This document provides information about quantum theory and atomic structure. It begins with goals and objectives for understanding concepts like the wave nature of light and quantum numbers. It then discusses electromagnetic radiation, Planck's quantization of energy, Bohr's model of the hydrogen atom, and the Schrodinger wave equation. It introduces quantum numbers like principal, angular momentum, and magnetic quantum numbers. It provides examples of determining orbital names and properties from quantum numbers. The key ideas are that energy and matter can behave as both particles and waves, and quantum theory was developed to explain atomic structure and spectra.
2. 7-2
Goals & Objectives
• See the following Learning Objectives on
pages 283 and 322.
• Understand these Concepts:
• 7.1-7, 9-12; 8.1-8.
• Master these Skills:
• 7.1-2, 5; 8.1-2.
3. 7-3
The Wave Nature of Light
Visible light is a type of electromagnetic radiation.
The wave properties of electromagnetic radiation are
described by three variables:
- frequency (n), cycles per second
- wavelength (l), the distance a wave travels in one cycle
- amplitude, the height of a wave crest or depth of a trough.
The speed of light is a constant:
c = n x l
= 3.00 x 108 m/s in a vacuum
7. 7-7
Sample Problem 7.1 Interconverting Wavelength and Frequency
PROBLEM: A dental hygienist uses x-rays (l= 1.00Å) to take a series of
dental radiographs while the patient listens to a radio station
(l = 325 cm) and looks out the window at the blue sky (l=
473 nm). What is the frequency (in s-1) of the
electromagnetic radiation from each source? (Assume that
the radiation travels at the speed of light, 3.00x108 m/s.)
PLAN: Use the equation c = nl to convert wavelength to frequency.
Wavelengths need to be in meters because c has units of m/s.
wavelength in units given
wavelength in m
frequency (s-1 or Hz)
use conversion factors
1 Å = 10-10 m
n =
l
c
8. 7-8
Sample Problem 7.1
SOLUTION:
For the x-rays: l = 1.00 Å x = 1.00 x 10-10 m
1 Å
10-10 m
= = 3.00 x 1018 s-1
n =
l
c 3.00 x 108 m/s
1.00 x 10-10 m
For the radio signal: = 9.23 x 107 s-1=n =
l
c 3.00 x 108 m/s
325 cm x
10-2 m
1 cm
For the blue sky: = 6.34 x 1014 s-1=n =
l
c 3.00 x 108 m/s
473 nm x
10-9 m
1 cm
11. 7-11
Energy and frequency
A solid object emits visible light when it is heated to about
1000 K. This is called blackbody radiation.
E = nhn
The color (and the intensity ) of the light changes as the
temperature changes. Color is related to wavelength and
frequency, while temperature is related to energy.
Energy is therefore related to frequency and wavelength:
E = energy
n is a positive integer
h is Planck’s constant
12. 7-12
Figure 7.6 Familiar examples of light emission related to
blackbody radiation.
Lightbulb filament
Smoldering coal Electric heating element
13. 7-13
The Quantum Theory of Energy
Any object (including atoms) can emit or absorb only
certain quantities of energy.
Energy is quantized; it occurs in fixed quantities, rather
than being continuous. Each fixed quantity of energy is
called a quantum.
An atom changes its energy state by emitting or
absorbing one or more quanta of energy.
DE = nhn where n can only be a whole number.
15. 7-15
Sample Problem 7.2
SOLUTION:
Calculating the Energy of Radiation from Its
Wavelength
PROBLEM: A cook uses a microwave oven to heat a meal. The
wavelength of the radiation is 1.20 cm. What is the energy
of one photon of this microwave radiation?
PLAN: We know l in cm, so we convert to m and find the frequency
using the speed of light. We then find the energy of one
photon using E = hn.
=E = hn =
hc
l
(6.626 x 10-34) J∙s)(3.00 x 108 m/s)
(1.20 cm)
10-2 m
1 cm
( )
= 1.66 x 10-23 J
18. 7-18
= RRydberg equation -
1
l
1
n2
2
1
n1
2
R is the Rydberg constant = 1.096776x107 m-1
Figure 7.9 Three series of spectral lines of atomic hydrogen.
for the visible series, n1 = 2 and n2 = 3, 4, 5, ...
19. 7-19
The Bohr Model of the Hydrogen Atom
1913
Bohr’s atomic model postulated the following:
• The H atom has only certain energy levels, which Bohr
called stationary states.
– Each state is associated with a fixed circular orbit of
the electron around the nucleus.
– The higher the energy level, the farther the orbit is
from the nucleus.
– When the H electron is in the first orbit, the atom is in
its lowest energy state, called the ground state.
20. 7-20
• The atom does not radiate energy while in one of its
stationary states.
• The atom changes to another stationary state only by
absorbing or emitting a photon.
– The energy of the photon (hn) equals the difference
between the energies of the two energy states.
– When the E electron is in any orbit higher than n = 1,
the atom is in an excited state.
21. 7-21
Figure 7.10 A quantum “staircase” as an analogy for atomic
energy levels.
22. 7-22
Figure 7.11 The Bohr explanation of three series of spectral lines
emitted by the H atom.
23. 7-23
Figure B7.1 Flame tests and fireworks.
copper 29Custrontium 38Sr copper 29Cu
The flame color is due to the emission
of light of a particular wavelength by
each element.
Fireworks display emissions similar to
those seen in flame tests.
Tools of the Laboratory
28. 7-28
The Wave-Particle Duality of Matter and Energy
Matter and Energy are alternate forms of the same entity.
E = mc2
All matter exhibits properties of both particles and
waves. Electrons have wave-like motion and therefore
have only certain allowable frequencies and energies.
Matter behaves as though it moves in a wave, and the
de Broglie wavelength for any particle is given by:
h
mu
l =
m = mass
u = speed in m/s
30. 7-30
Table 7.1 The de Broglie Wavelengths of Several Objects
Substance Mass (g) Speed (m/s) l (m)
slow electron
fast electron
alpha particle
one-gram mass
baseball
Earth
9x10-28
9x10-28
6.6x10-24
1.0
142
6.0x1027
1.0
5.9x106
1.5x107
0.01
25.0
3.0x104
7x10-4
1x10-10
7x10-15
7x10-29
2x10-34
4x10-63
31. 7-31
CLASSICAL THEORY
Matter
particulate,
massive
Energy
continuous,
wavelike
Since matter is discontinuous and particulate,
perhaps energy is discontinuous and particulate.
Observation Theory
Planck: Energy is quantized; only certain values
allowed
Blackbody radiation
Einstein: Light has particulate behavior (photons)Photoelectric effect
Bohr: Energy of atoms is quantized; photon
emitted when electron changes orbit.
Atomic line spectra
Figure 7.15
Major observations and theories
leading from classical theory to
quantum theory
32. 7-32
Since energy is wavelike,
perhaps matter is wavelike.
Observation Theory
deBroglie: All matter travels in waves; energy of
atom is quantized due to wave motion of
electrons
Davisson/Germer:
Electron beam is
diffracted by metal
crystal
Since matter has mass,
perhaps energy has mass
Observation Theory
Einstein/deBroglie: Mass and energy are
equivalent; particles have
wavelength and photons have
momentum.
Compton: Photon’s
wavelength increases
(momentum decreases)
after colliding with
electron
Figure 7.15 continued
QUANTUM THEORY
Energy and Matter
particulate, massive, wavelike
33. 7-33
Heisenberg’s Uncertainty Principle
Heisenberg’s Uncertainty Principle states that it is not
possible to know both the position and momentum of a
moving particle at the same time.
The more accurately we know the speed, the less
accurately we know the position, and vice versa.
D x∙m D u ≥
h
4p
x = position
u = speed
34. 7-34
Sample Problem 7.5
SOLUTION:
Applying the Uncertainty Principle
PROBLEM: An electron moving near an atomic nucleus has a speed
6x106 m/s ± 1%. What is the uncertainty in its position (Dx)?
PLAN: The uncertainty in the speed (Du) is given as ±1% (0.01) of
6x106 m/s. We multiply u by 0.01 and substitute this value
into Equation 7.6 to solve for Δx.
Du = (0.01)(6x106 m/s) = 6x104 m/s
Dx∙mDu ≥
h
4p
≥ 1x10-9 mDx ≥
h
4pmDu 4p (9.11x10-31 kg)(6x104 m/s)
6.626x10-34 kg∙m2/s
≥
35. 7-35
The Quantum-Mechanical Model of the Atom
The matter-wave of the electron occupies the space near
the nucleus and is continuously influenced by it.
The Schrödinger wave equation allows us to solve for
the energy states associated with a particular atomic
orbital.
The square of the wave function gives the probability
density, a measure of the probability of finding an
electron of a particular energy in a particular region of the
atom.
36. 7-36
The Shrodinger Atom (1926)
• Quantum Mechanical Model
– Wave equation is solved to obtain
– Wave function (orbital)
– Square of wave function 2 gives
probability of finding an electron in a
certain space
37. 7-37
The Schrödinger Equation
H = E
d2
dy2
d2
dx2
d2
dz2
+ +
8p2mQ
h2
(E-V(x,y,z)(x,y,z) = 0+
how y changes in
space
mass of
electron
total quantized
energy of the
atomic system
potential energy at
x,y,z
wave
function
39. 7-39
The Shrödinger Atom (1926)
• Wave function for each electron in an
atom contains four quantum numbers
– n Principle
– l Angular momentum
– ml Magnetic
– ms Spin
40. 7-40
Quantum Numbers and Atomic Orbitals
An atomic orbital is specified by three quantum numbers.
The principal quantum number (n) is a positive integer.
The value of n indicates the relative size of the orbital and therefore its
relative distance from the nucleus.
The magnetic quantum number (ml) is an integer with
values from –l to +l
The value of ml indicates the spatial orientation of the orbital.
The angular momentum quantum number (l) is an integer
from 0 to (n -1).
The value of l indicates the shape of the orbital.
41. 7-41
Quantum Numbers and The Exclusion Principle
An atomic orbital can hold a maximum of two electrons
and they must have opposing spins.
Each electron in any atom is described completely by a
set of four quantum numbers.
The first three quantum numbers describe the orbital, while the
fourth quantum number describes electron spin.
Pauli’s exclusion principle states that no two electrons in
the same atom can have the same four quantum numbers.
42. 7-42
Table 7.2 The Hierarchy of Quantum Numbers for Atomic Orbitals
Name, Symbol
(Property) Allowed Values Quantum Numbers
Principal, n
(size, energy)
Angular
momentum, l
(shape)
Magnetic, ml
(orientation)
Positive integer
(1, 2, 3, ...)
0 to n – 1
-l,…,0,…,+l
1
0
0
2
0 1
0
3
0 1 2
0
0-1 +1 -1 0 +1
0 +1 +2-1-2
43. 7-43
Sample Problem 7.6
SOLUTION:
Determining Quantum Numbers for an
Energy Level
PROBLEM: What values of the angular momentum (l) and magnetic
(ml) quantum numbers are allowed for a principal quantum
number (n) of 3? How many orbitals are allowed for n = 3?
PLAN: Values of l are determined from the value for n, since l can take
values from 0 to (n-1). The values of ml then follow from the
values of l.
For n = 3, allowed values of l are = 0, 1, and 2
For l = 1 ml = -1, 0, or +1
For l = 2 ml = -2, -1, 0, +1, or +2
There are 9 ml values and therefore 9 orbitals with n = 3.
For l = 0 ml = 0
PROBLEM: What values of the angular momentum (l) and magnetic
(ml) quantum numbers are allowed for a principal quantum
number (n) of 3? How many orbitals are allowed for n = 3?
44. 7-44
Sample Problem 7.7
SOLUTION:
Determining Sublevel Names and Orbital
Quantum Numbers
PLAN: Combine the n value and l designation to name the sublevel.
Knowing l, we can find ml and the number of orbitals.
n l sublevel name possible ml values # of orbitals
(a) 3 2 3d -2, -1, 0, 1, 2 5
0(b) 2 2s 0 1
(c) 5 1 5p -1, 0, 1 3
(d) 4 3 4f -3, -2, -1, 0, 1, 2, 3 7
PROBLEM: Give the name, magnetic quantum numbers, and number
of orbitals for each sublevel with the following quantum
numbers:
(a) n = 3, l = 2 (b) n = 2, l = 0 (c) n = 5, l = 1 (d) n = 4, l = 3
45. 7-45
Sample Problem 7.8
SOLUTION:
Identifying Incorrect Quantum Numbers
PROBLEM: What is wrong with each of the following quantum numbers
designations and/or sublevel names?
(a) A sublevel with n = 1 can only have l = 0, not l = 1. The only possible
sublevel name is 1s.
n l ml Name
(a)
(b)
(c)
1
4
3
1
3
1
0
+1
-2
1p
4d
3p
(b) A sublevel with l = 3 is an f sublevel, to a d sublevel. The name
should be 4f.
(c) A sublevel with l = 1 can only have ml values of -1, 0, or +1, not -2.
52. 7-52
The Shrödinger Atom (1926)
• Magnetic quantum number ml
– Value = –l to + l
– Properties of orbital:
orientation in space
53. 7-53
The Shrödinger Atom (1926)
• Spin quantum number ms
– Value = +1/2 or – 1/2
– Two electrons can occupy an orbital
one has spin +1/2, the other has spin – 1/2
54. 7-54
The Shrödinger Atom (1926)
• Subshell capacities in each shell (n)
– s (l=0) 1 orbital, 2 electrons
– p (l=1) 3 orbitals, 6 electrons (n≥2)
– d (l=2) 5 orbitals, 10 electrons (n≥3)
– f (l=3) 7 orbitals, 14 electrons (n≥4)
55. 7-55
Table 8.1 Summary of Quantum Numbers of Electrons in Atoms
Name Symbol Permitted Values Property
principal n positive integers (1, 2, 3, …) orbital energy (size)
angular
momentum
l integers from 0 to n-1 orbital shape (The l values
0, 1, 2, and 3 correspond to
s, p, d, and f orbitals,
respectively.)
magnetic ml
integers from -l to 0 to +l orbital orientation
spin ms
+½ or -½ direction of e- spin
56. 7-56
Orbitals: Order of Filling
• Lowest energy orbitals are filled first
• In subshells with multiple orbitals, start
filling at lowest ml.
• In subshells with multiple orbitals, each
orbital receives one electron (ms = +1/2)
until all are filled,
• then second electron electron goes in
(ms = –1/2).
61. 7-61
Some Notes on the d Subshell
and the Transition Metals
• There is special stability associated with
filled or half-filled subshells
• (n) s subshell fills first,
then (n–1) d subshell
• (n) s subshell empties first,
then (n–1) d subshell
62. 7-62
Some Notes on the d Subshell
and the Transition Metals
• Transition metals often form several
stable ions of differing charge:
• Fe 1s22s22p63s23p64s23d6
• Fe2+ 1s22s22p63s23p64s13d5
• Fe3+ 1s22s22p63s23p64s03d5
63. 7-63
Some Notes on the d Subshell
and the Transition Metals
• Some transition metals have anomalous
ground state electron configurations:
• 24Cr 1s22s22p63s23p64s13d5
• 29Cu 1s22s22p63s23p64s13d10
• 46Pd 1s22s22p63s23p64s23d104p65s04d10
68. 7-68
n l
#
as s, p, d, f
Electron Configurations and Orbital Diagrams
Electron configuration is indicated by a shorthand notation:
# of electrons in the sublevel
Orbital diagrams make use of a box, circle, or line for each
orbital in the energy level. An arrow is used to represent an
electron and its spin.
↑↓↑↓ ↑↓
70. 7-70
Building Orbital Diagrams
The aufbau principle is applied – electrons are always
placed in the lowest energy sublevel available.
The exclusion principle states that each orbital may
contain a maximum of 2 electrons, which must have
opposite spins.
H (Z = 1) 1s1
He (Z = 2) 1s2
1s
↑↓
1s
↑
71. 7-71
Building Orbital Diagrams
Hund’s rule specifies that when orbitals of equal energy
are available, the lowest energy electron configuration has
the maximum number of unpaired electrons with parallel
spins.
N (Z = 7) 1s22s22p3
↑↓
2s 2p
↑ ↑ ↑
72. 7-72
Sample Problem 8.1 Determining Quantum Numbers from
Orbital Diagrams
PLAN: Identify the electron of interest and note its level (n), sublevel,
(l), orbital (ml) and spin (ms). Count the electrons in the order in
which they are placed in the diagram.
PROBLEM: Write a set of quantum numbers for the third electron and
a set for the eighth electron of the F atom.
SOLUTION:
F (Z = 9) 1s22s22p3
For the 3rd electron: n = 2, l = 0, ml = 0, ms = +½
↑↓
2s 2p
↑↓ ↑ ↑↑↓
1s
For the 8th electron: n = 2, l = 1, ml = -1, ms = -½
73. 7-73
Partial Orbital Diagrams and
Condensed Configurations
A partial orbital diagram shows only the highest energy
sublevels being filled.
Al (Z = 13) 1s22s22p63s23p1
A condensed electron configuration has the element
symbol of the previous noble gas in square brackets.
Al has the condensed configuration [Ne]3s23p1
↑↓
3s 3p
↑
74. 7-74
Table 8.2 Partial Orbital Diagrams and Electron Configurations*
for the Elements in Period 3.
*Colored type indicates the sublevel to which the last electron is added.
75. 7-75
Electron Configuration and Group
Similar outer electron configurations correlate with
similar chemical behavior.
Elements in the same group of the periodic table have the
same outer electron configuration.
Elements in the same group of the periodic table exhibit
similar chemical behavior.
77. 7-77
Figure 8.9 Similar reactivities in a group.
Potassium reacting with water. Chlorine reacting with potassium.
78. 7-78
Figure 8.11 Orbital filling and the periodic table.
The order in which the orbitals are filled can be obtained directly
from the periodic table.
79. 7-79
Aid to memorizing sublevel filling order.
The n value is constant horizontally.
The l value is constant vertically.
n + l is constant diagonally.
80. 7-80
Categories of Electrons
Inner (core) electrons are those an atom has in common
with the pervious noble gas and any completed transition
series.
Outer electrons are those in the highest energy level
(highest n value).
Valence electrons are those involved in forming
compounds.
For main group elements, the valence electrons are the outer
electrons.
For transition elements, the valence electrons include the outer
electrons and any (n-1)d electrons.
81. 7-81
Sample Problem 8.2 Determining Electron Configurations
PROBLEM: Using the periodic table on the inside cover of the text (not
Figure 8.10 or Table 8.3), give the full and condensed
electron configurations, partial orbital diagrams showing
valence electrons only, and number of inner electrons for the
following elements:
(a) potassium
(K; Z = 19)
(b) technetium
(Tc; Z = 43)
(c) lead
(Pb; Z = 82)
PLAN: The atomic number gives the number of electrons, and the
periodic table shows the order for filling orbitals. The partial
orbital diagram includes all electrons added after the previous
noble gas except those in filled inner sublevels.
82. 7-82
Sample Problem 8.2
SOLUTION:
(a) For K (Z = 19)
[Ar] 4s1condensed configuration
1s22s22p63s23p64s1full configuration
There are 18 inner electrons.
partial orbital diagram ↑
4s 4p3d
83. 7-83
Sample Problem 8.2
SOLUTION:
(b) For Tc (Z = 43)
[Kr]5s24d5condensed configuration
1s22s22p63s23p64s23d104p65s24d5full configuration
There are 36 inner electrons.
partial orbital diagram ↑↓
5s 5p4d
↑ ↑ ↑ ↑ ↑
84. 7-84
Sample Problem 8.2
SOLUTION:
(c) For Pb (Z = 82)
[Xe] 6s24f145d106p2condensed configuration
1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p2full configuration
There are 78 inner electrons.
partial orbital diagram ↑↓
6s 6p
↑ ↑
85. 7-85
Exercises
• For the following electronic
configurations, give the four quantum
numbers, and the ground state element:
– 2p4 ________________________
– 3d6 ________________________
– 1s1 ________________________
– 4p2 ________________________
88. 7-88
Exercises
• For the following sets of quantum
numbers, give the electronic
configuration and the ground state
element:
– n l ml ms
– 3 1 0 +1/2 _______,_______
– 1 0 0 –1/2 _______, _______
– 4 2 2 +1/2 _______, _______