CHEMICAL
BONDING
Cocaine
Water
2
Chapter 9 Outline
•
•
•
•
Ionic -vs- Covalent
Bonding in molecules
Valance electrons
Lewis Structures
– Oxidation Numbers
– Formal Charges
• VESPR
Chemical Bonding
Problems and questions —
How is a molecule or polyatomic ion
held together?
Why are atoms distributed at strange
angles?
Why are molecules not flat?
Can we predict the structure?
How is structure related to chemical
and physical properties.
Forms of Chemical Bonds
• There are 2 extreme forms of
connecting or bonding atoms:
• Ionic—complete transfer of
electrons from one atom to
another
• Covalent—electrons shared
between atoms
• Most bonds are somewhere in
between.
Complete electron
Ionic
transfer from an
element of low IE (metal)
to an element of high EA (nonmetal)
Bonds
2 Na(s) + Cl2(g) ---> 2 Na+ + 2 Clresults in a metal/nonmetal or ionic compound.
Covalent Bonding
Covalent bond forms by the sharing of
VALENCE ELECTRONS, between
two nonmetals.
Valence Electrons
Electrons are divided between core and valence
electrons.
Na 1s2 2s2 2p6 3s1
Core = [Ne] and
valence = 3s1
Br
[Ar] 3d10 4s2 4p5
Core = [Ar] 3d10 and
valence = 4s2 4p5
.
35
: Br:
[Ar] 3d10 4s24p5
8
Valence Electrons
Covalent Bonding
The bond arises from the mutual attraction
of 2 nuclei for the same electrons.
A Bond is a balance of attractive and
repulsive forces.
Chemical Bonding
Objectives
Objectives are to understand:
1. e - distribution in molecules
and ions.
2. molecular structures.
3. bond properties and their
effect on molecular properties.
Electron
Distribution in
Molecules
• Electron distribution
is depicted with
Lewis electron
dot structures
• Electrons are
distributed as
shared or BOND
PAIRS and
unshared or LONE
PAIRS.
G. N. Lewis
1875 - 1946
Bond and Lone Pairs
Electrons are distributed as shared or
BOND PAIRS and unshared or
LONE PAIRS.
••
H
Cl
•
•
••
shared or
bond pair
lone pair (LP)
This is called a LEWIS
ELECTRON DOT structure.
13
Lewis Dot Electron Configs
Bond Formation
A bond can result from a “head-to-head” overlap
of atomic orbitals on neighboring atoms.
••
H
+
Cl
••
••
•
•
H
Cl
•
•
••
Overlap of H (1s) and Cl (3p)
This type of overlap places bonding
electrons in a MOLECULAR ORBITAL
along the line between the two atoms and
forms a SIGMA BOND ().
Rules of the Game
Observation that atoms want to
obtain a filled noble gas electron
configuration is called the
OCTET RULE
• There are exceptions to these
rules!!!
Building a Dot Structure
Ammonia, NH3
1. Decide on the central atom; never H.
Central atom is atom of lowest affinity for
electrons. Therefore, N is the central
atom.
2. Count valence electrons
H = 1 and N = 5
Total = (3 x 1) + 5
= 8 electrons / 4 pairs
Building a Dot Structure
3.
Form a sigma bond
between the central atom
and surrounding atoms.
H
N
H
H
4.
Remaining electrons go to
the central atom to form LONE
••
PAIRS to complete octet as
H N
needed.
3 BOND PAIRS and 1 LONE
H
PAIR. Note that N has a share
in 4 pairs (8 electrons), while
H shares 1 pair.
H
Sulfite ion, SO32Step 1. Central atom = S
Step 2. Count valence electrons
S= 6
3 x O = 3 x 6 = 18
Negative charge = 2
TOTAL = 26 e- or 13 pairs
Step 3. Form sigma bonds
Sulfite ion, SO32Step 1. Central atom = S
Step 2. Count valence electrons
S= 6
3 x O = 3 x 6 = 18
Negative charge = 2
TOTAL = 26 e- or 13 pairs
Step 3. Form sigma bonds
O
10 pairs of electrons are
now left.
O
S
O
Sulfite ion, SO32Remaining pairs become lone pairs,
first on outside atoms and then
remaining pairs on central atom.
•
•
•
•
O
••
O
••
-2
••
•
•
••
S
••
O ••
••
Each atom is surrounded by an octet of
electrons.
Carbon Dioxide, CO2
C
1. Central atom = _______
2. Valence electrons = __
16 or __
8 pairs
3. Form sigma bonds.
O
C
O
This leaves 6 pairs.
4. Place lone pairs on outer atoms.
••
••
•
•
O
••
C
O
••
•
•
Carbon Dioxide, CO2
4. Place lone pairs on outer atoms.
••
••
•
•
O
C
••
O
•
•
••
5. So that C has an octet, we form
DOUBLE BONDS between C and O.
••
••
•
•
O
••
C
O
••
•
•
•
•
O
••
C
The second bonding pair forms a pi
bond.
O
••
(p)
•
•
Double and even
triple bonds are
commonly
observed for C,
N, P, O, and S
C2F4
H2CO
SO3
Sulfur Dioxide, SO2
1. Central atom = S
2. Valence electrons = 18 or 9 pairs
••
•
•
O
••
••
••
S
O
•
•
••
3. Form pi (p) bond so that S has an octet
— but note that there are two ways of
doing this.
OR bring in
right pair
bring in
left pair
••
•
•
O
••
••
S
••
O
••
•
•
Sulfur Dioxide, SO2
OR bring in
right pair
bring in
left pair
••
•
•
••
••
O
S
••
O
•
•
••
This leads to the following structures.
•
•
O
••
••
S
••
O
••
•
•
•
•
••
••
O
S
••
O
••
••
These equivalent structures are called
RESONANCE STRUCTURES. The true
electronic structure is a HYBRID of the two.
Urea, (NH2)2CO
1. Arrangement, is C or O the central
atom?
2. Number of valence electrons = 24 e-
3. Draw sigma bonds.
O
C
O
H N
H
C
N
H
H
Urea, (NH2)2CO
4. Place remaining electron pairs
in the molecule.
••
••
O
••
H N
H
•
•
••
C
N H
H
5. Carbon needs an Octet, take from the Oxygen
Urea, (NH2)2CO
Exceptions to the Octet Rule
Occurs with B (the only 2nd period
exception) and elements of 3rd - 7th periods.
BF3
SF4
Boron Trifluoride
•
•
•
•
Central atom =
Valence electrons =
or electron pairs =
Assemble dot structure
••
•
•
F
•
•
••
•
•
F
B
••
•
•
F
••
•
•
The B atom has a share
in only 6 electrons (or
3 pairs). The B atom in
many molecules is
electron deficient.
Sulfur Tetrafluoride, SF4
• Central atom =
• Valence electrons = ___ or ___ pairs.
• Form sigma bonds and distribute
electron pairs.
••
•
•
••
F
••
••
•
•
F
••
S
••
F
••
••
F
••
•
•
•
•
5 pairs around the S
atom. A common
occurrence outside the
2nd period.
Draw Lewis Structures for:
N2
CH4
I2CO
HCOOH
NO2-
XeF4
Oxidation Number
• Oxidation Number is assigned based on
a set of rules.
• These rules are based on the Lewis
Structures of the compounds or ions.
• Oxidation Number = Group no.
- (no. assigned electrons)
- (no. of LP electrons)
Sample Problem
Formal Atom Charges
• Atoms in molecules often bear a
charge (+ or -).
• The predominant resonance structure
of a molecule is the one with charges
as close to 0 as possible.
• Formal charge = Group no.
- 1/2 (no. bond electrons)
- (no. of LP electrons)
Carbon Dioxide, CO2
Formal charge calculations
6 - (1/ 2)(4) - 4
••
•
•
O
••
C
4 - (1/ 2)(8) - 0
O
=
•
•
0
=
0
Carbon Dioxide, CO2
6 - (1/ 2)(2) - 6
••
•
•
O
C
6 - (1/ 2)(6) - 2
O
•
•
••
=
-1
C atom
charge is 0.
= +1
Which is the predominant resonance structure?
Carbon Dioxide, CO2
-0.73
-0.73
+1.46
Actual partial charges.
Boron Trifluoride, BF3
••
•
•
•
•
F
••
•
•
F
••
B
•
•
•
•
F
••
What if we form a B—F double
bond to satisfy the B atom octet?
Boron Trifluoride, BF3
••
•
•
F fc = 7 - 2 - 4 = +1
••
•
•
F
••
B fc = 3 - 4 - 0 = -1
•
•
•
•
F
••
• To have +1 charge on F, with
its very high affinity for
electrons, is not good.
• Negative charges are best
placed on atoms with high
affinity for electrons.
Thiocyanate ion, SCNWhich of three possible resonance
structures is the most important?
Calculate the formal charge for each
element.
Thiocyanate ion,
-0.16
-0.52
SCN
-0.32
Calculated partial charges
PRACTICE PROBLEM
Ox # = group # - lone pair electrons assigned electrons
F.C. = group # - lone pair electrons - 1/2
bonding electrons
Practice Problem
Determine oxidation numbers and formal
charges for the atoms in SO3-2.
MOLECULAR GEOMETRY
MOLECULAR GEOMETRY
VSEPR
• Valence Shell Electron
Pair Repulsion theory.
• Most important factor in
determining geometry is
relative repulsion between
electron pairs.
Molecule adopts
the shape that
minimizes the
electron pair
repulsions.
Figure 9.11
Figure 9.12
No. of e- Pairs
Around Central
Atom
2
Example
F—Be—F
Geometry
linear
180
F
3
F
planar
trigonal
B
F
120
H
4
C
H
109
tetrahedral
H
H
Notice where the central atom is
4 electron pair groups
Electron pair geometry
48
Molecular Shape name
Electron pair geometry
Figure 9.14
Molecular name
Brief Summary of VESPR
50
Electron Pair
Geometry
Molecular Geometry
2 electron pairs
Linear
Linear
3 electron pairs
Trigonal Planar
Trigonal Planar
Trigonal Planar
“V” Bent
4 electron pairs
Tetrahedral
Tetrahedral
Tetrahedral
Trigonal
Bipyramidal
Tetrahedral
Trigonal Pyramidal
“V” Bent
“look all 4 of them up”
Octahedral
“look all 4 of them up”
5 electron pairs
6 electron pairs
14 possible combinations….learn them all!
Structure Determination
by VSEPR
Ammonia, NH3
1. Draw electron dot structure
••
H N H
H
2. Count BP’s and LP’s around central
N atom = 4
(Called the number of structural pairs.)
Structure Determination
by VSEPR
Ammonia, NH3
H
1. Draw electron dot structure
2. Count BP’s and LP’s = 4
3. The 4 electron pairs are at the
corners of a tetrahedron.
H
H
N H
H
lone pair of electrons
in tetrahedral position
N
H
••
Structure Determination
by VSEPR
Ammonia, NH3
There are 4 electron pairs at the corners
of a tetrahedron.
lone pair of electrons
in tetrahedral position
••
H N H
H
N
H
H
H
The ELECTRON PAIR GEOMETRY is
tetrahedral.
Structure Determination
by VSEPR
Ammonia, NH3
The electron pair geometry is tetrahedral.
lone pair of electrons
in tetrahedral position
N
H
H
H
The MOLECULAR GEOMETRY or SHAPE
— the positions of the atoms —
is Trigonal PYRAMIDAL.
55
VESPR Step-by-step
Determination
1
2
Structure Determination
by VSEPR
Water, H2O
1. Draw electron dot structure
••
H O•• H
2. Count BP’s and LP’s = 4
Structure Determination
by VSEPR
Water, H2O
1. Draw electron dot structure
2. Count BP’s and LP’s = 4
3. The 4 electron pairs are at the
corners of a tetrahedron.
O
H
H
The electron pair
geometry is
TETRAHEDRAL.
••
H O•• H
Structure Determination
by VSEPR
Water, H2O
••
H O•• H
O
H
H
The molecular
geometry is
bent 109.5
The electron pair
geometry is
TETRAHEDRAL.
Structure Determination
by VSEPR
Formaldehyde, CH2O
1. Draw electron dot structure
•
•
H
O ••
C
H
Structure Determination
by VSEPR
Formaldehyde, CH2O
•
•
1. Draw electron dot structure
2. Count BP’s and LP’s = 3
H
(the double bond is treated as a
“lump” of electrons or one pair)
O ••
C
H
Structure Determination
by VSEPR
Formaldehyde, CH2O
•
•
O ••
1. Draw electron dot structure
H C
2. Count BP’s and LP’s = 3
3. There are 3 electron pairs are at the
corners of a planar triangle.
•
•
O ••
C
H
H
H
The electron pair
geometry is TRIGONAL
PLANAR with 120o bond
angles.
Structure Determination
by VSEPR
Formaldehyde, CH2O
•
•
O ••
The electron pair
geometry is
TRIGONAL PLANAR
C
H
H
The molecular
geometry is also
TRIGONAL PLANAR.
Structure Determination
by VSEPR
Methanol, CH3OH
1. Draw electron dot structure
H
••
H—C—O—H
••
1
H
2
2. Define bond angles 1 and 2
Structure Determination
by VSEPR
Methanol, CH3OH
1
H
H
C
O
H
2
Define bond angles 1 and 2
In both cases the atom
is surrounded by
4 electron pairs.
H
Structure Determination
by VSEPR
Acetonitrile, CH3CN
Define bond angles 1 and 2
Angle 1 = 109o
Angle 2 = 180o
One C is surrounded by 4 electron
“lumps” and the other by 2 “lumps”
Determine the
bond angles.
:
Phenylalanine,
an amino acid
120o
109o
109o
109o
120o
Compounds with 5 or 6 Pairs
Around the Central Atom
Compounds with 5 or 6 Pairs
Around the Central Atom
90
F
F
P
Trigonal bipyramid
F
120 5 electron pairs
F
F
90
6 electron pairs
F
F
S
F
F
Octahedral
F
F
90
69
STRUCTURES
WITH CENTRAL
ATOMS THAT DO
NOT OBEY THE
OCTET RULE
70
Deviations from the Octet Rule
Usually occurs with Group 3A elements
and with those of 3rd period and higher.
Consider boron trifluoride, BF3
••
F
•
•
•
•
••
•
•
F
••
B
•
•
•
•
F
••
71
Deviations from the Octet Rule
Consider boron trifluoride, BF3
••
The B atom is surrounded by
only 3 electron pairs.
Bond angles are
120o
Geometry
described as
planar trigonal
•
•
•
•
F
••
•
•
F
••
B
•
•
•
•
F
••
Sulfur Tetrafluoride, SF4
• Number of valence electrons =
• Central atom =
• Dot structure
Sulfur Tetrafluoride, SF4
• Number of valence electrons = 34
• Central atom = S
• Dot structure
••
••
•• F
••
••
••
••F S F••
••
••
•• F••
••
Sulfur Tetrafluoride, SF4
Number of valence electrons = 34
••
Central atom = S
••
•• F
••
••
Dot structure
••
•
•
F
•••
S
F•
••
•• F••
••
Electron pair geometry = ?
Sulfur Tetrafluoride, SF4
• Number of valence
electrons = 34
• Central atom = S
• Dot structure
••
••
•• F
••
•••
••
••F S F•
••
••
•• F••
••
Electron pair geometry
= trigonal bipyramid
(because there are 5
pairs around the S)
What is the Molecular Geometry?
Sulfur Tetrafluoride, SF4
Not all repulsions are equivalent.
L.P. - L.P. > L.P. - B.P. > B.P. - B.P.
Minimize lone pair bond pair repulsions at
the 90o angle.
90
••
F
S
F
F
F
120
Structural Determination
Determine the geometries for PF3,
NO2-, NO2+, ICl2-, ClF3, ClF5, and IBr4-,
giving VSEPR class, shape name, 3-D
diagram, and bond angles.
Solutions
Bond Properties
• What is the effect of bonding and
structure on molecular properties?
Buckyball in HIV-protease
Bond Order
• The number of bonds between a
pair of atoms.
H
single
BO = 1
1 
H
H
C
C
triple, BO = 3
1  and 2 p
C
N
double, BO = 2
1  and 1 p
Bond Order
Fractional bond orders occur in molecules with
resonance structures.
Consider NO2••
••
••
O
••
N
•• •
O•
••
••••
O
••
N
••
O
••
The N—O bond order = 1.5
Total # of e - pairs used for a type of bond
Bond order =
Total # of bonds of that type
3 e - pairs in N — O bonds
Bond order =
2 N — O bonds
Bond Order
Bond order is proportional to two
important bond properties:
(a) bond strength
(b) bond length
110 pm
745 kJ
414 kJ 123 pm
Bond Length
Bond length is the distance between
the nuclei of two bonded atoms.
Bond Length
Bond length
depends on size
of bonded atoms.
H—F
H—Cl
Bond distances measured
in Angstroms or pm where
1 A = 100 pm.
H—I
Bond Length
Bond length
depends on
bond order.
Different C-O Bond Lengths
C-C bond lengths
C-N bond lengths
Bond Strength
• Bond strength is measured by the energy
required to break a bond. See Table 9.9
C-C bond strengths
Bond Strength
BOND
H—H
C—C
C=C
CC
NN
STRENGTH (kJ/mol)
436
346
602
835
945
The GREATER the number of bonds
(bond order) the HIGHER the bond
strength and the SHORTER the bond.
Bond Strengths for O-O
Bond
HO—OH
••
O
••
••
O
O=O
Order
Length
1
147 pm
210 kJ/mol
128
394
?
121
498
•• • 1.5
O•
••
2
Strength
Using Bond Energies
DHorxn = S H (bonds broken) S H (bonds formed) or
DHorxn = energy input (+) +
energy output (-).
Net energy = DHrxn, bond =
energy required to break bond energy evolved when bonds are formed.
Using Bond Energies
Estimate the energy of the reaction
H—H + Cl—Cl ----> 2 H—Cl
H—H = 436 kJ/mol
Cl—Cl = 242 kJ/mol
H—Cl = 432 kJ/mol
Sum of H-H + Cl-Cl bond energies =
436 kJ + 242 kJ = +678 kJ
2 mol H-Cl bond energies = 864 kJ
Net = DH = +678 kJ - 864 kJ = -186 kJ
Using Bond Energies
Estimate the energy of the reaction
2 H—O—O—H ----> O=O + 2 H—O—H
Is the reaction exo- or endothermic?
Which is larger: energy req’d to break
bonds or energy evolved on making
bonds?
Using Bond Energies
2 H—O—O—H
----> O=O + 2 H—O—H
Energy required to break bonds:
break 4 mol of O—H bonds = 4 (463 kJ)
break 2 mol O—O bonds = 2 (146 kJ)
TOTAL ENERGY to break bonds = 2144 kJ
Using Bond Energies
2 H—O—O—H
----> O=O + 2 H—O—H
Energy evolved on making bonds:
make 1 mol of O=O bonds = 498 kJ
make 4 mol O—H bonds = 4 (463 kJ)
TOTAL ENERGY evolved on making bonds =
2350 kJ
Using Bond Energies
2 H—O—O—H
----> O=O + 2 H—O—H
Net energy = +2144 kJ - 2350 kJ = - 206 kJ
The reaction is exothermic!
More energy is evolved on
making bonds than is
expended in breaking
bonds.
Calculate DHorxn for the reaction
below using bond energies.
CH4 (g) + 2 O2 (g) ______> 2 H2O (g) + CO2 (g)
DH = [4(C-H) + 2(O=O)] - [4(H-O) + 2(C=O)]
DH = [4(413kJ) + 2(498kJ)] - [4(463kJ) + 2(732kJ)]
DH = -1166 kJ
More Sample Problems!
Molecular Polarity
Why are water molecules attracted
to a balloon that has a static
electric charge?
Bond Polarity
+
H
-
••
Cl ••
••
HCl is POLAR because it
has a positive end and
a negative end.
Polarity arises because
Cl has a greater share
in bonding electrons
than does H.
Bond Polarity
This model, calc’d using CAChe software,
shows that H is + (red) and Cl is - (yellow).
Calc’d charge is + or - 0.20.
This model shows that the electron density
is greater around Cl than around H.
+
H
-
••
Cl ••
••
Bond Polarity
Due to the bond polarity, the H—Cl
bond energy is GREATER than
expected for a “pure” covalent bond.
BOND
“pure” bond
real bond
ENERGY
339 kJ/mol calc’d
432 kJ/mol measured
Difference = 92 kJ. This difference is
proportional to the difference in
ELECTRONEGATIVITY,
Don’t confuse  with mole fraction from Ch 12
.
+
H
-
••
Cl ••
••
Bond Polarity
Partial charge on atom A =
Group Number of A
-
# lone pair electrons
-
A/(A+ B)( # bonding e- shared by A )
Electronegativity, 
 is a measure of the ability of
an atom in a molecule to
attract electrons to itself.
Concept proposed by
Linus Pauling
1901-1994
Linus Pauling, 1901-1994
The only person to receive two unshared
Nobel prizes (for Peace and Chemistry).
Chemistry areas: bonding, electronegativity,
protein structure
Electronegativity, 
See
Figure
9.10
• F has maximum .
• Atom with lowest  is the center atom
in most molecules.
• Relative values of  determine BOND
POLARITY (and point of attack on a
molecule).
Electronegativity, 
Figure 9.9
Basic Trend =
Bond Polarity
Which bond is more polar (or DIPOLAR)?
O—H
O—F

3.5 - 2.1
3.5 - 4.0
D
1.4
0.5
OH is more polar than OF
O—H
- +
O—F
+ -
and polarity is “reversed.”
Bond Polarity
Relative polarity in a bond is determined
by D , where the approximate % ionic
character is give by:
D = 1.0, 20%;
D = 1.5, 40%;
D = 2.0, 60%;
D = 2.5, 80%.
Molecular Polarity
Molecules—such as HCl and H2O— can be
POLAR (or dipolar).
They have a DIPOLE MOMENT. The polar HCl
molecule will turn to align with an electric field.
Molecular Polarity
The magnitude of the
dipole is given in
Debye units.
Named for Peter
Debye (1884 - 1966).
Received the 1936
Nobel prize for work
on x-ray diffraction
and dipole
moments.
Molecular Polarity
Molecules will be polar if
1. bonds are polar AND
2. the molecule is NOT “symmetric”
Symmetric molecules
Molecular Polarity
Unsymmetrical
Carbon Dioxide
• CO2 is NOT polar even though the CO bonds are
polar.
• CO2 is symmetrical.
Positive C
atom is reason
CO2 + H2O
gives H2CO3
Microwave
oven
Consequences
of H2O Polarity
Molecular Polarity
More Examples
F
B atom is
positive and
F atoms are
negative.
B
F
F
B—F bonds in BF3 are polar.
But molecule is symmetrical
and NOT polar
Molecular Polarity
H
B atom is
positive but
H & F atoms
are negative.
B
F
F
B—F and B—H bonds in HBF2
are polar. But molecule is
NOT symmetrical and is
polar
Polarity of Methane, CH4
H
C
H
H
H
Methane is symmetrical and is NOT polar.
Polarity of CH3F
F
C
H
H
H
C—F bond is very polar.
Molecule is not symmetrical
and so is polar.
Substituted Ethylene
• C—F bonds are MUCH more polar than
C—H bonds.
• Because both C—F bonds are on same
side of molecule, molecule is POLAR.
Substituted Ethylene
• C—F bonds are MUCH more polar than
C—H bonds.
• Because both C—F bonds are on opposing
ends of molecule, molecule is NOT
POLAR.
Practice Problems
1. Give the VSEPR class, bond angle, shape,
and polarity of KrF2, CO, and NO2-.
2. Calculate the heat of reaction for
C2H4 (g) + 3 O2 (g) --> 2 CO2 (g) + 2 H2O (g)
3. Calculate the carbon-carbon triple bond
energy in C2H2(g).
Heat of formation of C2H2 = 226 kJ/mole)
4. Determine the number of p and  bonds in
acetic acid.
5. Arrange in order of increasing polarity:
Sr-Br, P-F, S-F, Al-O
Practice Problems Answers
1. AX2E3, 1800, linear, nonpolar
AXE, 1800, linear, polar
AX2E, 1200, bent, polar
2. -1032 kJ
3. 818 kJ
4. 7  and 1 p
5. S-F, P-F, Al-O, Sr-Br,
Ionic Bonding
Ionic bonds form by the transferring of
VALENCE ELECTRONS, the
electrons at the outer edge of the atom.
Ionic Bonding
Na and F
Na 1s22s22p63s1
F 1s22s22p5
Na1+ 1s22s22p6
F1- 1s22s22p6
NaF
Ionic Bonding
Mg and O
Mg 1s22s22p63s2
O 1s22s22p4
Mg2+ 1s22s22p6
O2- 1s22s22p6
MgO
Ionic Bonding
Mg and F
Mg 1s22s22p63s2
F 1s22s22p5
F 1s22s22p5
Mg2+ 1s22s22p6
MgF2
F1- 1s22s22p6
F1- 1s22s22p6
Ionic Bonding
Na and O
Na 1s22s22p63s1
O 1s22s22p4
Na 1s22s22p63s1
Na1+ 1s22s22p63s1
Na1+ 1s22s22p63s1
O2- 1s22s22p6
Na2O
Covalent Bonding
Cl and F
Cl
.F
.
.
Cl . F
Covalent Bonding
S and F
.F
.S.
F
.. S .. F
Covalent Bonding
O and O
O
O
.
.
O .. O
Sample Problems
Draw dot structures for the following:
H2O
CO
NO21-1
H
-O- H
O
N
-O
-1
C
O
O- N
O
Draw Lewis Structures for:
N2
CH4
I2CO
HCOOH
NO2-
XeF4
N
N
Draw Lewis Structures for:
H
N2
CH4
I2CO
HCOOH
NO2-
XeF4
H
C
H
H
Draw Lewis Structures for:
N2
CH4
O
I2CO
HCOOH
NO2-
XeF4
I
C
I
Draw Lewis Structures for:
N2
CH4
I2CO
HCOOH
NO2-
XeF4
O
H
C
O
H
Draw Lewis Structures for:
N2
-
CH4
I2CO
O
N
O
HCOOH
NO2-
XeF4
-
O
O
N
O
Draw Lewis Structures for:
N2
CH4
I2CO
HCOOH
-
NO2
XeF4
..
Xe
..
Sample Problems
Determine the oxidation number for each
atom in the following:
CO2
SO32O = -2
O = -2
C=4
S=4
Sample Problems
Determine the formal charge for each atom
in the following:
CO
NO21O=1
single bond O = -1
C = -1
double bond O = 0
N=0
Structural Determination
PF3
PF3
AX3E
trigonal pyramidal
..
Structural Determination
NO2+
NO2+
linear
AX2
Structural Determination
-
ICl2
ICl2-
AX2E3
linear
180o
120o
..
Structural Determination
ClF3
ClF3
AX3E2
T - shaped
120o
Structural Determination
ClF5
ClF5
AX5E
square pyramidal
..
Structural Determination
-
IBr4
IBr4-
AX4E2
square planar
..
..
Calculate DHorxn for the reaction
below using bond energies.
C (g) + 2 Cl2 (g)
______>
CCl4 (g)
DH = [2(Cl-Cl)] - [4(C-Cl)]
DH = [2(242 kJ)] - [4(339 kJ)]
DH = - 872 kJ
Calculate DHorxn for the reaction
below using bond energies.
C (s) + 2 Cl2 (g)
______>
CCl4 (g)
DH = [ DHsubC + 2(Cl-Cl)] - [4(C-Cl)]
DH = [ 717 kJ + 2(242 kJ)] - [4(339 kJ)]
DH = -155 kJ
Calculate DHof for CH4 using
bond energies.
C (s) + 2 H2 (g) --> CH4 (g)
DHf = [DHsub C + 2(H-H)] - [4(C-H)]
DHf = [ 717 kJ + 2(436 kJ)] - [4(413kJ)]
DHf = - 63 kJ
Given DHof for C2H4 (g) is 52 kJ/mole.
Using bond energies, calculate the
carbon carbon bond energy in C2H4 (g).
2 C (s) + 2 H2 (g) -- > C2H4 (g)
DHf = [ 2(DHsubC) + 2(H-H) ] - [ 4(C-H) + 1(C=C) ]
52kJ = [ 2(717kJ) + 2(436kJ)] - [4(413kJ) + 1(C=C)]
(C=C) = 602 kJ
Bond Polarity
Ionic
Polar
Nonpolar
Covalent
Covalent
<-------------------------------------------------------------->
:F
Li
C : F
F : F