General Bonding Concepts
Chapter 8 Sections 1-12 only
Objectives
• Define 3 different types of bonds based on
electron distribution
• Use Coulomb's law equation to determine
attraction/repulsion
Chemical Bonds
• What is a bond?
• The forces that cause atoms to act in unison
• Why do bonds form?
• to use or achieve the lowest possible energy
• What is bond energy?
• the amount of energy required to BREAK the bond
Types of Bonds
• Ionic: Electrons are transferred
• Polar Covalent: Electrons are shared unequally
• Covalent: Electrons are shared equally
Ionic Bonding
What types of substances react to form ionic bonds?
Answer: Metals and non metals
Energy of interaction between a pair of ions can be
calculated using COULOMB'S LAW.
The formula:
 Q1Q2 
19
2.31 10 J  nm
E
 r 
Coulomb’s Law Equation
Constant:
Joules= energy unit,
nm = distance unit
2.3110
19
Charge on Ions
 Q1Q2 
J  nm
E
 r 
Distance between atoms
(in nm)
In Joules
Using Coulomb's Law
In NaCl the distance between ions is 0.276 nm,
What is the energy of this bond interaction?
What equation?
2.31 10
Set it up:
Answer:
2.3110
19
19
 Q1Q2 
J  nm
E
 r 
 (1)( 1) 
J  nm
E
 0.276 
E = -8.37x10-19 J
Coulomb's Law
• A Negative Coulomb's Law answer indicates
an attractive force
• A Positive Coulomb's Law answer indicates a
repulsive force
Coulomb’s Law Practice
Objectives
• Discuss relationship between bond energy and
bond length
• Discuss how electronegativity impacts bonding
• Determine if a bond will be polar, if so show
dipole moment
• Define bond, bond energy, bond length, dipole
moment
• Explain the three forces at work when atoms
bond
Relate Bond Energy and Bond Length
• Finish this statement:
• as bond length decreases bond energy…
INCREASES
Defend using Coulomb's Law:
2.31 10
19
 Q1Q2 
J  nm
E
 r 
What about two same type atoms?
• Example
H:H
• A bond will form if the energy of the aggregate is
lower than that of the separated ions or atoms
(p. 331)
Picture from:
chemkb.cs.iupui.edu
Bonding between same type ions
• These are covalent bonds.
• Bond Length is defined as the distance where
the lowest energy is achieved.
• Several forces are involved to establish bond
length
The Forces at Work
When same type atoms interact you get a covalent bond.
The following forces must be balanced for a covalent bond to occur:
Type of interaction:
What is this trying to do?
• Proton vs Proton
Repel
• Proton vs Electron
Attract
• Electron vs Electron
Repel
Key points of Covalent Bonding
• Shared Electrons
• Electrons shared in area between nuclei
• Bonds created to increase stability
Key points of Covalent Bonding
• Unequal sharing of electrons
• Based on electronegativity difference
• Electronegativity Differences and bond type
• 0-1 "covalent"
• 1-2 "polar covalent"
• >2 "ionic“
• Actually more like a sliding scale
Most Covalent
0
Most Ionic
4
Consequences of Unequal Sharing
• Dipole Moment: partial charges caused by the
unequal sharing of electrons
• Polar bond does not always equal polar
molecule
• Dipoles can cancel due to molecular structure
• Two ways to show dipoles:
HF
d d
HF
Assignment
• Page 383 # 24, 26, 30, 32
Show the bond polarity for the
following bonded atoms:
C-O
Se-S
P-H
Cl-I
H-Cl
Br-Te
Br-Br
O-P
Si-S
Objectives:
• Describe what it means for an atom/ion to be stable
• Use electron configurations to predict formulas of ionic
compounds
• Describe the relationship between parent atom size
and ion size for cations and anions
• Use a periodic trend to determine relative ion size in a
group
• Define an isoelectronic ion group
• Discuss the trend for ions size in an isoelectronic group
Stable Atoms
Quantum mechanical model has helped to show
that atoms in a stable compound have
achieved noble gas configuration (full energy
level) by either sharing electrons or by
forming ions by either the loss or gain of
electrons.
• Stable covalent compounds are achieved by
atoms sharing electrons so that both complete
their valence shells to attain noble gas
configuration
• Stable ionic compounds are formed when a nonmetal removes the valence electrons from the
metal so that it fills its valence shell to achieve
noble gas configuration, the metal reverts back to
the last noble gas configuration.
Predicting Ion Formation
You can use electron configurations to predict ion formation
Ca: [Ar]4s2
Ca wants to lose 2 electrons and achieve the configuration of Argon
O: [He] 2s22p4
O wants to gain 2 electrons and achieve the configuration of Neon
From this we predict:
The calcium ion
The oxide ion
Ca2+
O2-
The compound calcium oxide
CaO
Ion Prediction and Compound
Formation
What happens when aluminum and oxygen react?
• Look at electron configurations:
Al = [Ne] 3s23p1
O = [He] 2s22p4
• What does each atom "want"?
Al "wants" to lose 3 electrons,
O "wants" to gain 2 electrons
• What formula would you predict? Why?
• Al2O3
• all atoms are "satisfied"
Ion Size
• Important when determining stability,
structure, and properties of ionic solids and
aqueous ions.
• Impossible to precisely define ion size
• There are several ways to look at trends for
ion size
Ion Size and The Parent Atom
Cations are smaller than their parent atom.
Why?
• loss of electrons results in the loss of an entire
energy level which decreases the size of the
entity
Anions are larger than their parent atom.
Why?
• gain of electrons results in a fuller outer energy
level which increases the size of the entity
Ion Size and Periodic Groups
• Ion size increases down a group.
Why?
• as you move down a group, there are more
filled energy levels= larger ion
• There is not a trend for ion size across a period
like there is for other types of periodic trends.
Why?
• as you move across a period, there is a change
over from metals (cations) to non metals
(anions) so in the middle of a period, it will
change from smaller ions to large ones
Ion Size and Isoelectronic Ions
• Define Isoelectronic:
• ions containing the same number of electrons
• Example:
• O-2, F-1, Na+1, Mg+2, Al+3
• What is the electron configuration of each ?
They all have the configuration of NEON!
10 total electrons are in each ion!
• This is an isoelectronic set of ions
Ion Size and Isoelectronic Ions ...
continued
• The trend for isoelectronic ions is that ion size
decreases as atomic number (nuclear charge,
number of protons, z) increases.
Why??
• Isoelectronic ions by definition have the same
number of electrons, therefore, the greater the
number of protons (atomic number) the greater
the positive force drawing those electrons
towards the nucleus (making the ion smaller by
drawing the electrons closer)
Ion Size and Isoelectronic Ions...
Practice
• Arrange the following in order of decreasing ion
size
• Br-1, Rb+1, Se-2, Sr+2
Se-2, Br-1, Rb+1, Sr+2
• Choose the smallest ion from the following sets
• Li+1, Na+1, K+1, Rb+1
Li+1
• Ba+2, Cs+1, I-1, Te-2
Ba+2
Assignment:
• Page 383 # 35, 37, 39, 41
Answers:
#35 a. Sc+3 , b. Te-2 , c. Ce+4 and Ti+4, d. Ba+2
#37. La+3 , Ba+2 , Cs+1 , I-1 , Te-2
#39 a. Cu > Cu+1 > Cu+2
b. Pt+2 >Pd+2 >Ni+2
c. O-2 , O-1 , O
d. La+3 , Eu+3 , Gd+3 , Yb+3
e . Te-2 , I -1 , Cs+1 , Ba+2 , La+3
#41 a. Al2S3 – aluminum sulfide
b. K3N – potassium nitride
c. MgCl2 – magnesium chloride
d. CsBr – cesium bromide
Objectives:
• Define lattice energy
• Determine relative lattice energy among a
group of compounds
• Use lattice energy to calculate reaction
energy (5 step)
Lattice Energy
• Define Lattice Energy:
• the change in energy that takes place when
separated gaseous ions are packed together to form
an ionic solid
Na+1 (g) + Cl-1 (g) --> NaCl (s)
change in energy from
here
to here
is lattice energy !
The Textbooks's Perspective on Energy
• Viewed from the system's point of view
• Negative energy if the process is
exothermic
• Positive energy if the process is
endothermic
• This perspective gives lattice energy a
negative value!
Lattice Energy Calculation
• Lattice Energy is Calculated by a modified form of
Coulomb's Law.
1 2
Q Q 
LatticeEne rgy  k 

 r 
Where k is a constant that depends on the structure of
the solid and the electron configurations of the ions
Note: you will NOT be expected to calculate lattice energy,
just to know how it is calculated and to use lattice energy
to calculate total reaction energy
Estimating Relative Lattice Energy
 Q1Q 2 
LatticeEne rgy  k 

 r 
• Lattice energy increases as ion charge (Q1 and
Q2) increases
• Lattice energy decreases as the distance
between ions (r) increases
Practice:
Which compound in the following pairs has the
most exothermic lattice energy? Why?
• NaCl or KCl
• NaCl : smaller r
• Mg(OH)2 or MgO
• MgO : larger Q
• LiF or LiCl
• LiF : smaller r
• Fe(OH)2 or Fe(OH)3
• Fe(OH)3 : larger Q
• NaCl or Na2O
• Na2O : Larger Q
• MgO or BaS
• MgO : smaller r
Ionic Solid Formation Energy
Calculation
• Li (s) + 1/2 F2 (g) --> LiF (s) Note: must have balanced equation
• Use lattice energy to determine the energy change
experienced in this reaction.
• The calculation is broken into steps based on states of
matter
•
•
•
•
•
Sublimation
Ionization
Dissociation
Ion formation
Solid formation (lattice energy**)
** remember to use lattice energy you need separate,
gaseous ions. The other 4 steps just get you to this
point.
An Example
Step
Process
Energy change
Sublimation
Li (s) --> Li (g)
161 kJ/mol
Ionization
Li (g) --> Li+1 + 1 e-
520 kJ/mol
Dissociation
1/2 F2 (g) --> F (g)
154 kJ/mol
Ion Formation
F (g) + e- --> F-1 (g)
-328 kJ/mol
Solid Formation
Li+1(g) + F-1 (g) --> LiF (s)
-1047 kJ/mol
Total Energy
Li (s) + 1/2 F2 (g) --> LiF (s) -617 kJ
the negative sign tells that energy is being released (exothermic)
Assignment:
• Page 383-4 # 44, 45, 46, 49
Answers:
A better definition of ionic compounds:
• Any compound that conducts electricity when
melted is ionic
• Avoids confusion when polyatomic ions, which
are often held together covalently are part of
the compound
Use of Models
• Models attempt to explain how nature operates
• Models are human inventions: Models do not equal
reality
• Models can be wrong. They are oversimplifications.
Exceptions deal with items that do not meet
standards due to oversimplifications.
• Need to understand strengths and weaknesses
• When models are wrong, we can learn
Using Bond Energy to Calculate
Reaction Energy
• Bond Energy depends on the environment
• Example:
– C-H bond in HCCl3 = 380 kJ/mol
– C-H bond in C2H6 = 410 kJ/mol
• Average Bond Energy is still useful
Objectives:
•
•
•
•
Discuss a "bond" as a model
Explain why models are used in science
Discuss 5 properties of models
Define single, double and triple bonds- # of e- shared,
bond energy & length
• Use bond energies to calculate ΔH
• Explain the relationship between number of shared
electrons and bond strength and/or bond length
• Write the expression for ΔH when using bond energy to
calculate the enthalpy of the reaction
Ways to share electrons
• Single Bonds
C-C
• Double Bonds
C=C
- single pair of electrons shared
- two total shared electrons
- two pairs of electrons shared
- four total shared electrons
• Triple Bonds - three pairs of electrons shared
C=C
- six total shared electrons
Using bond energy to calculate Enthalpy
• H2 (g) + F2 (g) --> 2 HF (g)
• This reaction requires us to break a H-H bond
and a F-F bond on the reactant side of the
reaction It also requires that we make two H-F
bonds on the product side of the reaction.
• Breaking bonds is an endothermic processpositive numbers
• Making bonds is an exothermic processnegative numbers
• Average bond energy Table 8.4 is on page 351.
How it is done . . .
• Enthalpy change for a reaction is calculating by
the addition of the energy required to break old
bonds and the energy released by the formation
of new bonds.
• ΔH = ΣD (bonds broken) - ΣD (bonds formed)
• Note: you are actually adding the energy from
"bonds formed", however, since these are
releasing energy the sign on the numbers is
negative leading to the negative sign in the
equation.
Back to the problem . . .
•
•
•
•
H2 (g) + F2 (g) --> 2 HF (g)
Breaking one mol H-H @ 432 kJ/mol = 432 kJ
Breaking one mol F-F @ 154 kJ/mol = 154 kJ
Making two moles H-F @ 565 kJ/mol = 1130 kJ
• Bonds broken total= 586 kJ
• Bonds formed total= 1130 kJ
• 586 - 1130 = -544 kJ
• Average bond energy Table 8.4 is on page 351.
Try this one on your own . . .
carbon is central atom
• CH4 + 2 Cl2 + 2 F2 --> CF2Cl2 + 2HF + 2HCl
• Answer: -1194 kJ
Assignment:
• Page 384 # 53,55,57,59,61
Answers
Objectives
•
•
•
•
•
•
•
•
Explain the Localized Energy Model
Define lone pair
Define bonding pair
Define Lewis structure
Define duet rule
List elements that obey the duet rule
Define Octet rule
List elements that obey the octet rule
Objectives
• Draw Lewis structures for atoms
• Draw Lewis structures for molecules
• Explain how there are exceptions to the octet
rule
• List some elements that are exceptions to the
octet rule Draw Lewis structures for
compounds where elements exceed the octet
rule
The Localized Electron Model
**Assumes that a molecule is composed of
atoms that are bound together by sharing
pairs of electrons using the atomic orbitals of
the bound atoms
The LE Model Continued
• Electrons are assumed to be localized
on an atom or in the space between
two atoms.
• Lone Pair : Localized on one atom
• Bonding Pair: Found in space
between two atoms
Three Parts of LE Model
• Description of valence electron arrangement
with Lewis Structures
• Prediction of Molecular Geometry using
VSEPR Model
• Description of the type of atomic orbitals used
by atoms to share electrons.
• For this unit we are only going to deal with the first
part, the second and third will be worked with next
unit.
Lewis Structures
• Shows how valence electrons are arranged
among atoms in a molecule.
• Most important requirement for a stable
compound is the achievement of noble gas
configuration.
Two Basic Rules
• Duet Rule: a stable molecule is formed when
two electrons are shared
• Octet Rule: a stable molecule is formed when
electrons are shared so that each atom is
surrounded by eight electrons
Steps for Writing Lewis Structures
1. Sum the VALENCE electrons for all atoms in
the molecule, it does not matter how many
come from each, just the total
2. Use a pair of electrons to form a bond
between each pair of bond atoms
3. Arrange remaining atoms to satisfy the duet
rule for hydrogen and the octet rule for
second row elements.
An Example:
Water: H2O
1. Sum valence electrons: Total= 8 valence electrons
H= 1 valence electron
H= 1 valence electron
O= 6 valence electrons
2. Use a pair to form bond between atoms
3. Arrange remaining to satisfy duet/octet rule
Another Example:
Carbon Dioxide: CO2
1. Sum valence electrons
Total: 16 valence electrons
C= 4 valence electrons
O= 6 valence electrons
O= 6 valence electrons
2. Use pairs to bond atoms
3. Arrange remaining to meet duet/octet
Assignment
Page 385 # 67 and 68
Answers
Violations of the Octet Rule
• Be, B are often electron deficient
• Third row and beyond elements are able to
exceed the octet rule by placing extra
electrons in their "d" orbitals
Examples:
BF3
SF6
PCl5
Assignment
Page 385 69, 71,72
Answers
Objectives
• Define Resonance
• Define Resonance structure
• Explain why odd electron molecules cannot be shown
using the Localized Energy Model
• Define formal charge
• Explain the “problem” with oxidation state
assignments
• Explain how formal charge is different from oxidation
state
• Use formal charge to determine the most probable
Lewis structure for a molecule
Resonance
• Occurs when there is more than one equivalent
and valid Lewis structure for a molecule.
• Is necessary to compensate for the incorrect
assumption that electrons are localized
• In reality, electrons are delocalized- meaning
that they move around the entire molecule
• LE model is still useful, so the exception of
resonance is added to accommodate certain
molecules rather than tossing the whole model
An example:
• NO3-1
Another Example
• NO2-1
Assignment
Page 385 # 73,74,75
Answers
Odd - Electron Molecules
• Only a few molecules form containing odd
numbers of electrons
• Examples: NO and NO2
• The Localized Electron Model only works
with PAIRS of electrons so this model is not
able to accomodate odd electron molecules
Formal Charge
• Formal charge works with nonequivalent
Lewis structures often found in molecules
and polyatomic ions that have atoms that can
exceed the octet rule.
• Charges are utilized to determine the most
appropriate structure(s)
Charge Determination
• Oxidation states:
Both shared electrons count for the more
electronegative atom
• Results in exaggerated charge estimates
• Not good for determining proper Lewis
structures
Charge Determination
• Formal Charge:
The difference between the number of valence
electrons on the free atom and the number of
valence electrons assigned to the atom in the
molecule
• When calculating formal charge:
Lone pair electrons belong to the atom on which
they are located.
Shared pair electrons are split equally among
sharing atoms.
Rules for working with Formal Charge:
1. Atoms in molecules want to have formal
charges as close to zero as possible.
2. Negative formal charges should occur on the
more electronegative atom.
An Example:
• SO4-2
Another Example
• XeO3
• Hint: there are 8 possible Lewis structures
Assignment
• Page 386 # 81 and 82
Answers
End of Unit 1
• Chapter 8 Sections 1-12.