chapter 8 bonding and molecular structure · chapter 8 bonding and molecular structure ... chemical...

Jeffrey Mack

California State University,

Sacramento

Chapter 8 Bonding and Molecular Structure

• Things we must consider:

• What holds the atoms in a molecule or ionic

compound together?

• Why are atoms in molecules often distributed

at strange angles?

• Why are molecules three dimensional?

• Can we predict the structure of a compound?

• How does structure relate to chemical and

physical properties?

Chemical Bonding

Chemical bond: attractive force holding two or more atoms

together

Covalent bond: a sharing electrons between the atoms.

non–metal / non–metal bonds.

Ionic bond: electrostatic in nature, transfer of

electrons from a metal to a nonmetal.

(cation + anion)

Metallic bond: attractive force holding pure metals

together.

Cations in a “sea” of electrons.

Chemical Bonding

Ionic Bonding: complete transfer of 1 or more electrons from one atom to another.

(metals and non-metals)

Covalent Bonding: valence electrons shared

between atoms.

(non-metals and non-

metals)

Most bonds are somewhere in between.

Two Extreme Forms of Connecting or Bonding Atoms

Ionic Bonding results when an electron or electrons are

transferred from one atom to another.

The transfer results in each attaining an octet or Noble gas

electron configuration.

Na + Cl

3s1 3s23p5

Na+ Cl

e–

2s22p6

3s23p6

Na+ Cl-

[Ne] [Ar]

Noble gas electron

configurations

Ionic Bonding

• There are many examples of compounds

having covalent bonds, including the gases in

our atmosphere (O2, N2, H2O, and CO2),

common fuels (CH4), and most of the

compounds in your body.

• Covalent bonding is also responsible for the

atom-atom connections in polyatomic ions.

Covalent Bonding

A covalent bond results from a overlap of

valence orbitals on neighboring atoms.

Valence Electrons and Lewis Symbols for Atoms:

Covalent Bond Formation

• The American chemist Gilbert Newton

Lewis (1875–1946) introduced a useful

way to represent the valence shell

electrons of an atom.

• The element’s symbol represents the

nucleus including the core electrons.

• Up to four valence electrons, indicated

by dots, are placed one at a time

around the symbol.

• If any valence electrons remain, they

are paired with ones already there or

shared electrons up to a total of eight.

G. N. Lewis

1875 - 1946

Covalent Bonding & Lewis Structures

Notice that the group numbers of the main group elements represents the number of valence electrons (dots) for each element.

Lewis Dot Symbols

• Electrons are divided between core and valence electrons

B 1s2 2s2 2p1

• Core = [He] , valence = 2s2 2p1

Br [Ar] 3d10 4s2 4p5

Core = [Ar] 3d10 , valence = 4s2 4p5

Valence Electrons

• Covalent bonding, a bond results when one or more

electron pairs are shared between two atoms.

• The electron pair bond between the two atoms of an H2

molecule is represented by a pair of dots or a line.

• The representation of a molecule in this fashion is

called a Lewis electron dot structure or just a Lewis

structure in honor of G. N. Lewis.

H–H H

H-atom: H2 molecule:

H H

Lewis Dot Structures & the Octet Rule

• The number of unpaired valence electrons gives a general

indication as to the number of bonds an atom will likely form.

• Hydrogen has only one electron and therefore can only

make one covalent bond.

• Gr 7A has only one unpaired electron, so it generally forms

one bond.

• Gr 6A had two unpaired electrons, thus the likelihood of two

bonds. ... and so on.

Lewis Dot Structures & the Octet Rule

F

When other covalent species form, there are

additional electron pairs that do not participate in

bonding.

These are called “lone pairs” (lp)

& three lone pairs (octet)

H + H F

one bonding pair

hydrogen fluoride: HF

“single bond”

H– F

Lewis Dot Structures & the Octet Rule

• No. of valence electrons of an atom = Group number

• For Groups 1A-4A, no. of bond pairs = group number

• For Groups 5A -7A, BP’s = 8 - Grp. No.

• Hydrogen can only form one bond and never has lone

pairs.

• Boron often forms only 3 bonds.

• Elements in the 3rd period and beyond can exceed the

octet rule.

• Some stable molecules can form with an odd number of

electrons.

The Octet Rule

N N : :

Triple bond!

Example: N2

Each nitrogen atom needs 8 electrons to

complete an octet…

But each nitrogen has only 5 valence electrons!

As a result, the electrons must be shared.

N

N

There are 10

electrons

available

N2 needs 16

electrons (28) There are 10 valence electrons present

Therefore 6 electrons must be shared.

The remaining 4 electrons are the lone pairs

1. In general, for a poly atomic molecule the atom that

is the lowest and to the left on the periodic table

will be in the center. (lowest Electron Affinity)

2.Determine the total number of valence electrons in

the molecule or ion.

• In a neutral molecule, this number will be the sum

of the valence electrons for each atom.

• For an anion, add the number of electrons equal to

the negative charge.

• For a cation, subtract the number of electrons equal

to the positive charge.

Drawing Lewis Electron Dot Structures

3.Place one pair of electrons between each pair of

bonded atoms to form single bonds.

4.Use the remaining pairs to form more bonds or add

in as lone pairs to complete the octet of each atom.

5. If the central atom has fewer than eight electrons at

this point, change one or more of the lone pairs on

the terminal atoms into a bonding pair between the

central and terminal atom to form a multiple bond.

• Hydrogen atoms will never have lone pairs or

multiple bonds!

Drawing Lewis Electron Dot Structures

Ammonia, NH3

1. Decide on the central atom; Central atom is

generally the atom of lowest affinity for electrons &

never hydrogen.

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 Lewis Dot Structure

3. Form a single bonds between the central atom and each surrounding atom.

H H

H

N

H ••

H

H

N 4. The remaining electrons form

LONE PAIRS to complete octet as

needed.

3 BOND PAIRS and 1 LONE PAIR.

Note that N has a share in 4 pairs (8 electrons), while

H shares 1 pair.

Building a Lewis Dot Structure

Step 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 bonds

10 pairs of electrons are

now left.

Sulfite Ion, SO32−

4. The remaining pairs become lone pairs, first on outside atoms and then on central atom.

O O

O

S ••

•

•

••

••

•• ••

••

•

•

•

•

•

•

Each atom is surrounded by an octet

of electrons.

Sulfite Ion, SO32−

1. Central atom = _______

2. Valence electrons = __ or __ pairs

3. Form bonds.

4. Place lone pairs on outer atoms.

This leaves 6 pairs.

Carbon Dioxide, CO2

4. Place lone pairs on outer atoms.

The second bonding pair forms a double bond.

5. So that C has an octet, we shall form

DOUBLE BONDS between C and O.

Carbon Dioxide, CO2

C H O

4 + 2 × 1 + 6 = 12 electrons available

Carbon is the least electronegative,

hydrogen can form only one bond.

C

Formaldehyde, CH2O

Formaldehyde, CH2O

C H O

4 + 2 × 1 + 6 = 12 electrons available

Carbon is the least electronegative,

hydrogen can form only one bond.

C H H

O

C H O

4 + 2 × 1 + 6 = 12 electrons available

Carbon is the least electronegative,

hydrogen can form only one bond.

C H H

O

Formaldehyde, CH2O

C H O

4 + 2 × 1 + 6 = 12 electrons available

Carbon is the least electronegative,

hydrogen can form only one bond.

C H H

O

Formaldehyde, CH2O

C H O

4 + 2 × 1 + 6 = 12 electrons available

Carbon is the least electronegative,

hydrogen can form only one bond.

C H H

O

Formaldehyde, CH2O

C

O H H

: :

Additional

Stuctures: C H H

O

Formaldehyde, CH2O

C

O H H

: :

Additional

Stuctures: C H H

O

Formaldehyde, CH2O

C O H H

: :

C

O H H

: :

Additional

Stuctures: C H H

O

Formaldehyde, CH2O

correct structure

Not this one…

C O H H

: :

or this one…

C

O H H

: :

Additional

Stuctures: C H H

O

Formaldehyde, CH2O

correct structure

Not this one…

C O H H

: :

or this one…

nature likes symmetry!

C

O H H

: :

Additional

Stuctures: C H H

O

Formaldehyde, CH2O

correct structure

Not this one…

C O H H

: :

or this one…

nature likes symmetry!

You will see why the first structure is favored by formal

chargers later.

Double and even triple bonds are commonly observed for C, N, P, O, and S

H2CO

C2F4

SO3

Double & Triple Bonds

Lewis Structures of Common Oxoacids & Their Anions

Molecules and ions having the same number of valence electrons and the same Lewis structures are said to be isoelectronic

Common Isoelectric Molecules & Ions

• The formal charge is the charge that would reside on an

atom in a molecule or polyatomic ion if we assume that all

bonding electrons are shared equally.

• The formal charge for an atom in a molecule or ion is

calculated based on the Lewis structure of the molecule or

ion, using:

• NVE = the number of valence electrons in the uncombined

atom (and equal to its group number in the periodic table).

• LPE = number of lone pair electrons on an atom.

• BE = number of bonding electrons around an atom.

FC = NVE [LPE + ½ BE]

Atom Formal Charges in Covalent Molecules & Ions

Hydrogen Fluoride: H F

H F Number of valence

electrons:

Number of bonding

electrons:

Number of lone pair

electrons:

Formal charge:

1 7

2 2

0 6

2FC(H) 1 0

2

0

2FC(F) 7 6

2

H F

0

0 0

FC = NVE [LPE + ½ BE]

Calculating the Formal Charge on Each Atom in a Covalent Molecule

When the sum of the formal charges on the atoms in

a molecule equals the expected overall charge on the

molecule, the Lewis structure is valid.

HF is expected to be a neutral compound, and the

formal charges validate this Lewis dot structure.

H F 0 0 + = 0

Calculating the Formal Charge on Each Atom in a Covalent Molecule

• •

•

• O O C • •

•

•

O-atom: FC = 6 – [4 + ½ 4] = 0

C-atom: FC = 4 – [0 + ½ 8] = 0

0 + 0 + 0 = 0

Carbon Dioxide, CO2

[ :C N: ]–

FC(C)

–1

FC(N)

0

The negative charge resides on the carbon atom.

As expected the overall charge is 1

64 2

2

65 2

2

Use Formal Charges to Predict Which Atom Carries the Negative Charge on CN−

C

O H H

: :

C H H

O

C O H H

: :

0 0 0

0 0 0 +2

-2

not favorable!

0 0

-1 +1

not favorable!

This one is

most

favorable!

Generally, the structure with

the lowest formal charges on

each atom or negative

charges on the atoms with

the highest EA’s are often

most favored.

CH2O Formal Charges

Consider the nitrate anion: The double bond does not

have to be on the vertical O-atom. There are three

equivalent structures that can be drawn.

We see that the double bond moves around or

“resonates” between the structures indicating that

the molecule exhibits what is called “Resonance” .

N

O

O O :

:

: :

:

:

:

:

–

: N

O

O O :

: :

: : :

– :

: N

O

O O : :

:

: :

–

Resonance

All three of the Lewis structures are equivalent

Resonance Structures.

Resonance stabilizes the energy (lowers) by

distributing electron density over the entire molecule.

The measured fHo is more exothermic than the

predicted value.

N

O

O O :

:

: :

:

:

:

:

–

: N

O

O O :

: :

: : :

– :

: N

O

O O : :

:

: :

–

Resonance

Identifying Resonance: Compounds that exhibit

resonance will have fractional bond orders.

: N

O

O O :

: :

:

: :

–

# of A-B bondsBond Order =

# of A-B links

4 N O bondsBO 1.333

3 N O links

Resonance

Certain elements can violate the octet rule. Boron

may form stable compounds with only 6 valence

electrons.

Compounds in the 3rd period and beyond can have

more than 8 electrons.

BF3 SF4

Exceptions to the Octet Rule

There are three classes of exceptions to the

octet rule:

1. Molecules with an odd number of electrons.

2. Molecules in which one atom has less than

an octet

3. Molecules in which one atom has more than

an octet.

Exceptions to the Octet Rule

Examples: Generally molecules such as ClO2, NO,

and NO2 have an odd number of electrons.

NO (nitrogen monoxide) Available electrons = 11

There must be an odd electron!

Molecular Orbital Theory will explain why NO is

stable later.

N O N O

115.5...?

2

or

Odd Numbers of Electrons

• Relatively rare.

• Molecules with less than an octet are typical for

compounds of Groups 1A, 2A, and 3A

Most typical example is BH3.

Available Electrons = 3 + 3(1) = 6 B

H

H H

There are only 6 electrons on the central atom.

Hydrogen may only form one bond.

The formal charges (all zero) support the drawn

structure.

Incomplete Octet

• Atoms from the 3rd period onwards can

accommodate more than an octet.

• Beyond the third period, the d-orbitals are low

enough in energy to participate in bonding and

accept the extra electron density.

Example: BrF3

Available Electrons = 7 + 3(7) = 28

Each fluorine atom can only share one electron.

What does this mean??

Central Atoms that Exceed the Octet Rule

Bromine has 7 valence electrons (4s24p5)

Fluorine also has 7 (2s22p5)

BrF3

Br has the lowest electronegativity so it will be the

central atom in the molecule.

Each F shares

one electron to

complete an

octet.

There are 10

total electrons in

the Br valence Br F F

Available Electrons = 48

Sulfur has 6 valence electrons, fluorine has seven.

If each F shares one electron with one of the six

from S, F completes its octet.

This will yield a compound with

S in the center having 6 F’s

bonded.

12 valence electrons around

the S

S

F F

F F

F

F

SF6

Compounds in Which an Atom Has More Than Eight Valence Electrons

Lewis structures tell us how atoms are connected in a

molecule:

bonds (bp) & lone pairs (lp) etc…

The 3–D shape of a molecule is however, determined

by its bond angles.

The Lewis structure

suggests that the H–C–

H bond angles are 90o &

that the molecule is flat.

H

H

C H H

90o

Molecular Shapes

Experimentally however, the H–C–H bond angles are

found to be 109.5

If this is the case, the molecule cannot be planar…

The methane molecule must be 3-D since the sum of

the angles is greater than 360 degrees!

H

H

C H H 90° 90°

90° 90°

90° + 90° + 90° +90° = 360°

109.5 + 109.5 + 109.5 + 109.5 =

438.0°

Molecular Shapes

To accommodate the 109.5o bond angles, the atoms

adopt a new 3-D geometry to minimize the repulsion

between the atoms:

• The molecular geometry of CH4 is said to be

“tetrahedral”

• The H-atoms fit at the corners of a regular

tetrahedron shape with the carbon at the center.

H

H

C H H C H H

H

H

becomes…

109.5o

Molecular Shapes

• In order to predict molecular shape, we assume the

valence electrons of each atom in the molecule

repel one another.

• When this occurs, the molecule adopts a 3D

geometry that minimizes this repulsion where:

This process is known as:

Valence Shell Electron Pair Repulsion (VSEPR)

theory.

bp-bp lp-lp > lp-bp >

Molecular Shapes

Valence Shell Electron Pair Repulsion

A molecule can be described in terms of the

distribution of the bonding atoms about the

central atom:

Molecular Geometry (MG)

A molecule can be described in terms of the

distribution of the bonding pair electrons (bp)

and lone pair electrons (lp) about the central

atom:

Electron Pair Geometry (EPG)

Molecular Shapes

Central Atoms Surrounded Only by Single-Bond Pairs

2 pairs

of

electrons

3 pairs

of

electrons

4 pairs

of

electrons

5 pairs

of

electrons

6 pairs

of

electrons

Electron Pair Geometries

G e ome tr y E x a m pl e

N o. of e - P ai r s Ar o u n d Ce nt r a l Atom

G e ome tr y E x a m pl e

N o. of e - P ai r s Ar o u n d Ce nt r a l Atom

180°

l i n e ar 2 F—Be—F

G e ome tr y E x a m pl e

N o. of e - P ai r s Ar o u n d Ce nt r a l Atom

180°

l i n e ar 2 F—Be—F

120°

planar trigonal

F F B

F

3

G e ome tr y E x a m pl e

N o. of e - P ai r s Ar o u n d Ce nt r a l Atom

180°

l i n e ar 2 F—Be—F

120°

planar trigonal

F F B

F

3

H

H H

H

tetrahedral

109°

C 4

Electron Pair Geometries

4 bonding pairs of electrons about the central atom, no lone

pairs on the central atom:

4 groups of electrons (all bonding)

Electron Pair Geometry (EPG) :

Tetrahedral

Molecular Geometry (MG):

Tetrahedral

When there are no lone pairs of

electrons in a molecule, the molecular

and electronic geometries are the same!

Central Atoms with Single-Bond Pairs & Lone Pairs

4 total pairs of electrons about the central atom, three bonding,

one lone pair on the central atom:

Electron Pair Geometry (EPG) :

Tetrahedral

Molecular Geometry (MG):

Trigonal Pyramidal

The lone pair forces the bonds to

reduce below the normal 109.5°.

Central Atoms with Single-Bond Pairs & Lone Pairs

Four groups of electrons, 2 bonding, 2 lone pairs on the central

atom:

Electron Pair Geometry (EPG) :

Tetrahedral

Molecular Geometry (MG):

Bent

The lone pairs force the bonds to

reduce below the normal 109.5°.

Central Atoms with Single-Bond Pairs & Lone Pairs

BF3

Molecular and Electron Pair geometries are

described as trigonal planar.

12 pairs of electrons available, F

forms only 1 bond

The B atom is surrounded by only

3 electron pairs.

Bond angles are 120°

•

• F

B

••

•

•

Compounds with Less Than an Octet

All based on trigonal

bipyramidal shape

Molecular Geometries for 5 Electron Pairs

P has 5 valence electrons

Each Cl-atom needs one electron to complete the

octet

P : : . P

Cl

Cl

Cl

Cl

Cl

Molecular Geometry : Trigonal

Bipyramidal

Electron Pair Geometry: Trigonal

Bipyramidal

Example: PCl5

S has 6 valence electrons

Each F-atom needs one electron to complete the

octet

S : : : S

F

F

:

F

F

Molecular Geometry : See-Saw

Electron Pair Geometry: Trigonal

Bipyramidal

Example: SF4

All are based on the 8-

sided octahedron

Molecular Geometries for 6 Electron Pairs

6 electron pairs

F

F

F F

F F

O c t a h e d r o n

9 0°

9 0°

S

Compounds with 5 or 6 Pairs Around the Central Atom

Three groups of electrons, all bonding,

no lone pairs on the central atom:

The lone pair of electrons forces the molecule into a

bent molecular geometry.

S

:

SO2

3 groups of electrons (2 bonds, 1 lp)

Electron Pair Geometry (EPG) Trigonal Planar

Molecular Geometry (MG): bent

Example: SO2

Recall that electronegativity () measures the relative

tendency for an atom to polarize a bond. It follows Zeff on

the periodic table.

Bond Polarity & Electronegativity

As the difference in Electronegativity increases

(), so does the ionic character of the bond.

Increasing covalent character

Bond Polarity & Electronegativity

• A difference in electronegativity of 0 indicates

that the bond is purely covalent. There is equal

sharing of electrons.

• A difference of electronegativity between 0.1 to

0.3 indicates that the bond is non-polar

covalent.

• A difference of electronegativity between 0.4 to

1.7 indicates that the bond is polar covalent.

• A difference of electronegativity that is greater

than 1.8 indicates that the bond is ionic.

Bond Polarity & Electronegativity

The polarity of the individual bonds in a molecule will

determine the overall polarity of a molecule.

All homonuclear

diatomic molecules

a non-polar.

I I :

iodine (I2)

Notice the symmetry of the

molecule:

• When divided, the top and

bottom as well as the left and

right are mirror images of one

another.

• One also knows the molecule

is non polar because the

bond is non polar. = 2.5 2.5 = 0

Overall Polarity in Molecules

H F

• Fluorine has a larger

electroegativity value

than hydrogen.

• This means that the

electrons in the bond

are skewed toward

the F-atom.

• The electrons shift

toward the F-atom This polarizes the molecule

more positive end

(+)

more negative end

()

Example: HF

The greater the difference in electronegativity () the

more polar the covalent bond.

H vs. F

4.0 – 2.2 = 1.8

C vs. F

4.0 – 2.5 = 1.5

O vs. F

4.0 – 3.5 = 0.5

So in terms of polarity,

H-F > C-F > O-F

Electronegativity: H vs. F

When a molecule possesses a net dipole moment, it

is polar.

H2O

The molecule

is therefore

polar.

The individual O-H

bond dipoles result

in a “net” or

overall

dipole moment for

the molecule.

Bond Polarity & Molecular Polarity

Notice also that the molecule’s symmetry can be broken along

either O-H bond axis:

This side…

does not look like this side!

Whenever there exists a line or plane of asymmetry,

the molecule is Polar!

Molecular Shape & Molecular Polarity

CO2

non–polar

Symmetry across all

bond axes.

Molecular Shape & Molecular Polarity

Dipole Moments & Molecular Polarity

• The order of a bond is the number of bonding

electron pairs shared by two atoms in a

molecule

• Bond orders may be 1, 2, and 3, as well as

fractional values.

• Bond strength increases with bond order.

• Bond length is inversely proportional to bond

order.

• Bond strength also increases with the difference

in electronegativity between two covalently

bonded atoms.

Bond Properties: Order, Length, Energy

Covalent bond strength increases with increasing

example: HCl bond is stronger than the HBr bond

Covalent bond strength increases with increasing

bond order.

example: O=O bond in stronger than O–O bond

triple > double > single

Bond Order: 3 2 1

Bond length decreases with increasing bond order

(Strength)

example: O=O bond is shorter than O–O bond

Therefore shorter!

Bond Properties: Order, Length, Energy

Double bond Single bond

Triple

bond

Acrylonitrile

Bond Order: # of Bonds Between a Pair of Atoms

Fractional bond orders occur in molecules with resonance structures.

Consider NO2-

The N—O bond order = 1.5

O=NO

: :

: :

:

Fractional Bond Order

• 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

H—I

Bond distances measured

in Angstrom units where

1Å = 10-10 m.

Bond Length

Bond length depends on bond order.

Bond distances measured

in Angstrom units where

1Å = 10-10 m.

Bond Length

Bond order is proportional to two important bond properties:

(a) bond strength

(b) bond length

745 kJ

414 kJ

110 pm

123 pm

Bond Length & Bond Strength

BOND Bond dissociation enthalpy (kJ/mol)

H—H 436

C—C 346

C=C 602

CC 835

NN 945

The GREATER the number of bonds (bond order)

the HIGHER the bond strength and the SHORTER

the bond.

Bond Strength

The energy required to break a covalent bond is

called the bond dissociation enthalpy, D.

For the Cl2 molecule, D(Cl–Cl) is given by H for the

reaction:

Cl2(g) 2Cl(g)

When more than one bond is broken:

CH4(g) C(g) + 4H(g) H = 1660 kJ

the bond enthalpy is a fraction of H for the

atomization reaction:

D(C–H) = ¼H = ¼(1660 kJ) = 415 kJ

Strengths of Covalent Bonds

Estimate the energy of the reaction

H—H(g) + Cl—Cl(g) 2 H—Cl(g)

Net energy = ∆rH =

= energy required to break bonds

energy evolved when bonds are made

H—H = 436 kJ/mol

Cl—Cl = 242 kJ/mol

H—Cl = 432 kJ/mol

Using Bond Dissociation Enthalpies

It takes energy to break bonds. (endothermic)

Energy is releases when bonds are made.

(exothermic)

If more energy is released than is required to break

the reactant bonds, then the over all reaction is

exothermic.

If more energy is required to break bonds than is

released when the product bonds form, then the over

all reaction is endothermic.

rH = (bonds broken) (bonds formed)

Energy Supplied ( H>0)

Energy Released ( H<0)Molecule (g) Molecular fragments (g)

Estimate the rH for the following reaction using

average bond dissociation enthalpies.

CH4(g) + 2O2(g) CO2(g) + 2H2O(l)

Estimate the rH for the following reaction using

average bond dissociation enthalpies.

First draw out the Lewis structures to determine

the number and types of bonds

CH4(g) + 2O2(g) CO2(g) + 2H2O(l)

Estimate the rH for the following reaction using

average bond dissociation enthalpies.

First draw out the Lewis structures to determine

the number and types of bonds

C

H

H

H H + 2 O=O O=C=O O H

H + 2

CH4(g) + 2O2(g) CO2(g) + 2H2O(l)

Estimate the rH for the following reaction using

average bond dissociation enthalpies.

First draw out the Lewis structures to determine

the number and types of bonds

C

H

H

H H + 2 O=O O=C=O O H

H + 2

Second, count up the bonds of each type:

CH4(g) + 2O2(g) CO2(g) + 2H2O(l)

Estimate the rH for the following reaction using

average bond dissociation enthalpies.

CH4(g) + 2O2(g) CO2(g) + 2H2O(l)

First draw out the Lewis structures to determine

the number and types of bonds

C

H

H

H H + 2 O=O O=C=O O H

H + 2

Second, count up the bonds of each type:

4 (C–H) + 2 (O=O) 2 (C=O) + 4 (O–H)

Now sum up the bonds broken and subtract the

bonds formed.

4 (C–H) + 2 (O=O) 2 (C=O) + 4 (O–H) – [ ] rH =

Now sum up the bonds broken and subtract the

bonds formed.

4 (C–H) + 2 (O=O) 2 (C=O) + 4 (O–H) – [ ] rH =

rH = {4 (413 kJ/mol) + 2 (495 kJ/mol)}

– { 2 (745 kJ/mol + 4 (463 kJ/mol) }

Now sum up the bonds broken and subtract the

bonds formed.

4 (C–H) + 2 (O=O) 2 (C=O) + 4 (O–H) – [ ] rH =

rH = {4 (413 kJ/mol) + 2 (495 kJ/mol)}

– { 2 (745 kJ/mol + 4 (463 kJ/mol) }

= –718 kJ/mol

Now sum up the bonds broken and subtract the

bonds formed.

4 (C–H) + 2 (O=O) 2 (C=O) + 4 (O–H) – [ ] rH =

rH = {4 (413 kJ/mol) + 2 (495 kJ/mol)}

– { 2 (745 kJ/mol + 4 (463 kJ/mol) }

= –718 kJ/mol

By way of comparison, rH from fHo = –890 kJ/mol

890 ( 718)% error = 100 19.3%

890