Chapter 6
Electronic Structure of Atoms
Electronic Structure
• Electronic structure—the arrangement and
energy of electrons
• 1st lets talk about waves
– Why?
– Extremely small particles have properties that
can only be explained by learning about waves
2
Waves
• Wavelength ()
– Distance between corresponding points on
adjacent waves is the
• Frequency ()
– Number of waves passing a given point per unit of
time is the
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3
Waves
• Frequency ()
– Waves traveling at the same
velocity
Ȣ
• Wavelength inversely proportional
to frequency
•  1/
– Units is s-1
– Measure by counting number of
waves passing through a certain
point
What is the frequency of the
figures if time is 1 second?
• Figure (a): 2 s–1
• Figure (b): 4 s–1
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4
Electromagnetic Radiation
• Understanding the nature of electromagnetic radiation
helps to understand the electronic structure of atoms
• Electromagnetic radiation
– Form of energy
– Called radiant energy or optical energy
– Different types
• Propagating (moving) forms of energy
• Wavelike characteristics
• Travel at the speed of light (c) is 3.00  108 m/s
– c = 
http://images.wisegeek.com/electromagnetic-spectrum-with-gamma-rays.jpg
5
Example Wavelength and Frequency
Calculate the wavelength (in nm) of the red light emitted by a barcode scanner
that has a frequency of 4.62 × 1014 s–1.
GIVEN  = 4.26 x 1014 s-1
Solution
You are given the frequency of the light
FIND wavelength
and asked to find its wavelength.
Use the equation that relates speed of
light to frequency and wavelength and
solve for wavelength.
You can convert the wavelength from
meters to nanometers by using the
conversion factor between the two (1
nm = 10–9 m).
6
The Nature of Energy
Wave nature of light does
not explain how an object
can glow when its
temperature increases
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7
The Nature of Energy—Quanta
• Max Planck explained
– Energy is quantized
• subdivide into small but measurable
increments
– Energy comes in packets called
quanta (singular: quantum)
– Energy can be gained or lost only
in whole number multiples of h
• h is Planck’s constant
• h = 6.626 x 10-34 J.s
• DE = h
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8
The Photoelectric Effect
• Observations about the
photoelectric effect
– Metal emits electrons upon
exposure to electromagnetic
radiation
– Each metal has a frequency at
which it ejects electrons
– At lower frequency, electrons are
not emitted
– At higher frequency
• Emittance of electrons increases with
light intensity
• Kinetic energy of emitted electrons
increases linearly with frequency of
light
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9
The Photoelectric Effect
• Albert Einstien
– Proposed electromagnetic
radiation is quantized
– View electromagnetic
radiation as particles called
photons
– Energy of each photon
• E = h
• E = hc/
– h is Planck’s constant, 6.626
 10−34 J∙s
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10
The Photoelectric Effect
• Albert Einstien
– Energy has dual nature of
light
• Has properties of wave and
particulate matter
• Known as wave-particle duality
– Energy has mass
• E =mc2
– Mass of photon
• m =h/c
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11
Example Photon Energy
A nitrogen gas laser pulse with a wavelength of 337 nm contains 3.83 mJ of
energy. How many photons does it contain?
SORT You are given wavelength
and total energy of a light pulse and
asked to find the number of photons it
contains.
GIVEN Epulse = 3.83 mJ
λ = 337 nm
FIND number of of photons
CONCEPTUAL PLAN
STRATEGIZE In the first part of
the conceptual plan, calculate the
energy of an individual photon from
its wavelength.
In the second part, divide the total
energy of the pulse by the energy of
each photon to determine the number
of photons in the pulse (because the
total energy of the pulse is equal to the
sum of the energies of each photon).
RELATIONSHIPS USED E = hc/λ
12
Example Photon Energy
(continued)
SOLVE To execute the first
part of the conceptual plan,
convert the wavelength to
meters and substitute it into
the equation to calculate the
energy of a 337-nm photon.
To execute the second part of
the conceptual plan, convert
the energy of the pulse from
mJ to J.
SOLUTION
Epulse =
Then divide the energy of the
pulse by the energy of a
photon to obtain the number
of photons.
13
The Wave Nature of Matter
• Louis de Broglie
– Theorized that matter
should exhibit wave
properties
– Demonstrated the
relationship between
mass and wavelength
•
The wave nature of light
is used to produce this
electron micrograph
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h
 = mv
14
Example De Broglie Wavelength
Calculate the wavelength of an electron traveling with a speed of 2.65 × 106 m/s.
SORT You are given the speed of
an electron and asked to calculate its
wavelength.
GIVEN ν = 2.65 × 106 m/s.
FIND λ
CONCEPTUAL PLAN
STRATEGIZE The conceptual
plan shows how the de Broglie
relation relates the wavelength of an
electron to its mass and velocity.
SOLVE Substitute the velocity,
Planck’s constant, and the mass of an
electron into de Broglie relation to
calculate the electron’s wavelength.
To correctly cancel the units, break
down the J in Planck’s constant into
its SI base units (1 J = 1 kg · m2/s2).
RELATIONSHIPS USED
λ = h/mv (de Broglie relation)
SOLUTION
15
Atomic Emissions
How do atoms and molecules emit energy?
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16
Continuous vs. Line Spectra
• Continuous spectrum
– Spectrum that contains
all wave-lengths in a
specified region of the
electromagnetic
spectrum
– Rainbow from a white
light source
• Line spectrum
– Only discrete
wavelengths is observed
– Each element has a
unique line spectrum
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17
The Hydrogen Spectrum
• There are four lines in the hydrogen spectrum
• What does this mean?
– Only certain energies are allowed for the electron
in the hydrogen atom
– Energy of the electron in the hydrogen atom is
quantized
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Concept Check
1. Why is it significant that the color emitted from the
hydrogen emission spectrum is not white?
–
Shows that it is not a continuous spectrum. If continuous we
would see a white light, which contains all colors.
2. How does the emission spectrum support the idea of
quantized energy levels?
–
Since only certain colors are observed, this means that only
certain energy levels are allowed. An electron can exist at one
level or another, and there are regions of zero probability in
between.
19
The Bohr Model
•
Niels Bohr adopted
Planck’s assumption
and explained these
phenomena in this
way:
1. Electrons in an atom
can only occupy
certain circular
orbitals
(corresponding to
certain energies)
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20
The Bohr Model
2. Electrons in
permitted orbits have
specific, “allowed”
energies
3. Energy is only absorbed
or emitted in such a
way as to move an
electron from one
“allowed” energy state
to another
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21
The Bohr Model
2. Electrons in
permitted orbits have
specific, “allowed”
energies
3. Energy is only absorbed
or emitted in such a
way as to move an
electron from one
“allowed” energy state
to another
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22
Bohr Model
• Limitations
– Only works for hydrogen
– Circular motion is not wave-like in nature
• Points that are incorporated into the current
atomic model include the following:
1) Electrons exist in discrete energy levels
2) Energy is involved in the transition of an electron
from one level to another
23
QUANTUM MECHANICAL MODEL
OF THE ATOM
24
The Uncertainty Principle
Heisenberg showed you
can not know both
position and momentum
of a particle
h
(Dx) (Dmv) 
4
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25
Quantum Mechanics
• Quantum mechanics
– Developed by Erwin Schrödinger
– Mathematical equation which
includes both the wave and
particle nature of matter
– Solution of Schrödinger’s wave
equation is designated with a
lowercase Greek psi ().
– 2, gives the electron density, or
probability of where an electron
is likely to be at any given time
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26
Quantum Numbers
• Quantum numbers
– describe the orbital
– Set of three
• Principal quantum number (n)
• Angular momentum quantum number (l)
• Magnetic quantum number (ml)
27
Principal Quantum Number (n)
• Principal quantum
number (n)
– Known as the electron shell
– Describes the energy level on
which the orbital resides
– Correspond to the values in the
Bohr model
– The values of n are integers ≥ 1
– As n increases the electron is
less tightly bound to the
nucleus
3rd shell
2nd shell
1st shell
Nucleus
28
Angular Momentum Quantum
Number (l)
• Angular momentum quantum number
– Called the subshell
– Defines the shape of the orbital
– Values of l are integers ranging from 0 to n − 1
– Use letter designations to communicate the
different values of l, therefore, the shapes and
types of orbitals
29
Magnetic Quantum Number (ml)
• 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
– Determines number of electron orbitals in each
subshell
• 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
30
Magnetic Quantum Number (ml)
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31
Example Quantum Numbers I
What are the quantum numbers and names (for example, 2s, 2p) of the
orbitals in the n = 4 principal level? How many n = 4 orbitals exist?
Solution
n = 4; therefore l = 0, 1, 2, and 3
You first determine the possible
values of l (from the given value of
n). You then determine the possible
values of ml for each possible value
of l. For a given value of n, the
possible values of l are 0, 1, 2, . . . ,
(n – 1).
For a given value of l, the possible
values of ml are the integer values
including zero ranging from –l to +l.
The name of an orbital is its principal
quantum number (n) followed by the
letter corresponding to the value l.
32
ORBITAL SHAPES AND ENERGIES
33
s Orbitals
• s orbitals
– Value of l is 0
– Spherical in shape
– Radius of the sphere increases with the value of n
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s Orbitals
• For an ns orbital
– Number of peaks is n
– Number of nodes is n – 1
• Node is where there is zero
probability of finding an
electron
• As n increases
– Electron density is more
spread out
– Greater probability of
finding an electron further
from the nucleus
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35
p Orbitals
• p orbitals
– Value of l for p orbitals is 1
– Composed of two lobes with a node between them
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36
d Orbitals
• d orbitals
– Value of l is 2
– Four of the five d
orbitals have four
lobes
– Fifth resembles a p
orbital with a
doughnut around
the center
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37
f Orbitals
• f orbitals
– Value of l is 3
– Seven
equivalent
orbitals in a
sublevel
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38
Energies of Orbitals—Hydrogen
• For hydrogen atom
– Orbitals on the same
energy level have the
same energy
• Called degenerate
orbitals
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39
Energies of Orbitals— Many-electron Atoms
• Atoms with more than
one electron
– Not all orbitals on the
same energy level are
degenerate
• As number of electrons
increase so does the
repulsion between them
• Energy levels start to
overlap in energy
– 4s is lower in energy
than 3d
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40
Energies of Orbitals
41
Spin Magnetic Quantum Number, ms
• Spin Magnetic Quantum
Number
– Fourth quantum number
– Two electrons in the
same orbital do not have
exactly the same energy
– Two allowed values, +½
and –½
– The “spin” of an electron
describes its magnetic
field
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42
Pauli Exclusion Principle
• Pauli Exclusion Principle
– No two electrons in the same atom can have exactly the
same energy
– No two electrons in the same atom can have identical sets
of quantum numbers
– Every electron in an atom must differ by at least one of the
four quantum number values: n, l, ml, and ms
43
STOP
44
Chapter 7 Section 1
PERIODIC TABLE
45
Development of the Periodic Table
Modern Periodic
Table
– Dmitri Mendeleev
– Lothar Meyer
– Independently
came to the same
conclusion about
how elements
should be grouped
46
Mendeleev and the Periodic Table
Chemists mostly credit Mendeleev
– Organize the table by chemical properties
– Predicted some missing elements and their expected
properties, i.e. germanium
47
Periodic Table
Elements are arranged in order of atomic number
48
Periodic Table
Representative
elements
• Elements in
the same
group have
similar
chemical
properties
Why?
49
ELECTRON CONFIGURATION
50
Subatomic Particles in an Atom
• In a neutral atom the
number of protons
equal the number of
electrons
• Aufbau principle
(protons and neutrons)
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– Electrons are added to
orbitals as protons are
added to an atom
51
Electron Configurations
5
4p
• Electron configuration explains how electrons
are distributed in an atom
• Each component consists of
– a number denoting the energy level
(principal quantum number, n)
– a letter denoting the type of orbital (angular
momentum quantum number, l)
– a superscript denoting the number of
electrons in those orbitals
• The most stable organization is the lowest
possible energy, called the ground state
52
How do you determine an atoms electron
configuration?
53
Electron Configuration
Use the periodic table
54
Electron Configuration
Subshell (angular momentum
quantum number)
s
p
d
f
Electron maximum
2
6
10
14
2
columns
6
columns
10
columns
14
columns
55
Electron Configuration
Subshell
s
p
d
f
Electron
maximum
2
6
10
14
s
p
d
f
56
Electron Configuration
The periodic table can be used to determine
subshell (angular momentum quantum number, l)
s
p
d
f
57
Electron Configuration
All elements located in a group have
distinguishing electron in the same subshell
s1
s2
s1 s2
p1 p2 p 3 p4 p5 p6
d1 d2 d3 d4 d5 d6 d7 d8 d9 d10
f1 f2 f3 f4 f5 f6 f7 f8 f9 f10 f11 f12 f13 f14
58
Electron Configuration
Shell number equals period number
1
2
3
4
5
6
7
1s
2s
3s
4s
5s
6s
7s
6
7
Shell number equals period
number minus one
3d
4d
5d
6d
1s
2p
3p
4p
5p
6p
7p
4f
5f
Shell number equals period number minus two
59
Electron Configurations
Hydrogen (Z =1):
1s1
Helium (Z =2):
1s2
Lithium (Z =3):
1s22s1
Beryllium (Z =4):
1s22s2
Boron (Z =5):
1s22s22p1
Carbon (Z =6):
1s22s22p2
Iron (Z =26):
1s22s22p6 3s23p64s23d6
Mercury (Z =80):
1s22s22p63s23p64s23d104p6
5s24d105p66s24f145d10
60
EXAMPLE ELECTRON CONFIGURATIONS
Write electron configurations for each element.
(a) Mg
(a) Magnesium has 12 electrons. Distribute two of
these into the 1s orbital, two into the 2s orbital,
six into the 2p orbitals, and two into the 3s
orbital.
SOLUTION
Mg
1s22s22p63s2
You can also write the electron configuration
more compactly using the noble gas core
notation. For magnesium, use [Ne] to represent
1s22s22p6.
61
Electron Configurations of Ions
• Cations: The
electrons are lost
from the highest
energy level (n
value)
 K+ is 4s1 (losing a
4s electron)
 Al3+ is 1s22s22p6 (losing two 3s and 1 3p electrons)
• Anions: The electron configurations are filled to
ns2np6
 P3– is 1s22s22p63s23p6 (gaining three electrons in 2p)
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62
EXAMPLE ELECTRON CONFIGURATIONS
Continued
SKILLBUILDER PLUS
Write electron configurations for each ion. (Hint: To determine the number of
electrons to include in the electron configuration of an ion, add or subtract electrons
as needed to account for the charge of the ion.)
(a) Al3+
(b) Cl–
Answers:
Subtract 1 electron for each unit of positive charge. Add 1 electron for each unit of
negative charge.
(a) Al3+
(b)
1s22s22p6
63
Electron Configuration
• The filled inner shell electrons are called core electrons
– Include completely filled d or f sublevels
• Electrons in the outer most shell are called valence
electrons
• Elements in the same group of the periodic table have
the same number of electrons in the outer most shell
64
EXAMPLE VALENCE ELECTRONS AND CORE ELECTRONS
Write an electron configuration for selenium and identify the valence electrons and
the core electrons.
SOLUTION
Write the electron configuration for selenium by determining the total number of
electrons from selenium’s atomic number (34) and distributing them into the appropriate
orbitals.
Se
1s22s22p63s23p64s23d104p4
The valence electrons are those in the outermost principal shell. For selenium, the
outermost principal shell is the n = 4 shell, which contains 6 electrons (2 in the 4s
orbital and 4 in the three 4p orbitals). All other electrons, including those in the 3d
orbitals, are core electrons.
65
EXAMPLE VALENCE ELECTRONS AND CORE ELECTRONS
Continued
SKILLBUILDER | Valence Electrons and Core Electrons
Write an electron configuration for chlorine and identify the valence electrons
and core electrons.
Answer:
66
Conceptual Checkpoint
Which element has the fewest valence electrons?
(a) B
(b) Ca
(c) O
(d) K
(e) Ga
Answer:
67
Conceptual Checkpoint
Below is the electron configuration of calcium:
Ca
1s22s22p63s23p64s2
In its reactions, calcium tends to form the Ca2+ ion. Which electrons are lost
upon ionization?
(a) the 4s electrons
(b) two of the 3p electrons
(c) the 3s electrons
(d) the 1s electrons
Answer:
68
Condensed Electron Configurations
• Condensed electron configuration
– Shorthand electron configuration which uses
brackets around a noble gas symbol and listing
only valence electrons
Atom
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Electron Configuration
Condensed Configuration
Helium (Z =2)
1s2
Lithium (Z =3)
1s22s1
[He]2s1
Beryllium (Z =4)
1s22s2
[He]2s2
Boron (Z =5)
1s22s22p1
[He]2s22p1
Carbon (Z =6)
1s22s22p2
[He]2s22p2
Oxygen (Z =8)
1s22s22p4
[He]2s22p4
69
Condensed Electron Configurations
Groups possess the same valence electron
configuration
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70
EXAMPLE WRITING ELECTRON CONFIGURATIONS FROM THE PERIODIC
TABLE
Write a condensed electron configuration for arsenic based on its position in the
periodic table.
SOLUTION
The noble gas that precedes arsenic in
the periodic table is argon, so the inner
electron configuration is [Ar]. Obtain
the outer electron configuration by
tracing the elements between Ar and
As and assigning electrons to the
appropriate orbitals. Remember that
the highest n value is given by the row
number (4 for arsenic). So, begin with
[Ar], then add in the two 4s electrons
as you trace across the s block,
followed by ten 3d electrons as you trace across the d block
(the n value for d subshells is equal to the row number minus one), and finally the
three 4p electrons as you trace across the p block to As, which is in the third column
of the p block:
71
EXAMPLE WRITING ELECTRON CONFIGURATIONS FROM THE PERIODIC
TABLE
Continued
The electron configuration is:
As
[Ar]4s23d104p3
SKILLBUILDER | Writing Electron Configurations from the
Periodic Table
Use the periodic table to determine the condensed electron configuration for
tin.
Answer:
72
• Electron configuration specifies subshell
occupancy for electrons (angular momentum
quantum number, l)
What about the magnetic quantum number, ml ?
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73
ELECTRON ORBITAL DIAGRAMS
74
Electron Orbital Diagrams
• Magnetic quantum number, ml
– Determines number of orbitals in
each subshell
• Electron orbital diagram
– Shows number of electrons in
occupied electron orbitals
– Each box in the diagram represents
one orbital
– Half-arrows represent the electrons
– Direction of the arrow represents
the relative spin of the electron
(electron spin quantum number, ms
– Each orbital only holds 2 electrons
75
Electron Orbital Diagrams: Hund’s Rule
• Hund’s rule
– Every orbital in a
subshell is singly
occupied before any
orbital is doubly
occupied
– All of the electrons in
singly occupied orbitals
have the same spin
– Lowest energy attained
when the number of
electrons with the
same spin is maximized
 For a set of orbitals in the same sublevel, there must be one
electron in each orbital before pairing and the electrons
have the same spin
76
EXAMPLE WRITING ORBITAL DIAGRAMS
Write an orbital diagram for silicon.
SOLUTION
Since silicon is atomic number 14, it has 14 electrons. Draw a box for each
orbital, putting the lowest-energy orbital (1s) on the far left and proceeding to
orbitals of higher energy to the right.
Distribute the 14 electrons into the orbitals, allowing a maximum of 2 electrons
per orbital and remembering Hund’s rule. The complete orbital diagram is:
77
Electron Orbital Diagrams
• Electron configuration specifies subshell
occupancy for electrons (angular momentum
quantum number, l)
• Electron orbital diagram specifies orbital
occupancy for electrons
78
Electron Configuration Anomalies
 Some irregularities
occur when there
are enough
electrons to half-fill
s and d orbitals on a
given row
79
Chromium as an Anomaly
• Electron configuration for chromium is
[Ar] 4s1 3d5
rather than the expected
[Ar] 4s2 3d4
• This occurs because the 4s and 3d orbitals
are very close in energy
• These anomalies occur in f-block atoms
with f and d orbitals, as well
80
The Periodic Table – Final Thoughts
1. Valence electrons that primarily determine an
atom’s chemistry
2. Electron configurations can be determined from the
periodic table
3. Certain groups in the periodic table have special
names
4. Basic division of the elements in the periodic table is
into metals and nonmetals
81
82
What is the shape of a p orbital?
A.
B.
C.
D.
E.
Spherical
Doughnut
Pyramidal
Dumbbell
Square
83
How many orbitals are associated with a given set of
d orbitals with the same principal quantum number?
A.
B.
C.
D.
E.
1
7
3
9
5
84
What is the maximum number of electrons allowed in each
f orbital?
A.
B.
C.
D.
E.
2
6
8
10
14
85
What is the maximum number of electrons in the n = 4
principal energy level?
A.
B.
C.
D.
E.
6
8
18
32
64
86
Which of the following electron configurations is NOT
possible?
A.
B.
C.
D.
E.
1s22s22p3
[Ar]4s24d104p4
[Ne]3s2
[He]2s22p63s2
[Kr]5s24d105p2
87
What is the electron configuration for the copper atom?
A.
B.
C.
D.
E.
1s22s22p63s23p64s23d7
[Ar]4s24d9
[Ar]4s13d10
1s22s22p2
[Ar]4s23d9
88
How many unpaired electrons are in an iron atom?
A.
B.
C.
D.
E.
6
20
4
1
0
89
Which of the following atoms has the greatest metallic
character?
A.
B.
C.
D.
E.
Chlorine
Tellurium (Te)
Indium (In)
Carbon
Aluminum
90
The electron configuration for calcium is 1s22s22p63s2.
Which electrons are lost upon ionization?
A.
B.
C.
D.
E.
The 3s electrons
The 2s electrons
Two of the 2p electrons
The 1s electrons
None of the above
91
The lowest energy state of a
hydrogen atom is called its
_______ state.
a.
b.
c.
d.
bottom
ground
fundamental
original
92
s orbitals are shaped like
a.
b.
c.
d.
four-leaf clovers.
dumbbells.
spheres.
triangles.
93
At a node, the probability of
finding an electron is ___ %.
a.
b.
c.
d.
0
1
50
100
94
What is the maximum number
of electrons that can occupy
the 5d subshell
a.
b.
c.
d.
2
6
10
12
95
Conceptual Checkpoint
Which pair of elements has the same total number of electrons in p orbitals?
(a) Na and K
(b) K and Kr
(c) P and N
(d) Ar and Ca
Answer:
96
Which of the following elements has the following electron
configuration 1s22s22p63s23p5?
A.
B.
C.
D.
E.
O
Cl
F
Ar
S
97
What is the electron configuration of the ground state of the
P3– ion?
A.
B.
C.
D.
E.
[Ar]4s23d104p1
[Ne]3s23p6
[He]2s22p63s2
[Ne]3s23p3
[Ne]3s2
98
What is the electron configuration for sulfur?
A.
B.
C.
D.
E.
[Ar]4s24p5
[Ne]3s23p4
1s22s22p63s23p4
1s22s22p63s23p5
Two of the above
99
Which of the following ions has the same electron
configuration as an argon atom?
A.
Br–
B.
C.
D.
E.
K+
S3–
P3+
Ca+
100
How many core electrons are in an arsenic (element 33)
atom?
A.
B.
C.
D.
E.
33
28
18
5
8
101
Which element forms compounds with formulas similar to
compounds containing iodine?
A.
B.
C.
D.
E.
Ar
S
Na
P
F
102
The electron configuration for bromine is [Ar]4s23d104p5.
List the number of core electrons and valence electrons,
respectively.
A.
B.
C.
D.
E.
18 and 17
20 and 15
10 and 25
28 and 7
30 and 5
103
The electron configuration of a
carbon atom is
a.
b.
c.
d.
[He]2s22p6.
[He]2s22p4.
[He]2s22p2.
[He]2s2.
104
The electron configuration of a
germanium atom is
a.
b.
c.
d.
[Ar]4s24p2.
[Ar]4s23d104p2.
[Kr]4s23d104p2.
[Kr]4s23d104p2.
105
The electron configuration of a
copper atom is
a.
b.
c.
d.
[Ar]4s23d9.
[Ar]4s13d10.
[Ar]4s23d10.
[Ar]4s23d7.
106
The valence electron
configuration of elements in
column 6A(16) of the Periodic
Table is
a.
b.
c.
d.
np6.
ns0np6.
ns2np4.
impossible to predict because
each element is unique.
107
Identify the specific element
that corresponds to
1s22s22p63s2
a.
b.
c.
d.
Sodium
Beryllium
Magnesium
Calcium
108
Identify the specific element
that corresponds to
[Kr]5s24d105p4
a.
b.
c.
d.
tin
antimony
tellurium
Iodine
109
The elements located in Group
VIIA (Group 17) on the periodic
table are called
a.
b.
c.
d.
alkali metals.
noble gases.
chalcogens.
halogens.
110
Identify the group of elements that
corresponds to the generalized
electron configuration
[noble gas] ns2np5
a.
b.
c.
d.
Group 5A
Group 6A
Group 7A
Group 8A
111
Identify the group of elements that
corresponds to the generalized
electron configuration
[noble gas] ns2(n-1)d2
a.
b.
c.
d.
Group 3B
Group 4B
Group 5B
Group 6B
112