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Homework Problems
Chapter 6 Homework Problems: 1, 6, 8, 16, 18, 30, 34, 41, 48, 51, 54,
58, 66, 68, 69, 76, 86, 91, 96, 108, 130
CHAPTER 6
Quantum Theory and the Electronic Structure
of Atoms
Pre-Quantum Physics
Scientists have generally believed that the behavior of objects in
the universe can be summarized in terms of a small number of
fundamental physical laws. By 1900, most scientists believed that the
major laws of physics had been found.
Conservation laws (conservation of mass, conservation of
energy)
Thermodynamics (First law, second law, third law)
Laws of motion (Newton’s laws)
Newton’s theory of gravity
Maxwell’s equations for electricity and magnetism.
Atomic theory
Light
Light is a general term used for electromagnetic radiation.
Although some scientists (like Newton) believed that light was a particle
phenomenon, by 1900 most scientists were convinced that light was a
wave phenomenon, for reasons discussed below.
Properties of Waves
The following general terms are used for waves.
wavelength () – The distance between successive peaks of the
wave (SI units are m)
frequency ()– The number of peaks that pass a given point per
unit time (SI units are s-1, sometimes called Hertz (Hz)).
Wavelength, Frequency, and Wave Velocity
There is a general relationship between the wavelength,
frequency, and velocity (c, the speed) of a wave.
 = c
where c is the speed of the wave (SI unit = m/s)
Light
Experimentally it is found that the speed at which light travels in
a vacuum is independent of the wavelength or frequency of the light.
For light in vacuum, c = 2.998 x 108 m/s.
If we know either the frequency or the wavelength, we can use
the relationship
 = c
to find the missing quantity.
Example: A sodium lamp emits yellow light at a wavelength  =
589.3 nm. What is the frequency of the light?
Example: A sodium lamp emits yellow light at a wavelength  =
589.3 nm. What is the frequency of the light?
Since  = c, it follows that  = c/, so
 = (2.998 x 108 m/s) = 5.087 x 1014 s-1
589.3 x 10-9 m
Wavelength and Intensity
The color of light (for visible light) depends on the wavelength of
the light, while the intensity (energy per unit time) of light depends on
the amplitude of the light.
Electromagnetic Spectrum
All light is fundamentally the same, and we often use the word
“light” to mean any electromagnetic radiation. However, it is convenient
to divide the spectrum into regions based on wavelength or frequency.
Wave Properties of Light
Classically, light was considered a wave phenomena. This was
based on experimental observations such as the interference pattern for
light observed in the two slit experiment.
Interference – The increase or decrease in amplitude that occurs
when two waves of the same wavelength are combined together, due to
constructive or destructive interference.
Constructive and Destructive Interference
When two waves with the same wavelength combine together
they can be in phase (peaks line up with one another) or out of phase
(peaks of one wave line up with troughs of the other wave), or
somewhere between the two. When the waves are in phase, adding the
waves together increases the amplitude. This is called constructive
interference. When the waves are out of phase, adding the waves
together decreases the amplitude. This is called destructive interference.
Two Slit Interference
In the two slit experiment monochromatic light passes through
two small openings in a barrier. A diffraction pattern is observed.
white spots - places where
light has illuminated the
film
dark spots - places where
no light has illuminated
the film
If light were a particle, then no diffraction pattern should occur. Instead,
there should be only two spots on the film corresponding to the two slits.
Particle Properties of Light
While it was generally accepted that light behaved as a wave,
experiments at the end of the 19th century gave evidence that under some
conditions light behaved like a particle.
1) Blackbody radiation (explained by Max Planck in 1900).
2) Photoelectric effect (explained by Albert Einstein in 1905).
monochromatic light - light of a single color (wavelength)
critical wavelength (0) – If  > 0, no electrons are detected
Einstein Theory For the Photoelectric Effect
Einstein explained the photoelectric effect as follows:
1) Light consists of particles (now called photons) which have an
energy given by the Planck expression
Ephoton = h = hc/
h = 6.626 x 10-34 J.s
2) There is a minimum energy (W, the binding energy or work
function for the metal) required for an electron to escape from the metal.
3) When an electron in the metal absorbs a photon, the energy of
the photon is transferred to the electron. There are two possibilities
a) If the energy is less than the work function for the metal, the
electron does not have enough energy to escape.
b) If the energy is greater than the work function for the metal,
the electron can escape and be detected.
0 then represents the boundary between these two regions
Predictions From Einstein’s Theory
Several predictions follow from Einstein’s theory for the
photoelectric effect.
1) The critical wavelength 0 is related to the binding energy W.
Ephoton = h = hc/0 = W
for minimum photon energy
0 = hc/W
2) If  < 0 then electrons are produced instantaneously, even for
dim light. If  > 0 no photons are produced.
3) The maximum kinetic energy of the ejected electrons can be
found from conservation of energy
Ephoton = h = KEmax + W
KEmax = maximum KE of electron
KEmax = h - W
Therefore, a plot of KEmax vs  should have a slope equal to h.
The x-intercept in the above plot can be used to find the work
function for the metal.
Ephoton = h = KEmax + W
The minimum value for KEmax is 0
h0 = W , the work function for the metal.
CONCLUSION – Light has both wave and particle properties.
Atomic Spectra (Experimental)
When a sample of an element is heated to high temperature it will
emit light. The observed light emission is called a line spectrum.
Atomic Spectra (Observed Results)
1) Light emission occurs only at particular values of wavelength.
2) Different elements emit light at different wavelengths (and so
this light emission can be used to identify the presence of an element in a
sample, and even determine the concentration of element present).
3) An element will absorb light at the same wavelengths at which
it emits light.
Hydrogen Spectrum
For most elements the pattern of wavelengths where light is
emitted or absorbed is complicated. For hydrogen, however, the pattern
fits a simple equation called the Rydberg formula
1 = RH

1 - 1
nf2 ni2
RH = 0.01097 nm-1
ni, nf = 1, 2, 3, …
ni > n f
Example: At what wavelength (in nm) will light be emitted for
the transition ni = 3 --> nf = 2?
Example: At what wavelength (in nm) will light be emitted for
the transition ni = 3 --> nf = 2?
1/ = (0.01097 nm-1) [ (1/22) - (1/32)] = 1.524 x 10-3 nm-1
=
1
(1.524 x 10-3 nm-1)
= 656.3 nm
ni = 6
5
4
3
The concept of energy levels can be used to explain the pattern of
light absorption and emission for atoms. For example, for hydrogen the
energy levels are given by the equation:
En = - 2.18 x 10-18 J
n2
n = 1, 2, …
Other atomic spectra can also be
explained in terms of energy levels,
though there is no simple formula
for their location as there is for
hydrogen.
However, there is no explanation for the origin of these energy
levels by classical physics.
Wave Properties of Matter
Since light has “particle-like” properties it is reasonable to
consider the possibility that matter might, under some conditions, have
“wave-like” properties. The first person to explore this idea was Louis
de Broglie, in 1924.
E = mc2 (Einstein)
E = hc (Planck)

If we set these equal to one another, then
mc2 = hc or  = hc = h

mc2
mc
For a particle, replace c (speed of light) with v (speed of the particle), to
get
deBroglie = h
the de Broglie wavelength
mv
Interpretation of de Broglie Wavelength
What does the de Broglie wavelength mean? It is interpreted to
mean the length scale at which particles can exhibit wave behavior.
X-ray diffraction
electron diffraction
Aluminum foil
Using the de Broglie Equation
Example: What is the speed of an electron whose de Broglie
wavelength is  = 0.100 nm (approximate spacing between particles
in a crystal)?
Example: What is the speed of an electron whose de Broglie
wavelength is  = 0.100 nm (approximate spacing between particles in a
crystal)?
Since deBroglie =
then
v=
h
mv
h
m(deBroglie)
=
(6.626 x 10-34 J.s)
(9.109 x 10-31 kg) (0.100 x 10-9 m)
= 7.27 x 106 m/s
This is approximately 2.4 % of the speed of light.
Electron diffraction was first observed experimentally in 1927 by
the American physicists Clinton Davisson and Lester Germer.
Summary
From the previous discussion we get the following important
points:
1) Light has both wave-like and particle-like properties, which
can be brought out by different experiments.
2) Matter also has both wave-like and particle-like properties,
which can also be brought out in different experiments.
3) Many systems, such as atoms, behave as if they can only have
certain values for energy (energy levels).
Quantum Mechanics
Quantum mechanics is the theory developed to account for the
above observations. It is based on solving the Schrodinger equation.
[ (- 2/2m) (d2/dx2) + V(x) ] n(x) = Enn(x)
Schrodinger
Wavefunction
equation
Energy
Probability
By solving the Schrodinger equation for a system we can find the
possible values for energy and the probability of finding the particles
making up the system at a particular location in space.
Uncertainty Relationship
One general property of systems described by quantum mechanics is that they must satisfy an uncertainty principle (Heisenberg, 1927).
(x) (p) = (x) (mv)  (h/4)
p = mv = momentum
What this means is that it is not possible to assign a definite position for
a particle in a system. All that can be given is the probability of finding
the particle at a particular location. This is why, for example, we
describe the electrons in an atom as a “cloud” of charge surrounding the
nucleus.
Hydrogen Atom
The solutions to the Schrodinger equation for the hydrogen atom
are given in terms of quantum numbers, a series of numbers used to label
the solutions and which describe the solutions. For an electron in a
hydrogen atom there are four quantum numbers, each which gives
information about the state the electron is in.
Quantum Numbers
The quantum numbers and their possible values are as follows.
n = 1, 2, 3, …
Principal quantum number. Determines the
energy, the average distance between the
electron and the nucleus, and orbital size.
 = 0, 1, …, (n-1)
Angular momentum quantum number.
Determines the shape of the electron cloud
(orbital shape).
m = 0, 1, 2, …,  
Magnetic quantum number. Determines the
orientation of the orbital.
ms =  1/2
Spin quantum number. Determines the
orientation of the electron spin.
Example
If n = 2, what are the possible values for ?
If  = 2, what are the possible values for m?
Are these possible sets of quantum numbers for an electron?
n = 2,  = 1, m = -1, ms = ½
n = 3,  = 0, m = 1, ms = ½
Example
If n = 2, what are the possible values for ?
Answer: Since  ranges in value from  = 0 up to  = (n – 1), the
possible values for  are 0 or 1.
If  = 2, what are the possible values for m?
Answer: Since m  = 0, ±1, …, ± , the possible values for m are
m = 2, 1, 0, -1, and -2.
Are these possible sets of quantum numbers for an electron?
n = 2,  = 1, m = -1, ms = ½
yes
n = 3,  = 0, m = 1, ms = ½
no
Angular Momentum (Orbital) Quantum Number
We use letters to designate the various electron orbitals, which
describe the shape of the region where the electron is most likely to be
found.

orbital
m
0
s orbital (1)
0
1
p orbital (3)
1, 0, -1
2
d orbital (5)
2, 1, 0, -1, -2
3
f orbital (7)
3, 2, 1, 0, -1, -2, -3
Note that the total number of orbitals corresponding to a
particular value of  is equal to the number of possible values of m, or
(2 + 1).
We usually label the orbitals by both their value for n and their
value for . So, for example, if n = 3 and  = 1, we have the 3p orbitals.
Relationship Between n and Energy
For the hydrogen atom the energy of the electron depends only on
the value of the quantum number n.
En = (- 2.18 x 10-18 J)
n2
n = 1, 2, 3, ...
Orbital Shape (s and p Orbitals)
For each value of  there is a set of orbitals with distinctive
shapes. These orbitals represent the region of space where it is most
likely to find the electron.
 = 0 (s-orbital) (1)
 = 1 (p-orbitals) (3)
Orbital Shape (d Orbitals)
 = 2 (d-orbital) (5)
Orbitals With Different Values of n
The value of n does not affect the shape of an orbital. Therefore
a 1s and a 2s orbital have the same shape, and the 2p and 3p orbitals have
the same shape.
There are two effects that n has on the orbitals:
1) The larger the value of n, the larger the orbital. This leads to
the electron being further away from the nucleus.
2) Orbitals have radial nodes, or distances from the nucleus
where the probability of finding the electron goes to zero. The number
of radial nodes is given by the relationship
# nodes = (n - ) - 1
Size and Nodes For s-Orbitals
2s
# radial nodes = 1
3s
# radial nodes = 2
Spin Quantum Number (ms)
The spin quantum number indicates whether the electron is
spinning clockwise or counterclockwise, and can take on two values, +1/2
and - 1/2.
Direct evidence for this quantum number was first found
experimentally by Stern and Gerlach.
Multi-electron Atoms
There are two differences between hydrogen and other atoms,
which have more than one electron.
1) Each electron in a multi-electron atom has its own set of four
quantum numbers. These electrons must obey the Pauli exclusion
principle.
2) The energy for an electron in a multi-electron atom depends on
both n and . We label the various energy levels for the electrons by their
value for n and the letter used to represent the value for .
Example:
n = 2,  = 0
2s orbital
n = 3,  = 0
3s orbital
n = 3,  = 2
3d orbital
This dependence of energy on both n and  is a consequence of
electron-electron repulsion.
Energy Order for Multi-electron Atoms
There are two general statements we can make about the ordering
of orbitals in terms of energy for multi-electron atoms:
1) For orbitals with the same value of , the larger the value of n
the higher the energy for the orbital.
Example
1s < 2s < 3s < 4s ….
3d < 4d < 5d < …
2) For orbitals with the same value for n, the larger the value for 
the higher the energy for the orbital.
Example
2s < 2p
4s < 4p < 4d < 4f
This is not enough to figure out the normal ordering of orbitals in terms
of energy.
Energy Ordering For Orbitals
The usual ordering of energy for orbitals is as indicated below.
Notice that the ordering is not strictly in terms of the quantum number n
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f ...
Mnemonic Device For Energy Ordering
We may use the following mnemonic device for the order in
energy of the orbitals. The order in which the labels are crossed out
below is the order of energies.
1s
2s
2p
3s
3p
3d
4s
4p
4d
4f
5s
5p
5d
5f …
6s
6p
6d
6f …
So 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f ...
The Pauli Exclusion Principle
Each electron in a multi-electron atom has its own set of four
quantum numbers. An early observation concerning these quantum
numbers was made by Wolfgang Pauli, and is called the Pauli Exclusion
Principle.
No two electrons in an atom or ion can have the same set of
four quantum numbers.
Note that as many as three of the quantum numbers can be the
same, but at least one quantum number must be different.
As we will see, the arrangement of elements in the periodic table
and their periodic properties are in a real sense a consequence of the
exclusion principle.
Rules For Adding Electrons to Atoms
There are three rules that must be followed when adding
electrons to a multielectron atom to find the lowest energy state (ground
state) of the atom.
1) Pauli principle - No two electrons can have the same set of
four quantum numbers.
2) Aufbau principle - Electrons add to the lowest energy available
orbital until that orbital is filled.
3) Hund’s rule - Electrons add in such a way as to make as many
of the electrons as possible “spin up” (ms = 1/2).
Electron configuration - A list of each electron containing orbital,
in order of energy. A superscript to the right of the orbital is used to
indicate how many electrons are present in the orbital.
n

m
(1 e-) 1
0
0
1/
2
1s1
He (2 e-) 1
0
0
1/
2
1s2
1
0
0
- 1/2
(3 e-) 1
0
0
1/
1
0
0
- 1/2
2
0
0
1/
Element
H
Li
ms
electron configuration
2
2
1s22s1
n

m
(8 e-) 1
0
0
1/
1
0
0
- 1/2
2
0
0
1/
2
0
0
- 1/2
2
1
1
1/
2
2
1
0
1/
2
2
1
-1
1/
2
2
1
1
- 1/2
Element
O
ms
electron configuration
2
2
Maximum number of electrons in an orbital
= 2(2 + 1)
1s22s22p4
s-orbital
2
p-orbitals
6
d-orbitals 10
f-orbitals
14
Orbital Filling Diagram
An orbital filling diagram is a picture where the orbitals for an
atom are indicated by lines (or boxes), and electrons by arrows pointing
up (“spin up”) or down (“spin down”). Such a diagram is useful in
identifying the total number of unpaired electron spins in an atom.
Example:
O(8 e-) 1s22s22p4
__ __ __ __ __
1s
2s
2p
2 unpaired electron spins
Notice that if we are only interested in identifying unpaired
electron spins we need only examine those orbitals that are partially
filled. All electron spins will be paired up for filled orbitals.
Also notice how we use Hund’s rule in filling the orbitals.
Magnetic Properties of Atoms
An atom placed in a magnetic field will be attracted or repelled
by the field depending on the number of unpaired electron spins in the
atom. There are two cases
diamagnetic - Weakly repelled by an external magnetic field.
Occurs when there are no unpaired electron spins.
paramagnetic - Strongly attracted by an external magnetic field.
Occurs when there are one or more unpaired electron spins. The number
of unpaired spins can be determined by measuring the strength of the
attraction felt by the atoms.
Examples
1) Give the electron configuration for Sn (50 e-).
2) Consider the following elements: Li, Be, N.
How many unpaired electron spins occur in atoms of each of the
following elements? Which atoms are diamagnetic and which are
paramagnetic?
1) Give the electron configuration for Sn (50 e-).
Filling order: 1s<2s<2p<3s<3p<4s<3d<4p<5s<4d<5p<6s<…
Sn (50e-) 1s22s22p63s23p64s23d104p65s24d105p2
2) Consider the following elements: Li, Be, N.
How many unpaired electron spins occur in atoms of each of the
following elements? Which atoms are diamagnetic and which are
paramagnetic?
1s
2s
2p
Li
1s22s1
__
__
__ __ __ 1 unpaired
Be
1s22s2
__
__
__ __ __ 0 unpaired
N
1s22s22p3
__
__
__ __ __ 3 unpaired
Li and N have at least one unpaired spin and so are paramagnetic.
Be has no unpaired electron spins and so is diamagnetic. Note that we
only need to look at the partially filled orbitals to count unpaired spins.
Shorthand Notation For Electron Configurations
Electron configurations can be very long. For example
Sn (50 e-) 1s22s22p63s23p64s23d104p65s24d105p2
We may use the following shorthand notation for cases where
there are a large number of electrons. We represent the electron
configuration as the configuration for the closest noble gas with a smaller
number of electrons than the atom, plus the additional electrons.
In the above case, we would say
Sn (50e-) [Kr]5s24d105p2
Example: Give the electron configurations for O (8 e-), Ti (22 e-),
and Te (52 e-). Give these using both the long notation and the shorthand
notation.
Example: Give the electron configurations for O (8 e-), Ti (22 e-),
and Te (52 e-). Give these using both the long notation and the shorthand
notation.
Filling order: 1s<2s<2p<3s<3p<4s<3d<4p<5s<4d<5p<6s<…
O (8 e-)
1s22s22p4 = [He]2s22p4
Ti (22 e-) 1s22s22p63s23p64s23d2 = [Ar]4s23d2
Te (52 e-) 1s22s22p63s23p64s23d104p65s24d105p4 = [Kr]5s24d105p4
For atoms with a large number of electrons the shorthand notation
saves time and space.
Core and Valence Electrons
We may divide the electrons in an atom into two categories.
Valence electrons - The outermost electrons in an atom. There
are two important cases.
1) For main group elements, the valence electrons are the electrons found in orbitals with the largest value for n.
2) For transition metals, the valence electrons are all of the
electrons beyond the noble gas configuration.
Core electrons - All other electrons in the atom.
Examples:
O(8e-) 1s2 2s2 2p4 = [He]2s22p4
6 valence, 2 core
Ti(22 e-) 1s2 2s2 2p6 3s2 3p6 4s2 3d2 = [Ar]4s23d2 4 valence, 18 core
Importance of Valence Electrons
To a good first approximation the properties of the elements in
the periodic table are determined by the valence electrons they possess.
Chemical bonding takes place almost exclusively by the transfer
or sharing of valence electrons.
The chemical properties of a main group element are primarily
determined by the number of valence electrons per atom. Atoms in the
same group in the periodic table have the same electron configuration for
their valence electrons.
For transition metals, the arrangement of valence electrons also
plays a major role in determining the chemical properties of the
elements, but in a more complicated manner than for the main group
elements.
The Periodic Table
The periodic table arranges atoms of elements in such a way that
each element in a particular group in the table has the same number of
valence electrons. Since it is the valence electrons that are primarily
responsible for chemical properties of an element, that means that
elements in the same group will have similar chemical properties.
Example: Group 2A
Be (4 e-)
1s22s2 = [He]2s2
Mg (12 e-)
1s22s22p63s2 = [Ne]3s2
Ca (20 e-)
1s22s22p63s23p64s2 = [Ar]4s2
Sr (38 e-)
1s22s22p63s23p64s23d104p65s2 = [Kr]5s2
Ba (56 e-)
1s22s22p63s23p64s23d104p65s24d105p66s2 = [Xe]6s2
All group 2A elements have 2 valence electrons.
So what similarities in properties do we expect?
For example, consider the ionic compounds group 2A elements
make with the nonmetal fluorine.
BeF2
beryllium fluoride
MgF2 magnesium fluoride
CaF2
calcium fluoride
SrF2
strontium fluoride
BaF2
barium fluoride
Of course, there are other metals that form compounds with the
same stoichiometry (CuF2, copper (II) fluoride, for example). But if we
look at a large number of chemical and physical properties for the group
2A elements we find that they are similar to one another and different
from elements in other groups.
Electron Configuration for the Elements
The electron configurations for the main group and transition
metals are shown below. Note the similarities for elements in the same
group.
Anomalous Electron Configurations
The rules we have given for predicting electron configurations for
atoms work most of the time. However, there are occasional cases where
the actual electron configuration is different from the predicted
configuration.
Example:
Cr(24 e-) [Ar]4s23d4
[Ar]4s13d5
predicted
actual
These deviations from prediction occur because the rules
developed for predicting electron configurations are based on
approximate solutions to Schrodinger’s equation. We detect these
anomalous configurations experimentally.
End of Chapter 6
“Louis de Broglie made the astonishing suggestion that if light is
a particle as well as a wave, perhaps electrons and other particles also
behave as waves. He proposed this in a 1924 PhD thesis that did not
impress his examiners, and would have failed without the endorsement
of Einstein.” - Lee Smolin, The Trouble With Physics
“It’s a particle; it’s a wave.” - Keanu Reeves
“We virtually ignore the astonishing range of scientific and
practical applications that quantum mechanics undergirds: today an
estimated 30 percent of the U.S. gross national product is based on
inventions made possible by quantum mechanics, from semiconductors
in computer chips to lasers in compact-disc players, magnetic resonance
imaging in hospitals, and much more.”
Max Tegmark and John Wheeler Scientific American, Feb. 2001.