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1/7/2013
Chemistry: Atoms First
Julia Burdge & Jason Overby
Chapter 3
Quantum Theory and
the Electronic
Structure of Atoms
Kent L. McCorkle
Cosumnes River College
Sacramento, CA
Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
3
Quantum Theory and the Electronic Structure of
Atoms
3.7 Quantum Numbers
Principal Quantum Number (n)
Angular Momentum Quantum Number (l)
Magnetic Quantum Number (ml)
Electron Spin Quantum Number (ms)
3.8 Atomic Orbitals
s Orbitals
p Orbitals
d Orbitals and other High-Energy Orbitals
Energies of Orbitals
3.9 Electron Configuration
Energies of Atomic Orbitals in Many-Electron Systems
The Pauli Exclusion Principle
Aufbau Principle
Hund’s Rule
General Rules for Writing Electron Configurations
3.10 Electron Configurations and the Periodic Table
Forms of Energy
Potential energy is the energy possessed by an object by virtue of its
position.
There are two forms of potential energy of great interest to chemists:
 Chemical energy is energy stored within the structural units of
chemical substances.
 Electrostatic energy is potential energy that results from the
interaction of charged particles.
3
Quantum Theory and the Electronic Structure of
Atoms
3.1 Energy and Energy Changes
Forms of Energy
Units of Energy
3.2 The Nature of Light
Properties of Waves
The Electromagnetic Spectrum
The Double-Slit Experiment
3.3 Quantum Theory
Quantization of Energy
Photons and the Photoelectric Effect
3.4 Bohr’s Theory of the Hydrogen Atom
Atomic Line Spectra
The Line Spectrum of Hydrogen
3.5 Wave Properties of Matter
The de Broglie Hypothesis
Diffraction of Electrons
3.6 Quantum Mechanics
The Uncertainty Principle
The Schrödinger Equation
The Quantum Mechanical Description of the Hydrogen Atom
3.1
Energy and Energy Changes
Energy is the capacity to do work or transfer heat.
All forms of energy are either kinetic or potential.
Kinetic energy (Ek) is the energy of motion.
 m is the mass of the object
 u is its velocity
One form of kinetic energy of particular interest to chemists is
thermal energy, which is the energy associated with the random
motion of atoms and molecules.
Energy and Energy Changes
Kinetic and potential energy are interconvertible – one can be
converted to the other.
Although energy can assume many forms, the total energy of the
universe is constant.
 Energy can neither be created nor destroyed.
 When energy of one form disappears, the same amount of
energy reappears in another form or forms.
 This is known as the law of conservation of energy.
Q1 and Q2 represent two charges separated by the distance, d.
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Units of Energy
Worked Example 3.1
The SI unit of energy is the joule (J), named for the English
physicist James Joule.
Calculate the kinetic energy of a helium atom moving at a speed of 125 m/s.
It is the amount of energy possessed by a 2-kg mass moving at a
speed of 1 m/s.
Strategy Use Ek = ½mu2 to calculate the kinetic energy of an atom. Note that
for units to cancel properly, giving Ek in joules, the mass of the helium atom
(4.003 amu) must be in kilograms. The factor for conversion of amu to g is
1.661×10-24 g/1 amu. Therefore, the mass of a helium atom in kilograms is:
Ek = ½ mu2 = ½(2 kg)(1 m/s)2 = 1 kg∙m2/s2 = 1 J
The joule can also be defined as the amount of energy exerted
when a force of 1 newton (N) is applied over a distance of 1 meter.
1J=1N·m
Because the magnitude of a joule is so small, we often express
large amounts of energy using the unit kilojoule (kJ).
4.003 amu ×
1 kg
1.661×10-24 g
×
= 6.649×10-27 kg
1 amu
1×103 g
Solution Ek = ½mu2
= ½(6.49×10-27 kg)(125 m/s)
= 5.19×10-23 kg∙m2/s2 = 5.19×10-23 J
Think About It We expect the energy of a single atom, even a fast-moving
one, to be extremely small.
1 kJ = 1000 J
3.2 The Nature of Light
Worked Example 3.2
How much greater is the attraction between charges of +2 and -2 than attraction
between charges of +1 and -1 if the opposite charges are separated by the same
distance.
QQ
Strategy Use Eel α 1d 2 to compare the magnitudes of the two Eel values.
Because the distance between charges in the same in both cases, we can solve for
the ratio of Eel values without actually knowing the distance. Both the distance
and the proportionality constant cancel in the solution.
Think About It Doubling both charges causes a four-fold
in the magnitude of the electrostatic energy
Setup increase
Eel(+2,-2) = c Q1Q2 , Q1 = +2 and Q2 = -2
d particles.
between
charged
Eel(+1,-1) = c Q1Q2 , Q1 = +1 and Q2 = -1
Visible light is only a small component of the continuum of radiant energy
known as the electromagnetic spectrum.
d
Solution
2×(-2)
d
1×(-1)
c× d
c×
=4
The attraction between charges of +2 and -2 is four times as large as the
attraction between charges +1 and -1.
Properties of Waves
All forms of electromagnetic
radiation travel in waves.
Waves are characterized by:
Wavelength (λ; lambda) – the
distance between identical
points on successive waves
The Nature of Light
The speed of light (c) through a vacuum is a constant:
c = 2.99792458×108 m/s
Normally rounded to, c = 3.00×108 m/s.
Speed of light, frequency and wavelength are related:
Frequency (ν; nu) – the number
of waves that pass through a
particular point in 1 second.
Amplitude – the vertical distance
from the midline of a wave to the
top of the peak or the bottom of
the trough.
 λ is expressed in meters
 v is expressed in reciprocal seconds (s−1)
 s-1 is also known as hertz (Hz)
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The Electromagnetic Spectrum
An electromagnetic wave has both an electric field component and
a magnetic component.
The electric and magnetic components have the same frequency and
wavelength.
The Double-Slit Experiment
When light passes through two closely spaced slits, an interference
pattern is produced.
Constructive
interference is a result
of adding waves that
are in phase.
Destructive
interference is a result
of adding waves that
are out of phase.
This type of interference is typical of waves and demonstrates the
wave nature of light.
3.3
Worked Example 3.3
A laser commonly used in the treatment of vascular skin lesions has a wavelength
of 532 nm. What is the frequency of this radiation?
Strategy We must convert the wavelength to meters and solve for frequency
using c = λν.
c
Setup Rearranging the equation to solve for frequency gives ν = λ . The speed
of light, c, is 3.00×108 m/s.
Solution
1×10-9 m
λ = 532 nm×
1 nm
ν=
= 5.32×10-7 m
3.00×108 m/s
= 5.64×1014 s-1
5.32×10-7 m
Think About It Make sure your units cancel properly. A common error in
this type of problem is neglecting to convert wavelength to nanometers.
Quantum Theory
The energy E of a single quantum of energy is
 h is called Planck’s constant: 6.63×10−34 J∙s
The idea that energy is quantized rather than continuous is like
walking up a staircase or playing the piano
 You cannot step or play anywhere (continuous), you can only
step on a stair or play on a key (quantized).
Quantum Theory
When a solid is heated, it emits electromagnetic radiation, known as
blackbody radiation, over a wide range of wavelengths.
The amount of energy given off at a certain temperature depends on
the wavelength.
Classical physics failed to completely explain the phenomenon.
 Assumed that radiant energy was continuous; that is, could be
emitted or absorbed in any amount.
Max Planck suggested that radiant energy is only emitted or
absorbed in discrete quantities, like small packages or bundles.
A quantum of energy is the smallest quantity of energy that can be
emitted (or absorbed).
Photons and the Photoelectric Effect
Albert Einstein used Planck’s theory
to explain the photoelectric effect.
Electrons are ejected from the surface
of a metal exposed to light of a certain
minimum frequency, called the
threshold frequency.
The number of electrons ejected is
proportional to the intensity.
Below the threshold frequency no
electrons were ejected, no matter how
bright (or intense) the light.
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Photons and the Photoelectric Effect
Photons and the Photoelectric Effect
Einstein proposed that the beam of
light is really a stream of particles.
Shining light onto a metal surface can
be thought of as shooting a beam of
particles – photons – at the metal
atoms.
These particles of light are now called
photons.
Each photon (of the incident light)
must posses the energy given by the
equation:
If the ν of the photons equals the
energy the binds the electrons in the
metal, then the light will have enough
energy to knock the electrons loose.
If we use light of a higher ν, then not
only will the electrons be knocked
loose, but they will also acquire some
kinetic energy.
Photons and the Photoelectric Effect
This is summarized by the equation
 KE is the kinetic energy of the
ejected electron
 W is the binding energy of the
electron
Worked Example 3.4
Calculate the energy (in joules) of (a) a photon with a wavelength of 5.00×104
nm (infrared region) and (b) a photon with a wavelength of 52 nm (ultraviolet
region). (c) Calculate the maximum kinetic energy of an electron ejected by the
photon in part (b) from a metal with a binding energy of 3.7 eV.
Strategy We must use c = λν and E = hν to determine the energy of each
photon. In part (c), we will use hν = KE + W to find the kinetic energy of an
ejected electron. The binding energy, given in eV, must be converted to J.
Solution
1×10-9 m
(a) 5.00×104 nm ×
= 5.00×10-5 m
1 nm
c
3.00×108 m/s
ν=
=
= 6.00×1012 s-1
λ
5.00×10-5 m
E = hν = (6.63×10-34 J∙s)(6.00×10-12 s-1) = 3.98×10-21 J
Worked Example 3.4 (cont.)
Calculate the energy (in joules) of (a) a photon with a wavelength of 5.00×104
nm (infrared region) and (b) a photon with a wavelength of 52 nm (ultraviolet
region). (c) Calculate the maximum kinetic energy of an electron ejected by the
photon in part (b) from a metal with a binding energy of 3.7 eV.
Think About It Remember that frequency and wavelength
Solution
-9 m
are inversely
Thus, as wavelength decreases,
1×10proportional.
(b) 52 nm ×
= 5.2×10-8 m
frequency 1and
nmenergy increase. Note that in part (c),
subtracting
the 8binding
energy made
a relatively small
c
3.00×10
m/s
15 s-1
ν=
=
=of5.8×10
-8 m
change
to the energy
the incident
photon. If the incident
λ
5.2×10
photon had been in the X-ray region of the spectrum, the
E = hν = (6.63×10-34 J∙s)(5.8×1015 s-1) = 3.8×10-18 J
difference between its energy and the kinetic energy of the
ejected electron would have been negligible.
1.602×10-19 J
= 5.9×10-19 J
(c) W = 3.7 eV ×
1 eV
KE = hv - W = 3.8×10-18 J - 5.9×10-19 J = 3.2×10-18 J
3.4
Bohr’s Theory of the Hydrogen Atom
Sunlight is composed of various color
components that can be recombined to
produce white light.
The emission spectrum of a substance can
be seen by energizing a sample of material
with some form of energy.
The “red hot” or “white hot” glow of an
iron bar removed from a fire is the visible
portion of its emission spectrum.
The emission spectrum of both sunlight and
a heated solid are continuous; all
wavelengths of visible light are present.
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Atomic Line Spectra
Bohr’s Theory of the Hydrogen Atom
Line spectra are the emission of light only at specific wavelengths.
Every element has its own unique emission spectrum.
The Line Spectrum of Hydrogen
Bohr’s Theory of the Hydrogen Atom
The Rydberg equation can be used to calculate the wavelengths of
the four visible lines in the emission spectrum of hydrogen.
Neils Bohr attributed the emission of radiation by an energized
hydrogen atom to the electron dropping from a higher-energy orbit
to a lower one.
As the electron dropped, it gave up a quantum of energy in the form
of light.
 R∞ is the Rydberg constant (1.09737317 x
107
m−1)
Bohr showed that the energies of the electron in a hydrogen atom
are given by the equation:
 λ the wavelength of a line in the spectrum
 n1 and n2 are positive integers where n2 > n1.
The Line Spectrum of Hydrogen
As an electron gets closer to the nucleus, n decreases.
 En is the energy
 n is a positive integer
The Line Spectrum of Hydrogen
Bohr’s theory explains the line spectrum of the hydrogen atom.
Radiant energy absorbed by the atom causes the electron to move
from the ground state (n = 1) to an excited state (n > 1).
En becomes larger in absolute value (but more negative) as n gets
smaller.
En is most negative when n = 1.
Called the ground state, the lowest energy state of the atom
For hydrogen, this is the most stable state
Conversely, radiant energy is emitted when the electron moves
from a higher-energy state to a lower-energy excited state or the
ground state.
The quantized movement of the electron from one
energy state to another is analogous to a ball moving
and down steps.
The stability of the electron decreases as n increases.
Each energy state in which n > 1 is called an excited state.
nf is the final state
ni is the initial state
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Bohr’s Theory of the Hydrogen Atom
Suppose an electron is initially in an excited state, ni.
During emission, the electron drops to a lower energy state, nf.
The energy difference between the initial and final states is
Bohr’s Theory of the Hydrogen Atom
Bohr’s Theory of the Hydrogen Atom
To calculate wavelength, substitute c/λ for ν and rearrange:
nf is the final state
ni is the initial state
3.5
Worked Example 3.5
Wave Properties of Matter
Calculate the wavelength (in nm) of the photon emitted when an electron
transitions from the n = 4 state to the n = 2 state in a hydrogen atom.
Louis de Broglie reasoned that if light can behave like a stream of
particles (photons), then electrons could exhibit wavelike properties.
Setup
According to de
Broglie, electrons
behave like standing
waves.
Think About It Look again at the line spectrum of
h = 6.63×10-34 J∙s and c = 3.00×108 m/s
hydrogen and make sure your result matches one of them.
Note that for an emission, ni, is always greater than nf, and
-18
Solution
1
the 1equation
gives2.18×10
a positive Jresult.
- 12
=
λ
(6.63×10-34 J∙s )(3.00×108 m/s)
22
4
= 2.055×106 m-1
λ = 4.87×10-7 m ×
1 nm
1×10-9 m
= 487 nm
Only certain
wavelengths are
allowed.
At a node, the
amplitude of the wave
is zero.
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Wave Properties of Matter
Worked Example 3.6
De Broglie deduced that the particle and wave properties are
related by the following expression:
 λ is the wavelength associated with the particle
 m is the mass (in kg)
 u is the velocity (in m/s)
The wavelength calculated from this equation is known as the de
Broglie wavelength.
Calculate the de Broglie wavelength of the “particle” in the following two cases:
(a) a 25-g bullet traveling at 612 m/s and (b) an electron (m = 9.109×10-31 kg)
moving at 63.0 m/s.
Setup Think About It While you are new at solving these
problems, always write out the units of Planck’s constant
(J∙s) as kg∙m2/s. This will enable you to check your unit
-34 J∙s, or 6.63×10-34 kg∙m2/s; Remember m must be expressed in
h = 6.63×10
cancellations
and detect common errors such as expressing
kg.
mass in grams rather than kilograms. Note that the
calculated wavelength of a macroscopic object, even one
Solution
as small
as a bullet, is extremely small. An object must be
1 kg
= 0.025 kg
(a) 25 g ×
at least
asgsmall as a subatomic particle in order for its
1000
wavelength to be large
enough for us to observe.
6.63×10-34 kg∙m2/s
h
=
λ =
= 4.3×10-35 m
(0.025 kg)(612 m/s)
mu
(b) λ =
3.6
Diffraction of Electrons
Experiments have shown that electrons do indeed possess wavelike
properties:
h
mu
=
6.63×10-34 kg∙m2/s
= 1.16×10-5 m
(9.109×10-31 kg)(63.0 m/s)
Quantum Mechanics
The Heisenberg uncertainty principle states that it is impossible to
know simultaneously both the momentum p and the position x of a
particle with certainty.
 Δx is the uncertainty in position in meters
 Δp is the uncertainty in momentum
X-ray diffraction pattern of
aluminum foil
Electron diffraction pattern of
aluminum foil.
Worked Example 3.7
An electron in a hydrogen atom is known to have a velocity of 5×106 m/s + 1
percent. Using the uncertainty principle, calculate the minimum uncertainty in the
position of the electron and, given that the diameter of the hydrogen atom is less
than 1 angstrom (Å), comment on the magnitude of this uncertainty compared to
the size of the atom.
Strategy The uncertainty in the velocity, 1 percent of 5×106 m/s, is Δu.
Calculate Δx and compare it with the diameter of they hydrogen atom.
Setup The mass of an electron is 9.11×10-31 kg. Planck’s constant, h, is
6.63×10-34 kg∙m2/s.
 Δu is the uncertainty in velocity in m/s
 m is the mass in kg
Worked Example 3.7
An electron in a hydrogen atom is known to have a velocity of 5×106 m/s + 1
percent. Using the uncertainty principle, calculate the minimum uncertainty in the
position of the electron and, given that the diameter of the hydrogen atom is less
than 1 Think
angstrom
(Å), comment
on theerror
magnitude
of this uncertainty
to
About
It A common
is expressing
the masscompared
of
the sizethe
of particle
the atom.in grams instead of kilograms, but you should
discover this inconsistency if you check your unit cancellation
Solution
carefully.
Remember
if =
one
uncertainty
is small, the other
6 m/s
4 m/s
Δu = 0.01
× 5×10that
5×10
must be large. The uncertainty principle applies in a practical
h
way only
Δx = to submicroscopic particles. In the case of a
4πobject,
∙ mΔu where the mass is much larger than that
macroscopic
of an electron, small uncertainties, relative to the size of the
-34 kg∙m2/s
6.63×10
object,
both
position and velocity.
Δxare
= possible for-31
> 1×10-9 m
4π(9.11×10 kg)(5×104 m/s)
The minimum uncertainty in the position x is 1×10-9 m = 10Å. The uncertainty
is 10 times larger than the atom!
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Quantum Mechanics
Quantum Mechanics
Erwin Schrödinger derived a complex mathematical formula to
incorporate the wave and particle characteristics of electrons.
The Schrödinger equation specifies
possible energy states an electron can
occupy in a hydrogen atom.
Wave behavior is described with the wave function ψ.
The probability of finding an
electron in a certain area of
space is proportional to ψ2 and
is called electron density.
The energy states and wave functions
are characterized by a set of quantum
numbers.
Instead of referring to orbits as in the
Bohr model, quantum numbers and
wave functions describe atomic orbitals.
3.7 Quantum Numbers
Quantum Numbers
Quantum numbers are required to describe the distribution of
electron density in an atom.
The principal quantum number (n) designates the size of the
orbital.
There are three quantum numbers necessary to describe an atomic
orbital.
Larger values of n correspond to larger orbitals.
The allowed values of n are integral numbers: 1, 2, 3 and so forth.
 The principal quantum number (n) – designates size
 The angular moment quantum number (l) – describes shape
 The magnetic quantum number (m l) – specifies orientation
Quantum Numbers
The value of n corresponds to the value of n in Bohr’s model of the
hydrogen atom.
A collection of orbitals with the same value of n is frequently called
a shell.
Quantum Numbers
The angular moment quantum number (l) describes the shape of
the orbital.
The values of l are integers that depend on the value of the
principal quantum number
The magnetic quantum number (m l) describes the orientation of
the orbital in space.
The values of ml are integers that depend on the value of the
angular moment quantum number:
– l,…0,…+l
The allowed values of l range from 0 to n – 1.
 Example: If n = 2, l can be 0 or 1.
l
0
1
2
3
Orbital designation
s
p
d
f
A collection of orbitals with the same value of n and l is referred to
as a subshell.
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Quantum Numbers
Worked Example 3.8
Quantum numbers designate shells, subshells, and orbitals.
What are the possible values for the magnetic quantum number (m l) when the
principal quantum number (n) is 3 and the angular quantum number (l) is 1?
Strategy Recall that the possible values of ml depend on the value of l, not on
the value of n.
Setup The possible values of ml are – l,…0,…+l.
Solution The possible values of ml are -1, 0, and +1.
Think About It Consult Table 3.2 to make sure your answer is correct. Table 3.2
confirms that it is the value of l, not the value of n, that determines the possible
values of ml.
Quantum Numbers
Quantum Numbers
The electron spin quantum number (m s ) is used to specify an
electron’s spin.
A beam of atoms is split by a magnetic field.
Statistically, half of the electrons spin clockwise, the other half spin
counterclockwise.
There are two possible directions of
spin.
Allowed values of ms are +½ and −½.
3.8
Quantum Numbers
To summarize quantum numbers:
All s orbitals are spherical in shape but differ in size:
principal (n) – size
1s < 2s < 3s
angular (l) – shape
Required to describe an atomic orbital
magnetic (ml) – orientation
principal quantum
number (n = 2)
principal (n = 2)
2px
Atomic Orbitals
related to the magnetic
quantum number (ml )
angular momentum (l = 1)
electron spin (ms) direction of spin
2s
angular momentum
quantum number (l = 0)
Required to describe an
electron in an atomic orbital
ml = 0; only 1 orientation
possible
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Atomic Orbitals
Atomic Orbitals
The p orbitals:
The d orbitals:
Five orientations:
Three orientations:
l = 1 (as required for a p orbital)
ml = –1, 0, +1
l = 2 (as required for a d orbital)
ml = –2, –1, 0, +1, +2
Energies of Orbitals
Worked Example 3.9
The energies of orbitals in the hydrogen atom depend only on the
principal quantum number.
List the Think
values of
n, l, and
ml for each
the orbitals
in a to
4dverify
subshell.
About
It Consult
theoffollowing
figure
your
answers.
Strategy Consider the significance of the number and the letter in the 4d
designation and determine the values of n and l. There are multiple values for ml,
which will have to be deduced from the value of l.
3s subshell
3;
l = 0)
3rd shell (n
(n ==3p
3)subshell
(n3d
= subshell
3; l = 1) (n = 3; l = 2)
Setup The integer at the beginning of the orbital designation is the principal
quantum number (n). The letter in an orbital designation gives the value of the
angular momentum quantum number (l). The magnetic quantum number (ml) can
have integral values of – l,…0,…+l.
Solution
2s
2ndsubshell
shell (n = 2)2p subshell (n = 2; l = 1)
(n = 2; l = 0)
4d
principal quantum
number, n = 4
angular momentum
quantum number, l = 2
Possible ml are -2, -1, 0, +1, +2.
3.9
Electron Configurations
Electron Configurations
The electron configuration describes how the electrons are
distributed in the various atomic orbitals.
If hydrogen’s electron is found in a higher energy orbital, the atom
is in an excited state.
In a ground state hydrogen atom, the electron is found in the 1s
orbital.
A possible excited state electron
configuration of hydrogen
Ground state electron
configuration of hydrogen
2s
1s
2p
2p
2p
1s1
2s1
number of electrons in
the orbital or subshell
angular momentum (l = 0)
The use of an up arrow indicates an electron
with ms = + ½
Energy
Energy
principal (n = 1)
2s
2p
2p
2p
1s
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Electron Configurations
Electron Configurations
The helium emission spectrum is more complex than the hydrogen
spectrum.
In a multi-electron atoms, the energies of the atomic orbitals are split.
There are more possible energy transitions in a helium atom
because helium has two electrons.
Splitting of energy levels refers to
the splitting of a shell (n=3) into
subshells of different energies
(3s, 3p, 3d)
Electron Configurations
Electron Configurations
According to the Pauli exclusion principle, no two electrons in an
atom can have the same four quantum numbers.
The Aufbau principle states that electrons are added to the lowest
energy orbitals first before moving to higher energy orbitals.
The ground state electron
configuration of helium
2p
2p
The ground state electron
configuration of Li
1s2
2s
2p
Quantum number
Principal (n)
1s
describes the 1s orbital
Angular moment (l)
Magnetic (ml)
describes the electrons in the 1s orbital
Electron spin (ms)
1
0
0
+½
1
0
0
‒½
Energy
Energy
2p
Li has a total of 3 electrons
2s
1s
2p
2p
1s22s1
The third electron must go in the
next available orbital with the
lowest possible energy.
The 1s orbital can only accommodate 2
electrons (Pauli exclusion principle)
Electron Configurations
Electron Configurations
The Aufbau principle states that electrons are added to the lowest
energy orbitals first before moving to higher energy orbitals.
The Aufbau principle states that electrons are added to the lowest
energy orbitals first before moving to higher energy orbitals.
Be has a total of 4 electrons
B has a total of 5 electrons
The ground state electron
configuration of Be
1s
2p
2p
1s22s2
2p
Energy
Energy
2p
2s
The ground state electron
configuration of B
2p
2p
1s22s22p1
2s
1s
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Electron Configurations
Electron Configurations
According to Hund’s rule, the most stable arrangement of electrons
is the one in which the number of electrons with the same spin is
maximized.
According to Hund’s rule, the most stable arrangement of electrons
is the one in which the number of electrons with the same spin is
maximized.
C has a total of 6 electrons
The ground state electron
configuration of C
N has a total of 7 electrons
1s22s22p2
2p
1s22s22p3
2p
2p
2s
The 2p orbitals are of equal energy, or degenerate.
1s
Put 1 electron in each before pairing (Hund’s rule).
Energy
Energy
2p
The ground state electron
configuration of N
2p
2p
2s
The 2p orbitals are of equal energy, or degenerate.
1s
Put 1 electron in each before pairing (Hund’s rule).
Electron Configurations
Electron Configurations
According to Hund’s rule, the most stable arrangement of electrons
is the one in which the number of electrons with the same spin is
maximized.
According to Hund’s rule, the most stable arrangement of electrons
is the one in which the number of electrons with the same spin is
maximized.
O has a total of 8 electrons
The ground state electron
configuration of O
F has a total of 9 electrons
1s22s22p4
1s
2p
1s22s22p5
2p
2p
Once all the 2p orbitals are singly occupied, additional
electrons will have to pair with those already in the
orbitals.
Energy
Energy
2p
2s
2s
1s
2p
2p
When there are one or more unpaired electrons, as
in the case of oxygen and fluorine, the atom is
called paramagnetic.
Electron Configurations
Electron Configurations
According to Hund’s rule, the most stable arrangement of electrons
is the one in which the number of electrons with the same spin is
maximized.
General rules for writing electron
configurations:
Ne has a total of 10 electrons
Energy
2p
2s
1s
2p
The ground state electron
configuration of F
The ground state electron
configuration of Ne
1) Electrons will reside in the available
orbitals of the lowest possible energy.
1s22s22p6
2) Each orbital can accommodate a
maximum of two electrons.
2p
When all of the electrons in an atom are paired, as
in neon, it is called diamagnetic.
3) Electrons will not pair in degenerate
orbitals if an empty orbital is available.
4) Orbitals will fill in the order indicated
in the figure.
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3.10
Worked Example 3.10
Electron Configurations and the Periodic Table
Write the electron configuration and give the orbital diagram of a calcium (Ca)
atom (Z = 20).
The electron configurations of all elements except hydrogen and
helium can be represented using a noble gas core.
Setup Because Z = 20, Ca has 20 electrons. They will
fill in according to the diagram at right. Each s subshell
can contain a maximum of two electrons, whereas each p
subshell can contain a maximum of six electrons.
The electron configuration of potassium (Z = 19) is
1s22s22p63s23p64s1.
Solution
Because 1s22s22p63s23p6 is the electron configuration of argon, we
can simplify potassium’s to [Ar]4s1.
Ca 1s22s22p63s23p64s2
The ground state electron configuration of K:
1s2 2s2
2p6
3s2
3p6
4s2
Think About It Look at the figure again to make sure you have filled the
orbitals in the right order and that the sum of electrons is 20. Remember
that the 4s orbital fills before the 3d orbitals.
1s22s22p63s23p64s1
[Ar]
[Ar]4s1
Electron Configurations and the Periodic Table
Electron Configurations and the Periodic Table
Elements in Group 3B through Group 1B are the transition metals.
Following lanthanum (La), there is a gap where the lanthanide
(rare earth) series belongs.
Electron Configurations and the Periodic Table
Electron Configurations and the Periodic Table
After actinum (Ac) comes the actinide series.
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Electron Configurations and the Periodic Table
Electron Configurations and the Periodic Table
There are several notable exceptions to the order of electron filling
for some of the transition metals.
There are several notable exceptions to the order of electron filling
for some of the transition metals.
 Chromium (Z = 24) is [Ar]4s13d5 and not [Ar]4s23d4 as
expected.
 Copper (Z = 29) is [Ar]4s13d10 and not [Ar]4s23d9 as expected.
The reason for these anomalies is the slightly greater stability of d
subshells that are either half-filled (d5) or completely filled (d10).
Cr
 Chromium (Z = 24) is [Ar]4s13d5 and not [Ar]4s23d4 as
expected.
 Copper (Z = 29) is [Ar]4s13d10 and not [Ar]4s23d9 as expected.
The reason for these anomalies is the slightly greater stability of d
subshells that are either half-filled (d5) or completely filled (d10).
[Ar]
Cu
4s
3d
3d
3d
3d
3d
[Ar]
4s
Greater stability with half-filled
3d subshell
3
Write the electron configuration for an arsenic atom (Z = 33) in the ground state.
The order of filling beyond the noble gas core is 4s, 3d,
and 4p. Fifteen electrons go into these subshells because
there are 33 – 18 = 15 electrons in As beyond its noble gas
core.
2
2
6
2
6
2
3
Solution
As
[Ar]4s23d104p3
Think About It Arsenic is a p-block element; therefore, we should
expect its outermost electrons to reside in a p subshell.
3
3d
3d
3d
3d
Greater stability with filled 3d
subshell
Worked Example 3.11
Setup The noble gas core for As is [Ar], where Z = 18
for Ar.
3d
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Chapter Summary: Key Points
Forms of Energy
The Nature of Light
Properties of Waves
The Electromagnetic Spectrum
The Double-Slit Experiment
Quantum Theory
Quantization of Energy
Photons and the Photoelectric
Effect
Bohr’s Theory of the Hydrogen
Atom
Atomic Line Spectra
The Line Spectrum of Hydrogen
Wave Properties of Matter
The de Broglie Hypothesis
Diffraction of Electrons
Quantum Mechanics
The Uncertainty Principle
The Schrödinger Equation
The Quantum Mechanical
Description of the Hydrogen Atom
Quantum Numbers (n,l,ml,ms)
Atomic Orbitals
s orbitals, p orbitals, d orbitals and
other High-Energy Orbitals
Chapter Summary: Key Points
Energies of Orbitals
Electron Configuration
Energies of Atomic Orbitals in Many-Electron Systems
The Pauli Exclusion Principle
The Aufbau Principle
Hund’s Rule
General Rules for Writing Electron Configurations
Electron Configurations and the Periodic Table
Group Quiz #3
• Calculate the energy of a photon that
has a wavelength of 35.6 nm (in the xray region). (Hint: Watch units!!!)
14