Chemistry: A Molecular Approach, 2nd Ed.
Nivaldo Tro
Chapter 7
The Quantum–
Mechanical
Model of the
Atom
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Electromagnetic Radiation
Electromagnetic Radiation or “Light” is
composed of two orthogonal vectors:
An electric wave and a magnetic wave.
Electromagnetic Radiation
• Light is characterized by its wavelength and
frequency.
Electromagnetic Radiation
wavelength
wavelength
The intensity of
Amplitude light is a function
of the wave’s
amplitude.
Node
Ultraviolet radiation
A point of zero
amplitude is called
a “node”.
Electromagnetic Radiation
• The frequency of light is represented by the
Greek letter “nu”, 
•  has units of “cycles per sec” or Hertz (s-1)
• All radiation obeys the relationship:
 = c
• Long wavelength, low frequency
• Short wavelength, high frequency
• Wavelength and frequency are inversely
proportional.
Electromagnetic Radiation
Wavelength has units of length: m… m… nm… pm
Frequency has units of inverse time: s-1 or Hz (hertz)
 (m)   (s–1) = c (m s–1)
c, the speed of electromagnetic radiation (light)
moving through a vacuum is: 2.99792458  108 m/s
c
n=
l
or
c
l=
n
Electromagnetic Radiation
Long wavelength = low frequency
Short wavelength = high frequency
increasing
frequency
increasing
wavelength
Electromagnetic Radiation
The visible region of the electromagnetic spectrum is
only a small portion of the entire spectrum.
Electromagnetic Radiation
Problem:
Visible red light has a wavelength () of 685 nm,
calculate the frequency.
1m
685 nm 
 6.85  10 7 m
1  109 nm
m
3.00  10
c
s  4.38  1014 sec 1
frequency ( ) 

-7
 6.85  10 m
8
Quantization of Energy
Max Planck (1858-1947) proposed that light waves existed as discrete
packets of energy, “quanta” in order to account for the “ultraviolet
catastrophe” predicted by classical physics. The “ultraviolet catastrophe”
arises from the classical theory for the energy emitted by an ideal blackbody governed by the Rayleigh-Jeans law. According
to classical physics, the intensity of emitted light
approaches infinity as the wavelength of the light
approaches zero, hence the
term catastrophe.
Quantization of Energy
An object can gain or lose energy by absorbing or
emitting radiant energy in QUANTA. A quanta of
energy is the smallest unit of energy that may be
exchanged between oscillators or emitted as
radiation. It is too small to be observed in the
classical world in which we live.
Energy of radiation is proportional to frequency
E = h·
h = Planck’s constant = 6.6262 x 10-34 J·s
Quantization of Energy
E = h·
Light with long  (low ) has a low Energy
Light with a short  (high ) has a high Energy
Practice – Calculate the wavelength of a
radio signal with a frequency of 100.7 MHz
Given:  = 100.7 MHz
Find: , (m)
Conceptual  (MHz)
Plan:
 (s−1)
 (m)
Relationships: ∙ = c, 1 MHz = 106 s−1
Solve:
Check: the unit is correct, the wavelength is
appropriate for radiowaves
Tro: Chemistry: A Molecular Approach,
13
Planck’s Law
E = h´ n
yields:
and
h´c
E=
l
c
n=
l
• As the frequency of light increases, the energy of
the photon increases
• As the wavelength of light increases, the energy of
the photon decreases.
Blue Light, (higher frequency) has more energy than
Red Light, with a lower frequency.
Energy of Radiation
Problem:
Calculate the energy of 1.00
mol of photons of red light at
685 nm units of kJ/mol.
The energy of a photon in
units of kJ/mol can be
determined by converting
units of wavelength to
frequency to energy using
Planck’s law and Avogadro's
number via dimensional
analysis.
Energy of Radiation
Problem:
Calculate the energy of 1.00 mol of photons of red
light at 685 nm units of kJ/mol.
Ephoton
h´c
= h´ n =
l
Energy of Radiation
Problem:
Calculate the energy of 1.00 mol of photons of red
light at 685 nm units of kJ/mol.
Ephoton
Ephoton
hc
=
=
l
h´c
= h´ n =
l
Energy of Radiation
Problem:
Calculate the energy of 1.00 mol of photons of red
light at 685 nm units of kJ/mol.
Ephoton
Ephoton
hc
=
=
l
h´c
= h´ n =
l
6.626  10 34
Js
´3.00 ´ 108 m
s
photon
685 nm ´
1m
109 nm
Energy of Radiation
Problem:
Calculate the energy of 1.00 mol of photons of red
light at 685 nm units of kJ/mol.
Ephoton
Ephoton
hc
=
=
l
h´c
= h´ n =
l
Js
´3.00 ´ 108 m
s 6.022 ´ 1023 photons
photon
´
mole
1m
685 nm ´ 9
10 nm
6.626  10 34
J
= 1.75 ´ 10
mole of photons
5
Energy of Radiation
Problem:
Calculate the energy of 1.00 mol of photons of red
light at 685 nm units of kJ/mol.
J
1kJ
kJ
1.75 ´ 10
´ 3 = 175
mole of photons 10 J
mole
5
The units cancel leaving J/mol of photon.
This is in the range of energies that can break
chemical bonds!
Practice – What is the frequency of radiation
required to supply 1.0 x 102 J of energy from 8.5 x
1027 photons?
Given: Etotal = 1.0 x 102 J, number of photons = 8.5 x
27
Find: 10
Conceptual
Plan:
Relationships:
Solve:

number
photons
Ephoton
E=h, Etotal = Ephoton∙# photons
Tro: Chemistry: A Molecular Approach,
21
 (s−1)
Photoelectric Effects
• Certain metals will release (eject)
electrons when light strikes the metal
surface.
• The energy of the light must exceed a
minimum or “threshold energy” for
this to occur.
• Any excess energy beyond this
minimum goes into the kinetic energy
of the ejected electron. (They fly
away with greater velocity)
A. Einstein
(1879-1955)
Photoelectric Effect
• Classical theory suggests that energy of
an ejected electron should increase with
an increase in light intensity.
• This however is not experimentally
observed!
• No ejected electrons were observed until
light of a certain minimum energy is
applied.
• Number of electrons ejected depends on
light intensity so long as the light is above
a minimum energy. (This “minimum
energy” is also the ionization energy of
the metal.)
A. Einstein
(1879-1955)
Photoelectric Effect
Experiment demonstrates the particle like nature of
light.
Photoelectric Effect
• Conclusion: There is a one-to-one
correspondence between ejected electrons
and light waves.
• This can only occur if light consists of
individual units called “PHOTONS” .
• A Photon is a packet of light of discrete
energy.
Photoelectric Effect
Problem:
You are an engineer that is designing a switch that works via
the photoelectric effect. The metal used requires 6.710-19
J/atom to eject electrons. Will the switch work with light of 540
nm or greater? Why or why not?
Photoelectric Effect
Problem:
You are an engineer that is designing a switch that works via
the photoelectric effect. The metal used requires 6.710-19
J/atom to eject electrons. Will the switch work with light of 540
nm or greater? Why or why not?
Solution:
Determine if the energy of the light is greater than the
minimum threshold energy. If so, then electrons will
be ejected, if not, then the switch will not work.
Photoelectric Effect
Problem:
You are an engineer that is designing a switch that works via
the photoelectric effect. The metal used requires 6.710-19
J/atom to eject electrons. Will the switch work with light of 540
nm or greater? Why or why not?
Ephoton
h´c
= h´ n =
l
Photoelectric Effect
Problem:
You are an engineer that is designing a switch that works via
the photoelectric effect. The metal used requires 6.710-19
J/atom to eject electrons. Will the switch work with light of 540
nm or greater? Why or why not?
Ephoton
m
6.626 ´ 10 J × s ´3.00 ´ 10
s
m
540 nm ´ 9
10 nm
-34
Ephoton =
h´c
= h´ n =
l
8
= 3.71019J
Photoelectric Effect
Problem:
You are an engineer that is designing a switch that works via
the photoelectric effect. The metal used requires 6.710-19
J/atom to eject electrons. Will the switch work with light of 540
nm or greater? Why or why not?
3.710-19J < 6.7 10-19J
Conclusion:
The energy of the light is below the minimum
threshold. No ejection of electrons will occur. The
incident light must have a   297 nm to eject
electrons. (Confirm this on your own.)
Atomic Line Emission Spectra and
Niels Bohr
Niels Bohr
(1885-1962)
Bohr is credited with the first modern
model of the hydrogen atom based
on the “line spectra” of atomic
emission sources.
He proposed a “planetary” structure
for the atom where the electrons
circled the nucleus in defined orbits.
In this model, the attractive
electrostatic forces of the electron
and nucleus were balanced by the
centripetal forces of the orbiting
electron.
Spectrum of White Light
When white light passes through a prism, all
the colors of the rainbow are observed.
Spectrum of Excited Hydrogen Gas
When the light from a discharge tube containing a
pure element (hydrogen in this case) is passed
through the same prism, only certain colors (lines) are
observed. Recall that color (wavelength) is related to
energy via Planck’s law.
Line Emission Spectra of Excited
Atoms
• Excited atoms emit light of only certain
wavelengths
• The wavelengths of emitted light are unique
to each individual element.
Atomic Spectra & Bohr Model
Bohr asserted that line spectra of elements indicated
that the electrons were confined to specific energy
states called orbits.
The orbits or energy levels
are “quantized” such that
only certain levels are
allowed. n = 1, 2, 3... 
+
The Bohr Model:
r n = n 2a o
ao = Bohr radius (53 pm)
Atomic Spectra & Bohr Model
Bohr asserted that line spectra of elements indicated
that the electrons were confined to specific energy
states called orbits.
+
The lines (colors)
corresponded to “jumps” or
transitions between the
levels.
Quantum Mechanical Explanation of Atomic
Spectra
• Each wavelength in the spectrum of an atom
corresponds to an electron transition between
orbitals
• When an electron is excited, it transitions from an
orbital in a lower energy level to an orbital in a
higher energy level
• When an electron relaxes, it transitions from an
orbital in a higher energy level to an orbital in a
lower energy level
• When an electron relaxes, a photon of light is
released whose energy equals the energy difference
between the orbitals
Tro: Chemistry: A Molecular Approach,
37
Origin of Line Spectra
The “Balmer”
series for the
hydrogen atom
is in the visible
region of the
spectrum.
A “series” of
transitions end
with a common
lower level.
Line Spectra of Other Elements
• Each element has a unique line spectrum.
• The lines indicate that the electrons can only make
“jumps” between allowed energy levels.
• Knowing the color (wavelength) on can determine the
magnitude of the energy gaps using Planck's Law.
Line Emission Spectra of Excited
Atoms
High E
Short 
High 
Low E
Long 
Low 
Visible lines in H atom spectrum are called the
BALMER series.
The Bohr Model of the Atom
The energy of each level is given by:
En
Rhc
=- 2
n
R = Rydberg constant (1.097  107 m-1)
h = Planck’s constant
(6.626 10-34Js)
c = speed of light
(2.997  108 ms-1)
n = the quantum level of the
electron (1, 2, 3…)
The sign of En is negative because the potential energy
between the electron and the nucleus is attractive.
Energy Levels
•The location of electrons in an
energy level is indicated by
assigning a number n. The
value of n can be 1, 2, 3, 4, etc.
•The higher the n value, the
higher is the energy of the
“shell” that that the electrons
occupy.
Each shell can be thought of as a
step on a ladder
n=
n=3
n=2
n=1
Energy Levels
The spacing between adjacent
levels is given by:
DE = En+1 - En
between n = 1 and 2:
3Rhc
DE =
= 0.75 ´ Rhc
4
n=
virtual
continuum
of levels
n=4
n=3
n=2
between n = 2 and 3:
5Rhc
DE =
= 0.14 ´ Rhc
36
n=1
(as n increases, the levels get closer together)
Energy Levels
Key terms and Vocabulary:
Ground State: The lowest energy level (n = 1)
Excited State: A subsequently higher energy
level. n = 2 is the “first excited state” and so
on.
Absorption: An electron moving from a lower
energy level to a higher energy level via
excitation.
Emission: An electron moving from a higher to
a lower energy level accompanied by the
release of a photon.
Energy Absorption/Emission
Since the gaps between
states get closer and
closer together with
increasing n, the
frequency of the light
emitted changes.
Problem:
Determination the
photon wavelength of a
transition between two
energy levels:
Problem:
Determination the photon wavelength of a transition
between two energy levels:
Rhc
En = - 2
n
Problem:
Determination the photon wavelength of a transition
between two energy levels:
Rhc
En = - 2
n
DE = Efinal - Einitial
Problem:
Determination the photon wavelength of a transition
between two energy levels:
Rhc
En = - 2
n
DE = Efinal - Einitial
Rhc æ Rhc ö
DE = - 2 - ç - 2 ÷
nfinal è ninitial ø
Problem:
Determination the photon wavelength of a transition
between two energy levels:
Rhc
En = - 2
n
DE = Efinal - Einitial
æ 1
Rhc æ Rhc ö
1 ö
DE = - 2 - ç - 2 ÷ = -Rhc ç 2 - 2 ÷
nfinal è ninitial ø
è nfinal ninitial ø
Problem:
Determination the photon wavelength of a transition
between two energy levels:
æ 1
1 ö
DE = -Rhc ç 2 - 2 ÷
è nfinal ninitial ø
Problem:
Determination the photon wavelength of a transition
between two energy levels:
æ 1
1 ö
DE = -Rhc ç 2 - 2 ÷
è nfinal ninitial ø
h´c
E=
l
Problem:
Determination the photon wavelength of a transition
between two energy levels:
æ 1
1 ö
DE = -Rhc ç 2 - 2 ÷
è nfinal ninitial ø
h´c
E=
l
h´c
DE =
lphoton
Problem:
Determination the photon wavelength of a transition
between two energy levels:
æ 1
1 ö
DE = -Rhc ç 2 - 2 ÷
è nfinal ninitial ø
h´c
E=
l
h´c
DE =
lphoton
h´c
lphoton (meters) =
DE
Problem:
Determination the photon wavelength of a transition
between two energy levels:
æ 1
1 ö
DE = -Rhc ç 2 - 2 ÷
è nfinal ninitial ø
h´c
E=
l
h´c
DE =
lphoton
h´c
lphoton (meters) =
DE
take the absolute
value to assure a
positive wavelength.
The sign of E tells the direction:
(+) indicated absorption (-) indicates emission
Problem:
Determination the photon wavelength of a transition
between two energy levels:
R ´ h ´ c = 2.179 ´ 10-18 J
æ 1
1 ö
DE = -Rhc ç 2 - 2 ÷
nfinal = 5
è nfinal ninitial ø
DE = -2.179 ´ 10
-18
æ 1 1ö
J ´ç 2 - 2 ÷
è 5 2 ø energy
ninitial = 2
n =5
in
DE = 4.576 ´ 10-19 J
n =2
The value of E is positive because this is an absorption.
Problem:
Determination the photon wavelength of a transition
between two energy levels:
Watch your math…
æ 1
1 ö
ç n2 - n 2 ÷
è final
initial ø
=
1 1
- = 0.04 - 0.25 = -0.21
25 4
æ
ö
1
ç n 2 - n2 ÷
è final
initial ø
1
= +0.048
25 - 4
Problem:
Determination the photon wavelength of a transition
between two energy levels:
h´c
DE =
lphoton
h´c
λphoton (m)  DE
photon
=
= 4.576 ´ 10-19 J
6.626 1034 J  s
´ 2.997 ´ 108 m
4.576 1019 J
9
10 nm
 4.340 10 m ´
=
1m
7
434.0 nm
s
Particle-Wave Duality: A Prelude to
Quantum Mechanics
L. de Broglie
(1892-1987)
Louis de Broglie in
response to Planck &
Einstein’s assertion that
light was “particle-like”
(photon) stated that small
particles should exhibit a
characteristic wavelength.
Particle-Wave Duality: A Prelude to
Quantum Mechanics
E = mc 2
L. de Broglie
(1892-1987)
hn = mc 2
hn
= mc = p (momentum)
c
n 1
since
=
c l
Conclusion:
Light waves have mass,
particles have a wavelength.
h
h
= p or l =
l
p
Problem:
What is the wavelength associated with an 80g tennis
ball (d = 8 cm) moving at 115 miles per hour?
h
l =
p
 (tennis ball) = 2 x 10 –34 m
l(ball)
= 2.5 ´ 10-33
diameter(ball)
The ball is a bajillion times the
size of the wavelength !!!
No wonder we can’t see it.
Compare that to an electron moving at the same velocity:
 (electron) = 1.4 x 10
–5
m
l(electron)
= 3.2 ´ 109
diameter(electron)
By golly the electron looks a little like a wave !!!
Determinacy vs. Indeterminacy
• According to classical physics, particles move
in a path determined by the particle’s velocity,
position, and forces acting on it
– determinacy = definite, predictable future
• Because we cannot know both the position
and velocity of an electron, we cannot predict
the path it will follow
– indeterminacy = indefinite future, can only predict
probability
• The best we can do is to describe the
probability an electron will be found in a
particular region using statistical functions
Tro: Chemistry: A Molecular Approach,
62
Trajectory vs. Probability
Tro: Chemistry: A Molecular Approach,
63
Uncertainty Principle
W. Heisenberg
1901-1976
• According to Heisenberg the limits
of quantum mechanics state that it
is impossible to determine
simultaneously both the position
and velocity of an electron or any
other particle with any great degree
of accuracy or certainty.
• Therefore an electron is both a
particle and a wave simultaneously.
Dx × mDv ³ h
Wave or Quantum Mechanics
E. Schrodinger
1887-1961
• Taking on the ideas of Bohr, de
Broglie and Heisenberg, Irwin
Schrödinger proposed that matter
can be described as a wave.
• In this theory, the electron is
treated as both a wave and a
particle.
• An electron is described by a Wave
Function “” that completely
defines a system of matter.
Schrödinger’s Equation
• Schödinger’s Equation allows us to calculate the
probability of finding an electron with a particular
amount of energy at a particular location in the atom
• Solutions to Schödinger’s Equation produce many
wave functions, 
• A plot of distance vs. 2 represents an
orbital, a probability distribution map of a
region where the electron is likely to be
found
Tro: Chemistry: A Molecular Approach,
66
Solutions to the Wave Function, 
• Calculations show that the size, shape, and orientation
in space of an orbital are determined to be three
integer terms in the wave function
– added to quantize the energy of the electron
• These integers are called quantum numbers
– principal quantum number, n
– angular momentum quantum number, l
– magnetic quantum number, ml
Tro: Chemistry: A Molecular Approach,
67
Wave motion: wave length and nodes
“Quantization” in a standing wave
Wave Functions, Ψ
• The mental picture of an electron corresponds to a wave
superimposed upon the radial trajectory of a particle orbiting
the nucleus.
• The position of an electron is found from the solutions to the
Schrödinger Wave Equation which predict the “probability”
in a region space where the electron is likely to be found.
• The orbitals are really Probability Distributions
Types of Orbitals
• The solutions to the
Schrödinger equation
yields the probability in
3-dimensons for the
likelihood of finding and
electron about the
nucleus.
• It is these probability
functions that give rise
to the familiar hydrogenlike orbitals that
electrons occupy.
Principal Quantum Number, n
• Characterizes the energy of the electron in a
particular orbital
– corresponds to Bohr’s energy level
• n can be any integer  1
• The larger the value of n, the more energy the
orbital has
• Energies are defined as being negative
– an electron would have E = 0 when it just escapes
the atom
• The larger the value of n, the larger the orbital
• As n gets larger, the amount of energy between
orbitals gets smaller
Tro: Chemistry: A Molecular Approach,
71
Principal Energy Levels in Hydrogen
Tro: Chemistry: A Molecular Approach,
72
Angular Momentum Quantum Number, l
• The angular momentum quantum number determines
the shape of the orbital
• l can have integer values from 0 to (n – 1)
• Each value of l is called by a particular letter that
designates the shape of the orbital
–s orbitals are spherical
–p orbitals are like two balloons tied at the
knots
–d orbitals are mainly like four balloons tied at
the knot
–f orbitals are mainly like eight balloons tied at
the knot
Tro: Chemistry: A Molecular Approach,
73
Magnetic Quantum Number, ml
• The magnetic quantum number is an integer that
specifies the orientation of the orbital
–the direction in space the orbital is aligned
relative to the other orbitals
• Values are integers from −l to +l
–including zero
–gives the number of orbitals of a particular
shape
»when l = 2, the values of ml are −2, −1, 0,
+1, +2; which means there are five orbitals
with l = 2
Tro: Chemistry: A Molecular Approach,
74
Quantum Numbers & Electron
Orbitals
n defines the Principal energy level “shell”
There are n “sub–shells” for each n – level corresponding to l
if “n” equals:
n=1
n=2
n=3
“l” can have values of:
l=0
l=0&1
l = 0, 1 & 2
Each l is divided into (2l + 1) ml “orbitals” separated by
orientation.
if “l” equals: “ml” can have values of:
l=0
ml = 0
l=1
ml = 0, ±1
l=2
ml = 0, ±1, ±2
Quantum Numbers & Electron
Orbitals
Each “l” within an “n-level” represents a sub-shell.
Each “l” sub-shell is divided into ml degenerate
orbitals, where ml designates the spatial orientation
of each orbital.
l=0
l=1
l=2
l=3
Type of orbital
“s” sub-shell (sharp)
“p” sub-shell (Principal)
“d” sub-shell (diffuse)
“f” sub-shell (fine)
# of orbitals
1
3
5
7
each subshell contains 2l+1 orbitals
Energy Shells and Subshells
Tro: Chemistry: A Molecular Approach,
77
Types of Orbitals
s orbital
p orbital
d orbital
s-Orbitals
• l = 0, ml = 0
• 2l+1 = 1
• one s-orbital that extends in a radial manner from
the nucleus forming a spherical shape.
Spherical Nodes
2 s orbital
• All s-orbitals have “n-1” spherical
nodes.
• A 1s-orbital has none, a 2s has 1 and
so on.
• These nodes represent a sphere of
zero probability for finding an electron
1s, 2s, 3s-Orbitals
p-Orbitals
When n = 2, then l = 0 and 1
Therefore, in 2nd shell there are
2 types of orbitals, i.e. 2 subshells
For l = 0
ml = 0 (s-orbital)
For l = 1
ml = -1, 0, +1
2(1) + 1 or 3 p-orbitals
Each p-orbital is
characterized by a
“nodal plan”
p-Orbitals
The three degenerate p-orbitals spread out on the x,
y & z axis, 90° apart in space.
2px- & 3px-Orbital
d-Orbitals
When n = 3, l = 0, 1, 2 resulting in 3 sub-shells
within the shell.
For l = 0, ml = 0
s-sub-shell with a single s-orbital
For l = 1, ml = -1, 0, +1
p-sub-shell with 3 p-orbitals
For l = 2, ml = -2, -1, 0, +1, +2
d-sub-shell with 5 d-orbitals
2(2)+1 = 5
d-Orbitals
s-orbitals have no nodal
planes (l = 0)
p-orbitals have one nodal
plane (l = 1)
d-orbitals therefore have
two nodal planes (l = 2)
3dxy, 3dxz, 3dyz-Orbital
2
2
3dx - y
& 3dz
2-Orbital
f-Orbitals
When n = 4, l = 0, 1, 2 & 3 resulting in 4 sub-shells within the
shell.
For l = 0, ml = 0
s-sub-shell with a single s-orbital
For l = 1, ms = -1, 0, +1
p-sub-shell with 3 p-orbitals
For l = 2, ms = -2, -1, 0, +1, +2
d-sub-shell with 5 d-orbitals
For l = 3, ms = -3, -2, -1, 0, +1, +2, +3
f-sub-shell with 7 f-orbitals
2(3)+1 = 7
f-Orbitals
One of the 7 possible f-orbitals.
All have 3 nodal planes.
Summary of Quantum Numbers
Arrangement of Electrons in Atoms
• Each orbital can accommodate no more than
2 electrons
• Since each electron is unique, we need a
way to distinguish the individual electrons in
an orbital from one another.
• This is done via the 4th quantum number,
“ms”.
Electron Spin
When an atom with one
unpaired electron is passed
through a magnetic field, the
path is altered into two
directions.
Detector
N
Source of electrons
This means that the electrons
have a magnetic moment!
S
Electron Spin
Since there were 2 pathways in the experiment, there
must be 2 spins affected by the magnetic field. One
spinning to the right, one spinning to the left.
Each “spin state” is assigned a quantum number
ms = ± ½
+ ½ for “spin up”
 ½ for “spin down”
Electron Spin
Electron Spin
Quantum
Number, ms
The experiment results indicate that electron has an
intrinsic property referred to as “spin.”
Two spin directions are given by
ms where ms = +1/2 and -1/2.
Electron Spin
• It is not that the electrons are actually
spinning on axis...
• Rather it is that the mathematics that
describe the electrons “looks” like they are
spinning on axis.
Electron Spin & Magnetism
• Diamagnetic Substances:
Are NOT attracted to a
magnetic field
• Paramagnetic Substances:
ARE attracted to a magnetic
field.
• Substances with unpaired
electrons are paramagnetic.
Measuring Paramagnetism
When a substance containing unpaired electrons is place into
a magnetic field, it is attracted to that magnetic field.
The effect is proportional to the number of unpaired electrons.
Electron Quantum Numbers
Principal (n) shells
Angular (l) sub-shells
Magnetic (ml) orbitals
Spin (ms) individual electrons
n = 1, 2, 3, 4, ...
l = 0, 1, 2, ... n - 1
ml = - l ... 0 ... + l
ms =  ½