Transcript Atomic Structure of Atoms

Lecture Presentation
Chapter 6
Electronic Structure
of Atoms
© 2012 Pearson Education, Inc.
John D. Bookstaver
St. Charles Community College
Cottleville, MO
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 ().
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of Atoms
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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A.
B.
C.
D.
1.0 m and 3.0  108 cycles/s
2.0 m and 6.0  108 cycles/s
0.5 m and 3.0  108 cycles/s
0.5 m and 6.0  108 cycles/s
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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 = 
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B.
C.
D.
An X-ray has a longer wavelength and thus a higher frequency.
An X-ray has a shorter wavelength and thus a higher frequency.
An X-ray has a shorter wavelength and thus a lower frequency.
An X-ray has a longer wavelength and thus a lower frequency. Electronic
Structure
of Atoms
Sample Exercise 6.1 Concepts of Wavelength and Frequency
Two electromagnetic waves are represented in the margin. (a) Which wave has the higher
frequency? (b) If one wave represents visible light and the other represents infrared radiation,
which wave is which?
Solution
Practice Exercise
If one of the waves in the margin represents
blue light and the other red light, which is
which?
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Sample Exercise 6.2 Calculating Frequency from Wavelength
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?
Solution
Practice Exercise
(a) A laser used in eye surgery to fuse detached retinas produces radiation with a wavelength of
640.0 nm. Calculate the frequency of this radiation.
(b) (b) An FM radio station broadcasts electromagnetic radiation at a frequency of 103.4 MHz
(megahertz; 1 MHz = 106s-1 ). Calculate the wavelength of this radiation. The speed of light is
2.998 × 108 m/s to four significant digits.
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A. Yes: X-rays travel at at faster speeds than
visible light.
B. No: both X-rays and visible light travel at the
speed of light, c.
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The Nature of Energy
The wave nature of light
does not explain how
an object can glow
when its temperature
increases.
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A. White-yellow or yellow area in center,
bottom and middle
B. Blue-green area at right bottom
C. Green-yellow at middle right
D. Black area at left
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The Nature of Energy
Max Planck explained it by assuming that
energy comes in packets called quanta.
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of Atoms
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.
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of Atoms
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
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of Atoms
A. Lack of gas molecules keeps inside of vacuum
tube clean.
B. Electrons cannot move through a gas.
C. Electrons would strike gas molecules and many
electrons would not readily arrive at the positive
terminal.
D. Gas molecules in tube would prevent observation
of any emission of electrons.
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A.
B.
C.
D.
5 x 10-3 J
3 x 10-36 J
6 x 10-30 J
6.63 x 10-34 J
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Sample Exercise 6.3 Energy of a Photon
Calculate the energy of one photon of yellow light that has a wavelength of 589 nm.
Solution
Practice Exercise
(a) A laser emits light that has a frequency of 4.69 × 1014s-1. What is the energy of one photon of
this radiation?
(b) If the laser emits a pulse containing 5.0 × 1017 photons of this radiation, what is the total
energy of that pulse?
(c) If the laser emits 1.3 × 10-2J of energy during a pulse, how many photons are emitted?
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A. Photon of infrared light
B. Photon of ultraviolet light
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of Atoms
The Nature of Energy
Another mystery in
the early twentieth
century involved the
emission spectra
observed from
energy emitted by
atoms and
molecules.
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of Atoms
The Nature of Energy
• For atoms and
molecules, one does
not observe a
continuous spectrum,
as one gets from a
white light source.
• Only a line spectrum
of discrete
wavelengths is
observed.
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A. Only certain orbits are allowed.
B. An electron can only exist in one “allowed”
energy state.
C. Electrons do not spiral into the nucleus.
D. Electrons can only change from one “allowed”
energy state to another.
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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
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The Nature of Energy
The energy absorbed or emitted
from the process of electron
promotion or demotion can be
calculated by the equation:
E = −hcRH (
1
1
n f2
ni2
)
where RH is the Rydberg
constant, 1.097  107 m−1, and ni
and nf are the initial and final
energy levels of the electron.
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A. Ultraviolet light, because a greater amount of
energy is emitted during the second electron
transition than for the one with emission of
visible light
B. Infrared light, because a greater amount of
energy is emitted during the second electron
transition than for the one with emission of
visible light
C. Ultraviolet light, because a smaller amount of
energy is emitted during the second electron
transition than for the one with emission of
visible light
D. Infrared light, because a smaller amount of
energy is emitted during the second electron
transition than for the one with emission of
visible light
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A. It neither emits nor absorbs energy.
B. It both emits and absorbs energy
simultaneously.
C. It emits energy.
D. It absorbs energy.
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Sample Exercise 6.4 Electronic Transitions in the Hydrogen Atom
Using Figure 6.12, predict which of these electronic transitions produces the spectral line having the
longest wavelength: n = 2 to n = 1, n = 3 to n = 2, or n = 4 to n = 3.
Solution
Practice Exercise
Indicate whether each of the following electronic
transitions emits energy or requires the absorption
of energy: (a) n = 3 to n = 1; (b) n = 2 to n = 4.
FIGURE 6.12 Energy states in the hydrogen
atom. Only states for n = 1 through n = 4 and n =
∞ are shown. Energy is released or absorbed
when an electron moves from one energy state
to another.
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The Wave Nature of Matter
• Louis de Broglie posited
that if light can have
material properties,
matter should exhibit
wave properties.
• He demonstrated that the
relationship between
mass and wavelength
was
h
 = mv
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Sample Exercise 6.5 Matter Waves
What is the wavelength of an electron moving with a speed of 5.97 × 106 m/s?
The mass of the electron is 9.11 × 10-31 kg.
Solution
Practice Exercise
Calculate the velocity of a neutron whose de Broglie wavelength is 500 pm. The mass of a neutron is
given in the table inside the back cover of the text.
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A. Yes, but not observable because the matter
wavelength is too small for our eyes to detect.
B. Yes and observable because matter waves of
macroscopic objects are in the visible region
of light.
C. Matter waves are not produced for
macroscopic objects and thus none are
observable.
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The Uncertainty Principle
Heisenberg showed that the
more precisely the momentum
of a particle is known, the less
precisely is its position is
known:
(x) (mv) 
h
4
In many cases, our uncertainty
of the whereabouts of an
electron is greater than the
size of the atom itself!
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B.
C.
D.
The charge of subatomic particles compared to
zero charge for macroscopic particles.
The small mass and size of subatomic particles
compared to the mass and size of macroscopic
particles.
The small volume occupied by subatomic particles
compared to the volume occupied by macroscopic
particles.
The slower speeds of subatomic particles
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compared to the speeds of macroscopic particles. Structure
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Quantum Mechanics
• Erwin Schrödinger developed a
mathematical treatment into which
both the wave and particle nature
of matter could be incorporated.
• This is known as quantum
mechanics.
• The wave equation is designated
with a lowercase Greek psi ().
• The square of the wave equation,
2, gives a probability density map
of where an electron has a certain
statistical likelihood of being at
any given instant in time.
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B.
C.
D.
Far from the nucleus
Along the z axis
Along the y axis
Near the nucleus
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Quantum Numbers
• Solving the wave equation gives a set of
wave functions, or orbitals, and their
corresponding energies.
• Each orbital describes a spatial
distribution of electron density.
• An orbital is described by a set of three
quantum numbers.
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of Atoms
Principal Quantum Number (n)
• The principal quantum number, n,
describes the energy level on which the
orbital resides.
• The values of n are integers ≥ 1.
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Electronic
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of Atoms
Angular Momentum Quantum
Number (l)
• This quantum number defines the
shape of the orbital.
• Allowed values of l are integers ranging
from 0 to n − 1.
• We use letter designations to
communicate the different values of l
and, therefore, the shapes and types of
orbitals.
Electronic
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Structure
of Atoms
Angular Momentum Quantum
Number (l)
Value of l
0
1
2
3
Type of orbital
s
p
d
f
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Electronic
Structure
of Atoms
Magnetic Quantum Number (ml)
• The magnetic quantum number describes the
three-dimensional orientation of the orbital.
• Allowed values of ml are integers ranging
from −l to l:
−l ≤ ml ≤ l
• Therefore, on any given energy level, there
can be up to 1 s orbital, 3 p orbitals, 5 d
orbitals, 7 f orbitals, and so forth.
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A.
B.
C.
D.
An orbital is composed of some integral number of
orbits.
An orbit is a well-defined circular path around the
nucleus while an orbital is a wave function that gives the
probability of finding the electron at any point in space.
An orbit is a well-defined circular path around the
nucleus while an orbital is the object (electron) that is
moving around the nucleus.
There is no difference between the definitions of the
terms “orbit” and “orbital.” They simply were proposed
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by different scientists.
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Magnetic Quantum Number (ml)
• Orbitals with the same value of n form a shell.
• Different orbital types within a shell are subshells.
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B.
C.
D.
One subshell labeled 4s
Two subshells labeled 4s and 4p
Three subshells labeled 4s, 4p and 4d
Four subshell labeled 4s, 4p, 4d and 4f
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A.
B.
C.
D.
Because –1/(2)2 and –1/(1)2 are of much greater
difference than –1/(3)2 and –1/(2)2 in Bohr’s equation
that describes the hydrogen atom.
Because the Rydberg in Bohr’s equation constant is
higher for n = 1 to n = 2 transitions.
Because the energy difference between the 1st and 2nd
energy level of hydrogen is smaller than the energy
difference between the 2nd and 3rd energy level.
Because the electron in the hydrogen atom never
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occupies energy levels of higher than n = 1.
of Atoms
Sample Exercise 6.6 Subshells of the Hydrogen Atom
(a) Without referring to Table 6.2, predict the number of subshells in the fourth shell, that is, for n
= 4. (b) Give the label for each of these subshells. (c) How many orbitals are in each of these
subshells?
Solution
Practice Exercise
(a) What is the designation for the subshell with n = 5 and l = 1? (b) How many orbitals are
in this subshell? (c) Indicate the values of ml for each of these orbitals.
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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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Maxima
4
4
3
3
Nodes
4
3
3
4
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s Orbitals
Observing a graph of
probabilities of finding an
electron versus distance
from the nucleus, we see
that s orbitals possess n
− 1 nodes, or regions
where there is 0
probability of finding an
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electron.
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Structure
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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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of Atoms
A. Electrons are moving slower at edges than in interior of the lobes
B. Greater number of electrons at edges than in interior of the lobes.
C. Probability of finding an electron at edges is lower than in interior
of the lobes
D. Number of electrons is disappearing near the edges of the lobes
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A.
B.
C.
D.
2px
2py
2pz
2p
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d Orbitals
• The value of l for a
d orbital is 2.
• Four of the five d
orbitals have 4
lobes; the other
resembles a p
orbital with a
doughnut around
the center.
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of Atoms
Energies of Orbitals
• For a one-electron hydrogen
atom, orbitals on the same
energy level have the same
energy.
• That is, they are
degenerate.
• As the number of electrons
increases, though, so does
the repulsion between them.
• Therefore, in many-electron
atoms, orbitals on the same
energy level are no longer
degenerate.
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B.
C.
D.
4g
4d
4p and 4d
4d and 4f
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B.
C.
D.
3
6
9
18
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B.
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3s < 3p < 3d
3s < 3d < 3p
3d < 3p < 3s
3p < 3s < 3d
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of Atoms
Spin Quantum Number, ms
• In the 1920s, it was
discovered that two
electrons in the same
orbital do not have
exactly the same energy.
• The “spin” of an electron
describes its magnetic
field, which affects its
energy.
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Electronic
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of Atoms
Spin Quantum Number, ms
• This led to a fourth
quantum number, the
spin quantum number,
ms.
• The spin quantum
number has only 2
allowed values: +1/2
and −1/2.
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Electronic
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of Atoms
Pauli Exclusion Principle
• No two electrons in the
same atom can have
exactly the same energy.
• Therefore, no two
electrons in the same
atom can have identical
sets of quantum
numbers.
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Electron Configurations
5
4p
• This term shows the
distribution of all
electrons in an atom.
• Each component
consists of
– A number denoting the
energy level,
– A letter denoting the type
of orbital,
– A superscript denoting
the number of electrons
in those orbitals.
Electronic
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Structure
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Orbital Diagrams
• Each box in the
diagram represents
one orbital.
• Half-arrows represent
the electrons.
• The direction of the
arrow represents the
relative spin of the
electron.
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of Atoms
Hund’s Rule
“For degenerate
orbitals, the
lowest energy is
attained when
the number of
electrons with
the same spin is
maximized.”
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of Atoms
Sample Exercise 6.7 Orbital Diagrams and Electron Configurations
Draw the orbital diagram for the electron configuration of oxygen, atomic number 8. How many unpaired
electrons does an oxygen atom possess?
Solution
Practice Exercise
(a) Write the electron configuration for phosphorus, element 15. (b) How many unpaired electrons
does a phosphorus atom possess?
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A. 5d
B. 6s
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Periodic Table
• We fill orbitals in increasing order of energy.
• Different blocks on the periodic table (shaded
in different colors in this chart) correspond to
different types of orbitals.
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Electronic
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of Atoms
Sample Exercise 6.8 Electron Configurations for a Group
What is the characteristic valence electron configuration of the group 7A elements, the halogens?
Solution
Practice Exercise
Which family of elements is characterized by an ns2np2 electron configuration in the outermost
occupied shell?
Electronic
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of Atoms
Sample Exercise 6.9 Electron Configurations from the Periodic Table
(a) Based on its position in the periodic table, write the condensed electron configuration for bismuth,
element 83. (b) How many unpaired electrons does a bismuth atom have?
Solution
Practice Exercise
Use the periodic table to write the condensed electron configuration for (a) Co (element 27), (b) Te
(element 52).
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Some Anomalies
Some irregularities occur
when there are enough
electrons to half-fill s and d
orbitals on a given row.
•
•
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For instance, the electron
configuration for copper is
[Ar] 4s1 3d10
rather than the expected
[Ar] 4s2 3d9.
This occurs because the 4s
and 3d orbitals are very
close in energy.
These anomalies occur in fblock atoms, as well.
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A.
B.
C.
D
All three elements have the same relative energies
for their nd and (n + 1)s orbitals.
Nothing can be concluded, because the three
elements each have different valence electron
configurations.
The nd orbitals are of lower energy than
(n + 1)s orbitals for all three elements.
The nd orbitals are of higher energy than
Electronic
Structure
(n + 1)s orbitals for all three elements.
of Atoms