Chapter 8 - Silberberg

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Chapter 8 Electron Configuration and Chemical
Periodicity
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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 Some Key Periodic Atomic Properties
8.5 The Connection Between Atomic Structure and
Chemical Reactivity
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Mendeleevs Periodic Law
Arranged the 65 known elements by atomic mass and
by recurrence of various physical and chemical
properties. The Periodic Table today is very
similar but arranged according to atomic number
(number of protons). The arrangement led to
families of elements with similar properties and
at the time allowed for the prediction and
properties of elements yet to be discovered.
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Predicted Properties of eka Silicon(E)
Actual Properties of Germanium (Ge)
Property
atomic mass
72amu
72.61amu
appearance
gray metal
gray metal
density
5.5g/cm3
5.32g/cm3
molar volume
13cm3/mol
13.65cm3/mol
specific heat capacity
0.31J/gK
0.32J/gK
oxide formula
EO2
GeO2
oxide density
4.7g/cm3
4.23g/cm3
sulfide formula and solubility
ES2 insoluble in H2O soluble in aqueous (NH4)2S
GeS2 insoluble in H2O soluble in aqueous (NH4)2S
ECl4 (lt1000C)
chloride formula (boiling point)
GeCl4 (840C)
chloride density
1.9g/cm3
1.844g/cm3
element preparation
reduction of K2EF6 with sodium
reduction of K2GeF6 with sodium
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Remember from Chapter 7 - Quantum Numbers and
Atomic Orbitals
An atomic orbital is specified by three quantum
numbers.
n the principal quantum number - a positive
integer
l the angular momentum quantum number - an
integer from 0 to n-1
ml the magnetic moment quantum number - an
integer from -l to l
The three quantum numbers are actually giving the
energy of the electron in the orbital and a
fourth q.n. is needed to describe a property of
electrons called spin. The spin can be clockwise
or counterclockwise. The spin q.n., ms can be
½ or - ½ .
The Pauli Exclusion Principle - No two electrons
in the same atom can have the same four q.n.
Since the first three q.n. define the orbital,
this means only two electrons can be in the same
orbital and they must have opposite spins.
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Table 8.2 Summary of Quantum Numbers of
Electrons in Atoms
Name
Symbol
Permitted Values
Property
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Factors Affecting Atomic Orbital Energies
Higher nuclear charge lowers orbital energy
(stabilizes the system) by increasing
nucleus-electron attractions.
Additional electron in the same orbital
(makes less stable)
An additional electron raises the orbital energy
through electron-electron repulsions.
Additional electrons in inner orbitals (makes
outer orbital less stable)
Inner electrons shield outer electrons more
effectively than do electrons in the same
sublevel.
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The effect of another electron in the same orbital
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The effect of other electrons in inner orbitals
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The effect of orbital shape
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Order for filling energy sublevels with electrons
Illustrating Orbital Occupancies
The electron configuration

of electrons in the sublevel
n
l
The orbital diagram
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empty
A vertical orbital diagram for the Li ground state
half-filled
filled, spin-paired
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SAMPLE PROBLEM 8.1
Determining Quantum Numbers from Orbital Diagrams
PLAN
Use the orbital diagram to find the third and
eighth electrons.
SOLUTION
The third electron is in the 2s orbital. Its
quantum numbers are
1/2
2
0
0
The eighth electron is in a 2p orbital. Its
quantum numbers are
2
1
-1
-1/2
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Orbital occupancy for the first 10 elements, H
through Ne.
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Hunds rule
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Condensed ground-state electron configurations in
the first three periods.
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A periodic table of partial ground-state electron
configurations
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The relation between orbital filling and the
periodic table
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General pattern for filling the sublevels
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SAMPLE PROBLEM 8.2
Determining Electron Configuration
(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
Ar 4s1
There are 18 inner electrons.
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SAMPLE PROBLEM 8.2
continued
(b) for Mo (Z 42)
1s22s22p63s23p64s23d104p65s14d5
Kr 5s14d5
There are 36 inner electrons and 6 valence
electrons.
(c) for Pb (Z 82)
1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p2
Xe 6s24f145d106p2
There are 78 inner electrons and 4 valence
electrons.
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Defining metallic and covalent radii
Knowing the Cl radius and the C-Cl bond length,
the C radius can be determined.
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Trends in the Periodic Table
Atomic radii of the main-group and transition
elements.
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Trends in the Periodic Table
Periodicity of atomic radius
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SAMPLE PROBLEM 8.3
Ranking Elements by Atomic Size
(a) Ca, Mg, Sr
(b) K, Ga, Ca
(c) Br, Rb, Kr
(d) Sr, Ca, Rb
PLAN
Elements in the same group decrease in size as
you go up elements decrease in size as you go
across a period.
SOLUTION
(a) Sr gt Ca gt Mg
These elements are in Group 2A(2).
(b) K gt Ca gt Ga
These elements are in Period 4.
(c) Rb gt Br gt Kr
Rb has a higher energy level and is far to the
left. Br is to the left of Kr.
(d) Rb gt Sr gt Ca
Ca is one energy level smaller than Rb and Sr.
Rb is to the left of Sr.
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Trends in the Periodic Table
Periodicity of first ionization energy (IE1)
Energy required to remove one outermost electron.
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First ionization energies of the main-group
elements
Trends in the Periodic Table
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SAMPLE PROBLEM 8.4
Ranking Elements by First Ionization Energy
(a) Kr, He, Ar
(b) Sb, Te, Sn
(c) K, Ca, Rb
(d) I, Xe, Cs
PLAN
IE increases as you proceed up in a group IE
increases as you go across a period.
SOLUTION
(a) He gt Ar gt Kr
Group 8A(18) - IE decreases down a group.
(b) Te gt Sb gt Sn
Period 5 elements - IE increases across a period.
(c) Ca gt K gt Rb
Ca is to the right of K Rb is below K.
(d) Xe gt I gt Cs
I is to the left of Xe Cs is further to the
left and down one period.
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Trends in the Periodic Table
The first three ionization energies of beryllium
(in MJ/mol)
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SAMPLE PROBLEM 8.5
Identifying an Element from Successive
Ionization Energies
PLAN
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.
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Electron affinities of the main-group elements
Trends in the Periodic Table
Electron Affinity Energy change to add one
electron. In most cases, EA negative (energy
released becauseelectron attracted to nucleus
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Trends in three atomic properties
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Trends in metallic behavior
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Main-group ions and the noble gas configurations
Trends in the Periodic Table
Properties of Monatomic Ions
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SAMPLE PROBLEM 8.6
Writing Electron Configurations of Main-Group Ions
(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
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Magnetic Properties of Transition Metal Ions
A species with unpaired electrons exhibits
paramagnetism. It is attracted by an external
magnetic field. Species with all paired es, not
attracted........diamagnetic
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SAMPLE PROBLEM 8.7
Writing Electron Configurations and Predicting
Magnetic Behavior of Transition Metal Ions
(a) Mn2(Z 25)
(b) Cr3(Z 24)
(c) Hg2(Z 80)
PLAN
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
paramagnetic
paramagnetic
not paramagnetic (is diamagnetic)
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Ionic vs. atomic radius
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SAMPLE PROBLEM 8.8
Ranking Ions by Size
(a) Ca2, Sr2, Mg2
(b) K, S2-, Cl -
(c) Au, Au3
PLAN
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 gt Ca2 gt Mg2
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.
(b) S2- gt Cl - gt K
(c) Au gt Au3
The higher the charge, the smaller the ion.
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End of Chapter 8