Chapters 8 and 9
Covalent Bonding and Molecular
Structures
Objectives
You will be able to:
1. Write a description of the formation of the covalent bond between two hydrogen
atoms to form a hydrogen molecule. Your description should include mention
of overlapping atomic orbitals to form molecular orbitals. It should also include
rough pictures of the atomic orbitals and the molecular orbitals.
2. Write a description of the assumptions made in the Linear Combination of
Atomic Orbitals (LCAO) approximation for the molecular orbital approach to
covalent bonding.
3. Write an explanation for why bonding molecular orbitals are more stable and
why antibonding molecular orbitals are less stable than the separate atomic
orbitals that form them.
4. Write a description of how the σ1s, σ*1s, σ2s, σ*2s, σ2p, σ*2p, π*2p, and π*2p are
formed from the 1s, 2s and 2p atomic orbitals.
5. Draw sketches of the σ1s, σ*1s, σ2p, σ*2p, π2p, and π*2p molecular orbitals.
6. Write an explanation for why the σ2p molecular orbital of O2 is lower energy
than the π2p molecular orbitals and why the σ*2p is higher energy than the π*2p.
7. Draw molecular orbital diagrams for H2, O2, F2, and Ne2.
8. From a molecular orbital diagram of a molecule, predict its bond order.
9. From a molecular orbital diagram for a molecule, predict whether it is
paramagnetic or diamagnetic.
10.Draw the molecular orbital diagrams for CO and NO and use them to predict
the bond order of each and to explain why each is relatively stable despite their
strange Lewis Structures.
11.Describe the advantages and disadvantages of scientific models.
12.Write a list of the basic assumptions of the valence‑bond model of covalent
bonding.
13.Write or identify the number of valence electrons for each of the Representative
elements.
14.With reference to the valence bond model, describe how the covalent bond forms
between fluorine atoms in F2.
15.Write or identify a description of the information given in a Lewis structure.
16.With reference to the valence bond model, describe how the covalent bond forms
between hydrogen atoms in H2.
17.Explain how the carbon atom in CH4 is able to form four equivalent covalent
bonds to hydrogen atoms with bond angles of 109.5°.
18.Describe (1) how sp3 hybrid orbitals form, (2) their shapes, and (3) how they are
arranged in space.
19.Write or identify a description of the tetrahedral electron group geometry.
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Covalent Bonding and Molecular Structures
20.For ethene (ethylene) molecules, C2H4: (1) explain how the carbon atoms are
able to form four equivalent covalent bonds to hydrogen atoms, (2) explain why
all the bond angles are about 120°, (3) with reference to the valence bond model,
explain how the double bond between carbon atoms forms, and (4) explain why
one bond in the double bond is weaker than the other.
21.Describe (1) how sp2 hybrid orbitals form, (2) their shapes, and (3) how they are
arranged in space.
22.Write or identify a description of the trigonal planar electron group geometry.
23.For ethyne (acetylene) molecules, C2H2: (1) explain how the carbon atoms are
able to form two equivalent covalent bonds to hydrogen atoms, (2) explain why
the bond angles are 180°, (3) with reference to the valence bond model, explain
how the triple bond between carbon atoms forms, and (4) explain why two
bonds in the triple bond are weaker than the other one.
24.Describe (1) how sp hybrid orbitals form, (2) their shapes, and (3) how they are
arranged in space.
25.Write or identify a description of the linear electron group geometry.
26.Explain how the nitrogen atom in NH3 is able to form three equivalent covalent
bonds to hydrogen atoms with bond angles of 107°.
27.Write or identify a description of the trigonal pyramid molecular geometry.
28.Explain how the nitrogen atom in NH4+ is able to form four equivalent covalent
bonds to hydrogen atoms with bond angles of 109.5°.
29.Explain how the oxygen atom in H2O is able to form two equivalent covalent
bonds to hydrogen atoms with bond angles of 105°.
30.Write or identify a description of the bent molecular geometry.
31.With reference to the valence bond model, explain how the triple bond forms in
the carbon monoxide, CO, molecule.
32.Explain how the boron atom in BF3 is able to form three equivalent covalent
bonds to fluorine atoms with bond angles of 120°.
33.Explain how the phosphorus atom in PF5 is able to form five equivalent covalent
bonds to fluorine atoms with bond angles of 90°, 120°, and 180°.
34.Write or identify a description of the trigonal bipyramid electron group geometry.
35.Explain how the sulfur atom in SF4 is able to form four covalent bonds to
fluorine atoms with bond angles of 90°, 120°, and 180°.
36.Write or identify a description of the see-saw molecular geometry.
37.Explain how the iodine atom in IF3 is able to form three covalent bonds to
fluorine atoms with bond angles of 90° and 180°.
38.Write or identify a description of the T-shaped molecular geometry.
39.Explain how the xenon atom in XeF2 is able to form two covalent bonds to
fluorine atoms with a bond angle of 180°.
40.Write or identify a description of the linear molecular geometry.
41.Explain how the sulfur atom in SF6 is able to form six equivalent covalent bonds
to fluorine atoms with bond angles of 90° and 180°.
42.Write or identify a description of the octahedral electron geometry.
43.Explain how the iodine atom in IF5 is able to form five equivalent covalent bonds
to fluorine atoms with bond angles of 90° and 180°.
44.Write or identify a description of the square pyramid molecular group geometry.
159
45.Explain how the xenon atom in XeF4 is able to form four covalent bonds to
fluorine atoms with bond angles of 90° and 180°.
46.Write or identify a description of the square planar molecular geometry.
47.Write an explanation for why the elements listed on the Table 8,9.3 can have each
of the bonding patterns listed.
48.Given a Lewis structure or enough information to get one, do the following: (1)
list the hybridization for each atom in the structure, (2) write a description of the
formation of each bond in terms of the overlap of atomic orbitals, (3) describe
the electron group geometry around each atom that has two or more atoms
attached to it, (4) draw the geometric sketch of the molecule, including bond
angles, and (5) describe the molecular geometry around each atom that has two
or more atoms attached to it.
49.Given a Lewis structure or enough information to get one, determine the formal
charge for each atom in the structure.
50.Given a formula for a molecule or polyatomic ion, draw a reasonable Lewis
structure that corresponds to the formula.
51.Given a Lewis structure or enough information to get one, do the following:
(1) identify whether it is best described as having resonance or not, (2) if can
be seen as having resonance, draw the Lewis Structures for all of the reasonable
resonance forms, and (3) draw the resonance hybrid for the formula.
52.Draw or identify cis and trans isomers for molecules with carbon-carbon double
bonds.
53. Given a structure for a triglyceride, identify it as saturated or unsaturated, and
identify whether it is more likely to be solid or liquid at room temperature.
54. Given a formula for a molecular compound, predict whether or not it represents
a polar molecule.
55.Write a description of how London forces come to be.
56.Write an explanation for why larger molecules have stronger London forces than
smaller molecules.
57.Write a description of how London forces play a part in the attractions between
polar molecules.
58.Given a formula for a pure substance, identify which of the following types of
substances it represents: ionic compound, metallic substance, polar molecular
compound with hydrogen bonds, polar molecular substance without hydrogen
bonds, or nonpolar molecular substance.
59.Identify carbon in the diamond form, C(dia), and silicon dioxide, SiO2 as
examples of network crystals.
60.Given a formula for a pure substance, identify which of the following attractions
is the predominant force holding its particles in the liquid and solid state: ionic
bonds, metallic bonds, covalent bonds, dipole-dipole attractions, hydrogen
bonds, or London forces.
61.Given chemical formulas for two substances that are about the same size but
have different types of attractions between the particles, predict which substance
would have the strongest attractions between the particles in the liquid and solid
state.
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Covalent Bonding and Molecular Structures
62.Given chemical formulas for two substances with the same type of attraction
between the particles but different sizes, predict which substance would have the
stronger attractions between the particles in the liquid and solid state.
63. Given descriptions of two compounds that differ in the percentages of their
structures that are polar and nonpolar (hydrophilic and hydrophobic), predict
their relative solubilities in water and in hexane.
64.Explain why amphetamine can pass through the blood brain barrier more easily
than epinephrine.
65.Explain why methamphetamine is often converted to methamphetamine
hydrochloride.
66.Convert between the definition and the term for the following words or phrases.
Chapters 8 and 9 Glossary
Molecular orbital A volume that contains a high percentage of the electron charge for
an electron in a molecule or a volume within which an electron in a molecule has
a high probability of being found.
Bonding molecular orbital Formed from in-phase interaction of two atomic orbitals.
This leads to an increase in negative charge between two nuclei where the atomic
orbitals overlap and leads to more +/- attraction between the negative charge
generated by the electrons and the nuclei.
Antibonding molecular orbital Formed from out-of-phase interaction of two atomic
orbitals. This leads to a decrease in negative charge between two nuclei where the
atomic orbitals overlap and leads to less +/- attraction between the negative charge
generated by the electrons and the nuclei.
Valence bond model The model for covalent bonding that includes the assumptions
that (1) only the highest-energy (valence) electrons participate in bonding, (2)
covalent bonds arise due to the overlap of atomic orbitals on adjacent atoms,
forming molecular orbitals, and (3) covalent bonds often form to pair unpaired
electrons.
Valence electrons The highest‑energy s and p electrons for an atom. The electrons that
are most important in the formation of chemical bonds. The number of valence
electrons for the atoms of an element is equal to the element’s A-group number on
the periodic table.
Electron-dot symbol A representation of an atom that consists of its elemental symbol
surrounded by dots representing its valence electrons.
Lone pair Two electrons that are not involved in the covalent bonds between atoms
but are important for explaining the arrangement of atoms in molecules. They are
represented by pairs of dots in Lewis structures.
Lewis structure A representation of a molecule that consists of the elemental symbol
for each atom in the molecule, lines to show covalent bonds, and pairs of dots to
indicate lone pairs.
Hybrid orbital An atomic orbital formed from a blend of atomic orbitals, e.g. four sp3
hybrid orbitals are formed from the blend of one s orbital and three p orbitals.
Bond angle The angle formed by straight lines (representing bonds) connecting the
nuclei of three adjacent atoms.
161
Electron group geometry A description of the arrangement of all the electron groups
around a central atom in a molecule or polyatomic ion, including the lone pairs.
Molecular geometry The description of the arrangement of all the atoms around a
central atom in a molecule or polyatomic ion. This description does not consider
lone pairs.
Isomers Compounds that have the same molecular formula but different molecular
structures.
Formal charge A tool for evaluating Lewis structures, calculated from the formula:
A-group number minus number of bonds minus number of electrons in lone
pairs.
Resonance structures Two or more Lewis structures for a single molecule or polyatomic
ion that differ in the positions of lone pairs and multiple bonds but not in the
positions of the atoms in the structure. It is as if the molecule or ion were able to
shift from one of these structures to another by shifting pairs of electrons from one
position to another.
Resonance The hypothetical switching from one resonance structure to another.
Resonance hybrid A structure that represents the average of the resonance structures
for a molecule or polyatomic ion.
Delocalized electrons Electrons that are shared among three or more atoms.
Delocalized pi system A system of overlapping p orbitals used to describe the bonding
in resonance hybrids.
Cis isomer A structure that has like groups on different carbons (which are linked by
a double bond) and on the same side of the double bond.
Trans isomer A structure that has like groups on different carbons (which are linked
by a double bond) and on different sides of the double bond.
Triglyceride A compound with three hydrocarbon groups attached to a three carbon
backbone by ester functional groups.
Saturated fat A triglyceride with single bonds between all of the carbon atoms.
Unsaturated fat A triglyceride that has one or more carbon carbon double bonds.
Hydrogenation A process by which hydrogen is added to an unsaturated triglyceride
to convert double bonds to single bonds. This can be done by combining the
unsaturated triglyceride with hydrogen gas and a platinum catalyst.
Noncovalent interaction All forces of attraction between particles other than covalent,
ionic, or metallic bonds.
Intermolecular forces Attractions between molecules.
Dipole-dipole attraction The intermolecular attraction between the partial negative
end of one polar molecule and the partial positive end of another polar molecule.
Hydrogen bond The intermolecular attraction between a nitrogen, oxygen, or fluorine
atom of one molecule and a hydrogen atom bonded to a nitrogen, oxygen, or
fluorine atom in another molecule.
London dispersion forces, London forces, or dispersion forces The attractions
produced between molecules by instantaneous and induced dipoles.
Hydrophilic (“water loving”): A polar molecule or ion (or a portion of a molecule or
polyatomic ion) that is attracted to water.
Hydrophobic (“water fearing”): A nonpolar molecule (or a portion of a molecule or
polyatomic ion) that is not expected to mix with water.
Lipid bilayer Cell membrane.
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Covalent Bonding and Molecular Structures
Molecular Orbital Theory
The goal of molecular orbital theory is to describe molecules in a similar way to how we
describe atoms, that is, in terms of orbitals, orbital diagrams, and electron configurations.
For example, to give you a glimpse at where we are headed, the following are orbital
diagrams for O2 and O.
∗
σ2p
π ∗2p
π 2p
π∗2p
π 2p
σ2p
∗
σ2s
σ2s
∗
σ1s
σ1s
O2
2s
1s
2p
O
Each line in the molecular orbital diagram represents a molecular orbital, which
is the volume within which a high percentage of the negative charge generated by the
electron is found. The molecular orbital volume encompasses the whole molecule. We
assume that the electrons would fill the molecular orbitals of molecules like electrons
fill atomic orbitals in atoms.
• The molecular orbitals are filled in a way that yields the lowest potential
energy for the molecule.
• The maximum number of electrons in each molecular orbital is two. (We
follow the Pauli exclusion principle.)
• Orbitals of equal energy are half filled with parallel spin before they begin to
pair up. (We follow Hund’s Rule.)
Before we continue with a description of a model used to generate molecular orbital
diagrams, let’s get a review of light and electron waves and how two waves can interact.
The wave description of light describes the effect that the light has on the space around
it. This effect is to generate an oscillating electric and magnetic fields. These fields can
vary in intensity, which is reflected in varying brightness of light.
163
Source
Wavelength, λ, the distance
between two peaks
Magnetic field,
perpendicular to electric field
Electric field,
perpendicular to
magnetic field
Figure 8,9.1
Electromagnetic
Waves
Radiant
energy
The wave description of the electron describes the variation in the intensity of
negative charge generated by the electron.
Nucleus, about 0.00001
the diameter of the atom
The negative charge is most
intense at the nucleus
and decreases in intensity
with distance outward.
Figure 8,9.2
1s Orbital
Light waves can interact in-phase, which leads to an increase in the intensity of the
light (brighter) and out-of-phase, which leads to a decrease in the intensity of the light
(less bright).
Figure 8,9.3
Wave Interference
In-phase
Brighter
Dark
Out-of-phase
Electron waves can also interact in-phase and out-of-phase. In-phase interaction
leads to an increase in the intensity of the negative charge. Out-of-phase interaction
leads to a decrease in the intensity of the negative charge.
One common approximation that allows us to generate molecular orbital diagrams
for some small diatomic molecules is called the Linear Combination of Atomic
Orbitals (LCAO) approach. The following assumptions lie at the core of this model.
• Molecular orbitals are formed from the overlap of atomic orbitals.
• Only atomic orbitals of about the same energy interact to a significant
degree.
• When two atomic orbitals overlap, they interact in two extreme ways
to form two molecular orbitals, a bonding molecular orbital and an
antibonding molecular orbital.
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Covalent Bonding and Molecular Structures
For example, our model assumes that two 1s atomic orbitals can overlap in two
extreme ways to form two molecular orbitals. One of the ways the atomic orbitals
interact is in-phase, which leads to wave enhancement similar to the enhancement of
two in-phase light waves. Where the atomic orbitals overlap, the in‑phase interaction
leads to an increase in the intensity of the negative charge in the region where they
overlap. This creates an increase in negative charge between the nuclei and an increase
in the plus‑minus attraction between the electron charge and the nuclei for the atoms
in the bond. The greater attraction leads to greater stability and lower potential energy.
Because electrons in the molecular orbital are lower potential energy than in separate
atomic orbitals, energy would be required to shift the electrons back into the 1s orbitals
of separate atoms. This keeps the atoms together in the molecule, so we call this orbital
a bonding molecular orbital. The molecular orbital formed is symmetrical about
the axis of the bond. Symmetrical molecular orbitals are called sigma, σ, molecular
orbitals. The symbol σ1s is used to describe the bonding molecular orbital formed from
two 1s atomic orbitals.
The second way that two atomic orbitals interact is out‑of‑phase. Where the atomic
orbitals overlap, the out-of‑phase interaction leads to a decrease in the intensity of the
negative charge. This creates a decrease in negative charge between the nuclei and a
decrease in the plus-minus attraction between the electron charge and the nuclei for
the atoms in the bond. The lesser attraction leads to lower stability and higher potential
energy. The electrons are more stable in the 1s atomic orbitals of separate atoms, so
electrons in this type of molecular orbital destabilize the bond between atoms. We call
this type of molecular orbital an antibonding molecular orbital. The molecular orbital
formed is symmetrical about the axis of the bond, so it is a sigma molecular orbital
with a symbol of σ*1s. The asterisk indicates an antibonding molecular orbital.
The following diagram shows the bonding and antibonding molecular orbitals
formed from the interaction of two 1s atomic orbitals.
165
Decreased negative charge between the nuclei
leads to decreased attractions between the
negative charge from the electrons and the
positively charged nuclei. This makes the
sigma 1s anitbonding molecular orbital higher
potential energy than the separate 1s atomic orbitals.
σ∗1s
out-ofphase
nuclei
in-phase
1s 1s
σ1s
Increased negative charge between the nuclei
leads to increased attractions between the
negative charge from the electrons and the
positively charged nuclei. This makes the
sigma 1s bonding molecular orbital lower potential
energy than the separate 1s atomic orbitals.
∗
σ1s
1s
1s
σ1s
Figure 8,9.4
Formation of Sigma 1s
Molecular Orbitals
When two larger atoms combine to form a diatomic molecules (such as O2, F2, or
Ne2), more atomic orbitals interact. The LCAO approximation assumes that only the
atomic orbitals of about the same energy interact. For O2, F2, and Ne2, the orbital
energies are different enough in energy so only orbitals of the same energy interact to
a significant degree.
Like for hydrogen, the 1s from one atom overlaps the 1s from the other atom to
form a σ1s bonding molecular orbital and a σ*1s antibonding molecular orbital. The
shapes would be similar to those formed from the 1s orbitals for hydrogen. The 2s
atomic orbital from one atom overlaps the 2s atomic orbital from the other atom
to form a σ2s bonding molecular orbital and a σ*2s antibonding molecular orbital.
The shapes of these molecular orbitals would be similar to those for the σ1s and σ*1s
molecular orbitals. Both σ2s and σ*2s molecular orbitals are higher energy and larger
than the σ1s and σ*1s molecular orbitals.
The p atomic orbitals of the two atoms can interact in two different ways, parallel or
end-on. The molecular orbitals are different for each type of interaction. The end-on
interaction between two 2px atomic orbitals yields sigma molecular orbitals, which are
symmetrical about the axis of the bond.
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Covalent Bonding and Molecular Structures
Figure 8,9.5
Formation of Sigma 2p
Molecular Orbitals
σ*2p
out-ofphase
2px
2px
nuclei
in-phase
σ2p
The two 2py atomic orbitals overlap in parallel and form two pi molecular orbitals. Pi
molecular orbitals are asymmetrical about the axis of the bond.
Figure 8,9.6
Formation of Pi 2p
Molecular Orbitals
π 2p
*
out-ofphase
nuclei
in-phase
2py 2py
π 2p
There is less overlap
between parallel p orbitals
than between two p orbitals
overlapping end-on.
The 2pz-2pz overlap generates another pair of π2p and π*2p molecular orbitals. The
2pz-2pz overlap is similar to the 2py-2py overlap. To visualize this overlap, picture all
of the orbitals in the image above rotated 90°, so that the axes that run through each
atomic and molecular orbitals are perpendicular to the paper. The molecular orbitals
formed have the same potential energy as the molecular orbitals formed from the
2py-2py overlap.
There is less overlap for the parallel atomic orbitals. When the interaction is in-phase,
less overlap leads to less electron charge enhancement between the nuclei. This leads
to less electron charge between the nuclei for the pi bonding molecular orbital than
for the sigma bonding molecular orbital. Less electron character between the nuclei
means less plus-minus attraction, less stabilization, and higher potential energy for the
pi bonding molecular orbital compared to the sigma bonding molecular orbital.
When the interaction is out-of-phase, less overlap leads to less shift of electron
charge from between the nuclei. This leads to more electron charge between the nuclei
for the pi antibonding molecular orbital than for the sigma antibonding molecular
orbital. More electron charge between the nuclei means more plus-minus attraction
and lower potential energy for the pi antibonding molecular orbital compared to the
sigma antibonding molecular orbital.
167
The expected molecular orbital diagram from the overlap of 1s, 2s, and 2p atomic
orbitals is as follows. We will use this diagram to describe O2, F2, Ne2, CO, and NO.
π ∗2p
∗
σ2p
π 2p
π∗2p
π 2p
σ2p
∗
σ2s
σ2s
∗
σ1s
σ1s
We assume that the electrons would fill the molecular orbitals of molecules like
electrons fill atomic orbitals in atoms. We use the following procedure when drawing
molecular orbital diagrams.
• Determine the number of electrons in the molecule. We get the number
of electrons per atom from their atomic number on the periodic table.
(Remember to determine the total number of electrons, not just the valence
electrons.)
• Fill the molecular orbitals from bottom to top until all the electrons
are added. Describe the electrons with arrows. Put two arrows in each
molecular orbital, with the first arrow pointing up and the second pointing
down.
• Orbitals of equal energy are half filled with parallel spin before they begin
to pair up.
We describe the stability of the molecule with bond order.
bond order = 1/2 (#e− in bonding MO’s − #e− in antibonding MO’s)
We use bond orders to predict the stability of molecules.
• If the bond order for a molecule is equal to zero, the molecule is unstable.
• A bond order of greater than zero suggests a stable molecule.
• The higher the bond order is, the more stable the bond.
We can use the molecular orbital diagram to predict whether the molecule is
paramagnetic or diamagnetic. If all the electrons are paired, the molecule is diamagnetic.
If one or more electrons are unpaired, the molecule is paramagnetic.
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Chapter 8 & 9
Covalent Bonding and Molecular Structures
EXAMPLES:
1. The molecular orbital diagram for a diatomic hydrogen molecule, H2, is
∗
σ1s
σ1s
• The bond order is 1.
Bond Order =
1 2 (2
- 0) = 1
• The bond order above zero suggests that H2 is stable.
• Because there are no unpaired electrons, H2 is diamagnetic.
2. The molecular orbital diagram for a diatomic helium molecule, He2, shows the
following.
∗
σ1s
σ1s
• The bond order is 0 for He2.
Bond Order =
1 2 (2
- 2) = 0
• The zero bond order for He2 suggests that He2 is unstable.
• If He2 did form, it would be diamagnetic.
3. The molecular orbital diagram for a diatomic oxygen molecule, O2, is
π ∗2p
∗
σ2p
π 2p
π∗2p
π 2p
σ2p
∗
σ2s
σ2s
∗
σ1s
σ1s
• O2 has a bond order of 2.
Bond Order =
1 2 (10
- 6) = 2
• The bond order of two suggests that the oxygen molecule is stable.
• The two unpaired electrons show that O2 is paramagnetic.
169
4. The molecular orbital diagram for a diatomic fluorine molecule, F2, is
π ∗2p
∗
σ2p
π 2p
π∗2p
π 2p
σ2p
∗
σ2s
σ2s
∗
σ1s
σ1s
• F2 has a bond order of 1.
Bond Order =
1 2 (10
- 8) = 1
• The bond order of one suggests that the fluorine molecule is stable.
• Because all of the electrons are paired F2 is diamagnetic.
5. The molecular orbital diagram for a diatomic neon molecule, Ne2, is
π ∗2p
∗
σ2p
π 2p
π∗2p
π 2p
σ2p
∗
σ2s
σ2s
∗
σ1s
σ1s
• Ne2 has a bond order of 0.
Bond Order =
1 2 (10
- 10) = 0
• The bond order zero for Ne2 suggests that Ne2 is unstable.
• If Ne2 did form, it would be diamagnetic.
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Chapter 8 & 9
Covalent Bonding and Molecular Structures
We can describe diatomic molecules composed of atoms of different elements in
a similar way. The bond between the carbon and oxygen in carbon monoxide is very
strong despite what looks like a strange and perhaps unstable Lewis Structure.
C
O
The plus formal charge on the more electronegative oxygen and the minus formal
charge on the less electronegative carbon would suggest instability. The molecular
orbital diagram predicts CO to be very stable with a bond order of three.
π ∗2p
∗
σ2p
π 2p
π∗2p
π 2p
σ2p
∗
σ2s
σ2s
∗
σ1s
σ1s
We predict the nitrogen monoxide molecule to be unstable according to the Lewis
approach to bonding.
N
O
The unpaired electron and the lack of an octet of electrons around nitrogen would
suggest an unstable molecule. NO is actually quite stable. The molecular orbital
diagram predicts this by showing the molecule to have a bond order of 2.5.
π ∗2p
∗
σ2p
π 2p
π∗2p
π 2p
σ2p
∗
σ2s
σ2s
∗
σ1s
σ1s
171
Figure 8,9.7
Bonding in F2
Electrons in a sigma molecular orbital
lead to a sigma bond.
nuclei
2px
2px
σ2p
Figure 8,9.8
Bonding in H2
Electrons in a sigma molecular orbital
lead to a sigma bond.
nuclei
1s
1s
σ1s
Formation of sp3 hybrid orbitals
One 2s and three 2p atomic orbitals blend
2s
2px
2py
2pz
to form four equivalent sp3 hybrid orbitals, which are arranged in a
tetrahedral geometry with angles of 109.5°.
sp3
109.5°
109.5°
109.5°
sp3
sp3
109.5°
sp3
sp3
sp3
sp3
109.5°
sp3
Figure 8,9.9
Formation of sp3
Hybrid Atomic Orbitals
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Chapter 8 & 9
Covalent Bonding and Molecular Structures
Overlap of atomic orbitals in methane, CH4
H
109.5°
C
H
H
H
H 1s
sp3
109.5°
109.5°
C
Figure 8,9.10
Bonding in CH4
C
sp3
109.5°
1s
H
H
C
1s
sp3
sp3
1s
H
109.5°
C
Four C-H sigma bonds due to sp3-1s overlap
Formation of sp2 hybrid orbitals
One 2s and two 2p atomic orbitals blend
2s
2px
2py
2pz
to form three equivalent sp2 hybrid orbitals, which are arranged
in a trigonal planar geometry with angles of 120°. This leaves
one 2p atomic orbital unhybridized. The axis through the 2p orbital
is 90° from the axes through the sp2 hybrid orbitals.
sp2
sp2
2p
90°
120°
120°
sp2
120°
2
2p
sp2
Figure 8,9.11
Formation of sp2
Hybrid Atomic Orbitals
sp2
sp2
173
Figure 8,9.12
Bonding in C2H4
Overlap of atomic orbitals in ethene (ethylene), C2H4
1s
H
sp2
2p
2p
sp2
sp2
sp2
1s
H
sp2
sp2
C
C
(with stylized orbitals)
(with stylized orbitals)
2p
2p
sp2
sp2
One C-C sigma bond due to
sp2-sp2 overlap and one C-C
pi bond due to 2p-2p overlap
1s
H
1s
H
sp2
1s
1s sp2
sp2
sp2
4 C-H sigma bonds due to
sp2-1s overlap
1s
1s
174
Chapter 8 & 9
Covalent Bonding and Molecular Structures
Figure 8,9.13
Bonding in C2H2
Overlap of atomic orbitals in ethyne (acetylene), C2H2
2px
2py
2pz
2py
1s
sp
1s
sp
H
sp
C
sp
(with stylized orbitals)
H
C
(with stylized orbitals)
2pz
2pz
sp
C
C
C
C sp
One C-C sigma
bond due to
sp-sp overlap
C
sp
C
One C-C pi
bond due to
2py-2py overlap
1s
sp
H
Two C-H sigma
bonds due to
sp-1s overlap
C
H
Figure 8,9.14
Bonding in NH3
C
One C-C pi
bond due to
2pz-2pz overlap
1s
2py
2py
Overlap of atomic orbitals in ammonia, NH3
lone pair
N
H
H
107°
H
The lone pair is more repulsive than the
bond groups, so the hydrogen atoms are
pushed closer together than we would
predict from sp3 hybridization.
The angle is about 107° instead of 109.5°.
107°
N
sp3
1s
H
N
H
1s
sp3
107°
1s
H
Three N-H sigma bonds due to sp3-1s overlap
sp3
N
107°
175
Figure 8,9.15
Bonding in NH4+
Overlap of atomic orbitals in ammonium, NH4+
H
109.5°
N
H
H
H
H 1s
sp3
109.5°
109.5°
N
N
sp3
109.5°
1s
H
H
1s
N
sp3
sp3
1s
H
109.5°
N
Four N-H sigma bonds due to sp3-1s overlap
Figure 8,9.16
Bonding in H2O
Overlap of atomic orbitals in water, H2O
H
O
lone
pairs
105°
H
The lone pairs are more repulsive than the
bond groups, so the hydrogen atoms are
pushed closer together than we would
predict from sp3 hybridization.
The angle is about 105° instead of 109.5°.
1s H
sp3
Two O-H sigma bonds
due to sp3-1s overlap
O
105°
O
105°
sp3
1s
H
176
Chapter 8 & 9
Covalent Bonding and Molecular Structures
Figure 8,9.17
Bonding in BF3
Overlap of atomic orbitals in boron trifluoride, BF3
F
sp3
sp2
sp2
120°
sp3
B
sp2
sp3
F
F
Three B-F sigma bonds due to sp2-sp3 overlap
Formation of sp3d hybrid orbitals
One s, three p, and one d atomic orbitals blend
3s
3p
3dz2
3p
3dxz
3p
3dyz
3dx2- y2
3dxy
to form five equivalent sp3d hybrid orbitals, which are arranged
in a trigonal bipyramid geometry with angles of 90°, 120°, and 180°.
This leaves four d atomic orbitals unhybridized.
axial
equatorial
90°
90°
120°
90°
equatorial
equatorial
90°
180°
90°
90°
180°
120°
90°
axial
sp3d
sp3d
90°
120°
Figure 8,9.18
Formation of sp3d
Hybrid Atomic Orbitals
sp3d
120°
sp3d
90°
120°
120°
90°
90°
120°
90°
sp3d
180°
120°
90°
177
Figure 8,9.19
Bonding in PF5
Overlap of atomic orbitals in phosphorus pentafluoride, PF5
sp3d-sp3 overlap
sp3d-sp3 overlap
FF
sp3d-sp3 overlap
F
120°
P
180°
90°
F
120°
F
sp3d-sp3 overlap
F
sp3d-sp3 overlap
Five P-F sigma bonds due to sp3d-sp3 overlap
Figure 8,9.20
Bonding in SF4
Overlap of atomic orbitals in sulfur tetrafluoride, SF4
sp3d-sp3 overlap
FF
F
120°
90°
180°
S
F
sp3d-sp3 overlap
sp3d-sp3 overlap
lone pair
sp3d-sp3 overlap
F
Four S-F sigma bonds due to sp3d-sp3 overlap
178
Chapter 8 & 9
Covalent Bonding and Molecular Structures
Figure 8,9.21
Bonding in IF3
Overlap of atomic orbitals in iodine trifluoride, IF3
sp3d-sp3 overlap
FF
sp3d-sp3 overlap
I
lone pairs
180°
90°
F
sp3d-sp3 overlap
F
3
Three I-F sigma bonds due to sp3d-sp3 overlap
Figure 8,9.22
Bonding in XeF2
Overlap of atomic orbitals in xenon difluoride, XeF2
FF
sp3d-sp3 overlap
180°
lone pairs
Xe
lone pair
sp3d-sp3 overlap
F
Two Xe-F sigma bonds due to sp3d-sp3 overlap
179
Figure 8,9.23
Formation of sp3d2
Hybrid Atomic Orbitals
Formation of sp3d2 hybrid orbitals
One s, three p, and two d atomic orbitals blend
3s
3px
3dz2
3py
3dxz
3pz
3dyz
3dx2- y2
3dxy
to form six equivalent sp3d2 hybrid orbitals, which are arranged in an
octahedral geometry with angles of 90° and 180°. This leaves three
d atomic orbitals unhybridized.
sp3d2
sp3d2
180°
90°
180°
180°
90°
90°
sp3d2
sp3d2
90°
90°
90°
sp3d2
sp3d2
180°
180°
180°
sp3d2
sp3d2
sp3d2
90°
sp3d2
180°
sp3d2
90°
sp3d2
Overlap of atomic orbitals in sulfur hexafluoride, SF6
sp3d2-sp3 overlap
sp3d2-sp3 overlap
FF
F
sp3d2-sp3 overlap
F
sp3d2-sp3 overlap
S
F
sp3d2-sp3 overlap
F
F
sp3d2-sp3 overlap
Six S-F sigma bonds due to sp3d2-sp3 overlap
Figure 8,9.24
Bonding in SF6
180
Chapter 8 & 9
Covalent Bonding and Molecular Structures
Figure 8,9.25
Bonding in IF5
Overlap of atomic orbitals in iodine pentafluoride, IF5
sp3d2-sp overlap
sp3d2-sp3 overlap
sp3d2-sp3 overlap
FF
F
F
sp3d2-sp3 overlap
I
F
lone
pair
F
sp3d2-sp3 overlap
Five I-F sigma bonds due to sp3d2-sp3 overlap
Figure 8,9.26
Bonding in XeF4
Overlap of atomic orbitals in xenon tetrafluoride, XeF4
sp3d2-sp3 overlap
sp3d2-sp3 overlap
lone
pair
F
F
sp3d2-sp3 overlap
Xe
F
lone
pair
F
sp3d2-sp3 overlap
Four Xe-F sigma bonds due to sp3d2-sp3 overlap
181
The following assumptions lie at the core of the valence‑bond model.
• Only the highest energy electrons participate in bonding.
• Covalent bonds usually form to pair unpaired electrons.
• Covalent bonds arise due to the overlap of atomic orbitals on adjacent
atoms.
When the columns on the periodic table are numbered by the A group convention,
the number of valence electrons in each atom of a representative element is equal to the
element’s group number on the periodic table.
One valence
electron
1
H
Number of valence
electrons equals the
A-group number
3A
5
B
4A
6
C
5A
7
6A
8
O
15
16
33
As
7A
9
N
P
8A
F
S
34
17
Cl
35
Se
Br
52
53
Te
I
2
He
10
Ne
18
Ar
36
Kr
Figure 8,9.27
Valence Electrons
54
Xe
Table 8,9.1 Most Common Covalent Bonding Patterns
Element
Frequency of
pattern
Number
of bonds
Number of
lone pairs
Example
H
always
1
0
H
B
usually
3
0
B
C
usually
4
0
rarely
3
1
C
usually
3
1
N
commonly
4
0
usually
2
2
commonly
1
3
O
rarely
3
1
O
usually
1
3
X
N, P, & As
O, S, & Se
F, Cl, Br, & I
C
or
C
or
N
O
or
O
C
182
Chapter 8 & 9
Covalent Bonding and Molecular Structures
Formal Charge = Group Number − Number of lines − Number of dots
Formal Charge = (The number of valence electrons necessary to be uncharged)
− (the number of electrons from bonds assuming equal sharing)
− (the number of electrons from lone pairs)
Sample Study
Sheet 8,9.1:
Explanation
of Bonding
Patterns Valence Bond
Model
Tip-off: You will be given a bonding pattern for a nonmetal atoms and asked to
explain how it is possible in terms of the valence bond model.
General Procedure
Step 1 Write, “Only the highest energy electrons participate in bonding.”
Step 2 Draw the orbital diagram for the valence electrons. Include the empty
d orbitals for the third period nonmetals and below.
Step 3 If there is a formal charge, add or subtract electrons.
a. Add one electron for -1.
b. Subtract one electron for +1.
Step 4 If necessary, promote one or more electrons from a pair to an empty
orbital to get the number of unpaired electrons equal to the number of bonds
to be explained.
Step 5 Rewrite the orbital diagram, showing the predicted hybrid orbitals.
Step 6 Write, “Covalent bonds form in order to pair unpaired electrons.”
Step 7 Indicate that the unpaired electrons form the bonds.
Step 8 Indicate that the paired electrons are lone pairs.
Exercise 8,9.1 - Explanation of Bonding Patterns
Explain the following bonding patterns
a. C – 4 bonds, no lone pairs, and no formal charge
b. N – 4 bonds, no lone pairs, and a +1 formal charge
c. O – 1 bond, 3 lone pairs, and a −1 formal charge
d. S – 4 bonds, 1 lone pair, and no formal charge
e. I – 5 bonds, 1 lone pair, and no formal charge
183
Tip-off: You will be given a Lewis structure, and you will be asked the following.
• What is the hybridization for each atom in the structure?
• Write a description of the formation of each bond in terms of the overlap of
atomic orbitals.
• Describe the electron group geometry around each atom that has two or
more atoms attached to it.
• Draw the geometric sketch of the molecule, including bond angles.
• Describe the molecular geometry around each atom that has two or more
atoms attached to it.
General Steps for predicting hybridization for each atom in a structure.
Step 1 Count the number of electron groups around each atom.
An electron group is any one of the following.
• single bond
• double or triple bond (counts as one group)
• lone pair
Step 2 Apply the following guidelines. See Table 8,9.2.
1 group - No hybridization
(Hydrogen atoms are the only atoms that have only one group
around them in a Lewis structure. The 1s orbital is the only occupied
atomic orbital for hydrogen atoms. Because there is no other orbital
with which to blend, there is no hybridization. Hydrogen atoms use
the 1s atomic orbital to form bonds.)
2 groups - sp hybridization
3 groups - sp2 hybridization
4 groups - sp3 hybridization
5 groups - sp3d hybridization
6 groups - sp3d2 hybridization
General Steps for describing the bonds in terms of the overlap of atomic orbitals.
Step 1 Describe single bonds that do not involve hydrogen as due to the
overlapping of two hybrid orbitals, one from each atom.
Step 2 Describe single bonds that involve hydrogen as due to the overlapping a
hybrid orbital with a 1s orbital for the hydrogen atom.
Step 3 Describe double bonds as one bond forming from the overlap of two
hybrid orbitals, one from each atom. Describe the second bond as due to the
overlap of two p orbitals, one from each atom.
Step 4 Describe triple bonds as one bond forming from the overlap of two
hybrid orbitals, one from each atom. Describe each of the other two bonds as
due to the overlap of two p orbitals, one from each atom.
Sample Study
Sheet 8,9.2:
From Lewis
Structures
184
Chapter 8 & 9
Covalent Bonding and Molecular Structures
To determine the electron group geometry around each atom that is attached to
two or more atoms, use the number of electron groups around each atom and the
guidelines found on Table 8,9.2.
Use one or more of the following geometric sketches shown on Table 8,9.2 for the
geometric sketch of you molecule.
To describe the molecular geometry around each atom that has two or more atoms
attached to it, count the number of bond groups and apply the guidelines found on
Table 8,9.2. (Single, double, and triple bonds all count as one bond group.)
Note that if all of the electron groups attached to the atom are bond groups
(no lone pairs), the molecular geometry is the same as the electron group
geometry.
185
Table 8,9.2
Hybridization and Geometry
e−
groups
2
Hybrid
Orbitals
sp
3
sp2
General Geometric Sketch
180°
120°
4
sp3
e− group
geometry
linear
bond
angles
180°
bond
groups
2
lone
pairs
0
molecular
geometry
linear
trigonal
planar
120°
3
0
trigonal
planar
tetrahedral
109.5°
2
1
bent
4
3
0
1
2
2
tetrahedral
trigonal
pyramid
bent
5
0
4
3
2
1
2
3
6
5
0
1
4
2
109.5°
5
sp3d
trigonal
bipyramid
90°
90°,
120°,
180°
180°
120°
120°
6
sp3d2
octahedral
180°
90°
90°,
180°
trigonal
bipyramid
see-saw
T-shaped
linear
180°
90°
octahedral
square
pyramid
square
planar
186
Chapter 8 & 9
Covalent Bonding and Molecular Structures
Table 8,9.3
Common Covalent Bonding Patterns for the Nonmetal Elements
Element
Number of bonds
Number of lone
pairs
Formal charge
H
always 1
0
0
C
usually 4
0
0
rarely 3
1
-1
usually 3
1
0
commonly 4
0
+1
usually 2
2
0
commonly 1
3
-1
rarely 3
1
+1
F
always 1
3
0
P and As
usually 3
1
0
commonly 5
0
0
commonly 4
0
+1
usually 2
2
0
commonly 4
1
0
commonly 6
0
0
commonly 1
3
-1
possibly 3
1
+1
usually 1
3
0
possibly 3
2
0
possibly 5
1
0
possibly 7
0
0
rarely 2
3
0
rarely 4
2
0
N
O
S and Se
halogens except F
Xe
Table 8,9.4
usually none
The Most Common Bonding Pattern for Each Nonmetal Element
Group
Number of Covalent Bonds
Number of Lone Pairs
4A (C)
4
0
5A (N, P, As)
3
1
6A (O, S, Se)
2
2
7A (halogens)
1
3
8A (noble gases)
none
4
187
O
O
O
O N O
O N O
O N O
+
+
+
Tip-off: You are asked to draw a resonance hybrid.
Sample Study
General Steps
Sheet 8,9.3:
• Draw the skeletal structure with solid lines for the bonds that are found in
all of the resonance forms.
• Where there is sometimes a multiple bond and sometimes not, draw a
dotted line.
Resonance
Hybrids
• Draw in all of the lone pairs that are found on every one of the resonance
forms. (Leave off the lone pairs that are on one or more resonance form but
not on all of them.)
• Put on full formal charges on those atoms that have the formal charge in all
of the resonance forms.
• Put on partial formal charges on the atoms that have formal charges in one
or more of the resonance forms but not in all of them.
The resonance hybrid for the nitrate polyatomic ion is below.
O
O N
A bond found in all
the resonance structures
A bond found in at least one
but not all the resonance structures
−
O
A lone pair found in
all the resonance structures
Figure 8,9.28
Resonance Hybrid
188
Chapter 8 & 9
Sample Study
Sheet 8,9.4:
Resonance
and
Resonance
forms
Covalent Bonding and Molecular Structures
Tip-off - You are asked to draw a Lewis structure for a molecule or a polyatomic ion,
and the structure drawn has the following form.
X Y Z
Z can have more than one lone pair.
X and Y can have lone pairs.
The X‑Y bond can be a triple bond. The Y‑Z bond can be a double bond.
Z cannot be F with one bond and three lone pairs or O with two bonds
and two lone pairs.
General Steps: Follow these steps when writing the resonance forms.
• Shift one of the lone pairs on one adjacent atom down to form a multiple
bond.
• Shift one of the multiple bonds up to form a lone pair. You might find it
useful to draw the arrows to indicate the hypothetical shift of electrons.
• Repeat this process for each adjacent atom with a lone pair.
X Y Z
Sample Study
Sheet 8,9.5:
Drawing Lewis
Structures
X Y Z
Tip-off: You are asked to draw a Lewis structure from a chemical formula for a
molecule or a polyatomic ion.
General Steps (As you become more experienced with the process of drawing
Lewis structures, you will see shortcuts that will allow you to bypass some or all of
the following steps.)
Step 1 Determine the number of valence electrons for the molecule or polyatomic
ion.
• For neutral molecules, the total number of valence electrons is the sum of
the valence electrons of each atom.
Remember that the number of valence electrons for any one of the
representative elements is equal to its group number, using the A group
convention for numbering the groups. For example, chlorine, Cl, is in
group 7A, so it has seven valence electrons.
Hydrogen has one valence electron.
• For polyatomic cations, the total number of valence electrons is the sum of
the valence electrons for each atom minus the charge.
• For polyatomic anions, the total number of valence electrons is the sum of
the valence electrons for each atom plus the charge.
189
Step 2 Draw a reasonable skeletal structure joining the atoms with single bonds.
One or more of the following guidelines might help you with this step.
• Try to set the atoms in space to gain the most common number of
bonds for each atom. Table 8,9.3 lists the most common bonding
patterns for the nonmetal elements.
• Apply the following guidelines in deciding what goes in the center of
your structure.
∗ Hydrogen and fluorine atoms are never in the center.
∗ Oxygen atoms are rarely in the center.
∗ The element with the fewest atoms in the formula is usually in
the center.
∗ The atom that is capable of making the most bonds is usually
in the center. See Table 8,9.3.
∗ The atom with the lowest electronegativity is often in the
center.
• Oxygen atoms rarely bond to other oxygen atoms.
• The molecular formula often reflects the structural formula.
• Carbon atoms commonly bond to other carbon atoms.
Step 3 Subtract 2 electrons from the total for each of the single bonds (lines)
described in Step 2 above.
Step 4 Try to distribute the remaining electrons as lone pairs to get eight total
electrons around each atom except hydrogen, beryllium, and boron.
When an atom has a total of eight electrons around it, we say it has an octet of
electrons. Atoms of the noble gases other than helium have an octet of electrons
around them. The orbital diagram and electron‑dot structure for the valence
electrons of neon show this.
2p
Ne
2s
The stability of the octet is reflected in the fact that atoms in reasonable Lewis
structures often have this octet.
∗ Carbon, nitrogen, oxygen and fluorine always have 8 electrons around
them in a reasonable Lewis structure.
∗ Beryllium and boron can have less than eight electrons but never more
than eight.
∗ The nonmetal elements below the second period (P, S, Cl, Se, Br, and
I) usually have eight electrons around them but often have more.
∗ The metalloids arsenic, As, and tellurium, Te, form covalent bonds in
similar ways to phosphorus, P, and sulfur, S, so they usually have eight
electrons around them but often have more.
∗ Hydrogen will always have two electrons total from its one bond.
190
Chapter 8 & 9
Covalent Bonding and Molecular Structures
Step 5 Do one of the following.
• If in Step 4 you were able to get eight electrons around each atom other
than those of hydrogen, beryllium, and boron and if you used all of the
remaining valence electrons, go to Step 6.
• If you have electrons remaining after each of the atoms other than those of
hydrogen, beryllium, and boron have their octet, you can put more than
eight electrons around the elements in the third period and below.
• If you do not have enough electrons to get octets of electrons around each
atom (other than hydrogen, beryllium, and boron), convert one lone pair
into a multiple bond for each 2 electrons that you are short.
∗ If you would need two more electrons to get octets, convert one lone
pair in your structure to a second bond between two atoms.
∗ If you would need four more electrons to get octets, convert two lone
pairs into bonds. This could mean creating a second bond in two
different places or creating a triple bond in one place.
∗ If you would need six more electrons to get octets, convert three lone
pairs into bonds.
∗ Etc.
Step 6 Check your structure.
• First check to see if all of the atoms have their most common bonding
pattern.
If each atom has its most common bonding pattern, your structure is a
reasonable structure. Skip Step 7, and proceed to Step 8.
• Determine the formal charge for each atom that has other than its most
common bonding pattern. (Atoms with their most common bonding
pattern always have a zero formal charge.)
Formal Charge = Group Number − Number of lines − Number of dots
∗ If all of the formal charges are zero, skip Step 7, and go to Step 8.
∗ If you have formal charges, continue with Step 7.
Step 7 Try to rearrange your structure to eliminate or at least reduce any formal
charges. (This step is unnecessary if there are no formal charges.)
• One way to eliminate formal charges is to return to Step 2 and try another
skeleton.
• Formal charges can also be eliminated by converting lone pairs into bonds.
This process leads to placing more than eight electrons around some
atoms. Here is where we hit controversy. Chemists are not in agreement
about whether it is better to emphasize the importance of octets in
reasonable Lewis structures or better to minimize formal charges.
191
Step 8: Once you have a reasonable Lewis structure, consider the possibility of
resonance.
• If resonance is possible, write all of the reasonable resonance structures for
the original Lewis structure.
• Draw the resonance hybrid from the resonance forms.
It is possible to generate more than one reasonable Lewis structure for a given
formula. When this happens, the following rules will help you decide which is most
likely.
• The more usual or common the bond is, the more stable the structure. See
Table 8,9.3.
• A structure with formal charges is less stable than one without formal
charges.
• “+” formal charges are more likely on the less electronegative elements and
“-” formal charges are more likely on the more electronegative elements.
It is possible to have two Lewis structures for the same molecular formula, which
are considered equally reasonable according to the above criteria. The substances
represented by the Lewis structures are isomers, molecules that have the same molecular
formula but different structural formulas. The dimethyl ether and ethanol are isomers
of C2H6O.
H
H
C
H
H
O
C
H
H
dimethyl ether
H
H
H
C
C
H
H
ethanol
O
H
192
Chapter 8 & 9
Covalent Bonding and Molecular Structures
THE FOLLOWING IS A SUMMARY OF THE STEPS FOR DRAWING
LEWIS STRUCTURES FROM CHEMICAL FORMULAS.
Step 1 Determine the number of valence electrons for the molecule or polyatomic
ion.
Step 2 Draw a reasonable skeletal structure joining the atoms with single bonds.
Step 3 Subtract 2 electrons from the total for each of the single bonds described in
Step 2 above.
Step 4 Try to distribute the remaining electrons as lone pairs to get eight total
electrons around each atom except hydrogen, beryllium, and boron.
Step 5 Do one of the following.
• If in Step 4 you were able to get eight electrons around each atom other
than those of hydrogen, beryllium, and boron and if you used all of the
remaining valence electrons, go to Step 6.
• If you have electrons remaining after the atoms other than those of
hydrogen, beryllium, and boron have their octet, you can put more than
eight electrons around the elements in the third period and below.
• If you do not have enough electrons to get octets of electrons around each
atom (other than those of hydrogen, beryllium, and boron), convert one
lone pair into a multiple bond for each 2 electrons that you are short.
Step 6 Check your structure.
Step 7 Try to rearrange your structure to eliminate or at least reduce any formal
charges.
Step 8 Once you have a reasonable Lewis structure, consider the possibility of
resonance. If resonance is possible, draw the reasonable resonance structures and the
resonance hybrid for the structure.
Exercise 8,9.2 - Drawing Lewis Structures
Draw a reasonable Lewis structure for each of the following. If the structure has
resonance, draw all of the reasonable resonance structures and the resonance hybrid.
a. CH3Br
h. C3H7OH
b. ClF3
i. HCO2−
c. CH2O
j. CH2CHF
d. CN−
k. HCO2H
e. CF3CHCl2
l. NH2COCH3
f. C2H6O
m. SO42−
g. C2H4F2
n. PO43−
193
Write chemical formulas from names.
Draw Lewis structures.
Predict molecular geometry.
Use electronegativities to predict bond polarity.
Predict molecular polarity
Identify types of substances
Predict types of attractions between particles.
Predict relative strengths of these attractions.
Predict relative melting points and boiling points.
Figure 8,9.29
Connections Between Skills This figure shows the connections between the skills covered
in earlier chapters and the skills presented in this chapter. To master the new skills, you need
to have mastered the old ones. Start at the top of the sequence and work your way down,
convincing yourself as you go along that you can do each task.
For a molecule to be polar, it must meet two criteria:
• It must contain polar bonds.
• It must have an asymmetrical distribution of its polar bonds.
Electrons shift toward the chlorine atom,
forming partial plus and minus charges.
δ+
Hydrogen attracts electrons less.
δ−
H
Cl
Chlorine attracts electrons more.
Figure 8,9.30
The HCl molecule.
The polar covalent H‑Cl bond (or bond dipole) leads to an overall separation of positive and negative
charge in the molecule. Therefore, the HCl molecule is a polar molecule, or a molecular dipole.
H δ+
δ− O
105°
H δ+
Asymmetrical and polar
F
δ−
δ+
B
δ−
120°
δ−
F
F
Symmetrical and nonpolar
194
Chapter 8 & 9
Sample Study
Sheet 8,9.6:
Predicting
Molecular
Polarity
Covalent Bonding and Molecular Structures
Tip-off – You are asked to predict whether a molecule is polar or nonpolar; or you
are asked a question that cannot be answered unless you know whether a molecule
is polar or nonpolar. (For example, you are asked to predict the type of attraction
holding the particles together in a given liquid or solid.)
General Steps
Step 1 Draw a reasonable Lewis structure for the substance.
Step 2 Identify each bond as either polar or nonpolar. (If the difference in
electronegativity for the atoms in a bond is greater than 0.4, we consider the bond
polar. If the difference in electronegativity is less than 0.4, the bond is essentially
nonpolar.)
• If there are no polar bonds, the molecule is nonpolar.
• If the molecule has polar bonds, move on to Step 3.
Step 3 If there is only one central atom, examine the electron groups around it.
• If there are no lone pairs on the central atom, and if all the bonds to the
central atom are the same, the molecule is nonpolar.
• If the central atom has at least one polar bond and if the groups bonded to
the central atom are not all identical, the molecule is probably polar. Move
on to Step 4.
Step 4 Draw a geometric sketch of the molecule.
Step 5 Determine the symmetry of the molecule using the following steps.
• Describe the polar bonds with arrows pointing toward the more
electronegative element. Use the length of the arrow to show the relative
polarities of the different bonds. (A greater difference in electronegativity
suggests a more polar bond, which is described with a longer arrow.)
• Decide whether the arrangement of arrows is symmetrical or asymmetrical
∗ If the arrangement is symmetrical and the arrows are of equal length,
the molecule is nonpolar.
∗ If the arrows are of different lengths, and if they do not balance each
other, the molecule is polar.
∗ If the arrangement is asymmetrical, the molecule is polar.
Exercise 8,9.3 - Predicting Molecular Polarity
Decide whether the molecules represented by the following formulas are polar or
nonpolar. (You may need to draw Lewis structures and geometric sketches in order to
do this.)
a. BCl3
e. CBr4
b. C3H8
f. CH3CO2H
c. CS2
g. Br2O
d. NH3
h. IF3
195
Tip-off – You are asked to predict the relative strengths of attractions between
particles of two substances, or you are asked a question that cannot be answered
unless you know those relative strengths. (For example, you are asked to compare
certain properties of substances, such as their relativemelting and boiling point
temperatures.)
General Steps Step 1 Determine the type of attraction between the particles using the following
steps.
Step 1(a) Classify each substance as either a metallic element, an element
composed of atoms, an element composed of molecules, an ionic compound, or
a molecular compound.
• Metallic elements have metallic bonds.
• The particles in elements that are composed of atoms (the noble gases)
and molecules (the other nonmetallic elements) are held together by
London forces.
• The ions in ionic compounds are held together by ionic bonds.
• Continue to Step 1(b) for molecular compounds.
Step 1(b) For molecular compounds, draw the Lewis structure for the molecule.
• If the Lewis structure contains an O‑H, N‑H, or H‑F bond, the attractions
that form when the substance condenses are hydrogen bonds enhanced by
London forces.
• For other molecular compounds, do Step 1(c).
Step 1(c): If there are no O‑H, N‑H, or H‑F bonds, determine the polarity of
the bonds.
• If there are no polar bonds, the molecules are nonpolar and form London
forces when they condense.
• If there is at least one polar bond, do Step 1(d).
Step 1(d): Predict whether the polar bonds are symmetrically or
asymmetrically arranged.
• If the distribution of polar bonds is symmetrical and their dipoles equal,
the molecules are nonpolar and form London forces when they condense.
• If the distribution of polar bonds is asymmetrical, or symmetrical with
unequal dipoles, the molecules are polar and form dipole‑dipole attractions
enhanced by London forces when they condense.
Sample Study
Sheet 8,9.7:
Predicting
Types of
Attractions
and Relative
Strengths of
Attractions
Between
Particles
196
Chapter 8 & 9
Covalent Bonding and Molecular Structures
Step 2 Although we cannot predict the relative strengths of attractions between all
particles, we can apply one of the following guidelines to predict the relative strengths
of attractions between some particles.
• For substances that contain particles of about the same size, the substances
with chemical bonds (ionic, covalent, or metallic) have stronger attractions
between particles than substances with intermolecular attractions (hydrogen
bonds, dipole‑dipole attractions, or London forces).
Chemical bonds are generally stronger than intermolecular
attractions.
• For molecular substances that contain molecules of about the same size,
substances with hydrogen bonds have stronger attractions between the
particles than substances with either dipole‑dipole attractions or London
forces, and substances with dipole‑dipole attractions have stronger
attractions between the particles than London forces.
Hydrogen bonds are generally stronger than dipole‑dipole
attractions, which are generally stronger than London forces.
• For molecular substances that have the same type of intermolecular
attraction, larger molecules form stronger mutual attractions.
Larger molecules tend to have stronger attractions.
Figure 8,9.31 Scheme for Predicting Types of Attractions
What kind of substance?
Metallic element
Metallic bonds
Diamond
Other nonmetallic elements
Covalent bonds
London forces
Ionic compound
Molecular compound
Ionic bonds
Do the molecules contain O-H, N-H, or H-F bonds?
No
Yes
Are any bonds polar?
No
London forces
Hydrogen bonds
Yes
Symmetrical or asymmetrical arrangement of polar bonds?
Asymmetrical
Dipole-dipole attractions
Symmetrical
London forces
197
Table 8,9.5
Solid and Liquid Form
Type of substance
Particles to
visualize
Metal
Cations in a
sea of electrons
Atoms
Carbon atoms
Molecules
Elements
Noble Gases
Carbon (diamond)
Other Nonmetal
Elements
Ionic
Compounds
Molecular
Compounds
Nonpolar
Molecular
Polar molecules
without H‑F, O‑H,
or N‑H bond
Molecules with
H‑F, O‑H or N‑H
bond
Network crystals
Examples
Au
Type of attraction
between particles in
solid or liquid
Metallic bond
Xe
C(dia)
H2, N2, O2,
F2, Cl2, Br2,
I2, S8, Se8, P4
NaCl
London forces
Covalent bonds
London forces
Hydrocarbons
London forces
Molecules
HCl
Dipole-Dipole forces
Molecules
HF, H2O,
alcohols, NH3
Hydrogen bonds
Atoms
SiO2
Covalent bonds
Cations and
Anions
Molecules
Ionic bond
Exercise 8,9.4 - Predicting Type of Attraction
What is the primary type of attraction that holds the particles of each of these
substances in the liquid or solid form.
a. Ir
e. NH4OH
b. AlF3
f. BF3
c. C2H6
g. CH2O
d. CH3NH2
h. C3H7OH
198
Chapter 8 & 9
Covalent Bonding and Molecular Structures
Exercise 8,9.5 - Predicting Types and Strengths of attractions
Between Particles
For each of the following substances, write the name for the type of particle that
forms its basic structure and the name of the primary type of attraction that can form
between these particles. From each pair of substances, choose the one that you expect
would have stronger interparticle attractions.
a. ethanol, C2H5OH (in alcoholic beverages)
dimethyl ether, CH3OCH3 (refrigerant and solvent for propellant sprays)
b. manganese, Mn (improves corrosion resistance and hardness when alloyed
with other metals)
carbon tetrachloride, CCl4 (used to add chlorine atoms to organic
compounds)
c. propane, C3H8 (fuel and aerosol propellant)
n-heptane, C7H16 (standard for octane rating – 100% n-hexane is zero
octane)
d. acetone, CH3COCH3 (solvent for paints and varnishes)
lithium nitrate, LiNO3 (used to make ceramics and in rocket propellants)
199
Hydrophobic and Hydrophilic Substances
Organic compounds are often polar in one part of their structure and nonpolar in
another part. The polar section, which is attracted to water, is called hydrophilic
(literally, “water loving”), and the nonpolar part of the molecule, which is not expected
to be attracted to water, is called hydrophobic (“water fearing”). If we need to predict
the relative water solubility of two similar molecules, we can expect the one with the
proportionally larger polar portion to have higher water solubility. We predict that the
molecule with the proportionally larger nonpolar portion will be less soluble in water.
As an example, we can compare the structures of epinephrine (adrenaline) and
amphetamine. Epinephrine is a natural stimulant that is released in the body in
times of stress. Amphetamine (sold under the trade name Benzedrine) is an artificial
stimulant that causes many of the same effects as epinephrine. The three –OH groups
and the N H bond in the epinephrine structure cause a greater percentage of that
molecule to be polar, so we predict that epinephrine would be more soluble in water
than amphetamine.
Nonpolar
OH
NHCH3
HO
NH2
Polar
Polar
Nonpolar
HO
amphetamine
epinephrine
The molecular representations above are known as line drawings. The corners, where
two lines meet, represent carbon atoms, and the end of any line that does not have
a symbol attached also represents a carbon atom. We assume that each carbon has
enough hydrogen atoms attached to yield four bonds total. Compare the line drawings
to the more complete Lewis structures below:
H
H
C
C
H
H
C
C
C
H
C
C
H
H
H
C
H
C H
H
NH2
amphetamine
HO
HO
C
C
H
OH
C
C
C
H
C
C
H
H
H
C
NHCH3
H
epinephrine
The difference in polarity between epinephrine and amphetamine has an important
physiological consequence. The cell membranes that separate the blood stream from
the brain cells have a nonpolar interior and a polar exterior. The more polar a molecule
is, the more likely it is to be attracted to the exterior of the cell membrane enough to be
blocked from moving into the cell. This tends to prevent polar substances in the blood
from moving into the brain tissue. Epinephrine is too polar to move from the blood
stream into the brain, but amphetamine is not. The stimulant effects of amphetamine
are in part due to its ability to pass through the blood brain barrier.
Methamphetamine is a stimulant that closely resembles amphetamine. With only
200
Chapter 8 & 9
Covalent Bonding and Molecular Structures
one polar N-H bond, it has very low water solubility, but it can be induced to dissolve
in water when converted to the much more polar, ionic form called methamphetamine
hydrochloride. On the street, the solid form of this illicit ionic compound is known as
crystal meth.
Nonpolar
CH3NH
methamphetamine
Nonpolar
Slightly polar
CH3NH2+ Cl −
Very polar
methamphetamine hydrochloride