CHEMISTRY
The Molecular Nature of Matter and Change 3rd Edition
CHAPTER 8 LECTURE NOTES
Electron Configuration and
Chemical Periodicity
Chem 150 - Ken Marr - Winter 2003
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
8-3
8.2 Characteristics of Many-Electron Atoms
Table 8.2 Summary of Quantum Numbers of Electrons in Atoms
Name
principal
angular
momentum
magnetic
spin
8-6
Symbol
Permitted Values
Property
n
positive
integers(1,2,3,…)
orbital energy (size)
l
integers from 0 to n-1
ml
ms
integers from -l to 0 to +l
+1/2 or -1/2
orbital shape (The l values
0, 1, 2, and 3 correspond to
s, p, d, and f orbitals,
respectively.)
orbital orientation
direction of e- spin
Figure 8.1
The 4th quantum number, ms
1.
2.
3.
Electrons can Spin in two directions (like a top)
ms = spin quantum number
» can only have two values
ms = -1/2 or +1/2
Pauli Exclusion Principle
» No 2 electrons can have the same 4 quantum
numbers
» i.e. Only two electrons per orbital
Application of the Pauli Exclusion Principle
1.
How many electrons can occupy a....
a.
b.
c.
d.
e.
2s subshell?
3d subshell?
shell with n = 3?
shell with n = 4?
subshell with n = 3 and l = 1?
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)
1. Additional electron in the same orbital
2. An additional electron raises the orbital energy through
electron-electron repulsions.
3. Additional electrons in inner orbitals
4. Inner electrons shield outer electrons more effectively
than do electrons in the same sublevel
8-8
Figure 8.4
The effect
of another
electron in
the same
orbital
8-9
Figure 8.5
The effect of
other
electrons in
inner orbitals
8-10
Figure 8.6
8-11
The effect of orbital shape
Figure 8.7
Order for filling energy sublevels with electrons
Illustrating Orbital Occupancies
The electron configuration
#
nl
of electrons in the
sublevel
as s,p,d,f
The orbital diagram (box or circle)
8-12
Figure 8.8
A vertical
orbital diagram
for the Li
ground state
Empty
orbitals of the
2p sublevel
Half-filled
orbital of the 2s
sublevel
Filled 1s orbital with two
spin-paired electrons
8-13
SAMPLE PROBLEM 8.1 Determining Quantum Numbers from Orbital Diagrams
PROBLEM
PLAN
Write a set of quantum numbers for the third electron and a
set for the eighth electron of the F atom.
Use the orbital diagram to find the third and eighth electrons.
9F
1s
SOLUTION
2s
2p
The third electron is in the 2s orbital. Its quantum numbers are
n= 2
l= 0
ml = 0
ms= +1/2
The eighth electron is in a 2p orbital. Its quantum numbers are
n= 2
8-14
l= 1
ml = -1
ms= -1/2
Electronic Structure of
Multi-electron Atoms
1.
Aufbau Principle
»
»
2.
Electrons 1st fill orbitals of lowest energy
Orbitals of a sublevel half fill before electrons start
to pair
Orbital Notation vs Electronic Configuration
»
Examples:
–
3.
Noble gases, Halogens, Alkali Metals, etc.
Abbreviated electronic Configurations
Memory Aid for
the order of
sublevel filling
Electronic Configurations and
the Periodic Table
1.
2.
3.
Use the periodic table to remember the order
of orbital filling
Sublevels found in each…
a. Group
b. Period
Write Abbreviated electron configurations for
a. Mg
b. Fe
Figure 8.13
The relation between orbital filling and the periodic table
8-22
Figure 8.9
8-15
Orbital occupancy for the first 10 elements
Valence Shell Configurations
1.
2.
Valence Shell = Outermost shell (E-level)
Write the valence shell configurations
(abbreviated format) for
a. Alkali metals
b. Halogens
c. Group VA
d. 26Fe, 39Y
Unexpected Electron
Configurations
1.
2.
3.
Fully filled and half filled sublevels offer
stability
s and d sublevels of the transition metals are
very close in energy
Predict which transition elements in the 4th
period have unexpected electronic
configurations.
»
Write them in abbreviated format
Periodic Trends in Atomic Radius
1.
2.
3.
Effective Nuclear Charge, Zeff
Zeff = (Atomic number) – (Number of Core Electrons)
Valence electrons do not feel the full effect of the
nucleus because of screening by core electrons
e.g. Group 1A: H  Fr
Explain what happens to atomic radius...
»
»
»
Down a group? Which is more important, n or Eeff?
Across a period? Which is more important, n or Eeff?
What about the transition metals?
C – Cl Bond
SAMPLE PROBLEM 8.3 Ranking Elements by Atomic Size
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:
(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.
8-28
Size of an Atom vs. its Ion
1.
How does the size of an atom compare
to that of its Cation?
e.g. K
2.
vs
K+
How does the size of an atom compare
to that of its Anion?
e.g. F vs
F-
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
->
(c) Au+ > Au3+
8-46
K+
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.
Ionization Energy
1.
Ionization Energy
» Definition
» Endothermic......Why?
2.
Equations for
» 1st Ionization Energy
» 2nd Ionization Energy
» 3rd Ionizationenergy
SAMPLE PROBLEM 8.4
Ranking Elements by 1st Ionization Energy
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:
(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:
(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
8-32
IE decreases down a group.
I is to the left of Xe; Cs is furtther to the left and down
one period.
Periodic Trends in Ionization Energy
1.
2.
3.
Across a Period....Explain why!
Down a Group......Explain why!
Compare the relative 1st, 2nd, 3rd, etc.
Ionization Energies for....
»
»
elements in period 2
elements in period 3
SAMPLE PROBLEM 8.5
Identifying an Element from Successive Ionization Energies
PROBLEM: Name the Period 3 element with the following ionization
energies (in kJ/mol) and write its electron configuration:
PLAN:
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.
8-34
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): Group 7A elements gain one electron to gain noble gas
structure: I = [Kr]5s24d105p5 + e-  I- = [Kr]5s24d105p6 (Isoelectronic with Xe)
(b) Potassium (Z = 19): Group 1A elements lose one electron to gain noble gas
structure:
K = [Ar]4s1  K+ = [Ar] + e- (K+ is isoelectronic with Ar)
(c) Indium (Z = 49): Group 3A can lose either one electron or three electrons to gain
stability In = [Kr]5s24d105p1 
In+ = [Kr]5s24d10 + e+
In =[Kr]5s24d105p1)  In3+= [Kr] 4d10 + 3e-
8-40
Magnetic Properties of Atoms
1.
2.
3.
Paramagnetism
» Due to the spinning of unpaired electrons
» Atoms are attracted to a magnet
Diamagetism
» Atoms have all electrons paired
» Atoms not attracted to a magnet
Electric Motors
» Generate a magnetic field by moving electrons in
curved paths within a coil
SAMPLE PROBLEM 8.7
Writing Electron Configurations and Predicting Magnetic
Behavior of Transition Metal Ions
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:
(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)
(b) Cr3+(Z = 24)
Mn = [Ar]4s23d5
Cr = [Ar]4s23d6
(c) Hg2+(Z = 80) Hg = [Xe]6s24f145d10
8-43
Mn2+ = [Ar] 3d5
Cr3+ = [Ar] 3d5
Hg2+ =[Xe] 4f145d10
paramagnetic
paramagnetic
diamagnetic
Electron Affinity
1.
Definition
»
»
2.
3.
Exothermic.......why?
The most nonmetallic elements have the most
negative electron affinities
Equation
How does electron affinity vary ....
»
»
Across a Period? Why?
Down a Group? Why?
Figure 8.21
Trends in three atomic properties
Increases
Increases
8-36
Ionization Energy