3. Figure 8.12: A periodic table illustrating the building-up order.

4. Mendeleev’s periodic table generally organized elements by increasing atomic mass and with similar properties in columns. In some places, there were missing elements whose properties he predicted.
When gallium, scandium, and germanium were isolated and characterized, their properties were almost identical to those predicted by Mendeleev for eka-aluminum, eka-boron, and eka-silicon, respectively.

5. Figure 8.14: Mendeleev’s periodic table.

6. Periodic law states that when the elements are arranged by atomic number, their physical and chemical properties vary periodically.
We will look in more detail at three periodic properties: atomic radius, ionization energy, and electron affinity.

7. Effective Nuclear Charge
Effective nuclear charge is the positive charge that an electron experiences from the nucleus. It is equal to the nuclear charge, but is reduced by shielding or screening from any intervening electron distribution (inner shell electrons).

8. Effective nuclear charge increases across a period
Effective nuclear charge increases across a period. Because the shell number (n) is the same across a period, each successive atom experiences a stronger nuclear charge. As a result, the atomic size decreases across a period.

9. Atomic Radius
While an atom does not have a definite size, we can define it in terms of covalent radii (the radius in covalent compounds).

11. Figure 8.17: Representation of atomic radii (covalent radii) of the main-group elements.

12. Atomic radius is plotted against atomic number in the graph below
Atomic radius is plotted against atomic number in the graph below. Note the regular (periodic) variation.

13. A representation of atomic radii is shown below.

14. Refer to a periodic table and arrange the following elements in order of increasing atomic radius: Br, Se, Te. 35 Br 34 Se 52 Te
Te is larger than Se.
Se is larger than Br.
Br < Se < Te

15. First Ionization Energy (first ionization potential)
The minimum energy needed to remove the highest-energy (outermost) electron from a neutral atom in the gaseous state, thereby forming a positive ion

16. Periodicity of First Ionization
Energy (IE1)
Like Figure 8-18

17. Fig. 8.15

18. Left of the line, valence shell electrons are being removed
Left of the line, valence shell electrons are being removed. Right of the line, noble-gas core electrons are being removed.

19. Identifying Elements by Its Successive Ionization Energies
Problem: Given the following series of ionization energies (in kJ/mol)
for an element in period 3, name the element and write its
electron configuration:
IE IE IE IE4
, , ,600
Plan: Examine the values to find the largest jump in ionization energy,
which occurs after all valence electrons have been removed.
Use the periodic table!
Solution:

20. Identifying Elements by Its Successive Ionization Energies
Problem: Given the following series of ionization energies (in kJ/mol)
for an element in period 3, name the element and write its
electron configuration:
IE IE IE IE4
, , ,600
Plan: Examine the values to find the largest jump in ionization energy,
which occurs after all valence electrons have been removed.
Use the periodic table!
Solution:
The largest jump in IE occurs after IE3 so the element has
3 valence electrons thus it is Aluminum ( Al, Z=13), its
electron configuration is :
1s2 2s2 2p6 3s2 3p1

21. Fig. 8.16

22. Ranking Elements by First Ionization Energy
Problem: Using the Periodic table only, rank the following elements in
each of the following sets in order of increasing IE!
a) Ar, Ne, Rn b) At, Bi, Po c) Be, Na, Mg d) Cl, K, Ar
Plan: Find their relative positions in the periodic table and apply trends!
Solution:

23. Trends
Going down a group, first ionization energy decreases.
This trend is explained by understanding that the smaller an atom, the harder it is to remove an electron, so the larger the ionization energy.

24. Generally, ionization energy increases with atomic number.
Ionization energy is proportional to the effective nuclear charge divided by the average distance between the electron and the nucleus. Because the distance between the electron and the nucleus is inversely proportional to the effective nuclear charge, ionization energy is inversely proportional to the square of the effective nuclear charge.

25. Small deviations occur between Groups IIA and IIIA and between Groups VA and VIA.
Examining the valence configurations for these groups helps us to understand these deviations:
IIA ns2
IIIA ns2np1
VA ns2np3
VIA ns2np4
It takes less energy to remove the np1 electron than the ns2 electron.
It takes less energy to remove the np4 electron than the np3 electron.

26. Electrons can be successively removed from an atom
Electrons can be successively removed from an atom. Each successive ionization energy increases, because the electron is removed from a positive ion of increasing charge.
A dramatic increase occurs when the first electron from the noble-gas core is removed.

27. Refer to a periodic table and arrange the following elements in order of increasing ionization energy: As, Br, Sb.
Sb is larger than As.
As is larger than Br. 35 Br 33 As 51 Sb
Ionization energies:
Sb < As < Br

28. Overall periodic trends
Chemistry 140 Fall 2002
Overall periodic trends
Note: Electronegativity has similar trend as electron affinity 28

29. Reactivity of the Alkali Metals
2Li(s) + 2H2O(l) → 2LiOH(aq) + H2(g)
Lithium video
2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)
Sodium video
2K(s) + 2H2O(l) → 2KOH(aq) + H2(g)
Potassium video
Trend?

30. More Sodium Reaction Videos
2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)
Prepping Na
100 g Na in one piece
150 g Na in small pieces

32. Electronic Configuration Ions
Na 1s 2 2s 2 2p 6 3s Na+ 1s 2 2s 2 2p 6
Mg 1s 2 2s 2 2p 6 3s Mg+2 1s 2 2s 2 2p6
Al 1s 2 2s 2 2p 6 3s 2 3p Al+3 1s 2 2s 2 2p 6
O 1s 2 2s 2 2p O s 2 2s 2 2p 6
F 1s 2 2s 2 2p F s 2 2s 2 2p 6
N 1s 2 2s 2 2p N s 2 2s 2 2p 6

34. Isoelectronic Atoms and Ions
H- 1 { He } Li+ Be+2
N O- 2 F- { Ne } Na+ Mg+2 Al+3
P- 3 S- 2 Cl- { Ar } K+ Ca+2 Sc+3 Ti+4
As- 3 Se- 2 Br- { Kr } Rb+ Sr+2 Y+3 Zr+4
Sb- 3 Te I- { Xe } Cs+ Ba+2 La+3 Hf+4

35. Trends when atoms form chemical bonds
Chemistry 140 Fall 2002
Trends when atoms form chemical bonds
Empirical Observation
“when forming ionic compounds, elements tend to lose or gain electrons to be more like the nearest noble gas”
Nonmetals tend to gain e-’s
Metals tend to lose e-’s 35

36. Are ions bigger or smaller than atoms?
Chemistry 140 Fall 2002
Are ions bigger or smaller than atoms?
Representative cation
Na → Na+ + e-
Representative anion
F + e- → F- 36

38. Cations Contract Anions Add
Trends in ion size
Cations are always smaller than parent atom
- decreased e- repulsion (clouds contract)
- if emptying valence shell, “n” decreases
Anions are always larger than parent atoms
- increased e- repulsion (clouds expand)
Cations
Contract
Anions
Add

39. Trends in atom & ion size
Chemistry 140 Fall 2002 39

40. Chemistry 140 Fall 2002
Trends in ion size 40

41. Ranking Ions According to Size
Problem: Rank each set of Ions in order of increasing size.
a) K+, Rb+, Na+ b) Na+, O2-, F - c) Fe+2, Fe+3
Plan: We find the position of each element in the periodic table and
apply the ideas of size:
i) size increases down a group, ii) size decreases across a period but
increases from cation to anion. iii) size decreases with increasing
positive (or decreasing negative) charge in an isoelectronic series.
iv) cations of the same element decreases in size as the charge increases.
Solution:
a) since K+, Rb+, and Na+ are from the same group (1A), they increase
in size down the group: Na+ < K+ < Rb+
b) the ions Na+, O2-, and F- are isoelectronic. O2- has lower Zeff than F-,
so it is larger. Na+ is a cation, and has the highest Zeff, so it is
smaller: Na+ < F- < O2-
c) Fe+2 has a lower charge than Fe+3, so it is larger: Fe+3 < Fe+2

42. Chapter #9 - Models of Chemical Bonding
9.1) Atomic Properties and Chemical Bonds
9.2) The Ionic Bonding Model
9.3) The Covalent Bonding Model
9.4) Between the Extremes: Electronegativity and Bond Polarity
9.5) An Introduction to Metallic Bonding

44. Sodium Chloride

45. Depicting Ion Formation with Orbital
Diagrams and Electron Dot Symbols - I
Problem: Use orbital diagrams and Lewis structures to show the
formation of magnesium and chloride ions from the atoms, and
determine the formula of the compound.
Plan: Draw the orbital diagrams for Mg and Cl. To reach filled outer
levels Mg loses 2 electrons, and Cl will gain 1 electron. Therefore we
need two Cl atoms for every Mg atom.
Solution: Mg +
Mg Cl-
2 Cl .. . .. .. .. . Cl . .. .. .. ..
Mg + .
Mg Cl .. ..

46. . .. . . .. .. .. .. .. . Depicting Ion Formation from Orbital
Diagrams and Electron Dot Symbols - II
Problem: Use Lewis structures and orbital diagrams to show the
formation of potassium and sulfide ions from the atoms, and determine
the formula of the compound.
Plan: Draw orbital diagrams for K and S. To reach filled outer orbitals,
sulfur must gain two electrons, and potassium must lose one electron.
Solution:
2 K +
2 K S - 2 S . .. . .
2 - K .. .. .. .. .. . + S
2 K S K

47. Three Ways of Showing the Formation of Li+ and F - through Electron Transfer

48. Lewis Electron-Dot Symbols for Elements in Periods 2 & 3

49. The Reaction between Na and Br to Form NaBr
The Elements
The Reaction!

50. Melting and Boiling Points of Some Ionic Compounds
Compound mp( oC) bp( oC)
CsBr
NaI
MgCl
KBr
CaCl >1600
NaCl
LiF KF
MgO

52. Figure 9.11: Potential-energy curve for H2.

53. Covalent Bonding in Hydrogen, H2

55. Figure 9.10: The electron probability distribution for the H2 molecule.

56. Covalent bonds animation
animation

57. For elements larger than Boron, atoms usually react to
develop octets by sharing electrons. H, Li and Be strive
to “look” like He. B is an exception to the noble gas paradigm.
It’s happy surrounded by 6 electrons so the compound BH3 is
stable.
Try drawing a Lewis structure for methane.

58. Draw Lewis dot structures for the halogens.
Notice that these all follow the octet rule!
Try oxygen and nitrogen.
These also follow the octet rule!

59. Bond Lengths and Covalent Radius

60. Figure 9.14: The HCl molecule.

61. Figure 9.12: Molecular model of nitro-glycerin.
What is the formula for this
compound?

62. Rules for drawing Lewis structures
1. Count up all the valence electrons
2. Arrange the atoms in a skeleton
3. Have all atoms develop octets (except
those around He)

63. Make some Lewis Dot Structures with other elements:
SiH4
H2O
NH3
CH2O
C2H6
C2H6O

64. Figure 9.9: Model of CHI3 Courtesy of Frank Cox.

65. Make some Lewis Dot Structures with other elements:
CH4
H2O
NH3
CH2O
C2H6
C2H6O

66. Look at all these structures and make some bonding rules:

68. Rules for drawing Lewis structures
1. Count up all the valence electrons
2. Arrange the atoms in a skeleton
3. Have all atoms develop octets (except
those around He)
4. Satisfy bonding preferences!

69. A model of ethylene.

70. A model of acetylene.

71. A model of COCl2.

74. The Relation of Bond Order, Bond Length and Bond Energy
Bond Bond Order Average Bond Average Bond
Length (pm) Energy (kJ/mol)
C O
C O
C O
C C
C C
C C
N N
N N
N N
Table 9.4

75. Conceptual Problem 9.103

76. Fig. 9.14

77. Figure 9.15: Electronegatives of the elements.

78. The Periodic Table of the Elements
2.1 He
0.9
1.5
2.0
2.5
3.0
3.5
4.0 Ne
Electronegativity
0.9
1.2
1.5
1.8
2.1
2.5
3.0 Ar
0.8
1.0
1.3
1.5
1.6
1.6
1.5
1.8
1.8
1.8
1.9
1.6
1.6
1.8
2.0
2.4
2.8 Kr
0.8
1.0
1.2
1.4
1.6
1.8
1.9
2.2
2.2
2.2
1.9
1.7
1.7
1.8
1.9
2.1
2.5 Xe
0.7
0.9
1.1
1.3
1.5
1.7
1.9
2.2
2.2
2.2
2.4
1.9
1.8
1.8
1.9
2.0
2.2 Rn
0.7
0.9
1.1
Ce Pr Nd Pm
Yb Lu
1.1
1.1
1.1
1.2
1.2
1.1
1.2
1.2
1.2
1.2
1.2
1.2
1.2
1.3
1.3
1.5
1.7
1.3
1.3
1.3
1.3
1.3
1.3
1.3
1.5
Th Pa U Np
No Lr

79. Fig. 9.16

80. Fig. 9.17

81. Determining Bond Polarity from Electronegativity Values
Problem: (a)Indicate the polarity of the following bonds with a polarity
arrow: O - H, O - Cl, C - N, P - N, N - S, C - Br, As - S
(b) rank those bonds in order of increasing polarity.
Plan: (a) We use Fig to find the EN values, and point the arrow
toward the negative end. (b) Use the EN values.
Solution:
a) the EN of O = 3.5 and of H = 2.1: O - H
the EN of O = 3.5 and of Cl = 3.0: O - Cl
the EN of C = 2.5 and of P = 2.1: C - P
the EN of P = 2.1 and of N = 3.0: P - N
the EN of N = 3.0 and of S = 2.1: N - S
the EN of C = 2.5 and of Br = 2.8: C - Br
the EN of As = 2.0 and of O = 3.5: As - O
b) C - Br < C - P < O - Cl < P - N < N - S < O - H < As - O
< < < < < < 1.5

82. Fig. 9.18

83. Percent Ionic Character as a Function of Electronegativity Difference (En)
Fig. 9.19

84. Lewis Structures of Simple MoleculesH H H ..
CH4 H C C .. O H
Methane H H
Ethyl Alcohol (Ethanol) .. .. .. .. C F .. O K+ .. .. Cl .. .. .. .. .. O .. O
CF4 ..
KClO3
Potassium Chlorate
Carbon Tetrafluoride

85. Resonance:
Delocalized Electron-Pair Bonding - I
Ozone : O3 .. .. .. .. .. O O .. .. .. .. .. .. O O .. O O II I
Resonance Hybrid Structure .. O .. .. .. .. O O
One pair of electron’s resonances
between the two locations!!

86. Delocalized Electron-Pair Bonding - II
Resonance:
Delocalized Electron-Pair Bonding - II H H C C C C H C C H H H H C C H C H C H H C C C H H C H C H C H H C C
Benzene
Resonance Structure H

87. Lewis Structures of Simple Molecules
Resonance Structures -III
Nitrate .. .. .. O .. N .. .. O .. .. O .. .. .. O .. .. O N .. .. N .. .. .. O .. .. O .. .. .. .. O O

88. Lewis Structures for Octet Rule Exceptions.. .. .. B Cl .. .. .. .. F .. .. .. .. F Cl .. .. F
Each chlorine atom has
8 electrons associated.
Boron has only 6!
Each fluorine atom has
8 electrons associated.
Chlorine has 10 electrons! . .. .. .. .. .. .. N .. .. Cl Be Cl .. .. .. O O
Each chlorine atom has
8 electrons associated. The
beryllium has only 4 electrons.
NO2 is an odd electron atom.
The nitrogen has 7 electrons.

89. Resonance Structures - Expanded Valence Shells.. .. .. .. .. .. .. .. .. .. .. .. .. .. F F F F .. .. .. .. .. .. F S F .. P .. F .. .. .. .. .. .. F .. .. .. .. F F .. F
p = 10e-
S = 12e-
Sulfur hexafluoride ..
Phosphorous pentafluoride .. .. O S H .. O S H .. ..
Resonance Structures .. .. .. .. ..
Sulfuric acid
S = 12e-

90. Lewis Structures of Simple Molecules
. .
. . -2
Sulfate O
. .
. .
Resonance Structures-V
. .
. . O S O
. .
. . -2
. .
. . O
o *
o o O
o o
Plus 4 others
for a total of 6
x o o o
o o x
. .
o * O
x o S O
o o x
o o -2
. . O
. .
o o
x x
. .
. .
o o O
o o O S O
. .
. .
o o
. . O
. .
. .
x = Sulfur electrons
o = Oxygen electrons

91. VSEPR: Valence Shell Electron Pair Repulsion:
A way to predict the shapes of molecules
Pairs of valence electrons want to get as far
away from each other as possible in
3-dimensional space.

92. Balloon Analogy for the Mutual Repulsion of Electron Groups
Two
Three Four Five Six
Number of Electron Groups

95. AX2 Geometry - Linear .. .. .. .. .. ..
Molecular Geometry =
Linear Arrangement Cl Be Cl
BeCl2
1800
Gaseous beryllium chloride is an example of a molecule in which the
central atom - Be does not have an octet of electrons, and is electron
deficient.
Other alkaline earth elements also have the same valence electron
configuration, and the same geometry for molecules of this type.
Therefore this geometry is common to group II elements. .. .. .. .. O C O
CO2
1800
Carbon dioxide also has the same geometry, and is a linear molecule,
but in this case, the bonds between the carbon and oxygens are double
bonds.

96. The Two
Molecular
Shapes of the
Trigonal Planar
Electron-Group
Arrangement

97. AX3 Geometry - Trigonal Planar.. .. .. .. .. ..
All of the boron Family(IIIA)
elements have the same
geometry. Trigonal Planar ! F F
BF3 B
Boron Trifluoride
1200 .. .. .. F
AX2E SO2 .. - .. .. .. .. O S O .. ..
NO3- ..
1200 .. N .. .. .. .. O O
The AX2E molecules have a pair of
Electrons where the third atom
would appear in the space around the
central atom, in the trigonal planar
geometry.
1200
Nitrate Anion

98. The Three
Molecular Shapes
of the Tetrahedral
Electron-Group
Arrangement

99. .. AX4 Geometry - Tetrahedral H Methane H 109.50 CH4 H C H C H H H H
All molecules or ions with four electron groups around a central atom
adopt the tetrahedral arrangement H H
109.50 N H ..
107.30
109.50 + H+ N H H
Ammonia is in a tetrahedral shape,
but it has only an electron pair in
one location, so the smaller angle!
all angles are
the same!
Ammonium Ion H

100. The Four Molecular
Shapes of the Trigonal
Bipyramidal Electron-
Group Arrangement

101. AX5 Geometry - Trigonal Bipyramidal.. .. .. .. .. .. F I .. .. ..
86.20 .. .. .. I
1800 Br .. F .. .. .. .. .. I .. F ..
AX3E2 - BrF3 ..
AX2E3 - I3- .. .. .. Cl .. .. Cl .. .. P Cl ..
AX PCl5 .. .. Cl .. .. .. .. Cl

102. The Three Molecular
Shapes of the
Octahedral
Electron-Group
Arrangement

103. AX6 Geometry - Octahedral.. .. .. .. .. .. .. .. .. .. F F .. .. .. .. F .. F .. .. F .. F .. .. .. .. S I .. .. .. .. F F .. .. .. .. F .. F .. .. .. F
AX5E
Iodine Pentafluoride
AX6
Sulfur Hexafluoride .. .. .. .. .. .. F .. F .. Xe .. .. .. .. .. F .. F
Xenon Tetrafluoride
Square planar shape

104. Using VSEPR Theory to Determine Molecular Shape
1) Write the Lewis structure from the molecular formula to see the
relative placement of atoms and the number of electron groups.
2) Assign an electron-group arrangement by counting all electron
groups around the central atom, bonding plus nonbonding.
3) Predict the ideal bond angle from the electron-group arrangement
and the direction of any deviation caused by the lone pairs or double
bonds.
4) Draw and name the molecular shape by counting bonding groups
and non-bonding groups separately.

106. Hybrid Orbital Model

108. The sp Hybrid Orbitals in Gaseous BeCl2

112. The sp3 Hybrid Orbitals in NH3 and H2O

113. The sp3d Hybrid Orbitals in PCl5

114. The sp3d2 Hybrid Orbitals in SF6
Sulfur Hexafluoride SF6

116. Figure 10.26: Sigma and pi bonds.

117. Figure 10.27: Bonding in ethylene.

119. Figure 10.28: Bonding in acetylene.

121. Restricted Rotation of -Bonded Molecules
A) Cis - 1,2 dichloroethylene B) trans - 1,2 dichloroethylene

122. Postulating the Hybrid Orbitals in a Molecule
Problem: Describe how mixing of atomic orbitals on the central atoms
leads to the hybrid orbitals in the following:
a) Methyl amine, CH3NH b) Xenon tetrafluoride, XeF4
Plan: From the Lewis structure and molecular shape, we know the
number and arrangement of electron groups around the central atoms,
from which we postulate the type of hybrid orbitals involved. Then we
write the partial orbital diagram for each central atom before and after
the orbitals are hybridized.

123. Postulating the Hybrid Orbitals in a Molecule
Problem: Describe how mixing of atomic orbitals on the central atoms
leads to the hybrid orbitals in the following:
a) Methyl amine, CH3NH b) Xenon tetrafluoride, XeF4
Plan: From the Lewis structure and molecular shape, we know the
number and arrangement of electron groups around the central atoms,
from which we postulate the type of hybrid orbitals involved. Then we
write the partial orbital diagram for each central atom before and after
the orbitals are hybridized.
Solution:
a) For CH3NH2: The shape is tetrahedral around the C and N atoms.
Therefore, each central atom is sp3 hybridized. The carbon atom has
four half-filled sp3 orbitals: 2s 2p
sp3
Isolated Carbon Atom
Hybridized Carbon Atom

124. .. The N atom has three half-filled sp3 orbitals and one filled with a
lone pair. 2s
sp3 2p .. H C N H H H H

125. b) The Xenon atom has filled 5 s and 5 p orbitals with the 5 d orbitals
empty.
Isolated Xe atom
5 d
5 s
5 p
Hybridized Xe atom:
sp3d2
5 d

126. b) continued:For XeF4. for Xenon, normally it has a full octet of
electrons,which would mean an octahedral geometry, so to make the
compound, two pairs must be broken up, and bonds made to the four
fluorine atoms. If the two lone pairs are on the equatorial positions,
they will be at 900 to each other, whereas if the two polar positions are
chosen, the two electron groups will be 1800 from each other. Thereby
minimizing the repulsion between the two electron groups. .. F F F F
1800 Xe Xe .. F F F F
Square planar

127. That's all, folks!!