Download presentation

Presentation is loading. Please wait.

Published byVincent Norman Modified over 4 years ago

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

2
Quantum Theory

3
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

4
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

5
7.1

6
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

7
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

8
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

9
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

10
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

11
= 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

12
l = 0 (s orbitals) l = 1 (p orbitals) 7.6

13
l = 2 (d orbitals) 7.6

14
= 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

15
m l = -1m l = 0m l = 1 m l = -2m l = -1m l = 0m l = 1m l = 2 7.6

16
= 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.

18
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

20
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

21
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

22
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

23
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

24
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

25
“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

26
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

27
Core electrons = electrons in inner shells; Valence electrons = electrons in outer shell

28
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.

29
Outermost subshell being filled with electrons 7.8

30
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

31
7.8

32
Paramagnetic unpaired electrons 2p Diamagnetic all electrons paired 2p 7.8 (not drawn into a magnetic field)(attracted by a magnetic field)

33
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

34
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

35
The reason for this irregularity is that a slightly stability is associated with the half-filled (d 5 ) and completely filled (d 10 ) subshells.

36
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

Similar presentations

© 2020 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google