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Chemistry I Notes Ch. 5
Electrons in Atoms
Big Idea #2
Electrons and the Structure of Atoms
CHEMISTRY
•What gives gas-filled lights their colors?
An electric current
passing through the gas
in each glass tube
makes the gas glow
with its own
characteristic color.
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& YOU
Atomic Emission Spectra
• Light has properties of both waves and particles.
• Light is a form of electromagnetic radiation –
electric and magnetic fields oscillating at right
angles to each other.
• Wave Characteristics
–
–
–
–
–
Amplitude – height of wave from origin to peak (m)
Wavelength -  -Distance between successive crests (m)
Frequency –  - cycles per second (Hz)
Speed – c- distance per unit time Light= 3x108 m/s
Speed = frequency x wavelength c=  
Light and Atomic Emission Spectra
The Nature of Light
• By the year 1900, there was
enough experimental evidence to
convince scientists that light
consisted of waves.
• The amplitude of a wave is the
wave’s height from zero to the
crest.
• The wavelength, represented by
 (the Greek letter lambda), is the
distance between the crests.
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Light and Atomic Emission Spectra
The Nature of Light
• The frequency, represented by  (the
Greek letter nu), is the number of wave
cycles to pass a given point per unit of
time.
• The SI unit of cycles per second is called
the hertz (Hz).
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Light and Atomic Emission Spectra
The Nature of Light
The product of frequency and wavelength
equals a constant (c), the speed of light.
c = 
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Light and Atomic Emission Spectra
The frequency () and wavelength () of
light are inversely proportional to each
other. As the wavelength increases, the
frequency decreases.
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The Nature
of Waves
Electromagnetic Waves
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Light and Atomic Emission Spectra
The electromagnetic spectrum consists of
radiation over a broad range of wavelengths.
Low energy
( = 700 nm)
Frequency  (s-1)
3 x 106
102
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Wavelength  (m)
High energy
( = 380 nm)
3 x 1012
3 x 1022
10-8
10-14
Atomic Spectra cont…
LIGHT: What Is It?
• Light Energy
– Atoms
• As atoms absorb energy, electrons jump out to a
higher energy level.
• Electrons release light when falling down to the lower
energy level.
– Photons - bundles/packets of energy released
when the electrons fall.
• Light: Stream of Photons
Light and Atomic Emission Spectra
Atomic Emission Spectra
When atoms absorb energy, their
electrons move to higher energy
levels. These electrons lose energy by
emitting light when they return to
lower energy levels.
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Atomic Spectra cont…
• Quantum – Restriction on the amount of energy
an object emits or absorbs (Max Planck)
• E= h where
– E is the energy of a photon in joules (j)
– h is Planck’s constant 6.62x10-34 j-s
–  is the frequency in hertz (Hz) or 1/s
• Energies absorbed or emitted by atoms are
quantized
• Photoelectric Effect- electrons ejected from the
surface of a metal when light shines on it.
– For every metal there is a minimum frequency of light
needed to release electrons. This quantum of energy is
called a photon. (Albert Einstein)
• Thus light has particle wave duality
Light and Atomic Emission Spectra
Atomic Emission Spectra
• The energy absorbed by an electron for it to move
from its current energy level to a higher energy level
is identical to the energy of the light emitted by the
electron as it drops back to its original energy level.
• The wavelengths of the spectral lines are
characteristic of the element, and they make up the
atomic emission spectrum of the element.
• No two elements have the same emission spectrum.
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Sample Problem 5.2
Calculating the Wavelength of Light
Calculate the wavelength of the
yellow light emitted by a
sodium lamp if the frequency of
the radiation is 5.09 × 1014 Hz
(5.09 × 1014/s).
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Sample Problem 5.2
1 Analyze List the knowns and the unknown.
Use the equation c =  to solve for the
unknown wavelength.
KNOWNS
frequency () = 5.09 × 1014 /s
c = 2.998 × 108 m/s
UNKNOWN
wavelength () = ? m
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Sample Problem 5.2
2 Calculate Solve for the unknown.
Write the expression that relates the
frequency and wavelength of light.
c = 
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Sample Problem 5.2
2 Calculate Solve for the unknown.
Rearrange the equation to solve for .
c = 
c
= 
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Solve for  by
dividing both sides

by :c
 = 
Sample Problem 5.2
2 Calculate Solve for the unknown.
Substitute the known values for  and c into
the equation and solve.
c
2.998  108 m/s
–7 m
=
=
=
5.89

10

5.09  1014 /s
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Problem
• Calculate the wavelength of ultra violet light with a
frequency of 2.00X1016 Hz?
c    3.00 10 m / s
8
c 3.00 10 m / s
110 nm
8
 
 1.5 10 m 
 15nm
16 1
 2.00 10 s
1m
8
9
Atomic Spectra cont…
• Types of spectra
– Line Spectra – contain only specific wavelengths
or colors – atoms emit these when their electrons
are excited.
– Continuous Spectrum – white light broken into all
its colors R-O-Y-G-B-I-V (rainbow)
– Absorption spectra – Black lines on the continuous
spectra of star light that represent fingerprints of
the elements present.
Light and Atomic Emission Spectra
Atomic Emission Spectra
A prism separates light into the colors it
contains. White light produces a rainbow
of colors.
Screen
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Light
bulb
Slit
Prism
Light and Atomic Emission Spectra
Atomic Emission Spectra
Light from a helium lamp produces
discrete lines.
Screen
Helium
lamp
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Slit
Prism
Refraction of White Light
The Line Spectrum of Hydrogen
The Quantum Concept and Photons
The Quantization of Energy
•German physicist Max Planck (1858–1947)
showed mathematically that the amount of
radiant energy (E) of a single quantum
absorbed or emitted by a body is proportional
to the frequency of radiation ().
E
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 or E = h
The Quantum Concept and Photons
The Quantization of Energy
•The constant (h), which has a value of
6.626  10–34 J·s (J is the joule, the SI unit of
energy), is called Planck’s constant.
E
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 or E = h
The Quantum Concept and Photons
The Photoelectric Effect
•Albert Einstein used Planck’s quantum theory
to explain the photoelectric effect.
In the photoelectric effect, electrons
are ejected when light shines on a
metal.
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The Quantum Concept and Photons
The Photoelectric Effect
•Not just any frequency of light will cause the
photoelectric effect.
• Red light will not cause potassium to eject
electrons, no matter how intense the light.
• Yet a very weak yellow light shining on
potassium begins the effect.
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The Quantum Concept and Photons
The Photoelectric Effect
•To explain the photoelectric effect,
Einstein proposed that light could be
described as quanta of energy that behave
as if they were particles.
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The Quantum Concept and Photons
The Photoelectric Effect
•These light quanta are called photons.
• Einstein’s theory that light behaves as a stream
of particles explains the photoelectric effect
and many other observations.
• Light behaves as waves in other situations; we
must consider that light possesses both
wavelike and particle-like properties.
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The Quantum Concept and Photons
The Photoelectric Effect
No electrons are ejected
because the frequency
of the light is below the
threshold frequency.
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If the light is at or above
the threshold frequency,
electrons are ejected.
If the frequency is
increased, the ejected
electrons will travel
faster.
Sample Problem 5.3
Calculating the Energy of a Photon
What is the energy of
a photon of
microwave radiation
with a frequency of
3.20 × 1011/s?
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Sample Problem 5.3
1 Analyze List the knowns and the unknown.
Use the equation E = h ×  to calculate
the energy of the photon.
KNOWNS
frequency () = 3.20 × 1011/s
h = 6.626 × 10–34 J·s
UNKNOWN
energy (E) = ? J
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Sample Problem 5.3
2 Calculate Solve for the unknown.
Write the expression that relates the
energy of a photon of radiation and the
frequency of the radiation.
E = h
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Sample Problem 5.3
2 Calculate Solve for the unknown.
Substitute the known values for  and h
into the equation and solve.
E = h = (6.626  10–34 J·s)  (3.20  1011/s)
= 2.12  10–22 J
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Sample Problem 5.3
3 Evaluate Does the result make sense?
Individual photons have very small
energies, so the answer seems
reasonable.
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What is the frequency of a photon
whose energy is 1.166  10–17 J?
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What is the frequency of a photon
whose energy is 1.166  10–17 J?
E = h
E
=
h
E 1.166  10–17 J
 = h = 6.626 X 10–34 J·s = 1.760  1016 Hz
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Problem
• What is the energy of a photon of Infrared light
with a wavelength of 1.00x10-5 m ?
First calculate frequency using c=λv then
calculate energy using E=hv
8
5
c    3.00  10 m / s  1.00  10 m  
3.00  108 m / s
13 1


3.00

10
s ( Hz )
5
1.00  10 m
E  h  6.62  10 34 J  S  3.00  1013 Hz
E  1.99  10 20 J
An Explanation of Atomic Spectra
•The light emitted by an electron
moving from a higher to a lower
energy level has a frequency directly
proportional to the energy change of
the electron.
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Quantum Mechanics
The Wavelike Nature of Matter
•Today, the wavelike properties of beams of
electrons are useful in viewing objects that
cannot be viewed with an optical microscope.
• The electrons in an electron
microscope have much smaller
wavelengths than visible light.
• These smaller wavelengths allow a
much clearer enlarged image of a
very small object, such as this pollen
grain, than is possible with an ordinary
microscope.
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Quantum Mechanics
•Classical mechanics adequately
describes the motions of bodies
much larger than atoms, while
quantum mechanics describes the
motions of subatomic particles and
atoms as waves.
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Quantum Mechanics
The Heisenberg Uncertainty Principle
The Heisenberg uncertainty principle
states that it is impossible to know both the
velocity and the position of a particle at the
same time.
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Quantum Mechanics
The Heisenberg Uncertainty Principle
The Heisenberg uncertainty principle
states that it is impossible to know both the
velocity and the position of a particle at the
same time.
• This limitation is critical when dealing with
small particles such as electrons.
• But it does not matter for ordinary-sized
objects such as cars or airplanes.
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Quantum Mechanics
• To locate an electron, you might strike it with a photon.
• The electron has such a small mass that striking it with a
photon affects its motion in a way that cannot be
predicted accurately.
• The very act of measuring the position of the electron
changes its velocity, making its velocity uncertain.
Before
collision: A
photon strikes
an electron
during an
attempt to
observe the
electron’s
position.
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After
collision: The
impact
changes the
electron’s
velocity,
making it
uncertain.
The Heisenberg uncertainty principle
states that it is impossible to
simultaneously know which two
attributes of a particle?
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The Heisenberg uncertainty principle
states that it is impossible to
simultaneously know which two
attributes of a particle?
velocity and position
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Revising the Atomic Model
• The Bohr Model of the Atom
– Electrons occupy specific energy levels in atoms. These
energy levels are identified by a principle quantum
number – n (1-7)
– When electrons occupy the lowest energy levels
available the atom is in its ground state.
– When electrons occupy higher energy levels than the
ground state the atom is in an excited state.
– When atoms absorb energy the electrons jump to
higher energy levels (excited state). When they drop
back down to the ground state these atoms emits light
of specific wavelength corresponding to the energy
change of the electron (emission spectra)
Energy Levels in Atoms
The Bohr Model
Each possible electron orbit in Bohr’s
model has a fixed energy.
• The fixed energies an electron can
have are called energy levels.
• A quantum of energy is the amount of
energy required to move an electron
from one energy level to another
energy level.
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Energy Levels in Atoms
The Bohr Model
The rungs on this ladder are somewhat
like the energy levels in Bohr’s model of
the atom.
• A person on a ladder
cannot stand between
the rungs. Similarly, the
electrons in an atom
cannot exist between
energy levels.
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Energy Levels in Atoms
The Bohr Model
The rungs on this ladder are somewhat
like the energy levels in Bohr’s model of
the atom.
• The energy levels in
atoms are unequally
spaced, like the rungs in
this unusual ladder. The
higher energy levels are
closer together.
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Revising the Atomic Model
• Louis De Broglie (1924) proposed that particles of
matter have wavelike character
• Heisenberg’s Uncertainty Principle – The position
and speed of an electron cannot be measured
simultaneously. If you measure one you can’t
know the other.
• So electrons are located based on the probability
of finding them in a region of space around an
atom (orbital).
The Quantum Mechanical Model
• Austrian physicist Erwin Schrödinger (1887–
1961) used new theoretical calculations and
experimental results to devise and solve a
mathematical equation describing the
behavior of the electron in a hydrogen atom.
• The modern description of the electrons in
atoms, the quantum mechanical model,
came from the mathematical solutions to the
Schrödinger equation.
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The Quantum Mechanical Model
• Like the Bohr model, the quantum
mechanical model of the atom restricts
the energy of electrons to certain
values.
• Unlike the Bohr model, however, the
quantum mechanical model does not
specify an exact path the electron takes
around the nucleus.
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The Quantum Mechanical Model
The quantum mechanical model
determines the allowed energies
an electron can have and how
likely it is to find the electron in
various locations around the
nucleus of an atom.
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The Quantum Mechanical Model
• In the quantum mechanical
model, the probability of
finding an electron within a
certain volume of space
surrounding the nucleus
can be represented as a
fuzzy cloudlike region.
• The cloud is more dense
where the probability of
finding the electron is high.
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Electron cloud
Electron Arrangement in Atoms
• Quantum Mechanical Model – Treats the electron
as a wave with quantized energy and explains the
behavior of atoms.
• Electrons occupy a cloud where the probability of
finding an electron is highest where the cloud is
densest.
• Orbitals – a region around the nucleus where an
electron with a given amount of energy is likely
to be found. Maximum capacity is 2 electrons.
Atomic Orbitals
• The energy levels of electrons in the
quantum mechanical model are labeled by
principal quantum numbers (n).
• These numbers are assigned the values n =
1, 2, 3, 4, and so forth.
• For each principal energy level greater than
1, there are several orbitals with different
shapes and at different energy levels.
• These energy levels within a principal
energy level constitute energy sublevels.
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Orbitals and Energy
• Electrons are found in primary energy levels
sublevels and orbitals
– Primary energy level – quantum number n where n = 17 (floor of hotel)
– Sublevels – Each primary energy level has the number of
sublevels equal to its n quantum number they are
labeled s, p, d, & f (wing of hotel)
•
•
•
•
s sublevels contain only 1 s orbital that can hold 2 electrons
p sublevels contain 3 p orbitals that can hold 6 electrons
d sublevels contain 5 d orbitals that can hold 10 electrons
f sublevels contain 7 f orbitals that can hold 14 electrons
Atomic Orbitals
For a given principal energy level
greater than 1, there is one s orbital, 3
p orbitals, and 5 d orbitals.
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Orbitals and Energy cont…
• Orbitals contain only 2 electrons
– s orbital – spherical shape 1 per energy level
– p orbital – dumbell shaped 3 per energy level
– d orbital – double dumbell shaped 5 per energy
level
– f orbital – very complex shapes 7 per energy
level
Electrons within an orbital have opposite
spin – Pauli Exclusion Principle
1s Orbital
The Boundary Surface Representations
of All Three 2p Orbitals
The Boundary Surfaces of All of the 3d
Orbitals
Representation of the 4f Orbitals in
Terms of Their Boundary Surfaces
Orbital Energies
Electron Capacities of Shells and Orbitals
Primary
energy
Level
1
2
3
Orbital
Type
s sp spd
4
5
spdf
spdf
6
7
spd sp
Electron
Capacity
2 2 6 2 6 10 2 6 10 14 2 6 10 14 2 6 10 2 6
Total
Electrons
2
8
18
32
32
18
8
Quantum Numbers and Electrons
• If the primary energy level is given as n=1-7
• Then the number of sublevels in that primary
energy level = n
• The number of orbitals in that energy level = n2
• The number of electrons in that energy level= 2n2
Calculate the maximum number of
electrons in the 5th principal energy
level (n = 5).
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Calculate the maximum number of
electrons in the 5th principal energy
level (n = 5).
The maximum number of electrons that can
occupy a principal energy level is given by
the formula 2n2. If n = 5, 2n2 = 50.
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5.2 Electron Arrangement in Atoms cont…
• Electron configurations are determined by
distributing the atom’s electrons into energy
levels, sublevels , and orbitals based on a set of
principles.
– Aufbau Principle –Electrons are added one at a time to
the lowest energy orbitals available until all electrons
are accounted for.
– Pauli Exclusion Principle – Orbitals hold a maximum of
2 electrons with opposite spin.
– Hund’s Rule – Electrons occupy equal-energy orbitals
so that a maximum number of unpaired electrons
results (college dorm rule). Fill electrons into sublevels
1 per orbital until all orbitals have 1 then double up.
– Remember the exceptions at Copper and Chromium
Electron Configurations
Aufbau Principle
6p
5d
6s
4f
5p
4d
Increasing energy
5s
4p
4s
3d
3p
3s
2p
2s
1s
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According to the aufbau principle,
electrons occupy the orbitals of lowest
energy first. In the aufbau diagram,
each box represents an atomic orbital.
Electron Configurations
Aufbau Principle
6p
5d
6s
4f
5p
4d
Increasing energy
5s
4p
4s
3d
3p
3s
2p
2s
1s
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The aufbau diagram shows the
relative energy levels of the various
atomic orbitals. Orbitals of greater
energy are higher on the diagram.
Electron Configurations
Aufbau Principle
6p
5d
6s
4f
5p
4d
Increasing energy
5s
4p
4s
3d
3p
3s
2p
2s
1s
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The range of energy levels within a
principal energy level can overlap
the energy levels of another
principal level.
Electron Configurations
Pauli Exclusion Principle
• Spin is a quantum mechanical property of
electrons and may be thought of as
clockwise or counterclockwise.
• A vertical arrow indicates an electron and its
direction of spin ( or ).
• An orbital containing paired electrons is
written as .
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Electron Configurations
Hund’s Rule
According to Hund’s rule, electrons
occupy orbitals of the same energy in a
way that makes the number of electrons
with the same spin direction as large as
possible.
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Electron Configurations
Hund’s Rule
Three electrons would occupy three
orbitals of equal energy as follows.
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Electron Configurations
Hund’s Rule
Three electrons would occupy three
orbitals of equal energy as follows.
Electrons then occupy each orbital
so that their spins are paired with the
first electron in the orbital.
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Electron Configurations
Look at the orbital filling diagram of the oxygen atom.
• An oxygen
atom contains
eight
electrons.
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Electron Configurations of Selected Elements
Element
1s
2s
2px 2py 2pz
3s
Electron
configuration
H
1s1
He
1s2
Li
1s22s1
C
1s22s22p2
N
1s22s22p3
O
1s22s22p4
F
1s22s22p5
Ne
1s22s22p6
Na
1s22s22p63s1
Electron Configurations
Look at the orbital filling diagram of the oxygen atom.
Electron Configurations of Selected Elements
• The 1s orbital
has two
electrons of
opposite spin.
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Element
1s
2s
2px 2py 2pz
3s
Electron
configuration
H
1s1
He
1s2
Li
1s22s1
C
1s22s22p2
N
1s22s22p3
O
1s22s22p4
F
1s22s22p5
Ne
1s22s22p6
Na
1s22s22p63s1
Electron Configurations
Look at the orbital filling diagram of the oxygen atom.
Electron Configurations of Selected Elements
• The 1s orbital
has two
electrons of
opposite spin.
• The 2s orbital
also has two
electrons of
opposite spin.
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Element
1s
2s
2px 2py 2pz
3s
Electron
configuration
H
1s1
He
1s2
Li
1s22s1
C
1s22s22p2
N
1s22s22p3
O
1s22s22p4
F
1s22s22p5
Ne
1s22s22p6
Na
1s22s22p63s1
Electron Configurations
Look at the orbital filling diagram of the oxygen atom.
• Each of the
three 2p orbitals
has one
electron. The
remaining
electron now
pairs with an
electron
occupying one
of the 2p
orbitals.
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Electron Configurations of Selected Elements
Element
1s
2s
2px 2py 2pz
3s
Electron
configuration
H
1s1
He
1s2
Li
1s22s1
C
1s22s22p2
N
1s22s22p3
O
1s22s22p4
F
1s22s22p5
Ne
1s22s22p6
Na
1s22s22p63s1
Electron Configurations
• A convenient shorthand method for showing
the electron configuration of an atom involves
writing the energy level and the symbol for
every sublevel occupied by an electron.
• You indicate the number of electrons
occupying that sublevel with a superscript.
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Electron Configurations
• For hydrogen, with one electron in a 1s
orbital, the electron configuration is
written 1s1.
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Electron Configurations
• For hydrogen, with one electron in a 1s
orbital, the electron configuration is
written 1s1.
• For oxygen, with two electrons in a 1s
orbital, two electrons in a 2s orbital, and
four electrons in 2p orbitals, the electron
configuration is 1s22s22p4.
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Electron Configurations
• For hydrogen, with one electron in a 1s
orbital, the electron configuration is
written 1s1.
• For oxygen, with two electrons in a 1s
orbital, two electrons in a 2s orbital, and
four electrons in 2p orbitals, the electron
configuration is 1s22s22p4.
Note that the sum of the superscripts
equals the number of electrons in the
atom.
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CHEMISTRY
& YOU
Explain why the correct electron
configuration of oxygen is
1s22s22p4 and not 1s22s22p33s1.
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Sample Problem 5.1
Writing Electron Configurations
The atomic number of phosphorus is 15.
Write the electron configuration of a
phosphorus atom.
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Sample Problem 5.1
1 Analyze Identify the relevant concepts.
Phosphorus has 15 electrons. There is
a maximum of two electrons per
orbital. Electrons do not pair up within
an energy sublevel (orbitals of equal
energy) until each orbital already has
one electron.
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When writing electron
configurations, the sublevels
within the same principal
energy level are written
together.
Sample Problem 5.1
2 Solve Apply the concepts to
this problem.
• Use the aufbau diagram to place
electrons in the orbital with the lowest
energy (1s) first.
1s
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Sample Problem 5.1
2 Solve Apply the concepts to
this problem.
• Use the aufbau diagram to place
electrons in the orbital with the lowest
energy (1s) first.
• Continue placing electrons in each orbital
with the next higher energy level.
1s
2s
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2p
3s
3p
4s
Sample Problem 5.1
2 Solve Apply the concepts to
this problem.
Write the electron configuration.
• The electron configuration of phosphorus
is 1s22s22p63s23p3.
• The superscripts add up to the number of
electrons.
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Electron Configurations
Exceptional Electron Configurations
• You can obtain correct electron configurations
for the elements up to vanadium (atomic
number 23) by following the aufbau diagram for
orbital filling.
• If you were to continue in that fashion, however,
you would assign chromium and copper the
following incorrect configurations.
Cr 1s22s22p63s23p63d44s2
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Cu 1s22s22p63s23p63d94s2
Electron Configurations
Exceptional Electron Configurations
• The correct electron configurations are as
follows:
Cr 1s22s22p63s23p63d54s1
Cu 1s22s22p63s23p63d104s1
• These arrangements give chromium a
half-filled d sublevel and copper a filled d
sublevel.
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Electron Configurations
Exceptional Electron Configurations
Some actual electron configurations
differ from those assigned using the
aufbau principle because although
half-filled sublevels are not as stable
as filled sublevels, they are more
stable than other configurations.
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What is the correct electron
configuration of a sulfur atom?
A. 1s22s22p43s23p6
B. 1s22s22p63s23p3
C. 1s22s22p63s23p4
D. 1s22s22p63s63p2
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What is the correct electron
configuration of a sulfur atom?
A. 1s22s22p43s23p6
B. 1s22s22p63s23p3
C. 1s22s22p63s23p4
D. 1s22s22p63s63p2
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4S
3P
3S
2P
2S
1S