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Welcome to the “quantum” world
Chapter 8
The behavior of “large”-scale matter
in our everyday life is predictable: the
motions of the planets around the
Sun, the falling of an apple to the
ground, or the collisions of balls on a
pool table all can be described
effectively using classical physics (or
mechanics).
Electrons in Atoms
Dr. Peter Warburton
[email protected]
http://www.chem.mun.ca/zcourses/1050.php
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Welcome to the “quantum” world
Welcome to the “quantum” world
Partly this is because these small objects
are constantly awash in a sea of
electromagnetic radiation which can affect
how the objects move and interact with
each other. Also, the simple act of
measuring things like the position and
speed of these objects becomes more
complicated, since we must scale the
atomic scale events to provide responses
we can detect in the “large” world.
However, as objects get smaller and
smaller, it turns out that the behavior of
some of the smallest pieces of the
universe, like atoms and the electrons,
protons and neutrons they are made of
becomes less predictable – classical
physics no longer works well in describing
what’s going on.
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1
Welcome to the “quantum” world
Electromagnetic radiation
Ultimately, it is unreasonable to
expect the small pieces of the
universe to behave like things do in
our “large” experience of the
universe.
We need a special way to deal with
the “small” parts of the universe –
quantum mechanics!
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This is a form of energy transmission via
the wave propagation of electric AND
magnetic fields.
Unlike waves in our every day experience
(sound, water, seismic) these waves do
not require a medium for propagation –
electromagnetic radiation can travel
through a vacuum that contains no matter!
5
Waves
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Waves
Wavelength λ – The distance
between adjacent peaks (crests) in
the wave
SI unit: the meter (m)
though we sometimes see the nonSI angstrom (Å) where 1 Å = 10-10 m
For electromagnetic radiation in
chemistry we often deal with
wavelengths ranging in length from
kilometers (103 m - radio waves) to
picometers (10-12 m – X-rays)
Waves transmit energy via a
cyclic motion, either of the
medium itself or of the electric and
magnetic fields.
Frequency ν – The number of
cycles or events that happen in a
given time period. Has units of
inverse time like s-1.
SI unit: 1 Hz (hertz) = 1 s-1
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Waves
Electromagnetic radiation
Amplitude A – The maximum height
a point is displaced from it’s “rest”
position as the wave propagates.
In a medium like the water of the
ocean, a larger amplitude means a
greater amount of water is displaced
– implying greater energy. In fact E
Ñ A2. However, we’ll see for
electromagnetic radiation the energy
is determined by the wavelength and
NOT the amplitude – one way where
the quantum world is different!
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Electromagnetic radiation
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Electromagnetic spectrum
Electromagnetic radiation is light!
Speed of light (c) in a vacuum:
c = 2.99792458 x 108 m s-1
Higher
frequency
means shorter
wavelength
Relationship of the electromagnetic waves
to the speed of light:
speed of light = frequency x wavelength
c=νxλ
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Problem
Problem answer
ν = 4.34 x 1014 s-1 = 4.34 x 1014 Hz
The light from red LEDs (light emitting
diodes) is commonly seen in many
electronic devices. A typical LED
produces 690 nm light. What is the
frequency of this light?
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Wave interference
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Constructive interference
If equal wavelength
waves are in phase,
crests line up perfectly
with crests, and the
added wave has twice
the amplitude as the
individual waves. This
is called constructive
interference.
Regardless of the type of wave,
when two waves interact with
each other, their behavior is
additive. However, the result of
the additive behavior can be quite
variable, depending on the phase
of the interacting waves.
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Destructive interference
Wave interference
If equal wavelength
waves are out of
phase, crests line up
perfectly with troughs,
and the added wave
has zero amplitude
compared to the
individual waves. This
is called destructive
interference.
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The visible spectrum
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The visible spectrum
We’ve seen
Speed of light (c) in a vacuum:
c = 2.99792458 x 108 m s-1
However, when light travels through a physical
medium (air, glass, water, etc.) the speed of light
is slightly less than in a vacuum, and each
wavelength of light will be refracted (bent from
a straight path) by a different amount that
depends on wavelength.
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What we think of as white light actually contains
light of all wavelengths. Refraction of the white
light through a prism or a raindrop will cause the
white light to separate into the individual
wavelengths, resulting in the continuous visible
spectrum of colors!
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Discontinuous spectra
Discontinuous spectra
What happens when we refract light of a specific
color? That specific color, like white light, may
be made up of more than one wavelength, but
there are many wavelengths that are not
components of the spectrum we get.
Since it is the atoms of
a substance that are
somehow responsible
for the wavelengths of
light we see, we often
call discontinuous
spectra by the name
atomic (or line)
spectra
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Atomic spectrum of helium
Atomic spectrum of hydrogen
Here the purple/pink light
of a helium discharge tube
is refracted through a
spectroscope and results
in the atomic spectrum of
helium we see below. We
see the purple color is
actually a combination of
specific wavelengths of
red, yellow, green, blue
and violet!
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When Johann Balmer looked at the atomic spectrum for
hydrogen, he was able to detect a pattern! Of the four
lines he was able to see in the visible range he was able
to deduce that the frequencies of the four lines could be
predicted by the equation
= 3.288110 − where n > 2
n=6
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n=5
n=4
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n=3
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Blackbody radiation
Blackbody radiation
Hot objects give off light
– we call this blackbody
radiation.
We see a hot object has
a specific colour based
on the distribution of
wavelengths of light it
emits. This distribution
changes with
temperature!
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Blackbody radiation
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Energy is quantized
Classical theory predicts
the intensity of radiation
increases indefinitely with
wavelength – the
ultraviolet catastrophe.
A system may only possess very
specific amounts of energy, not a
continuum!
To go between two of the allowed
states of the system required the
absorbtion or emission of a quantum
of energy.
Max Planck –
“Energy, like
matter, is
discontinuous.”
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Energy of matter is quantized
Other evidence of quantization
ε = nhν
Photoelectric effect – Light shining on the
surface of some metals causes the
ejection of electrons.
However, electrons are only ejected when
the frequency ν of the light is above some
threshold ν0, regardless of the intensity
of the light. This is not classical behaviour!
The number of electrons ejected does
depend on the light intensity.
where
n is a positive integer
h = 6.62607 x 10-34 J s
ν is frequency of absorbed or emitted
radiation
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Photoelectric effect (Einstein)
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Energy of a photon is quantized
E = hν
Light can be treated as particles called
photons. The energy of a photon
depends on the light’s frequency when
treated as a wave (see slide 30), while a
greater intensity of light implies a
greater number of photons.
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The slope of
this graph is
Planck’s
constant!
31
where
h = 6.62607 x 10-34 J s
Planck’s constant
ν is frequency
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Explaining the photoelectric effect
Explaining the photoelectric effect
When a photon hits an bound electron in
an atom, the electron can absorb the
photon energy.
If the photon energy is greater than the
work function (the amount of energy
required to “just” unbind the electron from
the atom), then the remaining energy goes
into determining the kinetic energy of the
now unbound electron.
To measure the kinetic energy of the
electron, we set up a stopping voltage Vs
between two metal plates. The voltage is
adjusted until the electron stops moving,
which happens when
Ek = ½ mev2 = eVs
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Explaining the photoelectric effect
Ek = ½ me
v2 =
34
Work function and threshold frequency
We saw there was a threshold ν0 in
the photoelectric effect which defined
the minimum frequency (and
therefore minimum energy of a
photon since E = hν) required to eject
the electron. Any energy above the
threshold is what is measured by the
stopping voltage.
eVs
Here me is the electron rest mass of
9.109 x 10-34 kg
and e is the electron charge
1.602 x 10-19 C
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Work function and threshold frequency
Work function and threshold frequency
The work function can be expressed
By energy conservation, the energy of the
photon must be used to overcome the
work function, and provide the kinetic
energy of the electron, so
Ew = hν0
where ν0 is the minimum frequency
of light required to free an electron,
which depends on the metal.
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Ew + Ek = hν
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Problem
Work function and threshold frequency
The minimum energy (the work function)
required to cause the photoelectric effect
in potassium metal is 3.69 x 10-19 J. Will
photoelectrons be produced when blue
light of wavelength 400 nm is shone on the
metal? If they are ejected, what is the
velocity of the electrons?
Therefore
Ek = hν - Ew
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Problem answer
Photochemistry
Electrons will be ejected by 400 nm light,
with a velocity of 1.68 x 107 m s-1. This is
about 5.6% of the speed of light.
Photons of light can provide enough
energy to break chemical bonds by
changing the distribution of electrons in the
molecule. We can treat hν as a reactant!
O2 + hν ↓ 2 O
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Problem
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Problem answer
Chlorophyll absorbs light energies of 3.056
x 10-19 J photon-1 and 4.414 x 10-19 J
photon-1. To what color, frequency and
wavelength do these absorptions
correspond, and can you use these results
to explain why chlorophyll appears green?
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E = 3.056 x 10-19 J photon-1
ν = 4.612 x 1014 s-1 and λ = 650.4 nm
Absorption of red light
and E = 4.414 x 10-19 J photon-1
ν = 6.662 x 1014 s-1 and λ = 450.3 nm
Absorption of blue light
When you absorb red and blue, the color of
visible light that is reflected (and seen!) is around
a wavelength of 550 nm, which is green!
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The Bohr atom
The Bohr atom
The problem with this
image is that an orbiting
electron is always
accelerating – and
accelerating charges
give off light (and
therefore lose energy!)–
the electron should lose
energy and “death spiral”
into the nucleus!
Often our vision of the
electrons in an atom
is one of the electrons
“orbiting” the nucleus
like the planets orbit
around the Sun. This
would be classical
mechanics behaviour.
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The Bohr atom
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The Bohr atom
Second, an electron is
only allowed to have of
fixed set of orbits called
stationary states, which
depend on the angular
momentum of the
electron, which depends
π, where n is the
on nh/2π
principal quantum
number, and can only
be a non-zero integer, so
n = 1, 2, 3, -
Niels Bohr proposed
a different model
with three main
properties:
First, the orbit of an
electron is circular
like in classical
physics.
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The Bohr hydrogen atom
The Bohr atom
Radius of an allowed orbit is rn = n2a0
a0 is the Bohr radius where a0 = 53 pm = 0.53 Å
Third, an electron can
only move from one
stationary state to
another. This transition
requires the absorption
or emission of a photon
with an energy
matching the difference
of the energy of the
electron in the two
stationary states.
If we treat the infinitely separated nucleus and
electron as zero in energy, the energy of each
orbit is
=
−
where RH (the Rydberg
constant) is
2.179 x 10-18 J
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Energy level diagram
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Electronic transitions in a hydrogen atom
The photon absorbed or emitted must
have an energy that exactly matches the
energy difference between the two
electronic states
See slide 24!
Second excited state
n=3
First excited state
n=2
∆E=hν=R H
Ground state n = 1
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1
1
−
ni2 nf 2
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Problem
Problem answer
The energy of an electron in a hydrogen
atom is -4.45 x 10-20 J. What energy level
does the electron occupy?
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n=7
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Problem
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Problem answer
λ = 486.5 nm
Determine the wavelength of light
absorbed in an electron transition from n =
2 to n = 4 in a hydrogen atom.
n=6
n=5
n=4
n=3
The Balmer series represents all
the transitions to or from the n = 2
state!
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Absorption spectra
Helium absorption spectrum
The line spectra we saw
earlier are emission
spectra, where light is
given off.
Now we’ve seen that light
can also be absorbed,
which means we can have
absorption spectra
where specific
wavelengths are absorbed
(but usually only from the
ground state).
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Ionization energy of hydrogen
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If we consider ions that have only one
electron like He+, Li2+, Be3+ P then we
can use the Bohr models to predict the
energy levels and the ionization energy of
the last electron
− =
1
1
− 2 = R H 1−0 =R H
2
ni
nf
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Ionization energy of hydrogen-like ions
To completely remove the electron from
a hydrogen atom is an example of
ionization.
The Bohr model allows us to predict the
ionization energy of an electron in a
hydrogen atom from the ground state ni =
1 to an unbound electron where nf = ∞
∆E=hν=R H
Helium was actually
discovered from the
absorption spectrum of
the Sun in 1868.
Norman Lockyer
predicted the existence
of helium as an
element 27 years
before it was isolated
on Earth!
where Z is the atomic number for the ion
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Problem
Ionization energy of hydrogen-like ions
− =
Determine the wavelength (and color if it
falls in the visible spectrum) of light emitted
in an electron transition from n = 5 to n = 3
in a Be3+ ion.
As the atomic number goes up, the ionization
energy increases because we are trying to
remove a negatively charged electron from a
nucleus with a total Z+ charge. The electron is
more tightly electrostatically bound to a nucleus
with larger positive charge!
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Problem answer
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Problem
λ = 80.13 nm. This is actually ultraviolet
light.
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The frequency of the n = 3 to n = 2
transition for an unknown hydrogen-like ion
occurs at a frequency 16 times that for
hydrogen. What is the identity of the ion?
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Problem answer
Inadequacies of the Bohr model
Since there is a Z2 dependence for
hydrogen-like ions, Z must be 4 for a 16
times increase in frequency (and energy
of transition). This means the ion is Be3+.
We only talk about the Bohr model for hydrogen
and hydrogen-like ions because the model
cannot handle more than one electron!
Two or more electrons in an atom, molecule or
ion will interact with each other, changing the
energy levels in comparison to a single electron
interacting with a nucleus.
A new theory was required!
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Building a new theory
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Wave-particle duality
Wave-particle duality - we saw in the
photoelectric effect that Einstein treated
light (which we think of waves) as
particles called photons.
Louis de Broglie proposed that matter and
energy actually have “dual” nature where
“Small particles of matter may at times
display wave-like properties.”
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Starting from Einstein’s equation telling us
that matter and energy are interconvertible
E = mc2
de Broglie related this to the energy of a
photon
E = hν
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Wave-particle duality
Wave-particle duality
This means
Since
λν = c or ν/c = 1/λ
hν = mc2
which we can rearrange to
then
hν/c = mc = p
p = h/λ = mu
where p is the momentum (mass
times velocity) of the photon
where u is the speed of any
particle
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de Broglie wavelength
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de Broglie wavelength
h/p = h/(mu) = λ
h/p = h/(mu) = λ
Any wave can be described as a particle in
motion, and any particle in motion can be
described with a wavelength.
However, we see from the equation, that
since h is very small, then the wavelength
for a particle must also be very small –
UNLESS it’s mass is very small as well!
Consider the problem we did on slides 40
and 41 where an electron was ejected
from a potassium atom with a speed of
1.68 x 107 m s-1. The moving electron can
be treated as a wave with
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λ = h/(meu) = 43.3 pm
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de Broglie wavelength
X-rays can be in the pm range!
We also saw on slide 50 that the Bohr
radius of an electron in the ground state of
a hydrogen atom is 53 pm!
If moving electrons have wavelengths in
the picometer range, we should be able to
use them to probe the structure of matter.
X-ray diffraction
Electron beam
diffraction
How? We’ll see soon!
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Boats and waves
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Scanning electron microscopy (SEM)
To observe an object requires the waves
interacting with the object to be similar in size, or
smaller than the object. If the waves are too big,
the small object gets “overwhelmed” by the wave.
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Using the wave-like
properties of a beam of
electrons, we can get
very detailed images of
small objects that we
cannot get using lenses.
Here we have a white
blood cell (pink) infected
by HIV viruses (blue) at
13400x magnification!
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More SEM
Building a new theory
Heisenberg uncertainty principle - the idea
that very small particles have wave-like
qualities means we can never be
absolutely certain about it’s behavior, since
a particle is seen as discrete and localized
and a wave is variable and “spread out”
(delocalized)
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Heisenberg uncertainty principle
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Analogy with camera shutter speed
Is the car moving? Short
time frame of the picture
means we can’t be
absolutely sure. We
know x but aren’t sure
about p.
Here, the car is definitely
moving. But the long
time frame of the picture
means we don’t know
exactly where it was.
We know p but not x.
If we try and describe a particle by two
variables (usually it’s position x and
momentum p), it turns out that if we are
dealing with a small enough particle, then
we cannot accurately measure both things
at the same time. The uncertainty of the
measurements ∆x and ∆p are related!
∆x∆p ≤ h/4π
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Wave packet
Types of waves
To “localize” a small
particle, we must
describe it as a group
of waves - a wave
packet. The more
waves we have, the
better we know x, but
the harder it is to know
p, and vice versa (less
waves to get p, but
don’t know x)
Traveling waves – like
ocean waves or sound
waves, the crests and
troughs move
through space
Standing waves – like
a plucked guitar string,
the crests and
troughs DO NOT
move through space
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Types of waves
n=1
λ = 2L/n
n=2
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Standing waves in a Bohr orbit
Standing waves have
a wavelength that is an
integer fraction of twice
the total length
Standing waves also
have nodes where the
displacement is zero.
The number of such
nodes is n + 1.
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Imagine a guitar
string of length 2L
wrapped around to
form a circle of
circumference 2L.
Standing waves in
the string must still
obey
λ = 2L/n!
n=3
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Allowed standing
wave
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Standing waves in a Bohr orbit
If the standing wave
doesn’t obey the
relationship, it will
destructively interfere
with itself, cancelling
out to zero. The
standing wave is not
allowed!
de Broglie vs Bohr
The electrons in an
atom set up standing
waves to achieve the
allowed stationary
states for each n.
Still not good enough!
A guitar string is 1D,
and we need a 3D wave
to describe the electron
in all of the space
around the nucleus.
Disallowed
standing wave
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3D description of an electron
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Particle in a 1D box
Before we can understand how
to describe an electron in 3D,
we need a better
understanding of how to
describe a standing wave in
1D. To do this we solve a
model called particle in a box.
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The standing wave for an
electron in 1D can be treated
exactly like the guitar string.
There must be nodes at each
end of the string, while the
wave must be described by
some sort of cyclic function,
like a sine function!
We give the cyclic function
description a special name –
the wavefunction!
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Particle in a 1D box
Particle in a 1D box
For a 1D box (along the x
direction) of size L the
wavefunctions ψn that describe
stationary states of a particle in
a 1D box can be described as
The energy of each stationary
state of a wave can be found
since (using de Broglie)
Ek = ½ mv2 = p2/2m = h2/2mλ2
For a particle in a 1D box
h2
n2 h2
Ek =
=
2
8mL2
2mλ
2
nπx
ψn = sin
L
L
where n = 1,2,3,P
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Particle in a 1D box
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Particle in a 1D box
h2
n2 h2
Ek =
=
2
8mL2
2mλ
n2 h2
Ek =
=
2
8mL2
2mλ
The energy of the particle can
never be zero (the zero-point
energy means never at rest!)
Smaller L (better knowledge
of x) implies larger Ek and
therefore a less certain p
Most importantly, the
energy of the particle in
the box is quantized,
based on the allowed
values for n!
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h2
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What is the wavefunction?
Wavefunction analogy
The wavefunction is a mathematical
description of how a particle will behave
under certain conditions (like a box of a
certain size). Ultimately, it must contain
ALL the information of particle behavior
under ALL conditions.
However, the wavefunction is not
something physical we can measure.
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Consider all of the information available to
us through the internet. We certainly can’t
know all of it in our own heads, but under
the appropriate circumstances (a Google
search), we can find the information we
want.
We’ll see this analogy extend further in a
few slides.
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Particle in a 1D box
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Particle in a 1D box
Since the particle
behaves like a wave,
we can not exactly
say where the
particle is, but we
can evaluate the
probability of finding
the particle at a given
point in the box.
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We saw that the
energy of a
classical wave
depends on its
intensity, which
corresponds to
the square of
the amplitude.
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Particle in a 1D box
Particle in 3D box
The probability of
finding a particle is
the quantum
equivalent to the
intensity, and
depends on the
“square” of the
wavefunction ψ2
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In 3D space, we can treat the
wavefunction as a combination of three
1D wavefunctions, each with its own
principal quantum number n. When we do
this, we can get the energies of a particle
in a 3D box
&' (
ℎ + . / =
+
+
8* ,+ ,. ,/ 97
Problem
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98
Problem answer
What is the wavelength of the photon
emitted when an electron in a 1D box with
a length of 5.0 x 101 pm falls from the n = 5
level to the n = 3 level?
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λ = 0.52 nm
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Schrödinger equation
Schrödinger equation
We’ve seen that the wavefunction must
contain information about the particle in all
circumstances, and therefore we should be
able to use it to find many different
properties of the particle. To get the
specific property we’re interested in, we
apply an operator to the wavefunction which
should return the wavefunction multiplied by
an observable such as energy.
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101
Wavefunction of the H atom
0 ψ=Eψ
H
The operator H-hat is much like a Google
search of the internet. It “pulls out” the
property information we are interested in
from the large collection of information that
is the wavefunction.
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102
Wavefunction of the H atom
When we saw the energy
solutions of a particle in a
3D box, we did them in
terms of Cartesian
coordinates (x, y, z). It
turns out the math is more
conveniently done in what
is called spherical polar
coordinates (r, θ, φ).
φ
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Schrödinger equation –
In spherical polar
coordinates, we
define the electrons
position in an atom by
its distance r from the
nucleus, as well as
two angles θ and φ to
some arbitrary line.
103
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104
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Wavefunction of the H atom
Wavefunction of the H atom
Solutions of the Schrödinger
equation for the hydrogen
atom define orbitals that
describe the motion of the
electron in the H atom (not
Bohr orbits). These orbitals
will have features that
depend on both the radial
wavefunction and the angular
wavefunction.
This allows us to break the
wavefunction down into two
parts:
radial wavefunction R(r)
angular wavefunction Y(θ, φ)
This results in
ψ(r, θ, φ) = R(r) Y(θ, φ)
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105
Quantum numbers
106
Quantum numbers
We’ve already seen the principal quantum
number n (based on the angular momentum
of the electron).
The name is interesting, though. It implies
there must be other quantum numbers for
the electron to help describe other aspects
of the electron in the atom. It turns out
solutions for the Schrödinger equation for the
hydrogen atom require two more quantum
numbers! However, they are connectedP
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107
Principal (angular momentum) quantum
number n where
n = 1, 2, 3, P
Orbital angular momentum quantum
number l where
l = 0, 1, 2, 3, P, n-1
Magnetic quantum number ml where
ml = -l, (-l+1), P, -2, -1, 0, 1, 2, l-1, l
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Problems
Problem answers
a) Yes. For n = 3 the values of l can be
either 0, 1 or 2. For ANY of these
values of l, ml can be 0, since ml must
fall in the range of –l to +l in all cases.
b) If n = 3 then the values of l can be 0, 1
or 2. But since ml must fall between –l
and +l, a ml value of 1 eliminates l = 0
as a possibility, but leaves l = 1 or l = 2
as allowed values.
a) Can an orbital have the quantum
numbers n = 3, l = 0 and ml = 0?
b) For an orbital with n = 3 and ml = 1,
what is (are) the possible value(s) of l?
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109
110
Shells and subshells
Physical interpretation of the quantum numbers
The quantum numbers tell us something
about the orbitals the electrons move
within
n determines the energy and average
(most probable) distance from the nucleus
l determines the angular shape of the
orbital
ml determines the orientation (direction) of
the orbital
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Since we see the quantum numbers
are tied together, it makes sense to
logically group them together in some
way.
Orbitals that have the same value of n
are said to belong to the same shell.
Orbitals that have the same n and l are
said to belong to the same subshell.
111
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112
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Subshells
Subshells
We saw that l can take integer values
ranging from 0 to n-1, and therefore for a
given shell (given value of n) there can be
a variable number of subshells.
When n = 1 l can only be 0 (1 subshell)
When n = 2 l can be 0 or 1 (2 subshells)
When n = 3 l can be 0, 1 or 2 (3 subshell)
When n = 4 l can be 0, 1, 2 or 3 (4 subshell)
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113
Total orbitals in a subshell
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114
Total orbitals in a subshell
Since ml values can range from –l
to +l, for each subshell, there will
exist 2l+1 orbitals of a given
name in a subshell. These
should be similar in shape (same
l) but different in their direction
(different ml)
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We very quickly see the number of allowed
subshells in a shell equals the principal
quantum number for the shell. Based on
this, we give orbitals special names to
describe the shape of the subshell
When l = 0 (s orbital)
For higher values
When l = 1 (p orbital)
of l the names
When l = 2 (d orbital)
continue as g, h,
i, j, k, When l = 3 (f orbital)
115
There can only be 1 s orbital when l = 0
since ml can only be 0
There can be 3 p orbitals when l = 1
since ml = -1, 0 or 1
There can be 5 d orbitals when l = 2
since ml = -2, -1, 0, 1 or 2
There can be 7 f orbitals when l = 3 since
ml = -3, -2, -1, 0, 1, 2 or 3.
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116
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Orbital energies
Orbital energies
The energies of an electron in the orbitals in
each subshell for a hydrogen atom are given by
the energy of the shell (E depends on n only)!
En = -RH (1/n2)
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If there is no “favorite” direction determined by
an external electric or magnetic field, this means
each orbital of a subshell has the same energy.
We call orbitals at the same energy degenerate.
117
What do orbitals tell us?
118
Radial vs angular wavefunctions
We’ve already seen it’s
more convenient to
break the wavefunction
into radial and angular
parts. The radial part
tells us something about
the most probable
distance that we will
find an electron from the
nucleus when it is in a
given orbital.
The orbitals that result from the solutions
of the Schrödinger equation for the
hydrogen atom represent the stationary
states for the electron described by the
appropriate set of quantum numbers for
the orbital we are interested in. The
“square” of the orbital will tell us the
probability of finding an electron within
some part of the orbital.
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119
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120
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Radial vs angular wavefunctions
s orbitals for hydrogen-like atoms
If we look at the table,
we see that the angular
wavefunction for any s
orbital, regardless of the
value of n, is a constant!
This tells us that no
direction is “special” for
an s orbital. No special
direction implies a
spherical shape for an
s orbital in 3D.
The angular
wavefunction tells
us something
about the
directions in
which an electron
is most likely to
be found, when it
is in a given orbital.
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121
1s orbital is when n = 1 and l = 0
122
s orbitals when n > 1
Here we see three different ways of representing the
probability of finding an electron in the 1s orbital. No direction
is “special”, so we see “spherical distributions”, but the radial
wavefunction tells us we are more likely to see the electron
close to the nucleus, and less likely to see it far away.
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123
Here we see the 1s, 2s and 3s
orbitals represented by contours
within which we are likely to find the
electron 95% of the time. Since the
angular wavefunction is always
constant for s orbitals, s orbitals are
always spherical.
The main differences is s orbitals
when n = 1 or n = 2 or n = 3 comes
from the radial part of the
wavefunction.
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124
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s orbitals when n > 1
s orbitals when n > 1
When we look at the radial
wavefunctions for the 1s, 2s and 3s
orbitals, we see two interesting
features involving σ, which depends
on the value of n and the distance r:
For 2s there is (2-σ)
For 3 s there is (6-6σ+σ2)
If we think carefully, we realize there
can be certain distances r where the
interesting feature can BECOME
ZERO in the radial wavefunction,
which implies a zero probability of
finding the electron at that distance!
125
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Radial nodes
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126
Angular wavefunction of p orbitals
Radial nodes!
This means, for s orbitals, there will
exist n – 1 radial nodes at specific
distances where the electron CAN
NEVER BE FOUND! It turns out that
regardless of orbital shape, there
will always be n - l – 1 radial nodes
for an orbital.
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For p orbitals, we see that the angular
wavefunction is not a constant. Depending on
the direction considered (x, y or z) we will see a
dependence on one or possibly two angles!
127
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128
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Angular (or planar) nodes
Angular wavefunction of p orbitals
More correctly, we see that we have
dependencies on the sine or cosine of an
angle(s). Because the sine or cosine of a given
angle can be ZERO in certain cases, this implies
there will be directions in a given p orbital where
the electron CANNOT be found (probability of 0)
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129
A 2p orbital
Here we see the angular part of p orbital lying
along the z axis. The “square” of the orbital, gives
a probability of finding an electron within this p
orbital. In either case we see we will NEVER find
the electron lying in the xy plane. This is an
angular (or planar) node for the orbital.
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130
Angular (or planar) nodes
We’ve seen that there are three allowed values for ml in
what we call a p orbital for which l = 1. That is ml = -1, 0 or
1. This implies there are three different p orbitals P one
in each direction, each with its own planar node to
consider. However, notice that regardless of the p orbital
we consider, there is zero chance of finding the electron at
the nucleus!
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131
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132
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Angular (or planar) nodes
Total number of nodes for an orbital
It turns out that the number of angular
nodes a given orbital has is given by l.
For s orbitals l = 0 so no planar nodes
For p orbitals l = 1 so one planar node
Fro d orbitals l = 2 so two planar nodes
and so onP
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We’ve seen the number of radial nodes for
an orbital is n - l – 1 and the total number
of angular nodes is l. Therefore the total
number of nodes of either type for a given
orbital is
n-l–1+l=n–1
133
d orbitals
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134
d orbitals
We cannot see d orbitals until n is equal or
greater than 3, and l = 2. If we consider
the 3d orbitals, there should be 5 possible
orbitals, each containing zero radial nodes
and two planar nodes.
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135
Once again we see the planar nodes mean
that we can never find a d orbital electron
at the nucleus, because there are two
planar nodes that pass through it!
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136
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The 4th quantum number – ms
d orbitals
The two planar nodes are easy to identify
in the first 4 orbitals seen below. What
about the last one? It turns out that in that
orbital, the two planar nodes are “wrapped
around” to give a “double cone”
Notice that table 8.1 gives the angular and
radial wavefunctions for hydrogen-like
atoms (with only 1 electron). It turns out
that the hydrogen line spectrum can’t be
completely explained unless we add
another concept to the mix:
electron spin
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137
Stern-Gerlach experiment
138
Electron spin
If we pass a beam of silver atoms through a nonuniform magnetic field, the beam splits into two
beams. This implies that the electrons in the atom
interact with the magnetic field (via their own
magnetic field caused by the electron spin), which
can either work with the external field, or against it
(two options).
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139
1. The electron spin
generates a
magnetic field.
2. A pair of electrons
with opposing spin
has no net
magnetic field.
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140
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Electron spin and ms
Electron spin and ms
3. The silver beam splits
because there are an odd
number of electrons – we
can pair 23 sets of
electrons, but we have one
unpaired electron left over.
4. The unpaired electron
can either be
spin up (ms = +½ or ↑) or
spin down (ms = -½ or ↓)
spin up (ms = +½ or ↑) or
spin down (ms = -½ or ↓)
Notice that the electron
spin is a property of the
electron and does not
depend on which orbital
the electron is found in. It
DOES NOT DEPEND on
any of the other three
quantum numbers.
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141
Shorthand notation
1s1 is when
n = 1, l = 0 and one unpaired electron
or 2p2
n = 2, l = 1 and two paired electrons
or 3d1
n = 3, l = 2 and one unpaired electron
:;<=>?@=A=BC>?DE?><ECFA
such as 1s1 or 2p2 or 3d1
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142
Shorthand notation
To describe any electron in an atom, we
can provide the first three quantum
numbers (we can’t say what spin we have
without an experiment, we can only tell if
the spins are paired or not). However, its
easier to use a shorthand notation.
ℎ122 3456782ℎ891
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143
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144
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Problem
Problem answers
Identify the error in each set of quantum
numbers given in the form (n, l, ml, ms)
below:
a) (2, 1, 1, 0)
b) (1, 1, 0, ½)
c) (3, -1, 1, -½)
d) (0, 0, 0, -½)
e) (2, 1, 2, ½)
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a)
b)
c)
d)
e)
145
Multielectron atoms
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146
Multielectron atoms
The orbitals of multielectron atoms are
somewhat different than those for the
hydrogen atom. Where the radial and
angular wavefunctions are mathematically
similar, in multielectron atoms, the larger
nuclear charge means that an orbital of a
given type in a given shell will be lower in
energy than the same orbital in a hydrogen
atom.
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(2, 1, 1, 0) – ms must be +½ or -½
(1, 1, 0, ½) – l must be n-1 or less
(3, -1, 1, -½) – l cannot be less than 0
(0, 0, 0, -½) – n cannot be less than 0
(2, 1, 2, ½) – ml must be between –l
and +l
147
Also, the orbitals of different types within
the same shell are no longer degenerate –
they differ in energy from each other! The
reason for this is simple. Hydrogen only
has one electron, and therefore each
orbital of a given shell only “sees” the
nucleus. In multielectron atoms, the
presence of other electrons creates effects
that will make the orbital energies nondegenerate.
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148
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Effects of many electrons
Electron penetration
Electron shielding of the nucleus – the
electrons found in lower shells are closer to the
nucleus. This means the electrons in higher
shells don’t “see” the full nuclear charge, but
rather they “see” an effective nuclear charge,
since the inner electrons shield part of the
nuclear charge. A lower effective nuclear charge
for a given orbital will increase the orbital energy
(less electrostatic attraction) and increase the
average distance an electron in that orbital will
be found from the nucleus.
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149
Radial probability distributions
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150
Radial probability distributions
To get the radial probability distribution, we
multiply the square of the orbital wavefunction by
4πr2
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s orbital electrons are better at
shielding the nucleus than electrons
in any other orbital type because s
electrons are most often found at the
nucleus, while the planar nodes of
other orbitals guarantee the
electrons will never be found at the
nucleus. To understand this
difference of electron penetration,
it is best to look at radial probability
distributions.
151
What we then see are the distances from the
nucleus where the electron is most probably
going to be found. Larger “humps” closer to the
nucleus indicate a greater electron penetration.
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152
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Non-degenerate orbitals
Multielectron atom energy diagrams
Now we can explain why the orbitals of a
given shell in a multielectron atom are nondegenerate. Consider the n = 2 shell. Any
electrons in the 2s orbital will penetrate
close to the nucleus and shield the 2p
electrons, increasing its orbital energy
compared to the 2s orbital. However, the
p orbital electrons do not penetrate, and do
not shield the 2s electrons.
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153
Electron configurations
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154
Electron configurations
Imagine you can build an atom. You can
start with a nucleus with a given atomic
number Z and add in electrons one at a time
until you have added the number of electrons
you want. This is the aufbau (building up)
process.
Z electrons will give you a neutral atom.
More than Z electrons will give you a negative
ion, and less than Z electrons will give you a
positive ion.
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Generally, more electrons means a more
complicated pattern of shielding, meaning the
orbital energies within each shell tend to “spread
out” more with increasing numbers of electrons.
155
As you place each electron into the atom
or ion, it seems reasonable that it will go
into the lowest energy orbital available,
placing it as close to the nucleus as the
other electrons present will allow.
However, every time you add in a new
electron, you are making the shielding on
the higher energy orbitals more
complicated. The orbitals of a given shell
become non-degenerate.
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156
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Electron configurations
The “rules” of electron configurations
In reality we cannot build atoms this way.
However, we can use this idea as a good way to
understand which orbitals we will find the
electrons of atoms or ions in. All we need to do
now is figure out the rules to describe the
complicated pattern of shielding and the effect it
has on orbital energies, so we know which
available orbital is the lowest energy orbital. We
call the pattern of electron filling the electron
configuration.
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1. Electrons fill the orbital of lowest energy that is
not already filled. This guarantees the lowest
possible energy for the atom.
2. The Pauli exclusion principle – no two electrons
in an atom can have exactly the same set of
quantum numbers. This implies an orbital can
hold AT MOST two electrons.
3. Hund’s rule – When degenerate orbitals of the
same type are available in a shell, a single
electron must be placed in each before we start
to pair the electrons.
157
Order of orbital filling
158
Three different notations
The complicated effect of
shielding of orbitals that
affects the energy of the
higher orbitals AND
THEREFORE THE
ORDER THE ORBITALS
ARE FILLED is seen here.
In general, the orbitals are
usually filled in the order
spdf notation (condensed)
1s22s22p2
Only shows the total electrons of a subshell, but not how they are spread
over degenerate orbitals.
spdf notation (expanded)
1s22s22px12py1
Not only shows the total electrons of a subshell, but also shows how they
are spread over degenerate orbitals (Hund’s rule).
orbital diagram
1s, 2s, 2p, 3s, 2p, 4s, 3d, 4p, 5s, P
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Much like expanded notation, where we see the filling of degenerate
orbitals, we now also show electron spin.
159
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160
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Orbital diagram
Three different notations
orbital diagram
spdf notation (condensed)
spdf notation (expanded)
1s22s22p2
1s22s22px12py1
orbital diagram
We see here that the spins in the
degenerate orbital filling are
parallel. It turns out that parallel
spins give the lowest energy atom.
Any non-parallel spins in degenerate
orbitals will give a higher energy!
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161
Electron configuration shortcuts
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162
Electron configuration shortcuts
Here we see why we said [Ar] is 1s22s22p63s23p6
that the orbital filling
usually follows the order
found on slide 159. There
are some exceptions,
such as Cr and Cu, which
show the importance of
parallel spins being lower
in energy. This tends to
happen in d orbitals
totalling 5 or 10 electrons.
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Each representation above represents the lowest
possible energy of a carbon atom (with 6 electrons) by
orbital filling following the “rules”. The lowest energy
representations are the ground-state electron
configuration. Orbital diagrams that “break the rules” of
filling represent excited-state electron configurations of
an atom that is higher in energy than it is in its ground
state.
163
Since the order of orbital [Ar] is 1s22s22p63s23p6
filling is the same for all
atoms, we can express
the orbital diagrams for
atoms with a great
number of electrons by
only looking at the
orbital filling that occurs
for electrons after a
certain point – the noble
gas atom orbital diagram
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164
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Problems
Problem answers
a) Configurations (a) and (c) are
equivalent
b) This is an excited state. The ground
state configuration would look like
The spins are parallel in the highest
energy degenerate orbitals
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165
The periodic table
166
Electron configurations by group
Niels Bohr emphasized that elements in
the same group (column) of the periodic
table have similar electron
configurations. This implies a similarity
of the distribution of the outermost valence
electrons, which leads to a similarity in
physical properties and chemistry of those
elements! We’ll see more laterP
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167
Because the order of filling rules
are known, we can emphasize on
the periodic table which block of
orbitals was the last to be filled by
the electrons for the element.
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168
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Problems
Problem answers
a) Identify the element having the electron
configuration 1s22s22p63s23p63d24s2
b) Represent the electron configuration of
iron with an orbital diagram
c) For an atom of Sn, indicate the number
of: electronic shells that are filled or
partially filled, the number of 3p
electrons, the number of 5d electrons,
and the number of unpaired electrons.
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169
a) Ti (see slide 163 for confirmation)
b) Done in class
c) Sn has 5 filled or partially filled shells,
has six 3p electrons, has zero 5d
electrons and has two unpaired
electrons.
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