Chapter 8

It is the energy required to break a bond.
It gives us information about the strength of a bonding interaction.

Formed from electrostatic attractions of closely packed, oppositely charged ions.
Formed when an atom that easily loses electrons reacts with one that has a high electron affinity .

E
= 2.31 10
 19
J nm ( Q Q / )
This is a statement of Coulomb’s Law where:
Q
1 and Q
2
= numerical ion charges r = distance between ion centers (in nm)
When E is positive (+), repulsion is indicated.
When E is negative (-), attraction is indicated.

08_130
+ +
H atom H atom
Sufficiently far apart to have no interaction
+ +
0
H atom H atom
The atoms begin to interact as they move closer together.
+ +
H
2 molecule
Optimum distance to achieve lowest overall energy of system
-458
0
(a) (b)
H H
H H
H H
0.074
Internuclear distance (nm)
(H

H bond length)
H H
Interaction of two hydrogen atoms.

An indication of bond strength and bond length
– Single bond: 1 pair of e- shared
•• ••
Ex: F
2
:
F-F
:
•• ••
Longest, weakest
 Double bond: 2 pairs of e- shared
Ex: O
2
O=O
 Triple bond: 3 pairs of e- shared
Ex: N
2
:
N ≡ N
:
Shortest, strongest
8

- covalent bonds are formed by sharing electrons between nuclei.
H
.
+
.
H ----> H-H
coordinate covalent bonds are bonds where both shared electrons originate on the same atom
H
3
N: + H + ----> H
3
N-H +

Polar covalent bond -- covalent bond in which the electrons are not shared equally because one atom attracts them more strongly than the other. A dipole moment exists.
Nonpolar covalent bond -- covalent bond in which the electrons are shared equally between both atoms. No dipole moment exists.

The ability of an atom in a molecule to attract shared electrons to itself.

= (H

X) actual

(H

X) expected

08_132
Increasing electronegativity
H
2.1
Li
1.0
Na
0.9
K
0.8
Rb
0.8
Cs
0.7
Fr
0.7
Be
1.5
Mg
1.2
Ca
1.0
Sr
1.0
Ba
0.9
Ra
0.9
Sc
1.3
Y
1.2
La-Lu
1.0-1.2
Ac
1.1
Ti
1.5
Zr
1.4
Hf
1.3
Th
1.3
V
1.6
Nb
1.6
Ta
1.5
Pa
1.4
Cr
1.6
Mo
1.8
W
1.7
U
1.4
Mn
1.5
Tc
1.9
Re
1.9
Np-No
1.4-1.3
Fe
1.8
Ru
2.2
Os
2.2
Co
1.9
Rh
2.2
Ir
2.2
Ni
1.9
Pd
2.2
Pt
2.2
Cu
1.9
Ag
1.9
Au
2.4
Zn
1.6
Cd
1.7
Hg
1.9
B
2.0
Al
1.5
Ga
1.6
In
1.7
Tl
1.8
(a)
Li
1.0
Na
0.9
K
0.8
Rb
0.8
Cs
0.7
Fr
0.7
Be
1.5
Mg
1.2
Ca
1.0
Sr
1.0
Ba
0.9
Ra
0.9
Sc
1.3
Y
1.2
La-Lu
1.0-1.2
Ac
1.1
Ti
1.5
Zr
1.4
Hf
1.3
Th
1.3
H
2.1
V
1.6
Nb
1.6
Ta
1.5
Pa
1.4
C
2.5
Si
1.8
Ge
1.8
Sn
1.8
Pb
1.9
N
3.0
P
2.1
As
2.0
Sb
1.9
Bi
1.9
O
3.5
S
2.5
Se
2.4
Te
2.1
Po
2.0
F
4.0
Cl
3.0
Br
2.8
I
2.5
At
2.2
Increasing electronegativity
Cr
1.6
Mo
1.8
W
1.7
Mn
1.5
Tc
1.9
Re
1.9
U
1.4
Np-No
1.4-1.3
Fe
1.8
Ru
2.2
Os
2.2
Co
1.9
Rh
2.2
Ir
2.2
Ni
1.9
Pd
2.2
Pt
2.2
Cu
1.9
Ag
1.9
Au
2.4
Zn
1.6
Cd
1.7
Hg
1.9
B
2.0
Al
1.5
Ga
1.6
In
1.7
Tl
1.8
C
2.5
Si
1.8
Ge
1.8
Sn
1.8
Pb
1.9
P
2.1
As
2.0
Sb
1.9
Bi
1.9
N
3.0
Te
2.1
Po
2.0
S
2.5
Se
2.4
O
3.5
Cl
3.0
Br
2.8
I
2.5
At
2.2
F
4.0
(b)
Pauling Electronegativity Values

% Ionic Character where x
A
(IC)
 x
A
 x
B

100 %
 x
A is the larger electronegativity and x
B is the smaller value.
Watch significant figures!!!
Ionic Bond % IC > 50 %
Polar Covalent % IC 5 - 50 %
Nonpolar Covalent % IC < 5 %

Three Possible Types of Bonds
Nonpolar Covalent
(Electrons equally shared.)
Polar Covalent
(Electrons shared unequally.)
Ionic
(Electrons are transferred.)

A molecule, such as HF, that has a center of positive charge and a center of negative charge is said to be polar , or to have a dipole moment .
H F

+


08_131
(a)
 
F
 
H
H
 
 
H
F
 
F
 
 
 
F
F
H
 
H
 

  H F  
 
H F  

  H F  
(b)
  H F  
  H F  
The Effect of an electric field on hydrogen fluoride molecules.

08_133
 
H

+
O
 
 
 
H
(a) (b)
Dipole Moment for the water molecule.

08_134
H
 
H
 
3
 
N
(a)
H
 
+
(b)



Dipole moment for the ammonia molecule.

Does nitrogen trichloride exhibit a dipole moment?
Yes. It has three nonpolar bonds but, also, has a lone pair of electrons which makes it assymetrical and therefore, polar.

08_151
Nonpolar molecule--zero dipole moment.

Table 8.2 on page 356 in Zumdahl.

Achieving Noble Gas Electron
Configurations (NGEC)
Two nonmetals react: They share electrons to achieve NGEC.
A nonmetal and a representative group metal react (ionic compound): The valence orbitals of the metal are emptied to achieve NGEC. The valence electron configuration of the nonmetal achieves NGEC.
See Table 8.3 on page 361 in Zumdahl.

08_136
Li

Be

(0.60)
60
(0.31)
31
Na

Mg

Al

(0.95)
95
(0.65)
65
(0.50)
50
K

Ca

Ga

O

F

(1.40)
140
S

(1.36)
136
Cl

(1.84)
184
Se

(1.81)
181
Br

(1.33)
133
(0.99)
99
(0.62)
62
Rb

Sr

In

Sn

Sb

(1.98)
198
Te

(1.95)
195
I

(1.48)
148
(1.13)
113
(0.81) (0.71) (0.62)
81 71 62
(2.21)
221
(2.16)
216
Sizes of ion related to position on the periodic table.

Ions containing the same number of electrons
(O 2

, F

, Na + , Mg 2+ , Al 3+ )
O 2

> F

> Na + > Mg 2+ > Al 3+ largest smallest

The change in energy when separated gaseous ions are packed together to form an ionic solid .
M + ( g ) + X

( g )

MX( s )
Lattice energy is negative (exothermic) from the point of view of the system.

1. Sublimation of the solid metal
M( s )

M( g ) [endothermic]
2. Ionization of the metal atoms
M( g )

M + ( g ) + e

[endothermic]
3. Dissociation of the nonmetal
1 /
2
X
2
( g )

X( g ) [endothermic]

(continued)
4. Formation of X
X( g ) + e

 ions in the gas phase:

X

( g ) [exothermic]
5. Formation of the solid MX
M + ( g ) + X

( g )

MX( s ) [quite exothermic]

Lattice Energy = k
( Q Q / )
Q
1
, Q
2
= charges on the ions r = shortest distance between centers of the cations and anions

Mg 2+
08_139
( g ) + O 2( g )
737 Electron affinity
Mg 2+ ( g ) + O( g )
Mg 2+ ( g ) +
1
2
O
2
( g ) 247
-3916
Lattice energy
2180
Ionization energy
MgO( s )
Mg( g ) +
Mg( s ) +
1
2
1
2
O
2
( g )
O
2
( g )
-602
150
Overall energy change
Na( g ) + F( g )
Na + ( g ) +
1
2
F
2
( g ) 77
-328
495
109
Na( g
Ionization energy
) +
Na(s) +
1
2
1
2
F
F
2
2
( g )
( g )
Na + ( g ) + F ( g )
-923
-570
NaF( s )
Electron affinity
Lattice energy fluoride and magnesium oxide.

Bond breaking requires energy (endothermic).
Bond formation releases energy (exothermic).

H =

D ( bonds broken )
 
D ( bonds formed ) energy required energy released
Draw the Lewis Structure for each reactant and product before doing any calculations!

Single, Double, & Triple Bonds
Single bonds -- one shared pair of electrons.
Double bonds -- two shared pairs of electrons.
Triple bonds -- three shared pairs of electrons.
See bond energy Tables 8.4 & 8.5 on pages 372 in Zumdahl.

 A mount of energy required to break a particular bond between two elements in gaseous state. Given in kJ/mol.
Remember, breaking a bond always requires energy!
 Bond enthalpy indicates the “strength” of a bond.
 Bond enthalpies can be used to figure out

H rxn .
Ex: CH
4
(g) + Cl
2
(g) → CH
3
Cl (g) + HCl (g)

H rxn
= ?
 1 C-H & 1 Cl-Cl bond are broken (per mole)
 1 C-Cl & 1 H-Cl bond are formed (per mole)

H rxn
≈ 
(H bonds broken
) -

(H bonds formed
)
Note: this is the “opposite” of Hess’ Law where

H rxn
=

H products
–  h reactants
Bond Enthalpy link
32

Ex: CH
4
(g) + Cl
2
(g) → CH
3
Cl (g) + HCl (g)

H rxn
= ?
Bond Ave

H/mol Bond Ave

H/mol
C-H
H-Cl
C-C
413
431
348
Cl-Cl
C-Cl
C=C
242
328
614

H rxn
≈ 
(H bonds broken
) -

(H bonds formed
)

H rxn
≈ [(4(413) + 1(242)] – [1(328) + 3(413)+1(431)]

H rxn
≈
-104 kJ/mol

H rxn
= -99.8 kJ/mol (actual)
Note: 2 C-C
≠ 1 C=C
2(348) = 696 kJ
≠ 614 kJ
33

Ex: CH
4
(g) + Cl
2
(g) → CH
3
Cl(g) + HCl(g)

H rxn
=?
*CH
3
(g) + H(g) + 2 Cl(g)
H
Absorb E, break 1 C-H and 1 Cl-Cl bond
CH
4
(g) + Cl
2
(g)
Release E, form 1 C-Cl and 1 H-Cl bond

H rxn
CH
3
Cl (g) + HCl (g)

H rxn
=

(H bonds broken
) +

(- H bonds formed
)

H rxn
=

(H bonds broken
) -

(H bonds formed
)

Models are attempts to explain how nature operates on the microscopic level based on experiences in the macroscopic world.

A model does not equal reality.
Models are oversimplifications, and are therefore often wrong.
Models become more complicated as they age.
We must understand the underlying assumptions in a model so that we don’t misuse it.

A molecule is composed of atoms that are bound together by sharing pairs of electrons using the atomic orbitals of the bound atoms.

1. Description of valence electron arrangement (Lewis structure).
2. Prediction of geometry (VSEPR model).
3. Description of atomic orbital types used to share electrons or hold lone pairs.

Shows how valence electrons are arranged among atoms in a molecule.
Reflects central idea that stability of a compound relates to noble gas electron configuration.

Ionic Compounds
K

1
..
: Br
..
:

1
Covalent Compounds
..
..
: F
..
: F
..
:
In ionic compounds , electrons are transferred and ions are formed. In covalent compounds , electrons are shared to form a molecule.

Beryllium chloride -- BeCl
2
-- is electron deficient with four electrons. It forms a linear molecule.
Boron trifluoride -- BF
3
-- is electron deficient with six electrons. It forms a trigonal planar molecule.
See page 380 for the reaction between boron trifluoride and ammonia.

Expanded octet: occurs in molecules when the central atom is in or beyond the third period, because the empty 3d subshell is used in hybridization (Ch. 9)
PCl
5
SF
6
42

Comments About the Octet Rule
2nd row elements C, N, O, F observe the octet rule .
2nd row elements B and Be often have fewer than 8 electrons around themselves - they are very reactive.
3rd row and heavier elements CAN exceed the octet rule using empty valence d orbitals.
When writing Lewis structures, satisfy octets first , then place electrons around elements having available d orbitals .

• Sum the valence electrons from all the atoms.
• Use a pair of electrons to from a bond between each pair of bound atoms.
• Arrange remaining electrons to satisfy the duet rule for hydrogen and the octet rule for the second-row elements.

NO +
• 5 e + 6 e - 1 e = 10 e -
• [:N 
O:] +
• Each atom has an octet and is satisfied.

Formal Charge
Movie on Formal Charge
For each atom, the numerical difference between # of valence e- in the isolated atom and # of e- assigned to that atom in the Lewis structure.
To calculate formal charge:
1. Assign unshared e- (usually in pairs) to the atom on which they are found.
2. Assign one e- from each bonding pair to each atom in the bond. (Split the electrons in a bond.)
3. Then, subtract the e- assigned from the original number of valence e-.
#VALENCE e- in free atom
– #NON-BONDING e-
– ½(#BONDING e-)
FC

 Used to select most stable (and therefore most likely structure) when more than one structure are reasonable according to “the rules”.
 The most stable:
 Has FC on all atoms closest to zero
 Has all negative FC on most EN atoms.
 FC does not represent real charges; it is simply a useful tool for selecting the most stable Lewis structure.

Examples: Draw at least 2 Lewis structures for each, then calculate the FC of each atom in each structure.
1-
2
3
48

Occurs when more than one valid Lewis structure can be written for a particular molecule.
These are resonance structures . The actual structure is an average of the resonance structures called a resonance hybrid.
See the resonance structures for the nitrate ion on page 384 in Zumdahl.

 Equivalent Lewis structures that describe a molecule with more than one likely arrangement of e -
 Notation: use double-headed arrow between all resonance structures.
Ex: O
3
 Note: one structure is not “better” than the others. In fact, all resonance structures are wrong, because none truly represent the e- structure of the molecule. The “real” estructure is an “average” of all resonance structures.
50

NO
2
• contains 17 electrons.
• cannot satisfy the octet rule.
• a more sophisticated model is neededthe molecular orbital model.

The study of the threedimensional arrangement of atoms or groups of atoms within molecules and the properties which follow such arrangement.

V
S
E
P
R
The structure around a given atom is determined principally by minimizing electron pair repulsions.

1. Draw Lewis structure.
2. Put pairs as far apart as possible.
3. Determine positions of atoms from the way electron pairs are shared.( Parent
Geometry )
4. Determine the name of molecular structure from positions of the atoms.( Actual
Geometry )

Parent Geometry is electron pair arrangement about the central atom.
• linear
• trigonal planar
• tetrahedral
• trigonal bipyramidal
• octahedral
Actual Geometry is the arrangement of atoms about the central atom.
• linear
• bent
• trigonal pyramid
• seesaw
• T-shaped
• square pyramid
• square planar

08_142
Lone pair
N
H
H
N
(b)
H
(a)
Lone pair of electrons on the ammonia molecule.

08_143
(a)
O
Bonding pair
H
(b)
Lone pair
O
Lone pair
Bonding pair
H
O
H
H
(c)
Lone pairs on the water molecule.

08_144
P
Cl
Cl
Cl
P
Cl
Cl

Cl
Octahedral structure for phosphorus hexachloride.

08_145
Xe
Octahedral structure for xenon.

08_150
F
F
F
Xe
F
(a)
90 ° leads to the structure
F
F
F
Xe
F
F
F
Xe
F
180°
F leads to the structure
F
F
Xe
F
F
(b)
Parent and actual geometry for xenon tetrafluoride.

08_152
I
I
I
(a)
I
I
I
(b)
I
I
I
(c)
Three possible arrangements of the electron pairs in triiodide ion.

Table 8.6
Arrangements of Electron Pairs Around an Atom Yielding Minimum Repulsion
Number of
Electron Pairs
Arrangement of Electron Pairs Example
2 Linear
A
3 Trigonal planar
A
4
5
Tetrahedral
Trigonal bipyramidal
A
120°
A
90°
A
6 Octahedral

•
Determine the Lewis structure(s) for the molecule.
• For molecules with resonance structures, use any of the structures to predict the molecular structure.
•
Sum the electron pairs around the central atom to determine the parent geometry .
•
The arrangement of the pairs is determined by minimizing electron-pair repulsions.
(Actual
Geometry)

• Lone pairs require more space than bonding pairs since they are tightly attracted to only one nucleus. Lone pairs produce slight distortions of bond angles less than
120 o .