Chapter 7
Atomic Structure and Periodicity
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1
Overview
 Introduce Electromagnetic Radiation and The
Nature of Matter.
 Discuss the atomic spectrum of hydrogen and
Bohr model.
 Describe the quantum mechanical model of the
atoms and quantum numbers.
 Use Aufbau principle to determine the electron
configuration of elements.
 Highlight periodic table trends.
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2
Matter and Energy
Matter and Energy were two distinct
concepts in the 19th century.
Matter was thought to consist of particles,
and had mass and position.
Energy in the form of light was thought to
be wave-like.
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3
Physical Properties of Waves
Wavelength (l) is the distance between identical points on
successive waves.
Amplitude is the vertical distance from the midline of a
wave to the peak or trough.
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4
Properties of Waves
Frequency (n) is the number of waves that pass through a
particular point in 1 second (Hz = 1 cycle/s).
The speed (v or c) of the wave = l x n
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5
Electromagnetic Radiation
Electromagnetic radiation is the
emission and transmission of energy in
the form of electromagnetic waves.
Electromagnetic radiation travels through
space at the speed of light in a vacuum.
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Maxwell (1873), proposed that visible light consists of
electromagnetic waves.
Speed of light (c) in vacuum = 3.00 x 108 m/s
All electromagnetic radiation
lxn=c
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7
7.1
Electromagnetic Waves
Electromagnetic Waves have 3 primary
characteristics:
1.
Wavelength: distance between two
peaks in a wave.
2.
Frequency: number of waves per
second that pass a given point in space.
3.
Speed: speed of light is 2.9979  108
m/s.
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http://www.colorado.edu/physics/2000/waves_particles/wpwaves5.html
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Electromagnetic Wave Movie
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10
Wavelength and frequency can be interconverted.
n = c/l
n = frequency (s1)
l = wavelength (m)
c = speed of light (m s1)
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11
Electromagnetic Spectrum
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12
Plank
Studied
radiation
emitted
by
heated
bodies.
Results could not be
explained by the old
physics which stated that
matter can absorb and
emit any quantity of
energy.
Black Body Radiation
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13
Planck’s Constant
Transfer of energy is quantized, and can only
occur in discrete units, called quanta.
E = n hn =
hc
n
l
E = change in energy, in J
h = Planck’s constant, 6.626  1034 J s
n = frequency, in s1
l = wavelength, in m
n = integer = 1,2,3…
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14
S.E. 7.2
Calculate the frequency of red light of wave
length 4.50x102 nm.
c = υλ
or υ = c/λ
λ = 4.50 nm x 1m = 4.50x10-7 m
109 nm
υ = 2.9979x108 m/s = 6.66x1014 s-1 (Hz)
4.50x10-7 m
ΔE = h υ = 6.626x10-34J.s x 6.66x1014 s-1
= 4.41x10-19 J
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15
Particle Properties of Light
hn
Photoelectric Effect Solved
by Einstein in 1905
KE e-
Photon is a “particle” of light
hn = KE + BE
KE = hn - BE
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16
7.2
Diffraction
X-Ray Diffraction showed also that light
has wave properties.
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17
Figure
7.5:
(a) Diffraction occurs when electromagnetic
radiation is scattered from a regular array
of objects, such as the ions in a crystal of
sodium chloride. The large spot in the
center is from the main incident beam of X
rays. (b) Bright spots in the diffraction
pattern result from constructive interference
of waves. The waves are in phase; that is,
their peaks match. (c) Dark areas result
from destructive interference of waves. The
waves are out of phase; the peaks of one
wave coincide with the troughs of another
wave.
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Diffraction Movie
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Photoelectric Effect Movie
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20
Energy and Mass
Energy has mass
E = mc2 Einstein Equation
E = energy
m = mass
c = speed of light
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Energy and Mass
Ephoton =
mphoton
hc
l
h
=
lc
Radiation in itself is quantized
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22
Wavelength and Mass
de Broglie’s Equation
h
l =
mn
l = wavelength, in m
h = Planck’s constant, 6.626  1034 J s =
kg m2 s1
m = mass, in kg
v = speed, in ms1
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23
s.ex.7.3
Calculate the wavelength for an electron (mass = 9.11x10-31
kg) travelling at a speed of 1.0x107 m/s.
λ = h/mv = 6.626 x 10-34 kg m2/s
9.11x10-31 kg x 1.0x107 m/s
= 7.27x10-11 m
For the ball
λ = 6.626 x 10-34 kg m2/s
0.10 kg x 35 m/s
= 1.9x10-34 m
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24
What is the de Broglie wavelength (in nm)
associated with a 2.5 g Ping-Pong ball
traveling at 15.6 m/s?
l = h/mv
h in J•s m in kg u in (m/s)
l = 6.63 x 10-34 / (2.5 x 10-3 x 15.6)
l = 1.7 x 10-32 m = 1.7 x 10-23 nm
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25
A photon has a frequency of 6.0 x 104 Hz. Convert
this frequency into wavelength (nm). Does this frequency
fall in the visible region?
l
lxn=c
n
l = c/n
l = 3.00 x 108 m/s / 6.0 x 104 Hz
l = 5.0 x 103 m
l = 5.0 x 1012 nm
Radio wave
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26
Atomic Spectrum of Hydrogen
Continuous spectrum: Contains all the
wavelengths of light.
Line (discrete) spectrum: Contains only
some of the wavelengths of light.
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Figure 7.6:
(a) A continuous
spectrum
containing all
wavelengths of
visible light.
(b) The
hydrogen line
spectrum
contains only a
few discrete
wavelengths.
Atomic Spectrum of Hydrogen
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28
Bohr’s Model of
the Atom (1913)
1. e- can only have specific
(quantized) energy
values
2. light is emitted as emoves from one energy
level to a lower energy
level
En = -RH (
z2
n2
)
n (principal quantum number) = 1,2,3,…
RH (Rydberg constant) = 2.18 x 10-18J
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7.3
E = hn
E = hn
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ni = 3
ni = 3
ni = 2
nf = 2
Ephoton = E = Ef - Ei
1
Ef = -RH ( 2
nf
1
Ei = -RH ( 2
ni
1
E = RH( 2
ni
)
)
1
n2f
nnf f==11
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31
)
Calculate the wavelength (in nm) of a photon
emitted by a hydrogen atom when its electron
drops from the n = 5 state to the n = 3 state.
Ephoton = E = RH(
1
n2i
1
n2f
)
Ephoton = 2.18 x 10-18 J x (1/25 - 1/9)
Ephoton = E = -1.55 x 10-19 J
Ephoton = h x c / l
Ignore the (-) sign for
l and n
l = h x c / Ephoton
l = 6.63 x 10-34 (J•s) x 3.00 x 108 (m/s)/1.55 x 10-19J
l = 1280 nm
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The Bohr Model
Ground State: The lowest possible
energy state for an atom (n = 1).
Ionization: nf =  => 1/nf = 0 =>
E=0 for free electron.
Any bound electron has a negative
value to this reference state.
2
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33
7.5
Calculate the energy required to remove an electron
from hydrogen atom in its ground state.
n initial = 1
to
nfinal = ∞
ΔE = -2.178x10-18 J (1/ n2final - 1/ n2initial )
= -2.178x10-18 J (1/ ∞ - 1/12 )
= - 2.178x10-18 J ( 0 - 1) = 2.178x10-18 J
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34
To be well memorized
hc
E = hn =
l
P = mv
E = mc2
Ephoton = E = RH(
1
n2i
1
n2f
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)
35
Figure 7.9
The
Standing
Waves
Caused by
the
Vibration of
a Guitar
String
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Figure 7.10
The
Hydrogen
Electron
Visualized
as a
Standing
Wave
Around the
Nucleus
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Quantum Mechanics
Based on the wave properties of the atom
H  = E
 = wave function
H = mathematical operator
E = total energy of the atom
A specific wave function is often called an
orbital.
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38
Schrodinger Wave Equation
In 1926 Schrodinger wrote an equation that
described both the particle and wave nature of the eWave function (Y) describes:
1. energy of e- with a given Y
2. probability of finding e- in a volume of space
Schrodinger’s equation can only be solved exactly
for the hydrogen atom. Must approximate its
solution for multi-electron systems.
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Heisenberg Uncertainty Principle
h
x   mv 
4
x = position
mv = momentum
h = Planck’s constant
The more accurately we know a particle’s
position, the less accurately we can know its
momentum.
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Probability Distribution
of the wave function: Y2
 probability of finding an electron at a given
position
Radial probability distribution is the
probability distribution in each spherical
shell.
 SQUARE
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Schrodinger Wave Equation
Y = fn(n, l, ml, ms)
principal quantum number n
n = 1, 2, 3, 4, ….
distance of e- from the nucleus
n=1
n=2
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n=3
42
Distribution for
the Hydrogen 1s
Orbital in ThreeDimensional
Space (b) The
Probability of Find
the Electron at
Points Along a
Line Drawn From
the Nucleus
Outward in Any
Direction for the
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Figure 7.12: (a) Cross section of the hydrogen 1s
orbital probability distribution divided into
successive thin spherical shells. (b) The radial
probability distribution.
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Schrodinger Wave Equation
Y = fn(n, l, ml, ms)
angular momentum quantum number l
for a given value of n, l = 0, 1, 2, 3, … n-1
n = 1, l = 0
n = 2, l = 0 or 1
n = 3, l = 0, 1, or 2
l=0
l=1
l=2
l=3
s orbital
p orbital
d orbital
f orbital
Shape of the “volume” of space that the e- occupies
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Quantum Numbers (QN)
1.
Principal QN (n = 1, 2, 3, . . .) - related to size and
energy of the orbital.
2.
Angular Momentum QN (l = 0 to n  1) - relates to
shape of the orbital.
3.
Magnetic QN (ml = l to l) - relates to orientation
of the orbital in space relative to other orbitals.
4.
Electron Spin QN (ms = +1/2, 1/2) - relates to the
spin states of the electrons.
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Pauli Exclusion Principle
In a given atom, no two electrons can have
the same set of four quantum numbers (n, l,
ml, ms).
Therefore, an orbital can hold only two
electrons, and they must have opposite
spins.
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Each orbital can take a maximum of two
electrons and a minimum of Zero electrons.
Zero electrons does not mean that the
orbital does not exist.
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Degenerate
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Figure 7.13: Two
representations of
the hydrogen 1s,
2s, and 3s orbitals.
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Figure 7.14: Representation of the 2p orbitals.
(a) The electron probability distributed for a 2p
orbital. (b) The boundary surface representations
of all three 2p orbitals.
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2p
ml = -1
ml = 0
ml = 1
Degenerate Orbitals
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Figure 7.16: Representation of the
3d orbitals.
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3d
ml = -2
ml = -1
ml = 0
ml = 1
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ml = 2
55
Figure 7.20:
A comparison of the radial
probability distributions of the 2s and 2p orbitals.
P orbital is more diffuse
Zero probability to be
in the nucleus
Probability to be
in the nucleus
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Figure 7.21:
(a) The radial probability
distribution for an electron in a 3s orbital. (b) The
radial probability distribution for the 3s, 3p,
and 3d orbitals.
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How many 2p orbitals are there in an atom?
n=2
If l = 1, then ml = -1, 0, or +1
2p
3 orbitals
l=1
How many electrons can be placed in the 3d
subshell?
n=3
3d
l=2
If l = 2, then ml = -2, -1, 0, +1, or +2
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58
Energy of orbitals in a single electron atom
Energy only depends on principal quantum number n
n=3
n=2
En = -RH (
1
n2
)
n=1
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Energy of orbitals in a multi-electron atom
Energy depends on n and l
n=3 l = 2
n=3 l = 0
n=2 l = 0
n=1 l = 0
n=3 l = 1
n=2 l = 1
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“Fill up” electrons in lowest energy orbitals (Aufbau principle)
??
B
1s22s22p1
H 1 electron
1s1
C 6 electrons
B 5 electrons
Be 4 electrons
Li 3 electrons
Li 1s22s1
Be 1s22s2
He 2 electrons
He 1s2
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The most stable arrangement of electrons
in subshells is the one with the greatest
number of parallel spins (Hund’s rule).
Ne 1s22s22p6
F 1s22s22p5
O 1s22s22p4
N 1s22s22p3
C 1s22s22p2
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Figure 7.25: The electron configurations in the
type of orbital occupied last for the first 18
elements.
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Order of orbitals (filling) in multi-electron atom
1s < 2s < 2p < 3s < 3p
< 4s < 3d < 4p < 5s < 4d < 5p < 6s
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Figure 7.26: Electron configurations for
potassium through krypton.
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Figure 7.27:
The orbitals being filled for
elements in various parts of the periodic table.
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What is the electron configuration of Mg?
Mg 12 electrons
1s < 2s < 2p < 3s < 3p < 4s
1s22s22p63s2
2 + 2 + 6 + 2 = 12 electrons
Abbreviated as [Ne]3s2
[Ne]= 1s22s22p6
What are the possible quantum numbers for the
last (outermost) electron in Cl?
Cl 17 electrons
1s22s22p63s23p5
1s < 2s < 2p < 3s < 3p < 4s
2 + 2 + 6 + 2 + 5 = 17 electrons
Last electron added to 3p orbital
n=3
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or +1
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ms = ½ or -½
7.7
Valence Electrons
The electrons in the outermost principle
quantum level of an atom.
Atom
Valence Electrons
Ca
2
N
5
Br
7
Inner electrons are called core electrons.
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Broad Periodic Table
Classifications
Representative Elements (main group):
filling s and p orbitals (Na, Al, Ne, O)
Transition Elements: filling d orbitals (Fe,
Co, Ni)
Lanthanide and Actinide Series (inner
transition elements): filling 4f and 5f
orbitals (Eu, Am, Es)
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Figure 7.36:
Special
names for
groups in
the periodic
table.
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Outermost subshell being filled with electrons
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7.8
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Paramagnetic
unpaired electrons
2p
Diamagnetic
all electrons paired
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2p
73
Ionization Energy
The quantity of energy required
to remove an electron from the
gaseous atom or ion.
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X (g)
X+(g) + e-
I1 first ionization energy
X+ (g)
X2+(g) + e-
I2 second ionization energy
X2+ (g)
X3+(g) + e-
I3 third ionization energy
I1 < I2 < I3
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Al(g) → Al+ (g) + eAl+(g) → Al2+(g) + eAl2+(g) → Al3+(g) + eAl3+(g) → Al4+(g) + e-
I1 = 580 kJ/mole
I2 = 1815 kJ/mole
I3 = 2740 kJ/mole
I4 = 11,600 kJ/mol
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Periodic Trends
First ionization energy:
increases from left to right across a
period;
decreases going down a group.
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General Trend in First Ionization Energies
Increasing First Ionization Energy
Increasing First Ionization Energy
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Figure 7.31 Trends in Ionization Energies (kj/mol)
for the Representative Elements
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Figure 7.31: The values of first
ionization energy for the elements
in the first six periods.
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Exceptions
Be 1s2 2s2
B 1s22s22p1
Shielded Electron
N 1s22s2sp3
O 1s22s22p4
Repulsion to doubly electrons
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Electron Affinity
The energy change associated with
the addition of an electron to a
gaseous atom.
X(g) + e  X(g)
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X (g) + e-
X-(g)
F (g) + e-
X-(g)
H = -328 kJ/mol
O (g) + e-
O-(g)
H = -141 kJ/mol
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Figure 7.33: The electron affinity values for
atoms among the first 20 elements that form
stable, isolated X- ions.
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Exceptions
C-(g) can be formed easily while N-(g) cannot
be formed easily:
C1s22s2p3
Extra repulsion
2
2
4
N
1s 2s 2p
O-(g) can be formed because the larger
positive nucleus overcome pairing
repulsions.
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Periodic Trends
Atomic Radii:
decrease going from left to right
across a period;
increase going down a group.
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Figure 7.33 The Radious of an Atom (r) is Defined
as Half the Distance Between the Nuclei in a
Molecule Consisting of Identical Atoms
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Figure 7.35:
Atomic
radii (in
picometers)
for selected
atoms.
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Information Contained in the
Periodic Table
1.
2.
3.
4.
Each group member has the same valence
electron configuration (these electrons
primarily determine an atom’s chemistry).
The electron configuration of any
representative element.
Certain groups have special names (alkali
metals, halogens, etc).
Metals and nonmetals are characterized by
their chemical and physical properties.
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92