Ch 8 Bonding 128 slides.ppt

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Chapter 8

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Chapter 8

Table of Contents

8.1 Types of Chemical Bonds 8.2 Electronegativity

8.3 Bond Polarity and Dipole Moments

8.4 Ions: Electron Configurations and Sizes 8.5 Energy Effects in Binary Ionic Compounds 8.6Partial Ionic Character of Covalent Bonds 8.7The Covalent Chemical Bond: A Model

8.8Covalent Bond Energies and Chemical Reactions 8.9The Localized Electron Bonding Model

8.10Lewis Structures

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Chapter 8

Questions to Consider

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Section 8.1

Types of Chemical Bonds

What is meant by the term “chemical bond”?

• No simple, and yet complete, way to define this. • Forces that hold groups of atoms together and

make them function as a unit.

– An ionic bond is the attraction between the opposite charges in an ionic compound

– A covalent bond the the attraction of a

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Chapter 8

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Section 8.1

Why do atoms bond with each other to form compounds?

• Atoms firm bonds so that each individual atom in the bond can attain an octet

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Chapter 8

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Section 8.1

How do atoms bond with each other to form compounds?

•Ionic Bonding

–Electrons are transferred •Covalent Bonding

–Electrons shared equally

•Non Polar

–Electrons shared unequally

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Section 8.1

Energy of interaction between a single pair of ions

E = 2.31x10



19_{J nm}



Q1Q2 r



 

E = 2.31x10



19_{J nm}



 

1

 

1 0.276 nm

 



   8.37x10

19_J

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Section 8.1

Non Polar Covalent Bond

• We have seen that a bonding force develops when two different types of atoms react to form oppositely charged ions.

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Section 8.1

The Interaction of Two Hydrogen

Atoms

Fig. 8.1a

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Section 8.1

The Interaction of Two Hydrogen Atoms

• Two unfavorable potential energy terms – proton–proton repulsion

– electron–electron repulsion

• one favorable term, proton–electron attraction. • Under what conditions will the H₂ molecule be

favored over the separated hydrogen atoms • That is, what conditions will favor bond

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Section 8.1

Hydrogen Bond

• A bond will form (that is, the two hydrogen atoms will exist as a molecular unit) if the system can lower its total energy in the process.

– strong tendency in nature for any system to achieve the lowest possible energy

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Section 8.1

The Interaction of Two Hydrogen Atoms

The energy of this system is the sum of: potential energy that results from the attractions and repulsions among the charged particles

kinetic energy due to the motions of the electrons.

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Section 8.1

Non Polar Covalent Bond

• Equal sharing of electrons between atoms in a molecule.

• All diatomic elements H₂, N₂ O₂ etc

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Section 8.1

Polar Covalent Bond

• Unequal sharing of electrons between atoms in a molecule.

• Atoms have different electronegativities

• Results in a charge separation in the bond (partial positive and partial negative charge).

H



F

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Section 8.1

The Effect of an Electric Field on Hydrogen Fluoride Molecules

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Section 8.2

Electronegativity

• The ability of an atom in a molecule to attract shared electrons to itself.

– Only applies to covalent compounds – Ability to attract electrons

• Similar to the concept of electron affinity – Only applies to ionic compounds

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Section 8.2

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Section 8.2

• Electronegativites are assigned values

• Hydrogen was fixed at 2.1 and all other atoms were assigned relative to this values

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Section 8.2

• Expected bond energy for HBr.

• The measured H-Br bond energy is 0.7 higher than the expected value.

• Therefore Br has a value to 2.8

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Section 8.2

• On the periodic table, electronegativity generally increases across a period and decreases down a group.

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Section 8.2

Concept Check

What is the general trend for electronegativity across rows on theperiodic table?

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Section 8.2

Solution

What is the general trend for electronegativity across rows and down columns on the

periodic table?

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Section 8.2

Concept Check

What is the general trend for electronegativity down columns on the periodic table?

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Section 8.2

Solution

What is the general trend for electronegativity across rows and down columns on the

periodic table?

The electronegativity decreases going down a group because the size of the atoms

increases as you go down so when other electrons approach the larger atoms, the

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Section 8.2

Concept Check

In a bond between fluorine and iodine, which has more attraction for an electron? Why?

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Section 8.2

Solution

In a bond between fluorine and iodine, which has more attraction for an electron? Why?

The fluorine also has more attraction for an electron than does iodine. In this case the nuclear charge of iodine is greater, but the valence electrons are at a much higher

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Section 8.2

The Relationship Between Electronegativity and Bond Type

As the difference in electronegativity increases covalent bonds become more polar

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Section 8.2

Learning Exercise

Arrange the following bonds from most polar to least polar:

a) N–F O–F C–F

b) C–F N–O Si–F

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Section 8.2

Solution

Arrange the following bonds from most to least polar:

a) C–F, N–F, O–F

b) Si–F, C–F, N–O

c) B–Cl, S–Cl, Cl–Cl

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Section 8.2

Concept Check

Which of the following bonds would be the least polar yet still be considered polar

covalent?

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Section 8.2

Solution

Which of the following bonds would be the least polar yet still be considered polar

covalent?

Mg–O ionic

C–O 1.0

O–O 0

Si–O 1.7

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Section 8.2

Concept Check

Which of the following bonds would be the most polar without being considered ionic? Mg–O

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Section 8.2

Solution

Which of the following bonds would be the most polar without being considered ionic?

Mg–O ionic

C–O 1.0

O–O 0

Si–O 1.7

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Section 8.3

Bond Polarity and Dipole Moments

Dipole Moment

• Property of a molecule whose charge distribution can be represented by a center of positive

charge and a center of negative charge.

• Use an arrow to represent a dipole moment.

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Section 8.3

Electrostatic Potential Map

• Another way to represent charge distribution – Red indicates electron rich regions (-)

– Blue indicates electron poor regions (+)

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Section 8.3

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Section 8.3

Dipole Moment

a. Charge

distribution b. Dipole momentin an electric ﬁeld c. Electrostatic potentialdiagram

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Section 8.3

CO₂ – polar bonds - No Net Dipole Moment (Dipoles Cancel)

b. Opposing bond polarities cancel

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Section 8.4

Ions: Electron Configurations and Sizes

Stable Compounds

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Section 8.4

Electron Configurations in Stable Compounds

• When two nonmetals react to form a covalent bond, they share electrons in a way that

completes the valence electron configurations of both atoms.

• When a nonmetal and a representative-group

metal react to form a binary ionic compound, the

ions form so that the valence electron

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Section 8.4

Predicting Formulas of Ionic Compunds

• We have essentially already learned how to predict these formulas

• Metals form cations (to attain a noble gas configuration) – Charge is based on Group number for Groups 1 and 2 – For transition metals and larger metals charge can vary

(roman numeral in name)

• Nonmetals form anions (to attain a noble gas configuration)

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Section 8.4

Concept Check

Predict the formula for compounds formed between the following

Ca and Se

Mg and Br

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Section 8.4

Return to TOC

Solution

Predict the formula for compounds formed between the following

Ca and Se CaSe

Mg and Br MgBr₂

Pb (II) and O PbO

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Section 8.4

Sizes of Ions

• Cations

– Smaller than the parent ion – Electrons are being removed

– Group II cations smaller than Group I • Anions

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Section 8.4

Concept Check

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Section 8.4

Solution

The Be ion is half the size of the Li ion but the Mg ion is 2/3 the size of the Na ion and the Ca ion is 3/4 the size of the K ion. Explain this trend.

Both the Be and Li ions have 2 electrons but in Be they are being held by 4 protons and in Li only 3.

Both the Mg and Na ions have 10 electrons but in Mg they are being held by 12 protons and in Na only 11. The difference is much less

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Section 8.4

Isoelectronic Series

• Ions that all contain the same number of electrons

– O2– ,F– , Na+ , Mg2+, and Al3+

– All have the electron configuration of Ne – How do the sizes compare?

– Size depends on number of protons – Fewest protons – largest – oxygen (8)

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Section 8.4

Concept Check

Write the isoelectronic series that consists of a noble gas, an alkali metal, an alkaline earth metal and a halogen that all have 18 electrons (write the metals and the halogen as their most stable ions)

 _{What is the}_{electron configuration}_{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

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Section 8.4

Solution

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.

K+, Ca2+, Ar, Cl –

What is the electron configuration for each species? 1s2_2s2_2p6_3s2_3p6_. Determine the number of protons for each species.

K+_{has 19 protons, Ca}2+_{has 20 protons, Ar has 18 protons, and Cl}-_{has 17}

protons.

Rank the species according to increasing radius. Ca2+, K+, Ar, Cl

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Section 8.4

Periodic Table Allows Us to Predict Many Properties

• Trends for:

 Atomic size, ion radius, ionization energy, electronegativity

• Electron configurations

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Section 8.5

Energy Effects in Binary Ionic Compounds

• What are the factors that influence the stability and the structures of solid binary ionic

compounds?

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Section 8.5

Lattice Energy

• The change in energy that takes place when

separated gaseous ions are packed together to form an ionic solid.

• M+ (g) + X – (g)  MX (s)

• We can consider the energy changes in the formation of an ionic solid by considering the formation fo solid lithium fluoride from its

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Section 8.5

Formation of LiF

Li (s) + ½ F₂ (g)  LiF (s)

1. Li (s) Li (g) (sublimation)

2. Li (g) Li+_{(g) + e}–_(ionization)

3. ½ F₂ (g)  F (g) (dissociation)

4. F (g) + e – __F–_{(g) (formation of fluoride ions in the gas phase)}

5. Li+_{(g) F}– _(g)_{LiF (s) (lattice energy)}

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Section 8.5

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Section 8.5

Lattice Energy

• Can also be calculated from a modified form of Coulomb’s Law.

k = proportionality constant

Q₁ and Q₂ = charges on the ions

r = shortest distance between the centers of the cations and

anions

Everything else being equal the larger the values of Q (charges of the ions) the larger the lattice energies.

E = 2.31x10 19_{J nm} Q1Q2

r



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Section 8.5

Comparing

Energy Changes

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Section 8.6

Partial Ionic Character of Covalent Bonds

• How well can we tell the difference between an ionic bond and a polar covalent bond?

• There are probably no totally ionic bonds between discrete pairs of atoms.

• The evidence for this statement comes from

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Section 8.6

• None of the bonds reaches 100% ionic character even with compounds that have a large

electronegativity difference (when tested in the gas phase).

+

measured dipole moment of X Y

% ionic character of a bond = 100%

calculated dipole moment of X Y

 _ 



 

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Section 8.6

The relationship between the ionic character of a covalent bond and the electronegativity difference of the bonded atoms

These results are for the gas phase, where individual XY

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Section 8.6

Ionic Compound

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Section 8.7

The Covalent Chemical Bond: A Model

Models

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Section 8.7

Covalent bond in terms of Energy

• Bonds result from the tendency of a system to seek its lowest possible energy.

• 1652 kJ of energy is required to to dissociate a mole CH₄ into separate C and H atoms

– CH₄is lower in energy (more stable) – 1652 kJ/mole more stable

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Section 8.7

Covalent bond in terms of Energy

• It is important to note that the concept of a bond is a human invention.

• Bonds provide a method for dividing up the energy evolved when a stable molecule is formed from its component atoms.

• Thus in this context a bond represents a quantity of energy obtained from the overall molecular

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Section 8.7

Fundamental Properties of Models

1. A model does not equal reality.

2. Models are oversimplifications, and are therefore often wrong.

3. Models become more complicated and are modified as they age.

4. We must understand the underlying

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Section 8.8

Covalent Bond Energies and Chemical Reactions

Bond Energies

• To break bonds, energy must be added to the system (endothermic).

• To form bonds, energy is released (exothermic). • We can calculate ΔH as the sum of the energies

required to break old bonds minus the sum of the energies released in the formation of new bonds

ΔH = Σn×D (bonds broken) – Σn×D (bonds formed)

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Section 8.8

Learning Exercise

Predict ΔH for the following reaction:

Given the following information:

Bond Energy (kJ/mol)

C–H 413 C–N 305 C–C 347

3 3

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Section 8.8

Solution

Predict ΔH for the following reaction:

Given the following information:

Bond Energy (kJ/mol)

C–H 413 C–N 305 C–C 347 891

ΔH= [3(413) + 305 + 891] – [3(413) + 347 + 891] = – 42 kJ

CH₃N  C(g)  CH₃C  N(g)

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Section 8.9

The Localized Electron Bonding Model

Localized Electron Model

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Section 8.9

• Electron pairs are assumed to be localized on a particular atom or in the space between two

atoms:

 Lone pairs – pairs of electrons localized on

an atom

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Section 8.9

1. Description of valence electron arrangement (Lewis structure).

2. Prediction of geometry (VSEPR model).

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Section 8.10

Lewis Structures

Lewis Structure

• Shows how valence electrons are arranged among atoms in a molecule.

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Section 8.10

Duet Rule

• Hydrogen forms stable molecules where it shares two

electrons

Octet Rule

• Elements form stable molecules when surrounded by

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Section 8.10

Octet Rule

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Section 8.10

Single/Double/Triple Covalent Bonds

• Single - A covalent bond in which two atoms share one pair of electrons. H–H

• Double - A covalent bond in which two atoms share two pairs of electrons. O=C=O

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Section 8.10

Steps for Writing Lewis Structures

1. Sum the valence electrons from all the atoms. (Use the periodic table.)

Example: H₂O

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Section 8.10

2. Use a pair of electrons to form a bond between each pair of bound atoms.

Example: H₂O

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Section 8.10

3. Atoms usually have noble gas configurations. Arrange the remaining electrons to satisfy the octet rule (or duet rule for hydrogen).

Examples: H₂O, PBr₃, and HCN

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Section 8.10

1. Sum the valence electrons from all the atoms. 2. Use a pair of electrons to form a bond between

each pair of bound atoms.

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Section 8.10

Concept Check

Draw a Lewis structure for each of the following molecules:

PBr₃

F₂

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Section 8.10

Solution

Draw a Lewis structure for each of the following

PBr₃

F₂

P

Br

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Section 8.10

Concept Check

NH₃

CO₂

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Section 8.10

Solution

NH₃

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Section 8.11

Exceptions to the Octet Rule

• Boron tends to form compounds in which the boron atom has fewer than eight electrons around it (it does not have a complete octet).

BH₃ = 6e–

B

H

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Section 8.11

• When it is necessary to exceed the octet rule for one of several third-row (or higher) elements,

place the extra electrons on the central atom.

SF₄ = 34e– AsBr

5 = 40e–

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Section 8.11

Concept Check

BF₃

PCl₅

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Section 8.11

Solution

BF₃

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Section 8.11

Let’s Review

• C, N, O, and F should always be assumed to obey the octet rule.

• B and Be often have fewer than 8 electrons around them in their compounds.

• Second-row elements never exceed the octet rule.

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Section 8.11

Let’s Review

• When writing the Lewis structure for a molecule, satisfy the octet rule for the atoms first. If

electrons remain after the octet rule has been satisfied, then place them on the elements

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Section 8.12 Resonance

• More than one valid Lewis structure can be written for a particular molecule.

NO₃– = 24e–

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• Actual structure is an average of the resonance structures.

• Electrons are really delocalized – they can move around the entire molecule.

N

O

N

O

N

O

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Concept Check

Draw resonance Lewis structure for each of the following:

CO₂

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Solution

• Draw resonance Lewis structure for each of the following:

•

• CO2 •

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Concept Check

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Solution

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Formal Charge

• Used to evaluate nonequivalent Lewis structures.

• Atoms in molecules try to achieve formal charges as close to zero as possible.

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Formal Charge

• Formal charge = (# valence e– on free atom) – (#

valence e– assigned to the atom in the

molecule). • Assume:

 Lone pair electrons belong entirely to the atom in question.

 _{Shared electrons are divided equally between}

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Concept Check

Use Formal Charge to evaluate the three structures below. Based on formal charges, pick the least stable form of this ion.

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Solution

Use Formal Charge to evaluate the three structures below. Based on formal charges, pick the least stable form of this ion.

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Rules Governing Formal Charge

• To calculate the formal charge on an atom:

1. Take the sum of the lone pair electrons and one-half the shared electrons.

2. Subtract the number of assigned electrons from the number of valence electrons on the free, neutral atom.

3. The sum of the formal charges of all atoms in a given molecule or ion must equal the

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Concept Check

Consider the Lewis structure for POCl₃.

Assign the formal charge for each atom in the molecule.

P

Cl

O

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Solution

Consider the Lewis structure for POCl₃.

Assign the formal charge for each atom in the molecule.

P: 5 – 4 = +1 O: 6 – 7 = –1 Cl: 7 – 7 = 0

P

Cl

O

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Using Formal Charge to Evaluate Equivalent Lewis Structures

• If nonequivalent Lewis structures exist for a species,

• Atoms try to achieve a formal charge as close to zero as possible.

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Concept Check

Use formal charge to evaluate the following two equivalent structures for CO₂.

C

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Solution

Use formal charge to evaluate the following two equivalent structures for CO₂.

C

O

_C

O

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Concept Check

Draw formal charge to pick the best structure

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Solution

Draw formal charge to pick the best structure

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Section 8.13

Molecular Structure: The VSEPR Model

VSEPR Model

• VSEPR: Valence Shell Electron-Pair Repulsion. • The structure around a given atom is determined

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Section 8.13

Steps to Apply the VSEPR Model

1. Draw the Lewis structure for the molecule.

2. Count the electron pairs and arrange them in the way that minimizes repulsion (put the pairs as far apart as possible.

3. Determine the positions of the atoms from the way electron pairs are shared (how electrons are shared between the central atom and

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Section 8.13

Electron Pairs = Bonded Atoms

• BeCl₂ 180° Linear

• Electron pair arrangement

• Two pairs of electrons around the beryllium atom. To minimize electron electron repulsions place the pairs on opposite sides of the beryllium atom. Electron pair

arrangement is linear with a 180 degree bond angle.

• Molecular Structure

• Each electron pair on beryllium is shared with a chlorine atom. The molecular structure is linear with a

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Section 8.13

• BF₃ 120° Trigonal planar

• Three pairs of electrons around the boron atom. To

minimize electron electron repulsions place the pairs on 60 degrees apart. Electron pair arrangement is trigonal

planar (flat) with a 120 degree bond angle.

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Section 8.13

CH₄ 109.5° Tetrahedral

• Four pairs of electrons around the carbon atom. To

minimize electron electron repulsions place the pairs on 90 degrees apart. Electron pair arrangement is

tetrahedral with a 109.5 degree bond angle. Notice this

structure is not flat.

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Section 8.13

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Section 8.13

Electron Pairs ≠ Bonded Atoms

• What about molecules where the number of electron pairs is different from the number of bonded atoms

• Ammonia – 4 pairs of electrons – 3 bonded atoms

• Water – 4 pairs of electrons – only two bonded atoms

• Both of these have tetrahedral electron pair arrangement but what is the molecular structure. Neither is flat the

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Section 8.13

Electron pair arrangement vs Molecular Structure in NH₃

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Section 8.13

Electron pair arrangement vs Molecular Structure in H₂O

• (a) The tetrahedral arrangement of the four electron pairs around oxygen in the water molecule. (b) Two of the electron pairs are

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Section 8.13

Structures of Molecules That Have Four Electron Pairs

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Section 8.13

Arrangements of Electron Pairs and the Resulting Molecular Structures for Two, Three, and Four Electron Pairs

Electron Pair Arrangement vs

Molecular Structure

They are the same when the number of electron pairs around the central atom equals the number of

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Section 8.13

Concept Check

Determine the shape for each of the following molecules, and include bond angles:

HCN

(121)

Section 8.13

Solution

HCN HCN – linear, 180o

PH₃ PH₃ – trigonal pyramid, 109.5o

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Section 8.13

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Section 8.13

Arrangements of Electron Pairs and the Resulting Molecular Structures for Five and Six Electron Pairs

Bonded

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Section 8.13

Concept Check

SF₄

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Section 8.13

Solution

SF₄ 5 electron pairs/4 bonded

atoms – see saw, 90o, 120o

ICl₃ 5 electron pairs/3 bonded atoms T shaped 90o, 120o

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Section 8.13

Concept Check

True or false:

A molecule that has polar bonds will always be polar.

-If true, explain why.

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Section 8.13

Solution

True or false:

A molecule that has polar bonds will always be polar.

-If true, explain why.

-If false, provide a counter-example.

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Section 8.13

Concept Check

True or false:

Lone pairs make a molecule polar. -If true, explain why.

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Section 8.13

Solution

True or false:

Lone pairs make a molecule polar. -If true, explain why.

-If false, provide a counter-example.

•False, lone pairs do not always make a