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Electronic Structure of Atoms
Chapter 6
7-1
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Waves
• To understand the electronic structure of atoms,
one must understand the nature of electromagnetic
radiation.
• The distance between corresponding points on
adjacent waves is the wavelength ().
7-2
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Waves
• The number of waves
passing a given point per
unit of time is the
frequency ().
• For waves traveling at the
same velocity, the longer
the wavelength, the smaller
the frequency.
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Electromagnetic Radiation
• All electromagnetic radiation travels at the same velocity:
the speed of light (c), 3.00  108 m/s.
• Therefore,
c: speed of light (3.00  108 m/s)
c = 
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: wavelength (m, nm, etc.)
: frequency (Hz)
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Sample Problems
• The yellow light given off by a sodium vapor lamp used for public
lighting has a wavelength of 589 nm. What is the frequency of this
radiation?
• A laser used to weld detached retinas produces radiation with a
frequency of 4.69 x 1014 s-1. What is the wavelength of this
radiation?
c  
7-5
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Max Planck and Quantized Energy
• So an atom can change its energy by emitting
or absorbing quanta
• To understand quantization consider walking
up a ramp versus walking up stairs:
– For the ramp, there is a continuous change in
height whereas up stairs there is a quantized
change in height.
7-6
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The Nature of Energy
Max Planck explained it by assuming that energy
comes in packets called quanta.
7-7
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The Nature of Energy
• Einstein used this assumption to explain the
photoelectric effect.
• He concluded that energy is proportional to frequency:
E = h
where h is Planck’s constant, 6.626  10−34 J-s.
A change in an atom’s energy results in the gain or loss
of “packets” of fixed amounts energy called a
quantum.
7-8
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Quantum staircase.
7-9
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The Nature of Energy
• Therefore, if one knows the
wavelength of light, one can
calculate the energy in one
photon, or packet, of that
light:
c = 
E = h
7-10
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Sample Problem
Calculating the Energy of Radiation from Its
Wavelength
PROBLEM: A cook uses a microwave oven to heat a meal. The wavelength of
the radiation is 1.20cm. What is the energy of one photon of this
microwave radiation?
PLAN: After converting cm to m, we can use the energy equation, E = h
combined with  = c/ to find the energy.
SOLUTION:
E=
E = hc/
6.626X10-34J*s x 3x108m/s
1.20cm
10-2m
cm
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= 1.66x10-23J
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The Nature of Energy
•
Niels Bohr adopted Planck’s assumption and
explained these phenomena in this way:
1. Electrons in an atom can only occupy certain orbits
(corresponding to certain energies).
2. Electrons in permitted orbits have specific, “allowed”
energies; these energies will not be radiated from the
atom.
3. Energy is only absorbed or emitted in such a way as to
move an electron from one “allowed” energy state to
another; the energy is defined by
E = h
7-12
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Schrodinger Wave Equation
Quantum Numbers and Atomic Orbitals
An atomic orbital is specified by three quantum numbers.
n the principal quantum number - a positive integer, indicates the
relative size of the orbital or the distance from the nucleus.
The values of n are integers ≥ 1
l the angular momentum quantum number - an integer from 0 to n-1,
related to the shape of the orbital
ml the magnetic moment quantum number - an integer from -l to +l,
orientation of the orbital in the space around the nucleus 2l + 1
Ms spin quantum number ms = +½ or -½
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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
l=0
l=1
l=2
l=3
n = 1, l = 0
n = 2, l = 0 or 1
n = 3, l = 0, 1, or 2
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s orbital
p orbital
d orbital
f orbital
Value of l
0
1
2
3
Type of orbital
s
p
d
f
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s Orbitals
• The value of l for s orbitals is 0.
• They are spherical in shape.
• The radius of the sphere increases with the value
of n.
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p Orbitals
• The value of l for p orbitals is 1.
• They have two lobes with a node between them.
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Schrodinger Wave Equation
Y = fn(n, l, ml, ms)
magnetic quantum number ml
for a given value of l
ml = -l, …., 0, …. +l
if l = 1 (p orbital), ml = -1, 0, or 1
if l = 2 (d orbital), ml = -2, -1, 0, 1, or 2
orientation of the orbital in space
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7.6
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ml = -1
ml = -2
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ml = 0
ml = -1
ml = 0
ml = 1
ml = 1
ml = 2
7.6
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Schrodinger Wave Equation
Y = fn(n, l, ml, ms)
spin quantum number ms
ms = +½ or -½
ms = +½
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ms = -½
7.6
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Electron configuration is how the electrons are
distributed among the various atomic orbitals in an
atom.
number of electrons
in the orbital or subshell
1s1
principal quantum
number n
angular momentum
quantum number l
Gives you the shape of the subshell
Orbital diagram
H
1s1
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7.8
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“Fill up” electrons in lowest energy orbitals (Aufbau principle)
??
Be
Li
B5
C
3
64electrons
electrons
22s
222s
22p
12 1
BBe
Li1s1s
1s
2s
H
He12electron
electrons
He
H 1s
1s12
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7.9
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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).
Ne97
C
N
O
F
6
810
electrons
electrons
electrons
22s
222p
22p
5
246
3
Ne
C
N
O
F 1s
1s222s
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7.7
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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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7.7
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Outermost subshell being filled with electrons
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7.8
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7-25
Not in your book. This edition took it out from the book
7.8
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Energy of orbitals in a single electron atom
Energy only depends on principal quantum number n
n=3
n=2
n=1
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7.7
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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=3 l = 1
n=2 l = 1
n=1 l = 0
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7.7
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Schrodinger Wave Equation
Y = fn(n, l, ml, ms)
Existence (and energy) of electron in atom is described
by its unique wave function Y.
Pauli exclusion principle - no two electrons in an atom
can have the same four quantum numbers.
Each seat is uniquely identified (E, R12, S8)
Each seat can hold only one individual at a
time
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7.6
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The Hierarchy of Quantum Numbers for Atomic Orbitals
Name, Symbol
(Property)
Allowed Values
Quantum Numbers
Principal, n
Positive integer
(size, energy)
(1, 2, 3, ...)
1
Angular
momentum, l
0 to n-1
(shape)
0
0
0
0
Magnetic, ml
-l,…,0,…,+l
(orientation)
2
3
1
0
2
0
-1 0 +1
-1 0 +1
-2
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1
-1
0
+1 +2