum number) m l (magnetic quantum number) m s (spin quantum number) Allowed values: n = 1,2,3 …,n l = 0,1,2,…,(n-1) m l = (-l, …, 0. …+l) m s = +1/2, -1/2

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= 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 Schrodinger Wave Equation 7.6

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

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l = 2 (d orbitals) 7.6

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= 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 Schrodinger Wave Equation 7.6

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m l = -1m l = 0m l = 1 m l = -2m l = -1m l = 0m l = 1m l = 2 7.6

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= fn(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 Two possible spinning motions of an electron result in two different magnetic fields.

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Schrodinger Wave Equation = fn(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 7.6

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Which of the following sets of quantum numbers in an atom is acceptable? (a)(1, 0, +½, +½) (b) (3, 2, -2, -½) (c) (3, 3, 0, -½). (d) (4, 2, +3, +½) (e) (3, 0, +1, +½) Group practice problem of chap7:Q6 Allowed values: n = 1,2,3 …,n l = 0,1,2,…,(n-1) m l = (-l, …, 0. …+l) m s = +1/2, -1/2

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Energy diagram 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 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s

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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 7.8

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Electron Configurations Periods 1, 2, and 3 Three rules: 1.Electrons fill orbitals starting with lowest n and moving upwards (Aufbau principle: Fill up electrons in lowest energy orbitals ) 2. No more than two electrons can be placed in each orbital. No two electrons can fill one orbital with the same spin (Pauli exclusion principle: no two electrons in an atom can have the same four quantum numbers.) 3. For degenerate orbitals, electrons fill each orbital singly before any orbital gets a second electron (Hund’s rule: The most stable arrangement of electrons in subshells is the one with the greatest number of parallel spins ) Period 4 and Beyond the d orbitals begin to fill

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Which of the following violates the Pauli Exclusion Principle? (a)↑ ↑__ ↑↑ (b) ↑ ↑↓ ↑_ (c) ↑ ↑↓ ↓_ (d) ↑↓_ __ ↑ _ ↑_ ↑_ (e) ↑_ ↑__ ↑ _ ↓_ ↑↓_ Pauli exclusion principle: no two electrons in an atom can have the same four quantum numbers. No two electrons can fill one orbital with the same spin. ) Which of the following violates the Hund’s rule? (a)↑ ↑__ ↑↑ (b) ↑ ↑↓ ↑_ (c) ↑ ↑↓ ↓_ (d) ↑↓_ __ ↑ _ ↑_ ↑_ (e) ↑_ ↑__ ↑ _ ↓_ ↑↓_ Hund’s rule: The most stable arrangement of electrons in subshells is the one with the greatest number of parallel spins. For degenerate orbitals, electrons fill each orbital singly before any orbital gets a second electron. Practice problem of chap7:Q12,13

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“Fill up” electrons in lowest energy orbitals (Aufbau principle) H 1 electron H 1s 1 He 2 electronsHe 1s 2 Li 3 electronsLi 1s 2 2s 1 Be 4 electronsBe 1s 2 2s 2 B 5 electronsB 1s 2 2s 2 2p 1 ?? 7.9

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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 electronsN 1s 2 2s 2 2p 3 O 8 electronsO 1s 2 2s 2 2p 4 F 9 electronsF 1s 2 2s 2 2p 5 Ne 10 electronsNe 1s 2 2s 2 2p 6 7.7

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Core electrons = electrons in inner shells; Valence electrons = electrons in outer shell

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General rules for assigning electrons to atomic orbitals 1.Each shell or principle level of quantum number n contains n subshells. E.g. if n=2, there are two subshells (two values of l) of angular momentum quantum number. 2.Each subshell of quantum number l contains (2l+1) orbitals. E.g. if l=1, there are 3 p orbitals. 3.No more than two electrons can be placed in each orbital. 4.A quick way to determine the maximum number of electrons that an atom can have in principle level n is to use the formula of 2n 2.

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Outermost subshell being filled with electrons 7.8

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What is the electron configuration of Mg? Mg 12 electrons 1s < 2s < 2p < 3s < 3p < 4s 1s 2 2s 2 2p 6 3s 2 2 + 2 + 6 + 2 = 12 electrons 7.8 Abbreviated as [Ne]3s 2 [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 core electronsvalence electrons

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7.8

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Paramagnetic unpaired electrons 2p Diamagnetic all electrons paired 2p 7.8 (not drawn into a magnetic field)(attracted by a magnetic field)

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How many unpaired electrons are present in Al,N,Si,S? Unpaired electrons Al: 1s 2 2s 2 2p 6 3s 2 3p 11 N: 1s 2 2s 2 2p 33 Si: 1s 2 2s 2 2p 6 3s 2 3p 22 S: 1s 2 2s 2 2p 6 3s 2 3p 42 Practice problems of chap7:Q8,9

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Electron Configurations and the Periodic Table Shorthand way of writing electron configurations: Write the core electrons corresponding to the filled Noble gas in square brackets. Write the valence electrons explicitly. Example, P: 1s 2 2s 2 2p 6 3s 2 3p 3 but Ne is 1s 2 2s 2 2p 6 Therefore, P: [Ne] 3s 2 3p 3

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The reason for this irregularity is that a slightly stability is associated with the half-filled (d 5 ) and completely filled (d 10 ) subshells.

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8.2 ns 1 ns 2 ns 2 np 1 ns 2 np 2 ns 2 np 3 ns 2 np 4 ns 2 np 5 ns 2 np 6 d1d1 d5d5 d 10 4f 5f Ground State Electron Configurations of the Elements

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