Chemistry: Atoms First
Julia Burdge & Jason Overby
Chapter 6
Representing
Molecules
Kent L. McCorkle
Cosumnes River College
Sacramento, CA
Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6
Chemical Bonding I: Basic Concepts
6.1 The Octet Rule
Lewis Structures
Multiple Bonds
6.2 Electronegativity and Polarity
Electronegativity
Dipole Moment, Partial Charges and Percent Ionic
Character
6.3 Drawing Lewis Structures
6.4 Lewis Structures and Formal Charge
6.5 Resonance
6.6 Exceptions to the Octet Rule
Incomplete Octets
Odd Numbers of Electrons
Expanded Octets
6.1
The Octet Rule
According to the octet rule, atoms will lose, gain, or share electrons
in order to achieve a noble gas electron configuration.
••F •• F••
••
••
•F••
••
••
+
•• ••
•• ••
••F•
Each F counts both shared electrons
to “feel” as though it has a Ne
configuration.
Only valence electrons contribute to bonding.
F 1s22s22p5
valence electrons
The Octet Rule
Only two valence electrons participate in the formation of the F2
bond.
••
••
••
••
or
••F F••
••
••
••
••
••F •• F••
Pairs of valence electrons not involved in bonding are called lone
pairs.
••
••
••
••
••F •• F••
Lewis Structures
A Lewis structure is a representation of covalent bonding.
Shared electron pairs are shown either as dashes or as pairs of dots.
Lone pairs are shown as pairs of dots on individual atoms.
H with 2 e−
H with 2 e−
••
••
••
HOH
••
O with 8 e−
••
H O H
••
Shared electrons shown as
dashes (bonds)
Multiple Bonds
In a single bond, atoms are held together by one electron pair.
In a double bond, atoms share two pairs of electrons.
O with 8 e−
C with 8 e−
O with 8 e−
••
••
O C O
one shared pair of electrons
results in a single bond
2 shared pairs of electrons
result in double bonds
••
H O H
••
••
••
••
••
••
HOH
••
O=C= O
Multiple Bonds
A triple bond occurs when atoms are held together by three electron
pairs.
each N has 8 e−
3 shared pairs of electrons
result in a triple bond
N≡N
••
N
••
••
••
••
••
••
N
Multiple Bonds
Bond length is defined as the distance between the nuclei of two
covalently bonded atoms.
Multiple Bonds
Multiple bonds are shorter than single bonds.
For a given pair of atoms:
 triple bonds are shorter than double bonds
 double bonds are shorter than single bonds
N≡N
N=N
<
<
110 pm
124 pm
N−N
147 pm
Multiple Bonds
The shorter multiple bonds are also stronger than single bonds.
We quantify bond strength by measuring the quantity of energy
required to break it.
H2(g) → H(g) + H(g)
Bond energy = 436.4 kJ/mol
We will examine this in more detail in Chapter 10.
6.2
Electronegativity and Polarity
There are two extremes in the spectrum of bonding:
 covalent bonds occur between atoms that share electrons
 ionic bonds occur between a metal and a nonmetal and involve
ions
Bonds that fall between these extremes are polar.
In polar covalent bonds, electrons are shared but not shared equally.
M:X
Mδ+Xδ−
M+X−
Pure covalent bond
Neutral atoms held
together by equally
shared electrons
Polar covalent bond
Partially charged
atoms held together by
unequally shared
electrons
Ionic bond
Oppositely charged
ions held together by
electrostatic attraction
Electronegativity and Polarity
Electron density maps show the distributions of charge.
 Electrons spend a lot of time in red and very little time in blue.
Electrons are shared
equally
nonpolar covalent
Electrons are not shared
equally and are more
likely to be associated
with F
Electrons are not shared
but rather transferred
from Na to F
ionic
polar covalent
Electronegativity
Electronegativity is the ability of an atom in a compound to draw
electrons to itself.
Electronegativity and Polarity
Electronegativity varies with atomic number.
Electronegativity
There is no sharp distinction between nonpolar covalent and polar
covalent or between polar covalent and ionic.
The following rules help distinguish among them:
 A bond between atoms whose electronegativites differ by less
than 0.5 is general considered purely covalent or nonpolar.
 A bond between atoms who’s electronegativies differ by the
range of 0.5 to 2.0 is generally considered polar covalent.
 A bond between atoms whose electronegativities differ by 2.0
or more is generally considered ionic.
Worked Example 6.1
Classify the following bonds as nonpolar, polar, or ionic: (a) the bond in ClF, (b)
the bond in CsBr, and (c) the carbon-carbon double bond in C2H4.
Strategy Electronegativity values are: Cl (3.0), F (4.0), Cs (0.7), Br (2.8),
C (2.5). Use this information to determine which bonds have identical, similar,
and widely different electronegativities.
Solution
(a) The difference between the electronegativies of F and Cl is 4.0 – 3.0 = 1.0,
making the bond in ClF polar.
(b) In CsBr, the difference is 2.8 – 0.7 = 2.1, making the bond ionic.
(c) In C2H4, the two atoms are identical. (Not only are they the same element,
but each C atom is bonded to two H atoms.) The carbon-carbon double bond
is C2H4 is nonpolar.
Think About It By convention, the difference in electronegativity is always
calculated by subtracting the smaller number from the larger one, so the
result is always positive.
Dipole Moment, Partial Charges, and Percent Ionic Character
Regions where electrons
spend little time
••
The consequent charge separation can be
represented as:
 Deltas (δ) denote a partial positive or
negative charge.
••
H−F
••
δ+
δ−
••
H−F
••
••
An arrow is used to indicate the direction of
electron shift in polar covalent molecules.
Regions where electrons
spend a lot of time
Dipole Moment, Partial Charges, and Percent Ionic Character
A quantitative measure of the polarity of a bond is its dipole
moment (μ).
μ=Qxr
 Q is the charge.
 r is the distance between the charges.
 μ is always positive and expressed in debye units (D).
1 D = 3.336×10−30 C∙m
Dipole Moment, Partial Charges, and Percent Ionic Character
A quantitative measure of the polarity of a bond is its dipole
moment (μ).
μ=Qxr
Worked Example 6.2
Burns caused by hydrofluoric acid [HF(aq)] are unlike any other acid burns and
present unique medical complications. HF solutions typically penetrate the skin
and damage internal tissues, including bone, often with minimal surface damage.
Less concentrated solutions actually can cause greater injury than more
concentrated ones by penetrating more deeply before causing injury, thus
delaying the onset of symptoms and preventing timely treatment. Determine the
magnitude of the partial positive and partial negative charges in the HF molecule.
Strategy Solve for Q. Convert the resulting charge in coulombs to units of
electronic charge. According to Table 6.2, μ = 1.82 D and r = 0.92 Å for HF. The
dipole moment must be converted from debye to C∙m and the distance between
ions must be converted to meters.
3.336×10-30 C∙m
μ = 1.82 D ×
= 6.07×10-30 C∙m
1D
1×10-10 m
r = 0.92 Å ×
= 9.2×10-11 m
1Å
Worked Example 6.2 (cont.)
Solution In coulombs:
μ
6.07×10-30 C∙m
=
Q=
= 6.598×10-20 C
-11
r
9.2×10 m
In units of electronic charge:
6.598×10-20
1 eC×
= 0.41 e-19
1.6022×10 C
Therefore, the partial charges in HF are +0.41 and -0.41 on H and F, respectively.
••
H−F
••
-0.41
••
+0.41
Think About It Calculated partial charges should always be less than 1. If a
“partial” charge were 1 or greater, it would indicate that at least one electron
had been transferred from one atom to the other. Remember that polar bonds
involve unequal sharing of electrons, not a complete transfer of electrons.
Dipole Moment, Partial Charges, and Percent Ionic Character
Although the designations “covalent,” “polar covalent,” and “ionic”
can be useful, sometimes chemists wish to describe and compare
chemical bonds with more precision.
Comparing the calculated dipole moment with the measured values
gives us a quantitative way to describe the nature of a bond using
the term percent ionic character.
percent ionic character =
μ (observed)
×100%
μ (calculated assuming discrete charges)
Dipole Moment, Partial Charges, and Percent Ionic Character
The figure below demonstrates the relationship between percent
ionic character and the electronegativity difference in a
heteronuclear diatomic molecule.
Worked Example 6.3
Using data from Table 6.2, calculate the percent ionic character of the bond in HI.
Strategy Calculate the dipole moment in HI assuming that the charges on H and
I are +1 and -1, respectively; then calculate the percent ionic character. The
magnitude of the charges must be expressed in coulombs (1 e- = 1.6022×10-19
C); the bond length (r) must be expressed as meters (1 Å = 1×10-10 m); and the
calculated dipole moment should be expressed as debyes (1 D = 3.336×10-30
C∙m).
Setup From Table 6.2, the bond length in HI is 1.61 Å (1.61×10-10 m) and the
measured dipole moment of HI is 0.44 D.
Worked Example 6.3 (cont.)
Solution The dipole we would expect if the magnitude of the charges were
1.6022×10-19 C is
μ = Q × r = (1.6022×10-19 C) × (1.61×10-10 m) = 2.58×10-29 C∙m
Converting to debyes gives
2.58×10-29 C∙m ×
1D
= 7.73 D
3.336×10-30 C∙m
The percent ionic character of the H–I bond is
0.44 D
×100% = 5.7%.
7.73 D
Think About It A purely covalent bond (in a homonuclear diatomic molecule
such as H2) would have 0 percent ionic character. In theory, a purely ionic bond
would be expected to have 100 percent ionic character, although no such bond is
known.
6.3
Drawing Lewis Structures
Follow these steps when drawing Lewis structure for molecules and polyatomic ions.
1) Draw the skeletal structure of the compound. The least electronegative atom is
usually the central atom. Draw a single covalent bond between the central atom
and each of the surrounding atoms.
2) Count the total number of valence electrons present; add electrons for negative
charges and subtract electrons for positive charges.
3) For each bond in the skeletal structure, subtract two electrons from the total
valence electrons.
4) Use the remaining electrons to complete octets of the terminal atoms by placing
pairs of electrons on each atom. Complete the octets of the most electronegative
atom first.
5) Place any remaining electrons in pairs on the central atom.
6) If the central atom has fewer than eight electrons, move one or more pairs from
the terminal atoms to form multiple bonds between the central atom and
terminal atoms.
Drawing Lewis Structures
Draw the Lewis Structure for ClO3−.
Solution:
Step 1 Draw the skeletal structure of the compound. The least
electronegative atom is usually the central atom. Draw a single
covalent bond between the central atom and each of the surrounding
atoms.
chlorine’s electronegativity = 3.0
oxygen’s electronegativity = 3.5
Drawing Lewis Structures
Draw the Lewis Structure for ClO3−.
Solution:
Step 2 Count the total number of valence electrons present; add
electrons for negative charges and subtract electrons for positive
charges.
Cl: 1 x 7 = 7 e−
O: 3 x 6 = 18 e−
Charge:
1 e−
Total:
26 valence e−
Drawing Lewis Structures
Draw the Lewis Structure for ClO3−.
Solution:
Step 3 For each bond in the skeletal structure, subtract two
electrons from the total valence electrons.
3 bonds x 2 e− = 6 e−
26 – 6 = 20 e− left over
Drawing Lewis Structures
Draw the Lewis Structure for ClO3−.
Solution:
Step 4 Use the remaining electrons to complete octets of the
terminal atoms by placing pairs of electrons on each atom. Complete
the octets of the most electronegative atom first.
6 electrons used in single bonds
20 electrons must be placed in the
structure starting with terminal
atoms
The structure now has a total of 24
valence electrons of the available
26.
Drawing Lewis Structures
Draw the Lewis Structure for ClO3−.
Solution:
Step 5 Place any remaining electrons on the central atom.
The structure now has a total of 24
valence electrons of the available
26.
2 e− remain to be added to the
structure.
Lewis structures for polyatomic ions
are enclosed by square brackets.
Drawing Lewis Structures
Draw the Lewis Structure for ClO3−.
Solution:
Step 6 If the central atom has fewer than eight electrons, move one
or more pairs from the terminal atoms to form multiple bonds
between the central atom and terminal atoms.
All atoms already have an octet in this structure.
Drawing Lewis Structures
Worked Example 6.4
Draw the Lewis structure for carbon disulfide (CS2).
Setup
Step 1: C and S have identical electronegativities. We will draw the skeletal
structure with the unique atom, C, at the center.
Step 2: The total number of valence electrons is 16: 6 from each S atom and 4
from the C atom [2(6) + 4 = 16].
Step 3: Subtract 4 electrons to account for the bonds in the skeletal structure,
leaving us 12 electrons to distribute.
Step 4: Distribute the 12 remaining electrons as 3 lone pairs on each S atom.
Worked Example 6.4 (cont.)
Draw the Lewis structure for carbon disulfide (CS2).
Setup
Step 5: There are no electrons remaining after step 4, so step 5 does not apply.
Step 6: To complete carbon’s octet, use one lone pair from each S atom to make a
double bond to the C atom.
Solution
Think About It Counting the total number of valence electrons should be
relatively simple to do, but it is often done hastily and is therefore a potential
source of error in this type of problem. Remember that the number of valence
electrons for each element is equal to the group number of that element.
6.4
Lewis Structures and Formal Charge
Formal charge can be used to determine the most plausible Lewis
Structure when more than one possibility exists for a compound.
Formal charge = valence electrons – associated electrons
To determine associated electrons:
1) All the atom’s nonbonding electrons are associated with the
atom.
2) Half the atom’s bonding electrons are associated with the atom.
Lewis Structures and Formal Charge
Determine the formal charges on each oxygen atom in the ozone
molecule (O3).
••
•• ••
O =O− O
••
4 unshared + 4 shared = 6 e−
2
2 unshared + 6 shared = 5 e− 6 unshared + 2 shared = 7 e−
2
2
Valence e−
6
6
6
e− associated with atom
6
5
7
Difference (formal charge)
0
+1
−1
Worked Example 6.5
The widespread use of fertilizers has resulted in the contamination of some
groundwater with nitrates, which are potentially harmful. Nitrate toxicity is due
primarily to its conversion in the body to nitrite (NO2-), which interferes with the
ability of hemoglobin to transport oxygen. Determine the formal charges on each
atom in the nitrate ion (NO3-).
Strategy Follow the six steps to draw the Lewis structure of (NO3-). For each
atom, subtract the associated electrons from the valence electrons.
Setup
The N atom has five valence electrons and four associated electrons (one from
each single bond and two from the double bond). Each singly bonded O atom has
six valence electrons and seven associated electrons (six in three lone pairs and
one from the single bond). The doubly bonded O atom has six valence electrons
and six associated electrons (four in two lone pairs and two from the double
bond.)
Worked Example 6.5 (cont.)
Solution
The formal charges are as follows: +1 (N atom), -1 (singly bonded O atoms), and
0 (doubly bonded O atom).
Think About It The sum of formal charges (+1) + (-1) + (-1) + (0) = -1 is equal
to the overall charge on the nitrate ion.
Lewis Structures and Formal Charge
When there is more than one possible structure, the best arrangement
is determined by the following guidelines:
1) A Lewis structure in which all formal charges are zero is
preferred.
2) Small formal charges are preferred to large formal charges.
3) Formal charges should be consistent with electronegativities.
0
0
0
better structure (based
on formal charge)
••
••
Formal
charge
O≡C− O
+1
••
••
O=C= O
0
−1
Lewis Structures and Formal Charge
Based on formal charge, identify the best and the worst structures
for the isocyanate ion below:
Solution:
Step 1 Assign formal charges on each atom using the formula
Formal charge = valence electrons – associated electrons
Ve−
Ae−
FC
4
6
−2
5
4
+1
6
6
0
4
5
−1
5
4
+1
6
7
−1
4
7
−3
5
4
+1
6
5
+1
Lewis Structures and Formal Charge
Based on formal charge, identify the best and the worst structures
for the isocyanate ion below:
Solution:
Step 2 Determine the best and worst structure
FC
−2
+1
0
−1
+1
−1
−3
+1
+1
Best structure:
Worst structure:
Small formal charges
Large formal charges
Formal charges more
consistent with
electronegativities
Formal charges
inconsistent with
electronegativities
Worked Example 6.6
Formaldehyde (CH2O), which can be used to preserve biological specimens, is
commonly sold as a 37 percent aqueous solution. Use formal charges to
determine which skeletal arrangements of atoms shown here is the best choice for
the Lewis structure of CH2O.
Strategy The complete Lewis structures for the skeletons shown are:
Worked Example 6.6 (cont.)
Strategy The complete Lewis structures for the skeletons shown are:
In the structure on the left, the formal charges are as follows:
Both H atoms: 1 valence e- – 1 associated e- (from single bond) = 0
C atom: 4 valence e- – 5 associated e- (two in the lone pair, one from the single
bond, and two from the double bond) = -1
O atom: 6 valence e- – 5 associated e- (two from the lone pair, one from the single
bond, and two from the double bond) = +1
Formal charges 0
-1
+1
0
Worked Example 6.6 (cont.)
Strategy The complete Lewis structures for the skeletons shown are:
In the structure on the right, the formal charges are as follows:
Both H atoms: 1 valence e- – 1 associated e- (from single bond) = 0
C atom: 4 valence e- – 4 associated e- (one from each single bond, and two from
the double bond) = 0
O atom: 6 valence e- – 6 associated e- (four from the two lone pairs and two from
the double bond) = 0
Formal charges all zero
Worked Example 6.6 (cont.)
Solution Of the two possible arrangements, the structure on the left has an O
atom with a positive formal charge, which is inconsistent with oxygen’s high
electronegativity. Therefore, the structure on the right, in which both H atoms are
attached directly to the C atoms and all atoms have a formal charge of zero, is the
better choice for the Lewis structure of CH2O.
Think About It For a molecule, formal charges of zero are preferred. When
there are nonzero formal charges, they should be consistent with the
electronegativities of the atoms in the molecules. A positive formal charge on
oxygen, for example, is inconsistent with oxygen’s high electronegativity.
6.5 Resonance
A resonance structure is one of two or more Lewis structures for a
single molecule that cannot by represented accurately by only one
Lewis structure.
••
•• ••
O −O= O
••
••
•• ••
••
O =O− O
Resonance structures are a human invention.
Resonance structures differ only in the positions of their electrons.
Worked Example 6.7
High oil and gasoline prices have renewed interest in alternative methods of
producing energy, including the “clean” burning of coal. Part of what makes
“dirty” coal dirty is its high sulfur content. Burning dirty coal produces sulfur
dioxide (SO2), among other pollutants. Sulfur dioxide is oxidized in the
atmosphere to form sulfur trioxide (SO3), which subsequently combines with
water to produce sulfuric acid – a major component of acid rain. Draw all
possible resonance structures of sulfur trioxide.
Strategy Following the steps for drawing Lewis structures, we determine that a
correct Lewis structure for SO3 contains two sulfur-oxygen single bonds and one
sulfur-oxygen double bond.
But the double bond can be put in any one of three positions in the molecule.
Worked Example 6.7 (cont.)
Solution
Think About It Always make sure that resonance structures differ only in the
position of the electrons, not in the positions of the atoms.
6.6
Exceptions to the Octet Rule
Exceptions to the octet rule fall into three categories:
1) The central atom has fewer than eight electrons due to a
shortage of electrons.
H Be H
only 4 total valence
electrons in the system
 Elements in group 3A also tend to form compounds surrounded
by fewer than eight electrons.
 Boron, for example, reacts with halogens to form compounds
of the general formula BX3 having six electrons around the
boron atom.
Odd Numbers of Electrons
2) The central atom has fewer than eight electrons due to an odd
number of electrons.
••
••
••
•
O=N− O
17 valence electrons in the
system
Molecules with an odd number of electrons are sometimes referred
to as free radicals.
Worked Example 6.8
Draw the Lewis structure of chlorine dioxide (ClO2).
Strategy The skeletal structure is
This puts the unique atom, Cl, in the center and puts the more electronegative O
About
It ClO2 is used primarily to bleach wood pulp in
atoms Think
in terminal
positions.
the manufacture of paper, but it is also used to bleach flour,
disinfect
water,
and deodorize
certain
There are
a total drinking
of 19 valence
electrons
(7 from the
Cl andindustrial
6 from each of the
Recently,
has been
to for
eradicate
toxic
two O facilities.
atoms). We
subtract 4itelectrons
to used
account
the twothe
bonds
in the
mold
in homes
in15
New
Orleans
that wereasdamaged
by the
skeleton,
leaving
us with
electrons
to distribute
follows: three
lone pairs on
floodwaters
Katrina
in 2005. electron also on
each Odevastating
atom, one lone
pair on theof
ClHurricane
atom, and the
last remaining
the Cl atom.
Solution
Expanded Octets
3) The central atom has more than eight electrons.
Sulfur has 6 bonds
corresponding to 12 electrons
 Atoms in and beyond the third period can have more than eight
valence electrons.
 In addition to the 3s and 3p orbitals, elements in the third
period also have 3d orbitals that can be used in bonding.
Worked Example 6.9
Draw the Lewis structure of boron triiodide (BI3).
Strategy The skeletal structure is
There are a total of 24 valence electrons (3 from the B and 7 from each of the
three I atoms). We subtract 6 electrons to account for the three bonds in the
skeleton, leaving 18 electrons to distribute in lone pairs on each I atom.
Solution
Worked Example 6.9 (cont.)
Think About It Boron is one of the elements that does not always follow the
octet rule. Like BF3, however, BI3 can be drawn with a double bond in order to
satisfy the octet of boron. This gives rise to a total of four resonance structures.
Worked Example 6.10
Draw the Lewis structure of arsenic pentafluoride (AsF5).
Strategy The skeletal structure already has more than an octet around the As
atom.
Think About It Always make sure that the number of
There are
electrons
40 totalrepresented
valence electrons
in your
[5 from
finalAs
Lewis
(Group
structure
5A) andmatches
7 from each of
the fivethe
F atoms
total number
(Group 7A)].
of valence
We subtract
electrons
10 electrons
you aretosupposed
account fortothe five
bonds have.
in the skeleton, leaving 30 to be distributed. Next, we place three lone pairs
on each F atom, thereby completing all their octets and using up all the electrons.
Solution
Worked Example 6.11
Draw two resonance structures for sulfurous acid (H2SO3): one that obeys the
octet rule for the central atom, and one that minimizes the formal charges.
Determine the formal charges on each atom in both structures.
Strategy Begin by drawing the skeletal structure and counting the total number
of valence electrons. Use the steps outlined in Section 6.4 to draw the first
structure and reposition one or more lone pairs to adjust the formal charges for the
second structure.
Note that each hydrogen in an oxoacid is attached to an oxygen atom, and not
directly to the central atom. The total number of valence electrons is 26 (6 from S,
6 from each O, and 1 from each H).
Worked Example 6.11 (cont.)
Solution Following the steps we get the first structure:
-1
+1
From the
top O
atom, Itwhich
has three
lonesuch
pairs,
we reposition
Think
About
In some
species,
as the
sulfate ion,one
it islone pair to
create possible
a double to
bond
between too
O and
S todouble
get thebonds.
second
structure.with three
incorporate
many
Structures
and four double bonds to sulfur would give formal charges on S and
O that are inconsistent with the electronegativities of these elements.
In general, if you are trying to minimize formal charges by
expanding the central atom’s octet, only add enough double bonds to
Incorporating
theformal
doublecharge
bond on
results
in everyatom
atomzero.
having a formal charge of
make the
the central
zero.
Worked Example 6.12
Draw the Lewis structure of xenon tetrafluoride (XeF4).
Strategy Follow the steps for drawing Lewis structures. The skeletal structure is
There are 36 total valence electrons (8 from Xe and 7 from each of the four F
Think About It Atoms beyond the second period can
atoms). We subtract 8 electrons to account for the bonds in the skeleton, leaving
accommodate more than an octet of electrons, whether those
28 to distribute. We first complete the octets of all four F atoms. When this is
electrons are used in bonds or reside on the atoms as lone pairs.
done, 4 electrons remain, so we place two lone pairs on the Xe atom.
Solution
6
Chapter Summary: Key Points
Lewis Structures
Octet Rule
Bond Order
Electronegativity and Polarity
Dipole Moment
Percent Ionic Character
Drawing Lewis Structures
Formal Charge
Resonance Structures
Exceptions to the Octet Rule
Incomplete Octets
Odd Numbers of Electrons
Expanded Octets
Group Quiz #10
• Draw Lewis Structures for the following
substances:
• CO2
• CO
• HCN
• SO32-
61
Group Quiz #11
• Draw all possible Lewis Structures for:
•
•
•
•
•
POCl3
BF3
NO2PF5
SO3