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Chapter 8
Electron Configuration and
Chemical Periodicity
8-1
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Electron Configuration and Chemical Periodicity
8.1 Development of the Periodic Table
8.2 Characteristics of Many-Electron Atoms
8.3 The Quantum-Mechanical Model and the Periodic Table
8.4 Trends in Three Key Atomic Properties
8.5 Atomic Structure and Chemical Reactivity
8-2
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Figure 8.1
Observing the effect of electron spin.
The Stern-Gerlach experiment.
8-3
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Table 8.1 Summary of Quantum Numbers of Electrons in Atoms
Name
8-4
Symbol
Permitted Values
Property
principal
n
positive integers(1,2,3,…) orbital energy (size)
angular
momentum
l
integers from 0 to n-1
orbital shape (The l values
0, 1, 2, and 3 correspond to
s, p, d, and f orbitals,
respectively.)
magnetic
ml
integers from -l to 0 to +l
orbital orientation
spin
ms
+1/2 or -1/2
direction of e- spin
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Factors Affecting Atomic Orbital Energies
The Effect of Nuclear Charge (Zeffective)
Higher nuclear charge lowers orbital energy (stabilizes the
system) by increasing nucleus-electron attractions.
The Effect of Electron Repulsions (Shielding)
Additional electron in the same orbital
An additional electron raises the orbital energy through
electron-electron repulsions.
Additional electrons in inner orbitals
Inner electrons shield outer electrons more effectively than
do electrons in the same sublevel.
8-5
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Figure 8.2 The effect of orbital shape.
8-6
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Figure 8.3
Order for filling energy sublevels with
electrons.
Illustrating Orbital Occupancies
The electron configuration
#
of electrons in the sublevel
n l
as s,p,d,f
The orbital diagram (box or circle)
8-7
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TA pg. 241
A vertical orbital diagram for the Li ground state.
no color-empty
light - half-filled
dark - filled, spin-paired
8-8
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SAMPLE PROBLEM 8.1
PROBLEM:
Determining Quantum Numbers from Orbital
Diagrams
Write a set of quantum numbers for the third electron and a set
for the eighth electron of the F atom.
PLAN: Use the orbital diagram to find the third and eighth electrons.
9F
1s
2s
2p
SOLUTION: The third electron is in the 2s orbital. Its quantum numbers are
n= 2
l= 0
ml = 0
ms= + or -1/2
The eighth electron is in a 2p orbital. Its quantum numbers are
n= 2
8-9
l= 1
ml = -1, 0, or +1
ms= + or -1/2
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Table 8.2
8-10
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Figure 8.4
8-11
Condensed ground-state electron configurations in
the first three periods.
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Table 8.3
8-12
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Replace w/ Table 8.3 1e
8-13
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Figure 8.5
A periodic table of partial
ground-state electron
configurations.
8-14
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Figure 8.6
8-15
The relation between orbital filling and the
periodic table.
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SAMPLE PROBLEM 8.2
Determining Electron Configuration
PROBLEM:
Using the periodic table on the inside cover of the text (not Figure
8.12 or Table 8.4), give the full and condensed electrons
configurations, partial orbital diagrams showing valence electrons,
and number of inner electrons for the following elements:
(a) potassium (K: Z = 19) (b) molybdenum (Mo: Z = 42) (c) lead (Pb: Z = 82)
PLAN:
Use the atomic number for the number of electrons and the periodic
table for the order of filling for electron orbitals. Condensed
configurations consist of the preceding noble gas and outer electrons.
SOLUTION:
(a) for K (Z = 19)
1s22s22p63s23p64s1
full configuration
condensed configuration [Ar] 4s1
partial orbital diagram
4s1
8-16
There are 18 inner electrons.
3d
4p
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SAMPLE PROBLEM 8.2
continued
(b) for Mo (Z = 42)
full configuration 1s22s22p63s23p64s23d104p65s14d5
condensed configuration [Kr] 5s14d5
partial orbital diagram
There are 36 inner electrons
and 6 valence electrons.
5s1
4d5
5p
(c) for Pb (Z = 82)
full configuration
1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p2
condensed configuration
[Xe] 6s24f145d106p2
partial orbital diagram
There are 78 inner electrons
and 4 valence electrons.
8-17
6s2
6p2
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Figure 8.8
8-18
Defining metallic and covalent radii.
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Figure 8.9
Atomic radii of the maingroup and transition
elements.
8-19
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Figure 8.10
8-20
Periodicity of atomic radius.
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SAMPLE PROBLEM 8.3
PROBLEM:
Using only the periodic table (not Figure 8.15)m rank each set of
main group elements in order of decreasing atomic size:
(a) Ca, Mg, Sr
PLAN:
Ranking Elements by Atomic Size
(b) K, Ga, Ca
(c) Br, Rb, Kr
(d) Sr, Ca, Rb
Elements in the same group increase in size and you go down;
elements decrease in size as you go across a period.
SOLUTION:
(a) Sr > Ca > Mg
These elements are in Group 2A(2).
(b) K > Ca > Ga
These elements are in Period 4.
(c) Rb > Br > Kr
Rb has a higher energy level and is far to the left.
Br is to the left of Kr.
(d) Rb > Sr > Ca
Ca is one energy level smaller than Rb and Sr.
Rb is to the left of Sr.
8-21
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Figure 8.11
8-22
Periodicity of first ionization energy (IE1).
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Figure 8.12
8-23
First ionization energies of the main-group elements.
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Figure 8.13
The first three ionization energies of beryllium (in
MJ/mol).
For more data on sequential
ionization energies of the elements,
go to http://www.webelements.com or
click on the button below.
8-24
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SAMPLE PROBLEM 8.4
PROBLEM:
Using the periodic table only, rank the elements in each of the
following sets in order of decreasing IE1:
(a) Kr, He, Ar
PLAN:
Ranking Elements by First Ionization Energy
(b) Sb, Te, Sn
(c) K, Ca, Rb
(d) I, Xe, Cs
IE decreases as you proceed down in a group; IE increases as
you go across a period.
SOLUTION:
(a) He > Ar > Kr
Group 8A(18) - IE decreases down a group.
(b) Te > Sb > Sn
Period 5 elements - IE increases across a period.
(c) Ca > K > Rb
Ca is to the right of K; Rb is below K.
(d) Xe > I > Cs
I is to the left of Xe; Cs is furtther to the left and
down one period.
8-25
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Table 8.4
8-26
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SAMPLE PROBLEM 8.5
PROBLEM:
PLAN:
Identifying an Element from Successive
Ionization Energies
Name the Period 3 element with the following ionization energies
(in kJ/mol) and write its electron configuration:
IE1
IE2
IE3
IE4
IE5
1012
1903
2910
4956
6278
IE6
22,230
Look for a large increase in energy which indicates that all of the
valence electrons have been removed.
SOLUTION:
The largest increase occurs after IE5, that is, after the 5th valence
electron has been removed. Five electrons would mean that the
valence configuration is 3s23p3 and the element must be
phosphorous, P (Z = 15).
The complete electron configuration is 1s22s22p63s23p3.
8-27
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Figure 8.14
8-28
Electron affinities of the main-group elements.
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Figure 8.15
Trends in three atomic properties.
8-29
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Figure 8.16
8-30
Trends in metallic behavior.
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Figure 8.17
8-31
The trend in acid-base
behavior of element
oxides.
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Figure 8.18
8-32
Main-group ions and the noble gas configurations.
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SAMPLE PROBLEM 8.6 Writing Electron Configurations of Main-Group Ions
PROBLEM:
Using condensed electron configurations, write reactions for the
formation of the common ions of the following elements:
(a) Iodine (Z = 53)
(b) Potassium (Z = 19)
(c) Indium (Z = 49)
PLAN: Ions of elements in Groups 1A(1), 2A(2), 6A(16), and 7A(17) are usually
isoelectronic with the nearest noble gas.
Metals in Groups 3A(13) to 5A(15) can lose their np or ns and np
electrons.
SOLUTION:
(a) Iodine (Z = 53) is in Group 7A(17) and will gain one electron to be isoelectronic
with Xe: I ([Kr]5s24d105p5) + eI- ([Kr]5s24d105p6)
(b) Potassium (Z = 19) is in Group 1A(1) and will lose one electron to be isoelectronic
with Ar: K ([Ar]4s1)
K+ ([Ar]) + e(c) Indium (Z = 49) is in Group 3A(13) and can lose either one electron or three
electrons: In ([Kr]5s24d105p1)
In+ ([Kr]5s24d10) + e+
In ([Kr]5s24d105p1)
In3+([Kr] 4d10) + 3e-
8-33
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Figure 8.19
8-34
The Period 4 crossover in sublevel energies.
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Figure 8.20 Apparatus for measuring the magnetic behavior of a sample.
8-35
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SAMPLE PROBLEM 8.7
PROBLEM:
Use condensed electron configurations to write the reaction for the
formation of each transition metal ion, and predict whether the ion is
paramagnetic.
(a) Mn2+(Z = 25)
PLAN:
Writing Electron Configurations and Predicting
Magnetic Behavior of Transition Metal Ions
(b) Cr3+(Z = 24)
(c) Hg2+(Z = 80)
Write the electron configuration and remove electrons starting with
ns to match the charge on the ion. If the remaining configuration
has unpaired electrons, it is paramagnetic.
SOLUTION:
(a) Mn2+(Z = 25) Mn([Ar]4s23d5)
(b) Cr3+(Z = 24) Cr([Ar]4s23d6)
(c) Hg2+(Z = 80) Hg([Xe]6s24f145d10)
Mn2+ ([Ar] 3d5) + 2eCr3+ ([Ar] 3d5) + 3e-
paramagnetic
paramagnetic
Hg2+ ([Xe] 4f145d10) + 2enot paramagnetic (is diamagnetic)
8-36
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Figure 8.21
8-37
Depicting ionic radius.
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Figure 8.22
8-38
Ionic vs. atomic radii.
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SAMPLE PROBLEM 8.8
PROBLEM:
Ranking Ions by Size
Rank each set of ions in order of decreasing size, and explain your
ranking:
(a) Ca2+, Sr2+, Mg2+
PLAN:
(b) K+, S2-, Cl -
(c) Au+, Au3+
Compare positions in the periodic table, formation of positive and
negative ions and changes in size due to gain or loss of electrons.
SOLUTION:
(a) Sr2+ > Ca2+ > Mg2+
(b) S2- > Cl - > K+
(c) Au+ > Au3+
8-39
These are members of the same Group (2A/2) and
therefore decrease in size going up the group.
The ions are isoelectronic; S2- has the smallest Zeff and
therefore is the largest while K+ is a cation with a large Zeff
and is the smallest.
The higher the + charge, the smaller the ion.