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Chapter 3
Bonding: General Concepts
Chapter 3Table of Contents
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(3.1) Types of chemical bonds
(3.2) Electronegativity
(3.3) Ions: Electron configurations and sizes
(3.4) Energy effects in binary ionic compounds
(3.5) Partial ionic character of covalent bonds
(3.6) The covalent chemical bond: A model
(3.7) Covalent bond energies and chemical reactions
(3.8) The localized electron bonding model
Chapter 3Table of Contents
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(3.9) Lewis structures
(3.10) Exceptions to the octet rule
(3.11) Resonance
(3.12) Naming simple compounds
Chapter 3
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Questions to Consider
What is meant by the term “chemical bond”?
Why do atoms bond with each other to form compounds?
How do atoms bond with each other to form compounds?
Section 3.1Types of Chemical Bonds
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Representation of Molecular Structures
The force that holds atoms together is called a chemical bond
A covalent bond is formed when atoms share electrons
A collection of atoms makes a molecule
Molecules can be represented using:
Chemical formula
CO2
Structural formula
H—O—H
Section 3.1Types of Chemical Bonds
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Representation of Molecular Structures
Space-filling model: Indicates the relative sizes of the atoms as well as their relative orientation in the molecule
Ball and stick method: A three-dimensional model using spheres and rods
Section 3.1Types of Chemical Bonds
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Attributes of Molecules
Physical properties
Melting point
Hardness
Electrical and thermal conductivity
Solubility
Electric charge
Bond energy
Amount of energy needed to break a bond
Section 3.1Types of Chemical Bonds
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Types of Chemical Bonding
Ionic bonding: Occurs between an atom that easily loses electrons and an atom that has a high affinity for electrons
Ionic compounds are formed when metals react with non-metals
Section 3.1Types of Chemical Bonds
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Coulomb’s Law
It is used to calculate the energy of interaction between ions
E has units of joules
r is the distance between ions in nanometers
Q1 and Q2 are numerical ion charges
–19 1 2 = (2.31 × 10 J . nm)Q Q
Er
Section 3.1Types of Chemical Bonds
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Figure 3.2 - Interaction Between Two Identical Atoms
Section 3.1Types of Chemical Bonds
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Types of Chemical Bonding
Bond length is defined as the distance between two atoms where the energy is lowest
Based on diagram 3.2 (b):
Energy terms involved are net potential energy and kinetic energy
Zero point of energy is defined with the atoms at infinite separation
Energy rises steeply due to the repulsive forces when atoms are in close proximity
Section 3.1Types of Chemical Bonds
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Stability of Atoms in Bonding
Consider the H2 molecule
Electrons reside in the space between the two nuclei
Simultaneous attraction leads to increased stability compared to single H2 atoms
Increased attractive forces cause a decrease in potential energy of electrons
Section 3.1Types of Chemical Bonds
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Figure 3.3 - The Effect of an Electric Field on Hydrogen Fluoride Molecules
Section 3.1Types of Chemical Bonds
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Polar Covalent Bond
Unequal sharing of electrons between atoms in a molecule
Results in a charge separation in the bond
Partial positive and partial negative charge
Section 3.1Types of Chemical Bonds
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Concept Check
What is meant by the term “chemical bond?”
Why do atoms bond with each other to form molecules?
How do atoms bond with each other to form molecules?
Section 3.2Electronegativity
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What is Electronegativity?
The ability of an atom in a molecule to attract shared electrons to itself
Linus Pauling’s method of determining values of electronegativity
For a molecule HX, the relative electronegativities of the H and X atoms are determined by comparing the measured H—X bond energy with the “expected” H—X bond energy
H—H bond energy + X—X bond energyExpected H—X bond energy =
2
Section 3.2Electronegativity
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Determining Values of Electronegativity
Δ between actual and expected bond energies can be calculated by the formula
Δ = 0 if H and X have identical electronegativities
Shared electrons are closer to the X atom if the electronegativity of X is greater than X
The charge distribution is:
act exp = (H X) (H X)
+ –δ δH — X
Section 3.2Electronegativity
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Figure 3.4 - The Periodic Table
Section 3.2Electronegativity
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Table 3.1 - The Relationship Between Electronegativity and Bond Type
Section 3.2Electronegativity
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Interactive Example 3.1 - Relative Bond Polarities
Order the following bonds according to polarity: H—H, O—H, Cl—H, S—H, and F—H
Section 3.2Electronegativity
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Solution
The polarity of the bond increases as the difference in electronegativity increases. From the electronegativity values in Figure 3-4, the following variation in bond polarity is expected (the electronegativity value appears in parentheses below each element)
H—H < S—H < Cl—H < O—H < F—H(2.1) (2.1) (2.5) (2.1) (2.1) (2.1) (2.1)(3.0) (3.5) (4.0)
Section 3.2Electronegativity
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Concept Check
If lithium and fluorine react, which has more attraction for an electron? Why?
In a bond between fluorine and iodine, which has more attraction for an electron? Why?
Section 3.2Electronegativity
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Concept Check
What is the general trend for electronegativity across rows and down columns on the periodic table?
Explain the trend
Section 3.2Electronegativity
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Concept Check
Which of the following bonds would be the least polar yet be considered polar covalent?
Mg–O C–O O–O Si–O N–O
Section 3.2Electronegativity
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Figure 3.4 - The Periodic Table
Section 3.2Electronegativity
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Concept Check
Which of the following bonds would be the most polar without being considered ionic?
Mg–O C–O O–O Si–O N–O
Section 3.3Ions: Electron Configurations and Sizes
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Electron Configuration of Compounds
In a reaction between two nonmetals, electron sharing ensures complete valence electron configuration of both atoms Both nonmetals attain noble gas electron configurations
In a reaction between a nonmetal and a representative-group metal, a binary ionic compound is formed Ions are formed in such a way that both ions achieve noble
gas electron configuration
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Section 3.3Ions: Electron Configurations and Sizes
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Concept Check
Choose an alkali metal, an alkaline earth metal, a noble gas, and a halogen so that they constitute an isoelectronic series when the metals and halogen are written as their most stable ions.
What is the electron configuration for each species?
Determine the number of electrons for each species
Determine the number of protons for each species
Rank the species according to increasing radius
Rank the species according to increasing ionization energy
Section 3.3Ions: Electron Configurations and Sizes
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Section 3.3Ions: Electron Configurations and Sizes
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Predicting Formulas of Ionic Compounds
Ionic compound - Chemists’ term for compounds in solid state
Attraction between positive and negative ions is maximum
Caused by close proximity of the positive and negative ions, unlike those of compounds in the gaseous phase
Large attractive forces tend to increase the stability of a compound
Section 3.3Ions: Electron Configurations and Sizes
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Predicting Formulas of Ionic Compounds
Consider the valence electron configurations of oxygen and calcium
Ca [Ar]4s2
O [He]2s22p4
Electronegativity of oxygen is higher than that of calcium
The prediction made can be based on the fact that noble gas configurations are generally stable
Section 3.3Ions: Electron Configurations and Sizes
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Section 3.3Ions: Electron Configurations and Sizes
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Predicting Formulas of Ionic Compounds
An important fact that applies is that chemical compounds are always electrically neutral
In the reaction between calcium and oxygen:
There are equal numbers of calcium and oxygen ions
The empirical formula of the compound formed is CaO
Section 3.3Ions: Electron Configurations and Sizes
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Sizes of Ions
Ion size influences the structure and stability of ionic solids
Size is determined by measuring the distance between ion centers
Factors that influence ionic size
Size of the parent ion
Position in the periodic table
Isoelectric ions are ions that contain the same number of electrons
Section 3.3Ions: Electron Configurations and Sizes
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Figure 3.5 - Size of Ions with Relation to the Periodic Table
Section 3.3Ions: Electron Configurations and Sizes
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Interactive Example 3.3 - Relative Ion Size II
Choose the largest ion in each of the following groups:
a. Li+, Na+, K+, Rb+, Cs+
b. Ba2+, Cs+, I–, Te2–
Section 3.3Ions: Electron Configurations and Sizes
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Solution
a. The ions are all from Group 1A elements. Since size increases down a group (the ion with the greatest number of electrons is largest), Cs+ is the largest ion.
b. This is an isoelectronic series of ions, all of which have the xenon electron configuration. The ion with the smallest nuclear charge is largest.
Te2– ˃ I– ˃ Cs+ ˃ Ba2+
Z = 52 Z = 53 Z = 55 Z = 56
Section 3.3Ions: Electron Configurations and Sizes
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Section 3.4Energy Effects in Binary Ionic Compounds
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Lattice Energy
The change in energy that takes place when separated gaseous ions are packed together to form an ionic solid
It is mostly defined as the energy released when an ionic solid forms from its ions
It possesses a negative energy sign
Energy decreases
+ –M ( ) + X ( ) MX( )g g s
Section 3.4Energy Effects in Binary Ionic Compounds
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Energy Changes in the Formation of Lithium Fluoride
Consider the reaction
The reaction follows a certain sequence
Solid lithium undergoes sublimation
Energy of sublimation is 161 kJ/mol
Lithium atoms are ionized, forming lithium ions
520 kJ/mol of energy is released
2
1Li( ) + F ( ) LiF( )
2s g s
Li(s) Li(g)
+ –Li(g) Li (g) + e
Section 3.4Energy Effects in Binary Ionic Compounds
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Energy Changes in the Formation of Lithium Fluoride
Fluorine molecules undergo dissociation
F—F bonds in a half mole of F2 molecules are broken, forming a mole of fluorine atoms
154 kJ/mol of energy is required
Formation of F– ions occurs with an energy change of –328 kJ/mol
2
1 F (g) F(g)
2
– –F(g) + e F (g)
Section 3.4Energy Effects in Binary Ionic Compounds
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Energy Changes in the Formation of Lithium Fluoride
Gaseous ions of lithium and fluoride combine to form solid lithium fluoride
The overall process is shown in the table below
–Li (g) + F (g) LiF(s)
Section 3.4Energy Effects in Binary Ionic Compounds
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Figure 3.6 - Energy Changes Involved in the Formation of Lithium Fluoride
Section 3.4Energy Effects in Binary Ionic Compounds
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Lattice Energy Calculations
A modification of Coulomb's law is used
k is a proportionality constant that depends on the structure of the solid and the electronic configurations of the ions
Q1 and Q2 are the charges on the ions
r is the shortest distance between the centers of the anions and the cations
1 2Lattice energy = Q Q
kr
Section 3.4Energy Effects in Binary Ionic Compounds
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Energies Involved in the Formation of NaF(s) and MgO(s)
Energy released by Mg2+ and O2– ions is four times greater than the energy released by the combination of Na+ and F– ions, forming NaF
The energy needed to remove two electrons from the magnesium atom (2180 kJ/mol) is greater than the energy required to remove one electron from the sodium atom (495 kJ/mol)
737 kJ/mol of energy is required to add two electrons to the oxygen atom in the gas phase
Section 3.4Energy Effects in Binary Ionic Compounds
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Figure 3.8 - Formation of NaF(s) Vs MgO(s)
Section 3.5Partial Ionic Character of Covalent Bonds
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Formula for Percent Ionic Character of a Bond
There are probably no totally ionic bonds between discrete pairs of atoms
Calculations of the percent ionic character for bonds of various binary compounds in the gas phase
Percent ionic character of a bond can be calculated using the formula
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+ –
Measured dipole moment of X—Y 100%
Calculated dipole moment of X Y
Section 3.5Partial Ionic Character of Covalent Bonds
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Figure 3.10 - Relationship Between Ionic Character of a Covalent Bond and Electronegativity Difference of Bonded Atoms
Section 3.5Partial Ionic Character of Covalent Bonds
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Ionic Compounds
Problems faced in the identification of ionic compounds
No individual bonds are totally ionic
Many substances contain polyatomic ions
Operational definition of ionic compounds
Any compound that conducts an electric current when melted will be classified as ionic
Section 3.6The Covalent Chemical Bond: A Model
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Models
They are used to explain how nature operates on the microscopic level based on experiences in the macroscopic world
They are created by observing the properties of nature
The bonding model provides a framework to systematize chemical behavior
Molecules are viewed as collections of common fundamental components
Section 3.6The Covalent Chemical Bond: A Model
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Fundamental Properties of Models
A model does not equal reality
Models are oversimplifications and are therefore often wrong
Models become more complicated and are modified as they age
The underlying assumptions of a model must be properly understood
More is learnt when a model is wrong than when it is right
Section 3.7Covalent Bond Energies and Chemical Reactions
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Establishing the Sensitivity of Bonds to their Molecular Environment
Consider the stepwise decomposition of methane
4 3
3 2
2
Process Energy Required
CH ( ) CH ( ) + H( ) 435
CH ( ) CH ( ) + H( ) 453
CH ( ) CH( ) + H( )
g g g
g g g
g g g
425
CH( ) C( ) + H( ) 339
Total = 1652
1652 Average = = 413
4
g g g
Section 3.7Covalent Bond Energies and Chemical Reactions
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Establishing the Sensitivity of Bonds to their Molecular Environment
Establishing the sensitivity of bonds to their molecular environment
3
Measured C—H
bond energy
Molecule (kJ/mol)
HCBr
3
3
2 6
380
HCCl 380
HCF 430
C H 410
Section 3.7Covalent Bond Energies and Chemical Reactions
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Types of Bonds
Single Bond: A bond in which a single pair of electrons is shared
Double bond: A bond in which two pairs of electrons are shared
Triple bond: A bond in which three pairs of electrons are shared
Section 3.7Covalent Bond Energies and Chemical Reactions
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Table 3.3 - Average Bond Energies
Section 3.7Covalent Bond Energies and Chemical Reactions
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Bond Energies
Used to calculate approximate values for reactions
Consider the reaction
In the reaction above, two H—F bonds are formed by breaking one H—H bond and one F—F bond
Bonds can be broken when positive energy is added to the system
2 2H ( ) + F ( ) 2HF( )g g g
Section 3.7Covalent Bond Energies and Chemical Reactions
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Bond Energies
The formula for energy change
Σ = The sum of terms
D = Bond dissociation energy per mole of bonds
D is always positive
n = Moles of a particular type of bond
Energy required Energy released
Δ = (bonds broken) (bonds formed)E n D – n D
Section 3.7Covalent Bond Energies and Chemical Reactions
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Table 3.4 - Bond Lengths for Selected Bonds
Section 3.7Covalent Bond Energies and Chemical Reactions
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Interactive Example 3.4 - ΔE from Bond Energies
Using the bond energies listed in Table 3-3, calculate ΔE for the reaction of methane with chlorine and fluorine to give Freon-12 (CF2Cl2).
4 2 2 2 2CH ( ) + 2Cl ( ) + 2F ( ) CF Cl ( ) + 2HF( ) + 2HCl( )g g g g g g
Section 3.7Covalent Bond Energies and Chemical Reactions
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Solution
The idea here is to break the bonds in the gaseous reactants to give individual atoms and then assemble these atoms into the gaseous products by forming new bonds
The energy changes are combined to calculate ΔE
ΔE = energy required to break bonds – energy released when bonds form
The minus sign gives the correct sign to the energy terms for the exothermic processes
Energy Energy
required releasedReactants atoms products
Section 3.7Covalent Bond Energies and Chemical Reactions
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Solution
Reactant bonds broken
4CH : 4 mol C—H 4 mol413 kJ
× mol
2
= 1652 kJ
2Cl : 2 mol Cl—Cl 2 mol239 kJ
×mol
2
= 478 kJ
2F : 2 mol F—F 2 mol154 kJ
× mol
= 308 kJ
Total energy requied = 2438 kJ
Section 3.7Covalent Bond Energies and Chemical Reactions
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Solution
Product bonds formed
2 2CF Cl : 2 mol C—F 2 mol485 kJ
× mol
= 970 kJ
and
2 mol C—Cl 2 mol339 kJ
×mol
= 678 kJ
HF: 2 mol H—F 2 mol565 kJ
× mol
= 1130 kJ
HCl: 2 mol H—Cl 2 mol427 kJ
× mol
= 854 kJ
Total energy released = 3632 kJ
Section 3.7Covalent Bond Energies and Chemical Reactions
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Solution
Calculating ΔE
Since the sign of the value for the energy change is negative, this means that 1194 kJ of energy is released per mole of CF2Cl2 formed
ΔE = energy required to break bonds – energy released when bonds form
= 2438 kJ – 3632 kJ
= –1194 kJ
Section 3.8The Localized Electron Bonding Model
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Localized Electron (LE) Model
A molecule is composed of atoms that are bound together by sharing pairs of electrons using the atomic orbitals of the bound atoms
Pairs of electrons localized on a specific atom are called lone pairs
Electron pairs found in the space between the atoms are called bonding pairs
Section 3.8The Localized Electron Bonding Model
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Parts of the LE Model
Description of the valence electron arrangement in the molecule using Lewis structures
Prediction of the geometry of the molecule using the valence shell electron-pair repulsion (VSEPR) model
Description of the type of atomic orbitals used by the atoms to share electrons or hold lone pairs
Section 3.8The Localized Electron Bonding Model
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Lewis Structure
Shows how valence electrons are arranged among atoms in a molecule
Named after G.N. Lewis
Atoms achieving noble gas configurations are central to the formation of a stable compound
Section 3.8The Localized Electron Bonding Model
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Writing Lewis Structures
Only valence electrons are included
Dots represent electrons
Lewis structure for KBr
K has no valence electrons
Br– has a filled valence shell
Section 3.8The Localized Electron Bonding Model
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Writing Lewis Structures
The duet rule
A stable H2 molecule is formed by the combination of two hydrogen molecules
Sharing of electrons results in a filled valence shell
Section 3.8The Localized Electron Bonding Model
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Rules in Writing Lewis Structures
The octet rule
Elements form stable molecules when surrounded by eight electrons
Each fluorine atom has bonding pairs as well as lone pairs of electrons
Section 3.9Lewis Structures
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Problem Solving Strategy - Steps for Writing Lewis Structures
Sum the valence electrons from all the atoms
Use a pair of electrons to form a bond between each pair of bound atoms
Arrange the remaining electrons to satisfy the duet rule for hydrogen and the octet rule for the second-row elements
Section 3.9Lewis Structures
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Lewis Structure of Water
Stepwise approach
Sum the valence electrons for H2O
1 + 1 + 6 = 8 valence electrons
Using a pair of electrons per bond, draw the two O—H single bonds
H—O—H
A line replaces a pair of dots to indicate each pair of bonding electrons
H H O
Section 3.9Lewis Structures
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Lewis Structure of Water
Distribute the remaining electrons to achieve a noble gas configuration for each atom
Section 3.9Lewis Structures
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Interactive Example 3.5 - Writing Lewis Structures
Give the Lewis structure for each of the following
HF
N2
NH3
CH4
CF4
NO+
Section 3.9Lewis Structures
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Solution
Apply the three steps for writing Lewis structures
Lines indicate shared electron pairs
Dots indicate nonbonding pairs
Section 3.9Lewis Structures
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Solution
Section 3.9Lewis Structures
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Concept Check
Draw a Lewis structure for each of the following molecules:
H2
F2
HF
Section 3.9Lewis Structures
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Concept Check
Draw a Lewis structure for each of the following molecules:
NH3
CO2
CCl4
Section 3.10Exceptions to the Octet Rule
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Boron
Boron tends to form compounds in which the boron atom has fewer than eight electrons around it
It does not have a complete octet
The Lewis structure of boron fluoride
Section 3.10Exceptions to the Octet Rule
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Boron
Boron has only 6 electrons around it
Adding a double bond satisfies the octet rule
In stable compounds such as H3NBF4, boron has an octet of electrons
Section 3.10Exceptions to the Octet Rule
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Exceeding the Octet Rule
Observed only in elements in Period 3 of the periodic table and beyond
Consider the Lewis structure of sulfur hexafluoride
6 + 6 (7) = 48 electrons
Section 3.10Exceptions to the Octet Rule
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Interactive Example 3.6 - Lewis Structures that Violate Octet Rule I
Write the Lewis Structure for PCl5 Solution
Sum the valence electrons
5 + 5(7) = 40 electrons
Indicate single bonds between bound atomsP Cl
Section 3.10Exceptions to the Octet Rule
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Solution
Distribute the remaining electrons
30 electrons remain
Phosphorus, a third-row element, has exceeded the octet rule by two electrons
Section 3.10Exceptions to the Octet Rule
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Concept Check
Draw a Lewis structure for each of the following molecules:
BF3
PCl5 SF6
Section 3.10Exceptions to the Octet Rule
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Exceeding the Octet Rule in Certain Elements
When it is necessary to exceed the octet rule for one of several third-row (or higher) elements, assume that the extra electrons should be placed on the central atom
Calculating the valence electrons present in the triiodide ion (I3
–)
3(7) + 1 = 22 valence electrons
–1 chargeI
Section 3.10Exceptions to the Octet Rule
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Interactive Example 3.7 - Lewis Structures for Molecules that Violate the Octet Rule II
Write the Lewis structure for each molecule or ion
ClF3
XeO3
RnCl2 BeCl2 ICl4
–
Section 3.10Exceptions to the Octet Rule
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Solution
The chlorine atom accepts the extra electrons
All atoms obey the octet rule
Section 3.10Exceptions to the Octet Rule
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Solution
Radon, a noble gas in Period 6, accepts the extra electrons
Beryllium is electron-deficient
Iodine exceeds the octet rule
Section 3.11Resonance
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Resonance
More than one valid Lewis structure can be written for a particular molecule
Consider the nitrate ion
NO3– = 24 valence electrons
The correct description is given only by the superposition of all three structures
Section 3.11Resonance
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Resonance
It is the condition in which more than one valid Lewis structure can be written for a specific molecule
Atoms need to be connected in the same order
The resulting electron structure is derived from the average of the resonance structures
Represented by double-headed arrows
Section 3.11Resonance
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Interactive Example 3.8 - Resonance Structures
Describe the electron arrangement in the nitrite anion (NO2
–), using the localized electron model
Section 3.11Resonance
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Solution
Step 1 - Determine the valence electrons
5 + 2 (6) + 1 = 18
Step 2 - Use single bonds to indicate the structure
O—N—O
Step 3 - Distribute the remaining 14 electrons to produce Lewis structures
Section 3.11Resonance
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Odd-Electron Molecules
There are few molecules from non-metals that possess electrons in odd numbers
Nitric oxide (NO), emitted by automobiles, reacts with oxygen in the air to from nitrogen dioxide
A more sophisticated model than the localized electron model is required
The localized electron model is based on electron pairs
Section 3.11Resonance
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Concept Check
Draw a Lewis structure for each of the following molecules:
CO
CO2
CH3OH
OCN–
Section 3.11Resonance
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Formal Charge
It is used to estimate the charge on all possible nonequivalent Lewis structures
It is defined as the difference between the number of valence electrons on the free atom and the number of electrons assigned to the atom in the molecule
Formal charge = (number of valence electrons on a free atom) – (number of valence electrons assigned to the atom in the molecule)
Section 3.11Resonance
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Computing the Formal Charge of an Atom
Valence electrons are assigned to various atoms based on the following assumptions:
Lone pair electrons belong entirely to the atom in question
Shared electrons are divided equally between the two sharing atoms
The number of valence electrons is calculated
(Valence electrons)assigned = (number of lone electrons) + 1/2(number of shared electrons)
Section 3.11Resonance
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Formal Charge
Fundamental assumptions about formal charges in evaluating Lewis structures
Atoms in molecules try to achieve charges as close to zero as possible
Any negative formal charges are expected to reside on the most electronegative atoms
Section 3.11Resonance
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Rules Governing Formal Charge
To calculate the formal charge on an atom:
Take the sum of the lone pair electrons and one-half the shared electrons
Subtract the number of assigned electrons from the number of valence electrons on the free, neutral atom
Section 3.11Resonance
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Rules Governing Formal Charge
The sum of the formal charges of all atoms in a given molecule or ion must equal the overall charge on that species
If nonequivalent Lewis structures exist for a species, those with formal charges closest to zero and with any negative formal charges on the most electronegative atoms are considered to best describe the bonding in the molecule or ion
Section 3.11Resonance
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Interactive Example 3.9 - Formal Charges
Give possible Lewis structures for XeO3, an explosive compound of xenon. Which Lewis structure or structures are most appropriate according to the formal charges?
Section 3.11Resonance
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Solution
For XeO3 (26 valence electrons) we can draw the following possible Lewis structures:
Formal charges are indicated in parentheses
Section 3.11Resonance
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Solution
Based on the ideas of formal charge, it can be predicted that the Lewis structures with the lower values of formal charge would be most appropriate for describing the bonding in XeO3
Section 3.11Resonance
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Concept Check
Consider the Lewis structure for POCl3. Assign the formal charge for each atom in the molecule
P
Cl
Cl O
Cl
Section 3.12Naming Simple Compounds
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Binary Ionic Compounds (Type I)
They possess a cation in the formula followed by an anion
Rules in naming type I binary compounds
The cation is always named first and the anion second
A monatomic cation takes its name from the name of the element
A monatomic anion is named by taking the root of the element name and adding -ide
Section 3.12Naming Simple Compounds
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Table 3.5 - Common Monatomic Cations and Anions
Section 3.12Naming Simple Compounds
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Interactive Example 3.10 - Naming Type I Binary Compounds
Name each binary compound
CsF
AlCl3 LiH
Solution
CsF is cesium fluoride
AlCl3 is aluminum chloride
LiH is lithium hydride
Section 3.12Naming Simple Compounds
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Formulas from Names
In naming chemical compounds, the chemical formula is considered
The reverse process is also important
The formula of calcium chloride is written as CaCl2 Calcium forms two Ca2+ ions
Chloride is Cl–
So, two of these anions will be required to produce a neutral compound
Section 3.12Naming Simple Compounds
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Interactive Example 3.11 - Formulas from Names for Type I Binary Compounds
Given the following systematic names, write the formula for each compound:
Potassium iodide
Calcium oxide
Gallium bromide
Section 3.12Naming Simple Compounds
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Solution
Name Formula Comments
Potassium iodide KI Contains K+ and I–
Calcium oxide CaO Contains Ca2+ and O2–
Gallium bromide GaBr3 Contains Ga3+ and Br –
Must have 3Br– to balance charge of Ga3+
Section 3.12Naming Simple Compounds
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Binary Ionic Compounds (Type II)
Some metals form more than one type of positive ion
Results in the formation of more than one type of ionic compound with a given anion
The charge on the metal ion is specified using Roman numerals
Alternative for metals that form only two ions
Metals possessing ions with higher charge end in -ic
Metals possessing ions with lower charge end in -ous
Section 3.12Naming Simple Compounds
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Interactive example 3.12 - Naming Type II Binary Compounds
Given the following systematic names, write the formula for each compound:
Manganese (IV) oxide
Lead (II) chloride
Section 3.12Naming Simple Compounds
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Solution
Name Formula Comments
Manganese (IV) oxide MnO2 Two O2– ions (total charge 4–) are required by the Mn4+ ion [manganese (IV)]
Lead (II) chloride PbCl2 Two Cl– ions are required by the Pb2+ ion [lead (II)] for charge balance
Section 3.12Naming Simple Compounds
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Binary Ionic Compounds (Type II)
The Roman numeral naming system is used only in cases where more than one ionic compound forms between a given pair of elements
Elements that form only one cation are not identified by a Roman numeral
Group 1(A)
Group 2(A)
Aluminum
Silver
Section 3.12Naming Simple Compounds
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Figure 3.12 - Flowchart for Naming Binary Ionic Compounds
Section 3.12Naming Simple Compounds
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Interactive Example 3.13 - Naming Binary Compounds
Give the systematic name for each of the following compounds:
CoBr2
CaCl2 Al2O3
Section 3.12Naming Simple Compounds
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Solution
Formula
CoBr2 Cobalt (II) bromide Cobalt is a transition metal; the compoundname must have a Roman numeral. The two Br– ions must be balanced by a Co2+ ion.
CaCl2 Calcium chloride Calcium, an alkaline earth metal, forms only the Ca2+ ion. A Roman numeral is not necessary.
Al2O3 Aluminum oxide Aluminum forms only the Al3+ ion. A Roman numeral is not necessary.
Section 3.12Naming Simple Compounds
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Figure 3.13 - The Common Cations and Anions
Section 3.12Naming Simple Compounds
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Ionic Compounds with Polyatomic Atoms
Polyatomic ions are assigned specific names that must be memorized to name the compounds containing them
Oxyanions: Anions that contain an atom of a given element and different numbers of oxygen atoms
In a series with two oxyanions:
The oxyanion with lesser number of oxygen atoms ends in -ite
The oxyanion with higher number of oxygen atoms ends in -ate
Section 3.12Naming Simple Compounds
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Ionic Compound with Polyatomic Atoms
In a series with more than two oxyanions:
Hypo- is used as a prefix in the names of members with the fewest oxygen atoms
Per- is used as a prefix in the names of members with the most oxygen atoms
Section 3.12Naming Simple Compounds
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Interactive Example 3.14 - Naming Compounds Containing Polyatomic Ions
Give the systematic name for each of the following compounds:
Na2SO4
KH2PO4
Fe(NO3)3
Mn(OH)2
Na2SO3
Na2CO3
Section 3.12Naming Simple Compounds
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Section 3.12Naming Simple Compounds
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Solution
Formula Name Comments
Na2SO4 Sodium sulfate
KH2PO4 Pottasium dihydrogen sulfate
Fe(NO3)3 Iron (III) nitrate Transition metal—name must contain aRoman numeral. The Fe3+ ion balancesthree NO3
– ions
Mn(OH)2 Manganese (II) hydroxide Transition metal—name must contain aRoman numeral. The Mn2+ ion balances three OH– ions
Na2SO3 Sodium sulfite
Na2CO3 Sodium carbonate
Section 3.12Naming Simple Compounds
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Binary Covalent Compounds (Type III)
They are formed between two non-metals
Their names are similar to that of binary ionic compounds
Compound Systematic Name Common Name
N2O Dinitrogen monoxide Nitrous oxide
NO Nitrogen monoxide Nitric oxide
NO2 Nitrogen dioxide
N2O3 Dinitrogen trioxide
N2O4 Dinitrogen tetroxide
N2O5 Dinitrogen pentoxide
Section 3.12Naming Simple Compounds
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Interactive Example 3.15 - Naming Type III Binary Compounds
Name each of the following compounds:
PCl5 PCl3 SO2
Solution
Name Formula
PCl5 Phosphorus pentachloride
PCl3 Phosphorus trichloride
SO2 Sulfur dioxide
Section 3.12Naming Simple Compounds
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Figure 3.14 - Naming Binary Compounds
Section 3.12Naming Simple Compounds
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Figure 3.13 - Naming Chemical Compounds
Section 3.12Naming Simple Compounds
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Interactive Example 3.16 - Naming Various Compounds
Given the following systematic names, write the formula for each compound
Vanadium (V) fluoride
Dioxygen difluoride
Rubidium peroxide
Gallium oxide
Section 3.12Naming Simple Compounds
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Solution
Name Chemical Formula Comment
Vanadium (V) fluoride
VF5 The compound contains V5+ ions and requires five F– ions for charge balance
Dioxygen difluoride
O2F2 The prefix di- indicates two of each atom
Rubidium peroxide
Rb2O2 Since rubidium is in Group 1A, it forms only 1+ ions. Thus two Rb+ ions are needed to balance the 2– charge on the peroxide ion (O2
2–)
Gallium oxide Ga2O3 Since gallium is in Group 3A, like aluminum, it forms only 3+ ions. Two Ga3+
ions are required to balance the charge on three O2– ions
Section 3.12Naming Simple Compounds
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Acids
Molecules in which one or more H+ ions are attached to an anion
If the name of the anion ends in -ide, the acid is named with the prefix hydro- and the suffix -ic
The names of oxygen-containing anions are based on the root name of the anion, along with a suffix of -ic or -ous
If the name of the anion ends in -ate, the suffix -ic is added to the root name
If the name of the anion ends in -ite, the -ite is replaced by -ous
Section 3.12Naming Simple Compounds
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AcidsAcid Anion Name
HClO4 Perchlorate Perchloric acid
HClO3 Chlorate Chloric acid
HClO2 Chlorite Chlorous acid
HClO Hypochlorite Hypochlorous acid
Section 3.12Naming Simple Compounds
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The 6 strong acids are: (All the others are “weak”)
HCl
HBr
HI
HClO4
HNO3
H2SO4
Section 3.12Naming Simple Compounds
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Figure 3.16 - Naming Acids