General Chemistry

1. Bonding: General
Concepts
AP Chemistry Unit 8

2. Types of Chemical
Bonds

3. Ionic Bonds
Ionic Bonds are formed when an atom
that loses electrons relatively easily
reacts with an atom that has a high
attraction for electrons.
Ionic Compounds results when a metal
bonds with a nonmetal.

4. Bond Energy
Bond energy is the energy required to break a
bond.
The energy of interaction between a pair of ions
can be calculated using Coulomb’s law
r = the distance between the ions in nm.
Q1 and Q2 are the numerical ion charges.
E is in joules
E = (2.31x10-19
Jinm)
Q1Q2
r
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5. Bond Energy
When the calculated energy between
ions is negative, that indicates an
attractive force.
A positive energy is a repulsive energy.
The distance where the energy is
minimal is called the bond length.

6. Covalent Bonds
Covalent bonds form between
molecules in which electrons are shared
by nuclei.
The bonding electrons are typically
positioned between the two positively
charged nuclei.

7. Polar Covalent Bonds
Polar covalent bonds are an intermediate
case in which the electrons are not
completely transferred but form unequal
sharing.
A δ- or δ+ is used to show a fractional or
partial charge on a molecule with unequal
sharing. This is called a dipole.

8. Electronegativity

9. Electronegativity
Electronegativity is the ability of an atom in a
molecule to attract shared electrons to itself.
(electron love)
Relative electronegativities are determined by
comparing the measured bond energy with the
“expected” bond energy.
Measured in Paulings. After Linus Pauling the
American scientist who won the Nobel Prizes
for both chemistry and peace.

10. Electronegativity
Expected H-X bond energy=
H - H bond energy + X - X bond energy
2

11. Electronegativity
Electronegativity values generally increase
going left to right across the periodic table
and decrease going top to bottom.

12. Electronegativity and
Bond type

13. Bond Polarity and Dipole

14. Dipoles and Dipole
Moments
A molecule that has a center of positive
charge and a center of negative charge
is said to be dipolar or to have a dipole
moment.
An arrow is used to show this dipole
moment by pointing to the negative
charge and the tail at the positive
charge.

15. Dipoles and Dipole
Moments
Electrostatic potential
diagram shows
variation in charge.
Red is the most
electron rich region
and blue is the most
electron poor region.

16. Dipoles and Dipole
Moments

17. Dipoles and Dipole
Moments

18. Dipoles and Dipole
Moments

19. Dipoles and Dipole
Moments
Dipole moments are when opposing
bond polarities don’t cancel out.

20. Dipoles and Dipole
Moments

21. Example Problems
For each of the following molecules,
show the direction of the bond
polarities and indicate which ones have
a dipole moment: HCl, Cl2, SO3, CH4, H2S

22. HCl

23. Cl2

24. SO3

25. CH4

26. H2S

27. Ions: Electron
Configurations and Sizes

28. Electron Configurations
of Compounds
When two nonmetals react to form a
covalent bond, they share electrons in a way
that completes the valence electron
configurations of both atoms. That is, both
nonmetals attain noble gas electron
configurations.

29. Electron Configurations
of Compounds
When a nonmetal and a representative-group
metal react to form a binary ionic
compounds, the ions form so that the valence
electron configuration of the nonmetal
achieves the electron configuration of the
next noble gas atom and the valence orbitals
of the metal are emptied. In this way both
ions achieve noble gas electron
configurations.

30. Predicting Ionic
Formulas
To predict the formula of the ionic
compound, we simply recognize that the
chemical compounds are always electrically
neutral. They have the same quantities of
positive and negative charges.

31. Sizes of Ions
Size of an ion generally follows the same trend
as atomic radius. The big exception to this
trend is where the metals become nonmetals
and the ions switch charge.

32. Sizes of Ions
A positive ion is formed by removing one or
more electrons from a neutral atom, the
resulting cation is smaller than the neutral
atom.
Less electrons allow for less repulsions and
the ion gets smaller.

33. Sizes of Ions
An addition of electrons to a neutral atom
produces an anion that is significantly larger
than the neutral atom.
An addition of an electron causes additional
repulsions around the atom and therefore its
size increases.

34. Energy Effects in Binary
Ionic Compounds

35. Lattice Energy
Lattice energy is the change in energy that takes
place when separated gaseous ions are
packed together to form an ionic solid.
The lattice energy is often defined as the
energy released when an ionic solid forms
from its ions.
Lattice energy has a negative sign to show
that the energy is released.

36. Lattice Energy Example
Estimate the enthalpy of lithium fluoride and
the changes of energy and lattice energy
during formation:
1. Break down LiF into its standard state
elements (use formation reaction):
Li(s) + ½F2(g)  LiF(s)
Li+
(g) + F-
(g)  LiF(s)

37. Lattice Energy Example
Li(s) + ½F2(g)  LiF(s) Li+
(g) + F-
(g)  LiF(s)
2. Use sublimation and evaporation reactions to get
reactants into gas form (since lattice energy
depends on gaseous state). Find the enthalpies to
these reactions:
Li(s)  Li(g) 161 kJ/mol
Li(g) + ½F2(g)  LiF(s)

38. Lattice Energy Example
Li(g) + ½F2(g)  LiF(s) Li+
(g) + F-
(g)  LiF(s)
3. Ionize cation to form ions for bonding. Use
Ionization energy for the enthalpy of the reaction.
Li(g)  Li+
(g) + e- Ionization energy: 520 kJ/mol
Li+
(g) + ½F2(g)  LiF(s)

39. Lattice Energy Example
Li+
(g) + ½F2(g)  LiF(s) Li+
(g) + F-
(g)  LiF(s)
4. Dissociate diatomic gas to individual atoms:
½F2(g)  F(g) ½ Bond dissociation energy of F-F
= 154 kJ/ 2 = 77 kJ/mol
Li+
(g) + F(g)  LiF(s)

40. Lattice Energy Example
Li+
(g) + F(g)  LiF(s) Li+
(g) + F-
(g)  LiF(s)
5. Electron addition to fluorine is the electron affinity
of fluorine:
F(g) + e-  F-
(g) -328 kJ/mol
Li+
(g) + F-
(g)  LiF(s)

41. Lattice Energy Example
Li+
(g) + F-
(g)  LiF(s) Li+
(g) + F-
(g)  LiF(s)
6. Formation of solid lithium fluoride from the
gaseous ions corresponds to its lattice energy:
Li+
(g) + F-
(g)  LiF(s) -1047 kJ/mol

42. Lattice Energy Example
The sum of these five processes yields the overall reaction
and the sum of the individual energy changes gives the
overall energy change and the enthalpy of formation:
Li(s)  Li(g)
Li(g)  Li+
(g) + e-
½F2(g)  F(g)
F(g) + e-  F-
(g)
Li+
(g) + F-
(g)  LiF(s)
161 kJ
520 kJ
77 kJ
-328 kJ
-1047 kJ
Total = -617 kJ/mol

43. Lattice Energy

44. Lattice Energy
Lattice energy can be calculated with at form of
Coulomb’s law:
Q is the charges on the ions and r is the
shortest distance between the centers of the
cations and anions. k is a constant that
depends on the structure of the solid and the
electron configurations of the ions.
LatticeEnergy = k
Q1Q2
r
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45. Partial Ionic Character
of Covalent Bonds

46. Bond Character
Calculations of ionic character:
Even compounds with the maximum possible
electronegativity differences are not 100%
ionic in the gas phase. Therefore the
operational definition of ionic is any
compound that conducts an electric current
when melted will be classified as ionic.
Percent ionic character of a bond =
dipole moment of x - y
dipole moment of x+
yy
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47. Bond Character

48. The Covalent Chemical
Bond

49. Chemical Bond Model
A chemical bond can be viewed as forces that
cause a group of atoms to behave as a unit.
Bonds result from the tendency of a system
to seek its lowest possible energy.
Individual bonds act relatively independent.

50. Example
It takes 1652 kJ of energy required to break the
bonds in 1 mole of methane.
1652 kJ of energy is released when 1 mole of
methane is formed from gaseous atoms.
Therefore, 1 mole of methane in gas phase has
1652 kJ lower energy than the total of the
individual atoms.
One mole of methane is held together with 1652
kJ of energy.
Each of the four C-H bonds contains 413 kJ of
energy.

51. Example
Each of the four C-H bonds contains 413 kJ
of energy.
CH3Cl contains 1578 kJ of energy:
1 mol of C-Cl bonds + 3 mol (C-H bonds)=1578 kJ
C-Cl bond energy + 3 (413 kJ/mol) = 1578 kJ
C-Cl bond energy = 339 kJ/mol

52. Properties of Models
A model doesn’t equal reality; they are used
to explain incomplete understanding of how
nature works.
Models are often oversimplified and are
sometimes wrong.
Models over time tend to get over
complicated due to “repairs”.

53. Properties of Models
Remember that simple models often require
restrictive assumptions. Best way to use
models is to understand their strengths and
weaknesses.
We often learn more when models are
incorrect than when they are right.
Cu and Cr.

54. Covalent Bond Energies
and Chemical Reactions

55. Bond Energies
Bond energy averages are used for individual
bond dissociation energies to give
approximate energies in a particular bond.
Bond energies vary due to several reasons:
multiple bonds, 4 C-H bonds in methane
different elements in the molecule, C-H bond in
C2H6 or C-H bond in HCCl3

56. Bond Energy Example
CH4(g)CH3(g) + H(g)
CH3(g)CH2(g) + H(g)
CH2(g)CH(g) + H(g)
CH(g)C(g) + H(g)
435 kJ
453 kJ
425 kJ
339 kJ
Total 1652 kJ
Average 413 kJ

57. Bond Energy Example
HCBr3
HCCl3
HCF3
C2H6
380 kJ
380 kJ
430 kJ
410 kJ

58. Average Bond Energies

59. Bond Energy
A relationship also exists between the
number of shared electron pairs.
single bond – 2 electrons
double bond – 4 electrons
triple bond – 6 electrons

60. Bond Energy
Bond energy values can be used to calculate
approximate energies for reactions.
Energy associated with bond breaking have
positive signs
Endothermic process
Energy associated with forming bonds releases
energy and is negative.
Exothermic process

61. Bond Energy
A relationship exists between the number of
shared electron pairs and the bond length.
• As the number of electrons shared goes up the
bond length shortens.

62. Bond Energy

63. Bond Energy
ΔH = sum of the energies required to break old
bonds (positive signs) plus the sum of the
energies released in the formation of new bonds
(negative signs).
• D represents bond energies per mole and always
has positive sign
• n is number of moles
DH = Sn x D(bonds broken)- ån x D(bonds formed)

64. Bond Energy Example
H2(g) + F2(g) 2HF(g)
1 H-H bond, F-F bond and 2 H-F bonds
ΔH = DH-H + DF-F – 2DH-F
ΔH= (1mol x 432 kJ/mol) + (1mol x 154 kJ/mol)
– (2mol x 565 kJ/mol)
ΔH = -544 kJ

65. The Localized Electron
Bonding Model

66. Localized Electron
Model
The localized electron model assumes that a
molecule is composed of atoms that are
bound together by sharing pairs of electrons
using the atomic orbitals of the bound atoms.
Electrons are assumed to be localized on a
particular atom individually or in the space
between atoms.

67. Localized Electron
Model
Pairs of electrons that are localized on an
atom are called lone pairs.
Pairs of electrons that are found in the space
between the atoms are called bonding pairs

68. Localized Electron
Model
Three parts of the LE Model:
1. Description of the valence electron arrangement
in the molecule using Lewis structures.
2. Prediction of the geometry of the molecule using
VSEPR model
3. Description of the type of atomic orbitals used by
the atoms to share electrons or hold lone pairs.

69. Lewis Structures

70. Lewis Structures
The Lewis structure of a molecule show how the
valence electrons are arranged among the atoms
in the molecule.
Named after G. N. Lewis
Rules are based on observations of thousands of
molecules.
Most important requirement for the formation of
a stable compound is that the atoms achieve noble
gas electron configurations.

71. Lewis Structures
Only the valence electrons are included.
The duet rule: diatomic molecules can find
stability in the sharing of two electrons.
The octet rule: since eight electrons are
required to fill these orbitals, these elements
typically are surrounded by eight electrons.

72. Lewis Structure Steps
1. Sum the valence electrons from all the atoms.
Total valence electrons.
2. Use a pair of electrons to form a bond between
each pair of bound atoms.
3. Arrange the remaining electrons to satisfy the
duet rule for hydrogen and the octet rule for the
others.
a) Terminal atoms first.
b) Check for happiness

73. Examples
HF
N2
NH3
CH4
CF4
NO+

74. Exceptions to the Octet
Rule

75. Exceptions to the Octet
Rule
Incomplete: An odd number of electrons are
available for bonding. One lone electron is left
unpaired.
Suboctet: Less than 4 pairs of electrons are
assigned to the central atom
Suboctets tend to form coordinate
covalentbonds
BH3 + NH3

76. Exceptions to the Octet
Rule
Extended: The central atom has more than 4 pairs
of electrons.
At the third energy level and higher, atoms may
have empty d orbitals that can be used for
bonding.

77. General Rules
The second row elements C, N, O, and F always
obey the octet rule
The second row elements B and Be often have
fewer than eight electrons around them in their
compounds. They are electron deficient and very
reactive.
The second row elements never exceed the octet
rule, since their valence orbitals can only hold 8.

78. General Rules
Third-row and heavier elements often satisfy the
octet rule but can exceed the octet rule by using
their empty valence d orbitals.
When writing the Lewis structure for a molecule,
satisfy the octet rule for the atoms first. If
electrons remain after the octet rule has been
satisfied, then place them on the elements having
available d orbitals

79. Resonance

80. Resonance
Resonance is when more than on valid Lewis
structure can be written for a particular
molecule. The resulting electron structure of
the molecule is given by the average of these
resonance structures.

81. Resonance
The concept of resonance is necessary because
the localized electron model postulates that
electrons are localized between a given pair of
atoms. However, nature does not really operate
this way. Electrons are really delocalized- they
move around the entire molecule. The valence
electrons in a resonance structure distribute
themselves equally and produce equal bonds.

82. Formal Charge
Some molecules or polyatomic ions can have
several non-equivalent Lewis structures.
• Example: SO4
2-
Because of this we assign atomic charges to the
molecules in order to find the right structure.

83. Formal Charge
The formal charge of an atom in a molecule is
the difference between the number of
valence electrons on the free atom and the
number of valence electrons assigned to the
atom in the molecule
Formal charge = (# of valence electrons on
neutral ‘free atom’) – (# of valence electrons
assigned to the atom in the molecule)

84. Formal Charge
Assumptions on electron assignment:
Lone pair electrons belong entirely to the
atom in question.
Shared electrons are divided equally between
the two sharing atoms.

85. Formal Charge Example
SO4
2-: All single bonds
Formal charge on each O is -1
Formal charge on S is 2

86. Formal Charge Example
SO4
2-: two double bonds, two single
Formal charge on single bonded O is -1
Formal charge on double bonded O is 0
Formal charge on S is 0

87. Formal Charges
1. Atoms in molecules try to achieve formal charges
as close to zero as possible.
2. Any negative formal charges are expected to
reside on the most electronegative atoms.
If nonequivalent Lewis structures exist for a species,
those with formal charges closest to zero and
with any negative formal charges on the most
electronegative atoms are considered to best
describe the bonding in the molecule or ion.

88. Molecular Structure:
The VSEPR Model

89. VSEPR
Valence shell electron repulsion model is useful in
predicting the geometries of molecules formed
from nonmetals.
The structure around a given atom is
determined principally by minimizing electron
– pair repulsion.

90. VSEPR
From the Lewis structure, count the electron
pairs around the central atom.
Lone pairs require more room than bonding pairs
and tend to compress the angles between the
bonding pairs.
Multiple bonds should be counted as one
effective pair.
With a molecule with resonance, all structures
should yield the same shape.

91. Linear
180°
3-D linear

92. Trigonal Planer
120°
3-D trigonal planar
3-D trigonal w/lone pair

93. Tetrahedral
109.5°
3-D tetrahedral

94. Trigonal Pyramidal
107°
3-D tetrahedral 1 lone pair / trigonal pyramidal

95. Bent/V
104.5°
3-D tetrahedral 2 lone pair / bent

96. Tetrahedral Arrangements

97. Bipyramidal
Arrangements
trigonal bipyramidal
bipyramidal 1 lone pair / see saw
bipyramidal 2 lone pair / T shape
bipyramidal 3 lone pair / linear

98. Octahedral
Arrangements
octahedral
octahedral 1 lone pair / square pyramidal
octahedral 2 lone pair / square planar

99. Molecules without a
central atom
The molecular structure of more complicated
atoms can be predicted from the
arrangement of pairs around the center
atoms. A combination of shapes will result
that allows for minimum repulsion
throughout.

100. Molecules without a
central atom

102. The End