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Chapter 6
Quantum Theory and the Electronic
Structure of Atoms
6.1 The Nature of Light:
One of the ways that energy travels through space is
by electromagnetic radiation (light waves).
Waves have three primary characteristics:
1.
Wave length λ (Lambda):
It is the distance between two consecutive peaks
in a given wave.
2.
Frequency :
It is defined as the number of cycles (waves) per
second that pass a given point in space.
1Hz = 1s-1
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3. Amplitude:
It is the vertical distance from the midline of waves to
the top of the peak or the bottom of the trough
Speed of light c:
All types of radiation travel at the speed of light
(c = 2.9979 x 108 m/s or = 2.9979 x 1010 cm/s)
  1/ = c/
and
 = c
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4
Electromagnetic Spectrum
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Classification of electromagnetic radiation:
Radiation
Wave length in
meters
Gamma rays
10-12
X – rays
10-10
Ultraviolet
10-8
Visible
10-6
Infrared
10-4
Microwaves
10-2
FM
1
Radio Shortwave
102
AM
104
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Wave Calculation
The wavelength of a laser pointer is reported
to be 663 nm. What is the frequency of this
light?
 
c

10 9 m
  663 nm 
 6.63 10 7 m
nm
3.00 108 m/s
14 1


4.52

10
s
7
6.63 10 m
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Calculate the wavelength of light, in nm, of light
with a frequency of 3.52 x 1014 s-1.
 
c

3.00  108 m/s
7


8.52

10
m
14 1
3.52  10 s
9
10
nm
7
  8.52 10 m 
 852 nm
m
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Calculate the energy (in joules) of a photon with a
wavelength of 700.0 nm
109 m
  700.0 nm 
 7.00 107 m
nm
3.00 108 m/s
14 1


4.29

10
s
7
7.00 10 m
E  (6.63 1034 J  s)(4.29 1014 s 1 )
E  2.84 10 19 J
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Calculate the wavelength (in nm) of light with energy
7.83 x 1019 J per photon. In what region of the
electromagnetic radiation does this light fall?
7.83 10 19 J
15 1


1.18

10
s
34
6.63 10 J  s
3.00 108 m  s 1
7


2.53

10
m or 253 nm
15 1
1.18 10 s
(Ultraviolet region)
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6.3 Bohr’s Theory of the Hydrogen Atom:
• Planck’s theory along with Einstein’s ideas
not only explained the photoelectric effect,
but also made it possible for scientists to
unravel the idea of atomic line spectra
Atomic Line Spectra:
• Line spectra: emission of light only at specific
wavelengths
• Every element has a unique emission spectrum
• Often referred to as “fingerprints” of the element
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Emission Spectra of Hydrogen and Helium:
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Line Spectra of Hydrogen:
• Balmer (initially) and Rydberg (later) developed
the equation to calculate all spectral lines in
hydrogen
• The Rydberg equation:
1 1
  R  2  2 

 n2 n1 
1
R = 1.09737316 x 107 m-1
• Bohr’s contribution:
Bohr showed only valid energies for hydrogen’s
electron with the following equation:
E n  2.18  10
18
 1 
J 2 
n 
 
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and for an electronic transition from ni to nf
energy levels the energy change is given as:
 1
1 
E  2.18 10 J  2  2 
 n final n initial 
18
• Ground state: the lowest energy state of an
atom (n = 1)
• Excited state: each energy state in which n > 1
• Each spectral line corresponds to a specific
transition
• Electrons moving from ground state to higher
states require energy; an electron falling from a
higher to a lower state releases energy
• Bohr’s equation can be used to calculate the
energy of these transitions within the H atom
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Energy Transitions:
Calculate the energy needed for an electron to move
from n = 1 to n = 4:
E  2.18  10
 18
 1
1 
J  2  2 
 n final n initial 
1 1 
E  2.18  10 J  2  2 
4

1


18
E  2.04  10 18 J
E  2.04  10 18 J
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6.4 Wave Properties of Matter:
• Bohr could not explain why electrons were
restricted to fixed distances around the nucleus
• Louis de Broglie (1924) reasoned that if energy
(light) can behave as a particle (photon) then
perhaps particles (electrons) could exhibit wave
characteristics
• De Broglie proposed that electrons in atoms
behave as standing waves (like the wave created
when a guitar string is plucked)
h
 
mu
where,  = wavelength
m = mass (kg)
u = velocity (m/s)
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Calculate the de Broglie wavelength of the “particle”
in the following two cases:
1) A 25.0 g bullet travelling at 612 m/s:
6.63 10 34 kg  m 2 / s

 4.3 10 35 m *
(0.025 kg)(612 m/s)
2) An electron (mass = 9.109 x 10 31kg) moving
at 63.0 m/s Note: 1 Joule = 1 kg × m2/s2:

6.63 10 34 kg m 2 /s
5


1.16

10
m
31
(9.109 10 kg)(63.0 m/s)
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6.5 Quantum Mechanics:
• Scientists needed to understand exactly where
electrons are in an atom.
• Heisenberg’s uncertainty principle:
We cannot determine the position and the momentum
(i.e. velocity) of a particle at the same time
• Erwin Schrödinger, derived a complex mathematical
formula to incorporate wave and particle characteristics
and to determine the probabilty of finding an electron in
small volume in space around the nucleus.
• Quantum mechanics (wave mechanics) does not
allow us to specify exact location of electrons, we
can predict high probability of finding an electron
• Use the term atomic orbital to describe the electron’s
position of an electron within the atom
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6.6 Quantum Numbers:
• Each atomic orbital in an atom is characterized by a
unique set of three quantum numbers n, l, and ml
(obtained from solving Schrödinger’s wave equation)
1. Principal quantum number (n):
It determines the main energy level and:
n = 1, 2, 3, 4, ……
K L M N
2. Angular momentum quantum number (l):
It determines the shape of the orbital and:
l = 0, 1, 2, 3, …….. n-1
s p d f
3. Magnetic quantum number (ml):
It determines the orientation of the orbital and:
ml = -l, … 0, … +l
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Orbital
designation
l
1
0
1s
0
1
2
0
1
2s
2p
0
-1, 0, +1
1
3
3
0
1
2
3s
3p
3d
0
-1, 0, +1
-2, -1, 0, +1, +2
1
3
5
0
1
2
3
4s
4p
4d
4f
0
-1, 0, +1
-2, -1, 0, +1, +2
-3, -2,-1,0,+1,+2,+3
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Ml
Number
of orbitals
n
1
3
5
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Note:
1. In any principal energy level (n) there are n
subshells (l) from (0 to n-1).
2. In any subshell (l) there are (2l + 1) orbitals and
2(2l + 1) electrons.
3. Total number of electrons in any shell n is 2n2.
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• Electron spin quantum number (ms) - describes the
spin of an electron that occupies a particular orbital :
• Only two values: +1/2 or 1/2
• Electrons will spin opposite each other in the same
orbital
Which of the following are possible sets of quantum
numbers?
a)
1, 1, 0, +1/2
b) 2, 0, 0, +1/2
c)
3, 2, 2, 1/2
(l value not possible)
(possible)
(possible)
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6.7 Atomic Orbitals:
• Shapes of atomic orbitals
• s orbitals - spherical in shape
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• p orbitals - two lobes on opposite sides of
the nucleus:
• d orbitals - more variations of lobes:
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• f orbitals - complex shapes:
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6.8 Electron Configuration:
• Electron configuration is how electrons are
distributed in the various atomic orbitals.
• Ground state - electrons in lowest energy state
• Excited state - electrons in a higher energy
orbital
Pauli exclusion principle:

In a given atom no two electrons can have the
same set of four quantum numbers.

Since electrons in the same orbital can have the
same values of n, l and me they could have
different values of ms.

Since only two values of ms are allowed, an
orbital can only hold two electrons with different
spin.
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Hund’s role:
An atom in its ground state adopts a configuration
with the greater number of unpaired electrons with
the same spin quantum numbers (in degenerate
orbitals).
e.g: C6: 1s2, 2s2, 2p2
Three possibilities for 2s and 2p configuration:
1
2
3
Configuration 3 is lowest energy (ground) state
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6.9 Electron Configurations and the Periodic Table
The Aufbau (building up) Principle and the periodic
Table:

Electrons fill according to orbital energies (lowest to
highest)

As protons are added one by one to the nucleus to
build up the elements, electrons are similarly added
to hydrogen-like orbitals: H (Z = 1), He (Z = 2), Li (Z
= 3), Be (Z = 4), B (Z = 5), etc.

Electrons in the outermost principal quantum level
(highest n in a given atom) are known as valence
electrons (The inner electrons are known as core
electrons)

The elements in the same group (vertical column)
have the same valence electron configuration (e.g:
group 1A have ns1, group 2A have ns2, group 3 have
nd1, group 4 have nd2, group 7A have np5 and group
8A have np6configuration)
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
Groups nA (1A to 8A) are often called the maingroup or representative, elements (The group labels
1 to 8) indicate the total number of valence electrons
for the atoms in these groups.

Noble gas core configuration: can be used to
represent all elements but H and He:
Example (phosphorous P):
Z = 15: [1s22s22p6]3s23p3
[Ne] 3s23p3

As a result of the shielding effect the (n + 1)s orbitals
(e.g: 4s) always fill before the nd (e.g: 3d) orbitals:
e.g: Group 6 elements have ns1nd5 configuration
Cr: [Ar]4s13d5
(d orbitals are half filled)
Group 11 elements have ns1nd10 configuration
Cu: [Ar]4s13d10 (d orbitals are completely filled)
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• Remember Exceptions:
Cr (Z = 24)
and
Cu (Z = 29)
Cr = [Ar]4s13d5
and
Cu = [Ar]4s13d10
Reason: slightly greater stability associated with filled
and half-filled d subshells