1. Electron Configuration and
Chapter 8
Electron Configuration and
Chemical Periodicity

2. 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

3. Mendeleev’s 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.

4. Mendeleev’s Predicted Properties of Germanium (“eka Silicon”) and Its Actual Properties
Table 8.1
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/g*K
0.32J/g*K
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
chloride formula (boiling point)
ECl4; (<1000C)
GeCl4; (840C)
chloride density
1.9g/cm3
1.844g/cm3
element preparation
reduction of K2EF6 with sodium
reduction of K2GeF6 with sodium

5. 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.

6. Table 8.2 Summary of Quantum Numbers of Electrons in Atoms
Name
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

7. 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 (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.

8. The effect of another electron in the same orbital

9. The effect of other electrons in inner orbitals

10. The effect of orbital shape

11. 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

12. A vertical orbital diagram for the Li ground state
empty
A vertical orbital diagram for the Li ground state
half-filled
filled, spin-paired

13. SAMPLE PROBLEM 8.1
Determining Quantum Numbers from Orbital
Diagrams
PROBLEM:
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 =
l =
ml =
ms= 2
+1/2
The eighth electron is in a 2p orbital. Its quantum numbers are
n =
l =
ml =
ms= 2 1 -1
-1/2

14. Orbital occupancy for the first 10 elements, H through Ne.

15. Hund’s rule

16. Condensed ground-state electron configurations in the first three periods.

19. A periodic table of partial ground-state electron configurations

20. The relation between orbital filling and the periodic table

21. General pattern for filling the sublevels

22. 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)
condensed configuration
partial orbital diagram
full configuration
1s22s22p63s23p64s1
[Ar] 4s1
There are 18 inner electrons.
4s1

23. SAMPLE PROBLEM 8.2
continued
(b) for Mo (Z = 42)
condensed configuration
partial orbital diagram
full configuration
1s22s22p63s23p64s23d104p65s14d5
[Kr] 5s14d5
There are 36 inner electrons and 6 valence electrons.
5s1
4d5
(c) for Pb (Z = 82)
1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p2
condensed configuration
partial orbital diagram
full configuration
[Xe] 6s24f145d106p2
There are 78 inner electrons and 4 valence electrons.
6s2
6p2

24. Defining metallic and covalent radii
Knowing the Cl radius and the C-Cl bond length, the C radius can be determined.

25. Atomic radii of the main-group and transition elements.
Trends in the Periodic Table
Atomic radii of the main-group and transition elements.

26. Trends in the Periodic Table
Periodicity of atomic radius

27. 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
(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 > 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.

28. Periodicity of first ionization energy (IE1)
Trends in the Periodic Table
Periodicity of first ionization energy (IE1)
Energy required to remove one outermost electron.

29. First ionization energies of the main-group elements
Trends in the Periodic Table

30. SAMPLE PROBLEM 8.4
Ranking Elements by First 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
(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 > 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 further to the left and down one period.

31. The first three ionization energies of beryllium (in MJ/mol)
Trends in the Periodic Table
The first three ionization energies of beryllium (in MJ/mol)

32. 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:
IE1
IE2
IE3
IE4
IE5
IE6
1012
1903
2910
4956
6278
22,230
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.

33. Electron affinities of the main-group elements
Trends in the Periodic Table
Electron affinities of the main-group elements
Electron Affinity:
Energy change to add one electron.
In most cases, EA negative (energy released because electron attracted to nucleus

34. Trends in three atomic properties

35. Trends in metallic behavior

36. Main-group ions and the noble gas configurations
Trends in the Periodic Table
Main-group ions and the noble gas configurations
Properties of Monatomic Ions

37. 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) + e I- ([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-

38. 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 e’s, not attracted diamagnetic

39. 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)
(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:
(a) Mn2+(Z = 25) Mn([Ar]4s23d5) Mn2+ ([Ar] 3d5) + 2e-
paramagnetic
(b) Cr3+(Z = 24) Cr([Ar]4s23d6) Cr3+ ([Ar] 3d5) + 3e-
paramagnetic
(c) Hg2+(Z = 80) Hg([Xe]6s24f145d10) Hg2+ ([Xe] 4f145d10) + 2e-
not paramagnetic (is diamagnetic)

40. Ionic vs. atomic radius

41. SAMPLE PROBLEM 8.8
Ranking Ions by Size
PROBLEM:
Rank each set of ions in order of decreasing size, and explain your ranking:
(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+ > Ca2+ > 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- > Cl - > K+
(c) Au+ > Au3+
The higher the + charge, the smaller the ion.

42. End of Chapter 8