Chemistry: Atoms First
Julia Burdge & Jason Overby
Chapter 6
And 5.2
Representing
Molecules
Homework
Chapter 5: 3, 5, 7 and 9
Chapter 6: 23, 25, 29, 31, 35, 37, 39
41, 53, 57, 59, 73 and 79
Kent L. McCorkle
Cosumnes River College
Sacramento, CA
Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
The Lewis Theory
1. Valence electrons play a fundamental role in chemical bonding
2. Chemical bonding that results from the transfer of 1 or more
electrons from 1 atom to another leading to the formation of
positive and negative ions is called ionic
3. Chemical bonding that involves the sharing of electrons is
called covalent (non-polar and polar)
4. Electrons are transferred/ shared to the extent that each atom
acquires an especially stable electron configuration. Usually
the configuration of a noble gas (8 outer shell electrons)
5.2
Lewis Dot Symbols
When atoms form compounds, it is their valence electrons that
actually interact.
A Lewis dot symbol consists of the element’s symbol surround by
dots.
Each dot represents a valence electron.
•
•B•
Boron 1s22s22p1
3 valence electrons
•
B•
•
Lewis dot symbol for boron
•B•
•
•
•B
•
other reasonable Lewis dot
symbols for boron
Lewis Dot Symbols
Atoms combine in order to achieve a more stable electron
configuration.
Maximum stability results when a chemical species is isoelectronic
with a noble gas.
Na: 1s22s22p63s1
Na+: 1s22s22p6
10 electrons total,
isoelectronic with Ne
Cl: 1s22s22p63s23p5
Cl‒: 1s22s22p63s23p6
18 electrons total,
isoelectronic with Ar
Lewis dot symbols of the main group elements.
Dots are not paired until absolutely necessary.
•
•B•
1s22s22p1
•
•C•
•
1s22s22p2
••
•N•
•
1s22s22p3
5 valence electrons; first pair formed
in the Lewis dot symbol
Na •
For main group metals such as Na, the number of dots is
the number of electrons that are lost.
••
•O•
••
For nonmetals in the second period, the number of
unpaired dots is the number of bonds the atom can form.
Ions may also be represented by Lewis dot symbols.
Remember the charge
Na•
Na 1s22s22p63s1
Valence electron lost in the formation
of the Na+ ion.
••
•O•
••
O 1s22s22p4
Na+
Na+ 1s22s22p6
Core electrons not represented in the
Lewis dot symbol
••
•• O ••
••
2‒
O2‒ 1s22s22p6
Write Lewis dot symbols for (a) fluoride ion (F-), (b) potassium ion (K+), and
(c) sulfide ion (S2-).
Strategy Starting with the Lewis dot symbols for each element, add dots (for
anions) or remove dots (for cations) as needed to achieve the correct charge on
each ion. Don’t forget to include the appropriate charge on the Lewis dot symbol.
Solution (a)
(b) K+
(c)
Think About It For ions that are isoelectronic with noble gases, cations should
have no dots remaining around the element symbol, whereas anions should have
eight dots around the element symbol. Note, too, that for anions, we put square
brackets around the Lewis dot symbol and place the negative charge outside the
brackets. Because the symbol for a common cation such as the potassium ion
has no remaining dots, square brackets are not necessary.
6.3
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.
Write the Lewis structure of nitrogen trifluoride (NF3).
Step 1 – N is less electronegative than F, put N in center
Step 2 – Count valence electrons N - 5 (2s22p3) and F - 7 (2s22p5)
5 + (3 x 7) = 26 valence electrons
Step 3 – Draw single bonds between N and F atoms and complete
octets on N and F atoms.
Step 4 - Check, are # of e- in structure equal to number of valence e- ?
3 single bonds (3x2) + 10 lone pairs (10x2) = 26 valence electrons
F
N
F
F
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)
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
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
Bond length is defined as the distance between the nuclei of two covalently
bonded atoms.
Multiple bonds are shorter than single bonds.
N≡N
For a given pair of atoms:
110 pm
 triple bonds are shorter than double bonds
 double bonds are shorter than single bonds
N=N
<
N−N
<
124 pm
147 pm
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.
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.
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.
Formal Charges
Sometimes, while drawing Lewis structures, you don't have observational data to use.
Formal charges can be used to find out what to choose.
Formal charge = (# of Valence shell e- of an atom) - (# Bond pair e-) - (# of unshared e-)
For example, lets take the first incorrect drawing of sulfuric acid.
Formal charge on S = 6 - 4 - 0 = 2
Formal charge on H = 1 - 1 - 0 = 0
Formal charge on O = 6 - 2 - 4 = 0 (on the ones bonded to H)
Formal charge on O = 6 - 1 - 6 = -1 (on the isolated ones)
We can disregard the ones with 0 formal charge.
The ones that do have formal charge are the sulfur and the isolated oxygens.
Since the 2 oxygens are -1 and sulfur is 2, another bond goes from each oxygen
to the sulfur to cancel the formal charge. (2 + 2(-1) = 0)
When several Lewis structures are possible, those with the smallest formal
charges are the most stable and are preferred.
Two possible skeletal structures of formaldehyde (CH2O)
H
C
O
H
H
C
H
O
An atom’s formal charge is the difference between the
number of valence electrons in an isolated atom and the
number of electrons assigned to that atom in a Lewis
structure.
formal charge
on an atom in
a Lewis
structure
=
total number
total number
of valence
of nonbonding
electrons in electrons
the free atom
-
1
2
(
total number
of bonding
electrons
The sum of the formal charges of the atoms in a molecule
or ion must equal the charge on the molecule or ion.
)
H
-1
+1
C
O
formal charge
on an atom in
a Lewis
structure
H
=
C – 4 eO – 6 e2H – 2x1 e12 e-
2 single bonds (2x2) = 4
1 double bond = 4
2 lone pairs (2x2) = 4
Total = 12
total number
total number
of valence
of nonbonding
electrons in electrons
the free atom
formal charge
= 4 -2 -½ x 6 = -1
on C
formal charge
= 6 -2 -½ x 6 = +1
on O
-
1
2
(
total number
of bonding
electrons
)
H
H
0
C
formal charge
on an atom in
a Lewis
structure
0
O
=
C – 4 eO – 6 e2H – 2x1 e12 e-
2 single bonds (2x2) = 4
1 double bond = 4
2 lone pairs (2x2) = 4
Total = 12
total number
total number
of valence
of nonbonding
electrons in electrons
the free atom
formal charge
= 4 - 0 -½ x 8 = 0
on C
formal charge
= 6 -4 -½ x 4 = 0
on O
-
1
2
(
total number
of bonding
electrons
)
Formal Charge and Lewis Structures
1. For neutral molecules, a Lewis structure in which there
are no formal charges is preferable to one in which
formal charges are present.
2. Lewis structures with large formal charges are less
plausible than those with small formal charges.
3. Among Lewis structures having similar distributions of
formal charges, the most plausible structure is the one in
which negative formal charges are placed on the more
electronegative atoms.
Which is the most likely Lewis structure for CH2O?
H
-1
+1
C
O
H
H
H
0
C
0
O
6.4
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.
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
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.)
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.
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
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
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
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:
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
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
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.
Write the Lewis structure of the carbonate ion (CO32-).
Step 1 – C is less electronegative than O, put C in center
Step 2 – Count valence electrons C - 4 (2s22p2) and O - 6 (2s22p4)
-2 charge – 2e4 + (3 x 6) + 2 = 24 valence electrons
Step 3 – Draw single bonds between C and O atoms and complete
octet on C and O atoms.
Step 4 - Check, are # of e- in structure equal to number of valence e- ?
3 single bonds (3x2) + 10 lone pairs (10x2) = 26 valence electrons
Step 5 - Too many electrons, form double bond and re-check # of e-
O
C
O
2 single bonds (2x2) = 4
1 double bond = 4
O
8 lone pairs (8x2) = 16
Total = 24
The Double bond could be between C and any of the
oxygens………this is known as resonance
Resonance Structures
Sometimes, a single Lewis structure does not adequately
represent the true structure of a molecule.
Consider the carbonate ion, CO32-
carbon (C) has four valence electrons x 1 carbon = 4 eoxygen (O) has six valence electrons x 3 oxygens = 18 eThe ion has an overall negative two charge so we add 2 e- to
give a total of 24 e- to be placed in the Lewis structure.
Carbon is the central atom, the three oxygens are bound to it and
electrons are added to fulfill the octets of the outer atoms.
All the available electrons have been used but carbon is electron
deficient - it only has six electrons around it.
So………………………
we share a non-bonding electron pair on an oxygen with the
carbon to create a double bond and thereby fulfill carbon's octet.
becomes
or
becomes
or
becomes
Three equivalent Lewis structures (formal charges are included)
can be drawn for the carbonate ion. The true structure of the
carbonate ion is an average of the three resonance structures
connected by a double headed arrow and enclosed in brackets.
This can also be represented by a resonance hybrid where a
dashed line is used to indicate the delocalized electrons.
6.5
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.
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.
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.
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.
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
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.
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
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).
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.
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
6.6
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.
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.
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.
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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
6.2
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
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 is the ability of an atom in a compound to draw
electrons to itself.
Electronegativity varies with atomic number.
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
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