Chemistry
Second Edition
Julia Burdge
Lecture PowerPoints
Jason A. Kautz
University of Nebraska-Lincoln
8
Chemical Bonding I:
Basic Concepts
Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
1
8
Chemical Bonding I: Basic Concepts
8.1 Lewis Dot Symbols
8.2 Ionic Bonding
Lattice Energy
The Born-Haber Cycle
8.3 Covalent Bonding
Lewis Structures
Multiple Bonds
Comparison of Ionic and Covalent Compounds
8.4 Electronegativity and Polarity
Electronegativity
Dipole Moment and Partial Charges
8.5 Drawing Lewis Structures
8.6 Lewis Structures and Formal Charge
8.7 Resonance
8.8 Exceptions to the Octet Rule
Incomplete Octets
Odd Numbers of Electrons
Expanded Octets
8.9 Bond Enthalpy
8.1
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
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.
Boron
•
•B•
1s22s22p1
3 valence electrons
•
B•
•
Lewis dot symbol for boron
•B•
•
•
•B
•
other reasonable Lewis dot
symbols for boron
Lewis Dot Symbols
Lewis dot symbols of the main group elements.
Lewis Dot Symbols
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.
Lewis Dot Symbols
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
8.2 Ionic Bonding
Ionic bonding refers to the electrostatic attraction that holds oppositely
charge ions together in an ionic compound.
K•
•• ••I • + e−
••
•• ••I • + K •
••
K+ + e−
−
••
•• I ••
••
K+ +
•• ••I ••
••
−
The attraction between the cation and
anion draws them together to form KI
Ionic Bonding
Lattice energy is the amount of energy required to convert a mole of
ionic solid to its constituent ions in the gas phase.
−
+
+
−
NaCl(s)  Na+(g) + Cl−(g)
−
−
+
+
−
+
−
−
+
Hlattice = +788 kJ/mol
+
Ionic Bonding
The magnitude of lattice energy is a measure of an ionic compound’s
stability.
Lattice energy depends on the magnitudes of the charge and on the
distance between them.
F
Q1
Q1  Q2
d2


d
Q = amount of charge
d = distance of separation
Q2
Ionic Bonding
The magnitude of lattice energy is a measure of an ionic compound’s
stability.
Lattice energy depends on the magnitudes of the charge and on the
distance between them.
Ionic Bonding
Ionic Bonding
A Born-Haber cycle is a cycle that relates the lattice energy of an ionic
compound to quantities that can be measured.
Ionic Bonding
Calculate the lattice energy of rubidium iodide (RbI).
Solution:
Step 1: Combine the pertinent thermodynamic data.
Rb+(g) + e− + Cl(g)
IE1(Na)
Rb+(g) + Cl−(g)
EA1(Cl)
Energy
Rb(s) + I(g)
ΔH°[I(g)]
f
ΔH°[Rb(g)]
f
Rb(g) + ½I2(g)
Rb(s) + ½I2(g)
ΔH°[RbI(s)]
f
RbI(s)
Lattice energy
of RbI
Ionic Bonding
Calculate the lattice energy of rubidium iodide (RbI).
Solution:
Step 2: Use Hess’s law to calculate the lattice energy.
Rb+(g) + e− + Cl(g)
IE1(Na)
Rb+(g) + Cl−(g)
EA1(Cl)
Energy
Rb(s) + I(g)
ΔH°[I(g)]
f
ΔH°[Rb(g)]
f
Rb(g) + ½I2(g)
Rb(s) + ½I2(g)
Lattice energy
of RbI
ΔH°[RbI(s)]
f
RbI(s)
Lattice energy = ΔHf°[Rb(g)] + ΔHf°[I(g)] + IE1(Rb) + |ΔHf°[RbI(s)]| − EA(I)
Ionic Bonding
Calculate the lattice energy of rubidium iodide (RbI).
Solution:
Step 3: Using the data found in the textbook, calculate the lattice energy.
Step
Energy (kJ/mol)
ΔHf°[Rb(g)]
85.8
ΔHf°[I(g)]
106.8
IE1(Rb)
403
ΔHf°[RbI(s)]
−328
EA(I)
295
Lattice energy = ΔHf°[Rb(g)] + ΔHf°[I(g)] + IE1(Rb) + |ΔHf°[RbI(s)]| − EA(I)
Lattice energy = 85.8 kJ/mol + 106. kJ/mol + 403 kJ/mol + 328 kJ/mol − 295 kJ/mol
Lattice energy = 629 kJ/mol
8.3
Covalent Bonding
The Lewis theory of bonding states that a chemical bond involves atoms
sharing electrons.
H•
+
•H
H •• H
Each H counts both electrons to “feel”
as though it has a He configuration
When electrons are shared between atoms, a covalent bond results.
Each electron in the bond is attracted to both nuclei.
Covalent Bonding
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
Covalent Bonding
Pairs of valence electrons not involved in bonding are called lone pairs.
••
••
••
••
••F •• F••
Covalent Bonding
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 atoms.
H with 2 e−
H with 2 e−
••
••
••
HOH
••
O with 8 e−
••
H O H
••
Shared electrons shown as
dashes (bonds)
Covalent Bonding
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−
one shared pair of electrons
results in a single bond
••
H O H
••
O with 8 e−
••
••
••
••
••
••
••
HOH
••
C with 8 e−
O C O
2 shared pairs of electrons
result in double bonds
O=C= O
Covalent Bonding
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 triple bonds
N≡N
••
N
••
••
••
••
••
••
N
Covalent Bonding
Bond length is defined as the distance between the nuclei of two covalently
bonded atoms.
Covalent Bonding
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
1.10 Å
<
N=N
1.24 Å
<
N−N
1.47 Å
Multiple bonds are stronger than single bonds.
Covalent Bonding
There are two types of attractive forces in covalent molecules:
Intramolecular bonding force that holds the atoms together
Intermolecular forces between different molecules.
Intermolecular forces are weak
compared to intramolecular
forces.
Covalent compounds tend to
be gases, liquids, or low-melting
solids.
Covalent Bonding
8.4
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 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 and Polarity
Electronegativity is the ability of an atom in a compound to draw
electrons to itself.
Electronegativity and Polarity
Electronegativity is the ability of an atom in a compound to draw
electrons to itself.
Electronegativity and Polarity
There is no sharp distinction between the different types of bonds:
nonpolar covalent and polar covalent
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.
Electronegativity and Polarity
An arrow is used to indicate the direction of electron shift in polar covalent
molecules.
Regions where electrons
spend little time
δ+
••
H−F
••
δ−
••
••
••
H−F
••
Regions where electrons
spend a lot of time
Electronegativity and Polarity
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
Expressed in debye units (D)
1 D = 3.336 x 10−30 C•m
Electronegativity and Polarity
A quantitative measure of the polarity of a bond is its dipole moment (μ).
μ=Qxr
8.5
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.
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.
Complete the octets of the most electronegative atom first.
5) Place any remaining electrons on the central atom.
6) Complete the octet of the central atom by forming multiple bonds.
Drawing Lewis Structures
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.
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. 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.
Drawing Lewis Structures
Draw the Lewis Structure for ClO3−.
Solution:
Step 5 Complete the octet of the central atom by forming multiple bonds.
All atoms have an octet in this structure.
8.6
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
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
8.7 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
8.8
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
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
Exceptions to the Octet Rule
Exceptions to the octet rule fall into three categories:
3) The central atom has more than eight electrons.
Sulfur has 6 bonds corresponding
to 12 electrons
8.9
Bond Enthalpy
The bond enthalpy is the enthalpy change associated with breaking a
bond in 1 mole of gaseous molecule.
H2(g) → H(g) + H(g)
ΔH° = 436.4 kJ/mol
The enthalpy for a gas phase reaction is given by:
ΔH° = ΣBE(reactants) – ΣBE(products)
ΔH° = total energy input – total energy released
bonds broken
bonds formed
Bond Enthalpy
Bond enthalpy change in an exothermic reaction:
Bond Enthalpy
Bond enthalpy change in an endothermic reaction:
Bond Enthalpy
Calculate the enthalpy of reaction for the following:
CH4(g) + Br2(g) → CH3Br(g) + HBr(g)
ΔH° = ΣBE(reactants) – ΣBE(products)
Step 1: Draw the Lewis structures to determine what bonds should be
broken and what bonds are to be formed.
Bond Enthalpy
Step 2: Bonds to break: 4 C−H and 1 Br−Br
Bonds to form: 3 C−H, 1 C−Br, and 1 H−Br
Bond
Bond enthalpy
(kJ/mol)
C−H
414
Br−Br
192.5
C−Br
276
H−Br
366.1
ΔH° = ΣBE(reactants) – ΣBE(products)
ΔH° = [4BE(C−H) + 1BE(Br−Br)] – [3BE(C−H) + 1BE(C−Br) + 1BE(H−Br)]
ΔH° = [4(414 kJ/mol) + 1(192.5 kJ/mol)] – [3(414 kJ/mol) + 1(276 kJ/mol) + 1(366.1 kJ/mol)]
ΔH° = -36 kJ/mol
8
Chapter Summary: Key Points
Lewis Dot Symbols
Ionic Bonding
Lattice Energy
Born-Haber Cycle
Covalent Bonding
Octet Rule
Lewis Structures
Bond Order
Bond Polarity
Electronegativity
Dipole Moment
Drawing Lewis Structures
Formal Charge
Resonance Structures
Incomplete Octets
Odd Numbers of Electrons
Expanded Octets
Bond Enthalpy