Chemical Principles
Chapter 13 Bonding: General Concepts
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Chapter 13 Bonding: General Concepts
 13.1 Types of Chemical Bonds
 13.2 Electronegativity
 13.3 Bond Polarity and Dipole Moments
 13.4 Ions: Electron Configurations and Sizes
 13.5 Formation of Binary Ionic Compounds
 13.6 Partial Ionic Character of Covalent Bonds
 13.7 The Covalent Chemical Bond: A Model
 13.8 Covalent Bond Energies and Chemical Reactions
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Chapter 13 Bonding: General Concepts
 13.9 The Localized Electron Bonding Model
 13.10 Lewis Structures
 13.11 Resonance
 13.12 Exceptions to the Octet Rule
 13.13 Molecular Structure:The VSEPR Model
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Introduction
 Bonding and structure play a central role in determining the
course of all chemical reactions,
 Graphite is a soft, slippery material used as a lubricant in locks.
Diamond is one of the hardest materials known, valuable both as a
gemstone and in industrial cutting tools. Both composed solely of
carbon atoms, have such different properties?
 Silicon and carbon are next to each other in Group 4A on the
periodic table.

SiO2 is the empirical formula of silica, which is found in sand and
quartz, whereas carbon dioxide is a gas, a product of respiration.
 Why are they so different?
 The answer lies in the bonding within these substances.
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Introduction
Quartz grows in beautiful, regular
crystals.
Two forms of carbon:
graphite and diamond.
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13.1 Types of Chemical Bonds
 The bonding forces that produce this great thermal stability result
from the electrostatic attractions of the closely packed, oppositely
charged ions. This is an example of ionic bonding.
 an ionic compound results when a metal reacts with a
nonmetal.
 Coulomb’s law:
 in solid sodium chloride, where the distance between the centers
of the Na+ and Cl- ions is 276 picometers (0.276 nm), the ionic
energy per pair of ions is
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13.1 Types of Chemical Bonds
 The negative sign indicates an attractive force. That is, the ion pair
has lower energy than the separated ions. For a mole of pairs of
Na+ and Cl- ions, the energy of interaction is
 Note that this energy refers to a mole of Na+ and Cl- ion pairs in
the gas phase where a given pair is far from any other pair. In solid
sodium chloride, which contains a large array of closely packed Na+
and Cl- ions, where a given ion is close to many oppositely charged
ions, the energy associated with ionic bonding is much greater than
504 kJ/mol because of the larger numbers of interacting ions.
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13.1 Types of Chemical Bonds
Figure 13.1 (a) The interaction of two hydrogen atoms. (b) Energy profile as
a function of the distance between the nuclei of the hydrogen atoms. As the
atoms approach each other, the energy decreases until the distance reaches
0.074 nm (0.74 A) and then begins to increase again because of repulsions.
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13.1 Types of Chemical Bonds
 The distance at which the energy is minimum is called the
equilibrium internuclear distance or the bond length.
1.The energy terms involved are the potential energy that results
from the attractions and repulsions among the charged particles
and the kinetic energy caused by the motions of the electrons.
 2. The zero reference point for energy is defined for the atoms at
infinite separation.
 3. At very short distances the energy rises steeply because of the
great importance of the internuclear repulsive forces at these
distances.
 4. The bond length is the distance at which the system has minimum
energy, and the bond energy corresponds to the depth of the “well”
at9this distance.
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13.1 Types of Chemical Bonds
 The type of bonding we encounter in the hydrogen molecule and in
many other molecules in which electrons are shared by nuclei is
called covalent bonding.
 In covalent bonding two identical atoms share electrons equally. The
bonding results from the mutual attraction of the two nuclei for the
shared electrons.
 Between these extremes lie intermediate cases in which the atoms
are not so different that electrons are completely transferred but
are different enough so that unequal sharing results.
 These are called polar covalent bonds. An example of this type
of bond occurs in the hydrogen fluoride (HF) molecule.
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13.1 Types of Chemical Bonds
Figure 13.2
The effect of an electric field on hydrogen
fluoride molecules.
(a) When no electric field is present, the
molecules are randomly oriented.
(b) When the field is turned on, the
molecules tend to line up with their
negative ends toward the positive pole
and their positive ends toward the
negative pole.
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13.2 Electronegativity
 electronegativity: the ability of an atom in a molecule to attract
shared electrons to itself.
為合適幾何平均，而不是
算術平均？
 Pauling then obtained absolute electronegativity values for the
elements by assigning a value of 4.0 to fluorine (the element with
the highest electronegativity).
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13.2 Electronegativity
Figure 13.3
The Pauling electronegativity values. Electronegativity generally increases
across a period and decreases down a group.
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13.2 Electronegativity
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13.2 Electronegativity
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13.3 Bond Polarity and Dipole Moments
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13.3 Bond Polarity and Dipole Moments
 Solving for d gives 6.66 x 10-20 C.
 Since the charge on an electron is 1.60 x 10-19 C, each atom in HF
has a fractional charge of
 From this argument we might say that HF has 42% ionic bonding.
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13.3 Bond Polarity and Dipole Moments
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13.3 Bond Polarity and Dipole Moments
Figure 13.5
(a)The charge distribution in the water molecule. (b) The water molecule
in an electric field. (c) The electrostatic potential diagram of the water
molecule.
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13.3 Bond Polarity and Dipole Moments
Figure 13.6
(a)The structure and charge distribution of the ammonia molecule. The polarity
of the N-H bonds occurs because nitrogen has a greater electronegativity than
hydrogen. (b) The dipole moment of the ammonia molecule oriented in an
electric field. (c) The electrostatic potential diagram for ammonia.
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13.3 Bond Polarity and Dipole Moments
Figure 13.7
(a) The carbon dioxide molecule. (b) The opposed bond polarities cancel
out, and the carbon dioxide molecule has no dipole moment. (c) The
electrostatic potential diagram for carbon dioxide.
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13.3 Bond Polarity and Dipole Moments
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Example 13.2 Bond Polarity and
Dipole Moment
 For each of the following molecules, show the direction of the
bond polarities. Also indicate which ones have dipole moments:
HCl, Cl2, SO3 (planar), CH4 (tetrahedral), and H2S (V-shaped).
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Example 13.2 Solution
The HCl molecule: Because the electronegativity of chlorine (3.2)
is greater than that of hydrogen (2.2), the chlorine is partially
negative, and the hydrogen is partially positive.
The Cl2 molecule: Because the two chlorine atoms share the
electrons equally, no bond polarity occurs. The Cl2 molecule has
no dipole moment.
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Example 13.2 Solution
The SO3 molecule
However, the bond polarities
cancel, and the molecule has no
dipole moment.
The CH4 molecule
Since the bond polarities cancel, the
molecule has no dipole moment.
The H2S molecule:
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13.4 Ions: Electron Configurations and
Sizes
 (1) When two nonmetals react to form a covalent bond, they share
electrons in a way that both nonmetals attain noble gas electron
configurations. 完成八隅體
 (2) When a nonmetal and a representative group metal react to
form a binary ionic compound, the ions form so that the valence
electron configuration of the nonmetal is completed and the
valence orbitals of the metal are emptied. In this way both ions
achieve noble gas electron configurations.
 Predicting Formulas of Ionic Compounds
 MgO(s), which contains Mg2+ and O2- ions , is very stable, but the
isolated gas-phase ion pair Mg2+ … O2- ions is not energetically
favorable in comparison with the separate neutral gaseous atoms.
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13.4 Ions: Electron Configurations and
Sizes
 (a) In the solid state of an ionic compound, the ions are relatively
close together, and many ions are simultaneously interacting:
(b)
(a)
 (b) In the gas phase of an ionic substance, the ions would be
relatively far apart and would not contain large groups of ions:
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13.4 Ions: Electron Configurations and
Sizes
 The electronegativity of oxygen (3.4) is much greater than that of
calcium (1.0). Because of this large difference, electrons will be
transferred from calcium to oxygen to form oxygen anions and
calcium cations in the compound.
 Note that oxygen needs two electrons to fill its 2s and 2p valence
orbitals and to achieve the configuration of neon (1s22s22p6). And
by losing two electrons, calcium can achieve the configuration of
argon. Two electrons are therefore transferred:
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13.4 Ions: Electron Configurations and
Sizes
 Because aluminum has the configuration [Ne]3s23p1, it must lose
three electrons to form the Al3+ ion and thus achieve the neon
configuration.
 Since the compound must be electrically neutral, there must be
three O-2 ions for every two Al3+ ions, and the compound has the
empirical formula Al2O3.
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13.4 Ions: Electron Configurations and
Sizes
Figure 13.8 Size generally increases down a group. In a series of isoelectronic ions,
size decreases with increasing atomic number.
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13.5 Formation of Binary Ionic
Compounds
 lattice energy—the change in energy that takes place when
separated gaseous ions are packed together to form an ionic solid:
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13.5 Formation of Binary Ionic
Compounds
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Figure 13.9 The energy changes involved in the formation of solid lithium
fluoride from its elements
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13.5 Formation of Binary Ionic
Compounds
Figure 13.10 The structure of lithium fluoride. Represented by a balland-stick model. Note that each Li+ ion is Surrounded by six F- ions, and
each F- ion is surrounded by six Li+ ions. (b) Represented with the ions
shown as spheres. The structure is determined by packing the spherical
ions in a way that both maximizes the ionic attractions and minimizes
the ionic repulsions.
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13.5 Formation of Binary Ionic
Compounds
 Lattice Energy Calculations
 The energy released when the gaseous Mg2+ and O2- ions combine
to form solid MgO is much greater (more than four times greater)
than that released when the gaseous Na+ and F- ions combine to
form solid NaF.
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13.5 Formation of Binary Ionic
Compounds
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13.5 Formation of Binary Ionic
Compounds

Ionization energy: The energy required to remove two electrons
from the magnesium atom (735 kJ/mol for the first and 1445 kJ/mol for
the second, yielding a total of 2180 kJ/mol) is much greater than the
energy required to remove an electron from a sodium atom (495
kJ/mol).

Electron affinity: Energy (737 kJ/mol) is required to add two electrons
to the oxygen atom in the gas phase. Addition of the first electron is
exothermic (2141 kJ/mol), but addition of the second electron is quite
endothermic (878 kJ/mol). This latter energy must be obtained indirectly,
since the O2-(g) is not stable.

The answer lies in the lattice energy. Note that the lattice energy
for combining gaseous Mg2+ and O2- ions to form MgO(s) is 3000 kJ/mol
more negative than that for combining gaseous Na+ and F- ions to form
NaF(s).
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13.6 Partial Ionic Character of Covalent
Bonds
 Partial Ionic Character of Covalent Bonds
(a)
(b)
(c)
Figure 13.12
The three possible types of bonds: a covalent bond formed between
identical F atoms; the polar covalent bond of HF, with both ionic and
covalent components; and (c) an ionic bond with no electron sharing.
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13.6 Partial Ionic Character of Covalent
Bonds
Figure 13.13 The relationship between the ionic character of a covalent bond
and the electronegativity difference of the bonded atoms. The compounds
normally considered to be ionic in the solid phase are shown in red.
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13.7 The Covalent Chemical Bond: A
Model
 Chemical bonds can be viewed as forces that cause a group of
atoms to behave as a unit.
 Example 1 : CH4
 about 1652 kJ of energy is required to break a mole of methane
(CH4) molecules into separate C and H atoms.
 The energy of stabilization of CH4 is divided equally among the four
bonds to give an average C-H bond energy per mole of C-H bonds:
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13.7 The Covalent Chemical Bond: A
Model
 Example 2 : CH3Cl
 Experiments have shown that about 1578 kJ of energy is required
to break down 1 mole of gaseous CH3Cl molecules into gaseous
carbon, chlorine, and hydrogen atoms. The reverse process can be
represented as
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13.8 Covalent Bond Energies and
Chemical Reactions
 Example : CH4
 This example shows that the C-H bond is somewhat sensitive to its
environment. We use the average of these individual bond
dissociation energies even though this quantity only approximates
the energy associated with a C-H bond in a particular molecule.
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13.8 Covalent Bond Energies and
Chemical Reactions
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13.8 Covalent Bond Energies and
Chemical Reactions
 Bond Energy and Enthalpy
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13.8 Covalent Bond Energies and
Chemical Reactions
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13.9 The Localized Electron Bonding
Model

The localized electron (LE) model assumes that a molecule is
composed of atoms that are bound together by using atomic orbitals to
分子軌域
share electron pairs.

The LE model has three parts:

1. Description of the valence electron arrangement in the molecule
using Lewis structures (will be discussed in the next section).

2. Prediction of the geometry of the molecule, using the valence shell
electron pair repulsion (VSEPR) model (will be discussed in Section
13.13).

3. Description of the types of atomic orbitals used by the atoms to share
electrons or hold lone pairs (will be discussed in Chapter 14).
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13.10 Lewis Structures
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13.10 Lewis Structures
 The Lewis structure of a molecule represents the arrangement
of valence electrons among the atoms in the molecule.
 Lewis structures show only valence the electrons.
 Using dots to represent electrons
 No dots are shown in the K+ ion since it has no valence electrons.
The Br- ion is shown with eight electrons since it has a filled
valence shell.
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13.10 Lewis Structures
 By sharing electrons, each hydrogen in H2 has two electrons. This
gives each hydrogen a filled valence shell.
 The shared pair of electrons is called a bonding pair.
 Each fluorine atom also has three pairs of electrons not involved in
bonding. These are the lone pairs.
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13.10 Lewis Structures
 Neon does not form bonds since it already has an octet of valence
electrons (it is a noble gas).
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13.10 Lewis Structures
 Example 1: H2O
 Step 1: We sum the valence electrons for H2O as shown:
 Step 2: Using one pair of electrons per bond, we draw in the two
O-H single bonds:
 Note that a line instead of a pair of dots is used to indicate each
pair of bonding electrons.
 Step 3: We distribute the remaining electrons around the atoms to
achieve a noble gas electron configuration for each atom.
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13.10 Lewis Structures
 Example 2: CO2
 1. The total number of electrons. There are 16 valence electrons in
this structure, which is the correct number.
 2. The octet rule for each atom. Each oxygen has eight electrons,
but the carbon has only four. This cannot be the correct Lewis
structure.
 Now each atom is surrounded by 8 electrons, and the total number
of electrons is 16, as required. Thus the correct Lewis structure for
carbon dioxide has two double bonds.
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13.10 Lewis Structures
 Example 3: CN-
 After drawing a single bond (C-N), we distribute the remaining
electrons to achieve a noble gas configuration for each atom. Eight
electrons remain to be distributed.
 This structure is incorrect because C and N have only six electrons
each instead of eight.
 both carbon and nitrogen have eight electrons
Correct!!
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13.10 Lewis Structures
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13.11 Resonance
 Resonance is invoked when more than one valid Lewis structure
can be written for a particular molecule.
 The nitrate ion does not have one double and two single bonds—it
has three equivalent bonds. The nitrate ion does not exist as any of
the three extreme forms as indicated by the Lewis structures.
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Example 13.7 Resonance Structures
 Describe the electron arrangement in the nitrite anion (NO2-)
using the LE model.
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Example 13.7 Solution
5 + 2(6) +1 = 18 valence electrons.
The remaining 14 electrons (18 - 4) can be distributed to produce
these structures:
The electronic structure of the molecule is not correctly represented
by either resonance structure but by the average of the two.
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13.12 Exceptions to the Octet Rule
 Note that in this structure boron has only six electrons around it.
 The octet rule for boron
 since fluorine is so much more electronegative than boron, this
structure seems questionable.
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13.12 Exceptions to the Octet Rule
 The sum of the valence electrons for SF6 is 6 + 6(7) = 48 electrons
 The empty 3d orbitals on sulfur can be used to accommodate
extra electrons: The sulfur atom can have 12 electrons around it
by using the 3s and 3p orbitals to hold 8 electrons, with the extra
4 electrons in the formerly empty 3d orbitals.
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13.12 Exceptions to the Octet Rule
 Lewis Structures and the Octet Rule
 1. The second-row elements C, N, O, and F should always be
assumed to obey the octet rule.
 2. The second-row elements B and Be often have fewer than eight
electrons around them in their compounds. These electrondeficient compounds are very reactive.(沒有超八，因為不夠八個)
 3. The second-row elements never exceed the octet rule, since
their valence orbitals (2s and 2p) can accommodate only eight
electrons.(第二行不可能有超八)
 4. Third-row and heavier elements often satisfy the octet rule but
are assumed in the simplest model to exceed the octet rule by
using their empty valence d orbitals.(有可能有超八)
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13.12 Exceptions to the Octet Rule
 Lewis Structures and the Octet Rule
 5. When writing the Lewis structure for a molecule, first draw
single bonds between all bonded atoms, and then satisfy the octet
rule for all the atoms. If electrons remain after the octet rule has
been satisfied, place them on the elements having available d
orbitals (elements in the third period or beyond).
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Example 13.9 Lewis Structures for
Molecules That Violate the Octet Rule II
 Write the Lewis structure for each molecule or ion.
 a. ClF3
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b. XeO3
c. RnCl2
d. BeCl2
e. ICl4
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Example 13.9 Solution
a. ClF3
b. XeO3
e. ICl4
c. RnCl2
d. BeCl2
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13.12 Exceptions to the Octet Rule
 Hyperconjugation
 Hyperconjugation involves binding n atoms to a central atom using
fewer than n electron pairs around the central atom.
 For the cases of SF6 and PF5,
 There are two extreme points of view:
(1)
(1) The S and P atoms in these molecules exceed the octet rule,
placing the extra electrons in 3d orbitals,

(2) the S and P atoms in these molecules obey the octet rule,
resorting to hyperconjugation.
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13.12 Exceptions to the Octet Rule
Note that in each of these resonance structures, the sulfur atom
always has an octet of electrons around it. The “true” structure is a
composite of these equivalent resonance structures.
The overall bonding is described as a combination of covalent and
ionic contributions. The covalent contribution to the bonding involves
four electron pairs “spread out” over the six S–F bonds.
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13.12 Exceptions to the Octet Rule
 The ionic contribution to the bonding
 Note that for each Lewis structure two of the fluorines have 21
formal charges and the sulfur has a 12 formal charge,
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13.12 Exceptions to the Octet Rule
 Formal Charge
 How do we decide which of the many possible Lewis structures
best describes the actual bonding in molecules or polyatomic ions
often have many nonequivalent Lewis structures?
 One method involves estimating the charge on each atom in the
various possible Lewis structures and using these charges to select
the most appropriate structure(s).
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13.12 Exceptions to the Octet Rule
 1. Lone pair electrons belong entirely to the atom in question.
 2. Shared electrons are divided equally between the two sharing atoms.
 two of the possible Lewis structures for the sulfate ion, which has
32 valence electrons.
 One possible Lewis structure:
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13.12 Exceptions to the Octet Rule
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13.12 Exceptions to the Octet Rule
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13.12 Exceptions to the Octet Rule
 Second possible Lewis structure:
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13.12 Exceptions to the Octet Rule
 Second possible Lewis structure:
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13.12 Exceptions to the Octet Rule
 electrons will naturally flow from negatively charged parts of the
molecule to positively charged parts, thus minimizing the charges on
the atoms.
 Note that these structures exceed the octet rule, thus requiring the
sulfur to use its 3d orbitals to hold electrons.
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13.12 Exceptions to the Octet Rule
 The other school of thought argues for the primacy of the octet
rule and against the use of 3d orbitals. This position favors the
single-bonded resonance structure.
Which point of view is correct?
 One pertinent fact is that the sulfur–oxygen bonds in SO42- are
known by experiment to be shorter than expected for normal
single bonds.
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13.12 Exceptions to the Octet Rule
 This would seem to favor the resonance structures with double
bonds. However, one can also argue that the high formal charges
on the atoms in the single-bonded structure cause ionic
attractions that pull the atoms closer together.
 Thus both schools of thought can adequately explain the short
sulfur–oxygen bond lengths.
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13.12 Exceptions to the Octet Rule
 Rules Governing Formal Charge
 1. To calculate the formal charge on an atom:

• Take the sum of the lone pair electrons and one-half of the shared
electrons.

• Subtract the number of assigned electrons from the number of valence
electrons on the free, neutral atom to obtain the formal charge.
 2. The sum of the formal charges of all atoms in a given molecule or
ion must equal the overall charge on that species.
 3. If nonequivalent Lewis structures exist for a species containing
second-row atoms, 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.
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Example 13.10 Formal Charges
 Give possible Lewis structures for XeO3, an explosive compound
of xenon. Determine the formal charges of each atom in the
various Lewis structures.
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Example 13.10 Solution
For XeO3 (26 valence electrons), the following possible Lewis
structures (formal charges are indicated in parentheses):
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13.12 Exceptions to the Octet Rule
 The concept of formal charge is most often used to evaluate the
importance of various Lewis structures for molecules that exhibit
resonance. However, formal charge arguments also can be helpful in
predicting which, among a given group of atoms, is the central atom
in a simple molecule.
 Example: CO2
 why is carbon dioxide O-C-O rather than C-O-O?
 all atoms have formal charges of 0.
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13.12 Exceptions to the Octet Rule
 all the Lewis structures give unreasonable formal charges. Consider
the following possibilities, where the formal charges are listed
below each atom.
 None of these Lewis structures (with their resulting formal charges)
agrees with our observation that oxygen has a significantly greater
electronegativity than carbon. It doesn’t make sense that a
negatively charged carbon atom next to a positively charged oxygen
atom.
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13.13 Molecular Structure: The VSEPR
Model
 the valence shell electron-pair repulsion (VSEPR) model:
the structure around a given atom is determined principally by
minimizing electron-pair repulsions.
 (1) BeCl2, which has the Lewis structure
 the best arrangement places the pairs on opposite sides of the
beryllium atom at 180 degrees:
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13.13 Molecular Structure: The VSEPR
Model
 (2) BF3, which has the Lewis structure
 To minimize the repulsions, the electron pairs are farthest apart at
angles of 120 degrees:
trigonal planar
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13.13 Molecular Structure: The VSEPR
Model
 (3) CH4 the methane molecule, which has the Lewis structure
?
 The carbon atom and the electron pairs are centered in the plane
of the paper, and the angles between the pairs are all 90 degrees.
tetrahedral :
angles of 109.5
degrees
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13.13 Molecular Structure: The VSEPR
Model
Example: ammonia, NH3
Draw the Lewis structure.
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13.13 Molecular Structure: The VSEPR
Model
 Count the pairs of electrons, and arrange them to
minimize repulsions.
 The NH3 molecule has four pairs of electrons: three bonding pairs
and one nonbonding pair. From the discussion of the methane
molecule, we know that the best arrangement of four electron pairs
is a tetrahedral array, as shown below
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13.13 Molecular Structure: The VSEPR
Model
 Determine the positions of the atoms.
 Name the molecular structure.
 the name of the molecular structure is always based on the
positions of the atoms
 The molecular structure of ammonia is a trigonal pyramid
 (one triangular side is different from the other three), rather than
a85tetrahedron.
Zumdahl/Decoste | Chemical Principles, 8e, © 2017 Cengage Learning
13.13 Molecular Structure: The VSEPR
Model
 Determine the positions of the atoms.
 Name the molecular structure.
 the name of the molecular structure is always based on the
positions of the atoms
 The molecular structure of ammonia is a trigonal pyramid
 (one triangular side is different from the other three), rather than a
tetrahedron.
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Example 13.11 Prediction of Molecular
Structure I
 Describe the molecular structure of the water molecule.
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Example 13.11 Solution
The Lewis structure for water is
There are four pairs of electrons: two bonding pairs and two nonbonding pairs. To minimize repulsions, these are best arranged in a
tetrahedral array,
The atoms in the H2O
molecule form a V-shape
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13.13 Molecular Structure: The VSEPR
Model
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Figure 13.18 The bond angles in the CH4, NH3, and H2O molecules.
Note that the bond angle between bonding pairs decreases as the number
of lone pairs increases.
Zumdahl/Decoste | Chemical Principles, 8e, © 2017 Cengage Learning
13.13 Molecular Structure: The VSEPR
Model
 Lone pairs require more room than bonding pairs and tend to
compress the angles between the bonding pairs.
Figure 13.19
(a) In a bonding pair of electrons, the
electrons are shared by two nuclei.
(b) In a lone pair, since both electrons
must be close to a single nucleus, they
tend to take up more of the space
around that atom.
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13.13 Molecular Structure: The VSEPR
Model
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13.13 Molecular Structure: The VSEPR
Model
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13.13 Molecular Structure: The VSEPR
Model
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Model
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Model
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13.13 Molecular Structure: The VSEPR
Model
 The VSEPR Model and Multiple Bonds
 Example: three resonance structures of NO3-
 This planar structure is the one expected for three pairs of
electrons around a central atom.
 For the VSEPR model, multiple bonds count as one effective
electron pair.
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13.13 Molecular Structure: The VSEPR
Model
 Molecules Containing No Single Central Atom
 Example: methanol
(a)
(b)
(c)
Figure 13.22
The molecular structure of methanol. (a) The arrangement of electron pairs
and atoms around the carbon atom. (b) The arrangement of bonding and
lone pairs around the oxygen atom. (c) The molecular structure.
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13.13 Molecular Structure: The VSEPR
Model
 The VSEPR Model—How Well Does It Work?
 The VSEPR model is very simple with only a few rules to
remember, but it fails in a few instances.
 Example:
 phosphine (PH) has a Lewis structure analogous to NH3. However,
the bond angles of phosphine are actually 94 degrees, NOT about
107 degrees like the case of NH3.
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