Chemistry-140
Lecture 22
Chapter 9:
Bonding & Molecular Structure:
Fundamental Concepts
 Chapter Highlights
 bonding types
 Lewis symbols & octets
 ionic bonding & ionic lattices

covalent bonding & Lewis dot structures

resonance structures

breaking the octet rule & formal charge

bond order & bond energy

VSEPR molecular shape & polarity
Chemistry-140
Lecture 22
Ionic, Covalent & Metallic Bonding
 Ionic bond: Term given to the electrostatic (charge-based)
attractive forces which hold oppositely charged ions together
Li+
F-
Chemistry-140
Lecture 22
Ionic, Covalent & Metallic Bonding
 Covalent bond: The sharing of electrons between two atoms
that acts to hold the atoms together
H
:
H
Chemistry-140
Lecture 22
Ionic, Covalent & Metallic Bonding
 Metallic bond: Is found in metals. Atoms of the metal are
bound to several neighbors, holding the atoms together but
allowing electrons to move freely.
M
:::
M
:::
M
Chemistry-140
Lecture 22
Lewis Symbols
 Lewis symbols = electron-dot symbols
 Lewis symbols place one dot for each valence electron
around the symbol of the element.
For Example:
S: [Ne]3s23p4
S
Chemistry-140
Lecture 22
Lewis Octets
 Octet rule: Atoms tend to gain, lose, or share enough
electrons to become surrounded by eight valence electrons
Attain the closed shell configuration of a Group 18 inert gas
Ne: [He]2s22p6
Ne
Chemistry-140
Lecture 22
Ionic Bonding
 The ionic bond is formed when ions of opposite charge
(anions and cations) are attracted and held to one another
by electrostatic attractions
Na(g)
Na
+
Cl
+
Cl(g)
NaCl(g)
Na+
+
Cl
-
Chemistry-140
Lecture 22
Energetics of Forming a Na-Cl Ionic Bond
Na(g)
Cl(g) + e-
Na+(g) + e-
Cl-(g)
DE = +496 kJ/mol
DE = -349 kJ/mol
Therefore electron transfer costs 147 kJmol-1 !!!!!
Chemistry-140
Lecture 22
Energetics of Forming a Na-Cl Ionic Bond
But, Coulomb’s law...
Q1Q2 

E=k 

 d 
 where: Q1 and Q2 are the charges on the cation and anion, d is
the distance between the nuclei (sum of the ionic radii)
and k is a constant = 8.99 x 109 J-m/C2
 1.602 x 1019  - 1.602 x 1019  

E = (6.022 x 1023)(8.99 x 109) 
10
1.57  1.81 x 10


E = -411 kJ/mol
Chemistry-140
Lecture 22
Energetics of Forming a Na-Cl Ionic Bond
Therefore:
the overall DE for the reaction
Na(g)
+
Cl(g)
NaCl(g)
DE = (-411 + 147) kJ/mol = -264 kJ/mol
Chemistry-140
Lecture 22
Ionic Lattices
 In order to maximize the attractions among ions, ionic
solids exist in lattices, which are regularly repeating three-
dimensional arrays of ions
Chemistry-140
Lecture 22
Coordination Number
 Coordination number: the number of close contacts in the
lattice array (equals 6 for this Na+ ion below)
Chemistry-140
Lecture 22
Lattice Energies
 Lattice energy: the energy required to separate the
crystalline solid into the constituent gaseous ions.
NaCl(s)
Na+(g) + Cl-(g)
It is a measure of the stability of the crystalline state
Note: The lattice energy for NaCl(s) is -786 kJ/mol as
compared to -264 kJ/mol that we calculated!!
Chemistry-140
Lecture 22
Lattice Energies
In summary lattice energies:
 increase as charges of the ions increase
 increase as sizes of the ions decrease
Coulomb’s Law!!!
Q1Q2 

E=k 

 d 
 increase with increasing coordination number
Chemistry-140
Lecture 22
Born-Haber Cycle
Na+(g)
+
Cl-(g)
Step 4
Step 3
Step 5
Na(g)
+
Step 1
Na(s)
Cl(g)
Step 2
+
1/2 Cl2(g)
DHfo
NaCl(s)
Chemistry-140
Lecture 23
Chapter 9:
Bonding & Molecular Structure:
Fundamental Concepts
 Chapter Highlights

bonding types

Lewis symbols & octets

ionic bonding & ionic lattices
 covalent bonding & Lewis dot structures

resonance structures

breaking the octet rule & formal charge

bond order & bond energy

VSEPR molecular shape & polarity
Chemistry-140
Lecture 23
Covalent Bonding
 Covalent bond: a bond formed between two atoms by
sharing of electrons.
 Lewis Structures for H2 and Cl2
H
+
H
Cl + Cl
H H
= H-H
Cl Cl = Cl-Cl
Chemistry-140
Lecture 23
Covalent Single Bonds
 Different multiplicities of covalent bonds are possible.
Single bonds are covalent bonds in which one pair of
electrons is shared by the two atoms
Cl Cl
or
Cl-Cl
Chemistry-140
Lecture 23
Multiple Covalent Bonds
 Different multiplicities of covalent bonds are possible.
Double bonds are covalent bonds in which two pairs of
electrons are shared by the two atoms.
Triple bonds are covalent bonds in which three pairs of
electrons are shared by the two atoms
N
+
N
N N
or
=N
N=
=N
N=
N=N
_
N N
1.10 A
1.24 A
1.47 A
Chemistry-140
Lecture 23
Bond Polarity
 Recall: for a covalent bond the bonding electrons are equally
shared between two atoms.
 Recall: for an ionic bond the bonding electrons are separated
between the ions (electrostatic attraction).
 When sharing is not equal, the bond is called a polar bond.
 Equal sharing is sometimes referred to as a nonpolar bond.
Chemistry-140
Lecture 23
Electronegativity
 Electronegativity: the ability of an atom in a molecule to
attract electrons to itself. The higher an element's
electronegativity, the better it competes for electrons.
 Electronegativity is related to ionization energy and
electron affinity. The scale (Pauling scale) has no units
Chemistry-140
Lecture 23
Electronegativity Values
Chemistry-140
Lecture 23
Polar vs. Nonpolar Bonds
 Electronegativity difference between two atoms of a bond is
related to the polarity of the bond. The greater the
electronegativity difference, the more polar the bond.
> 2.0 = ionic,
< 0.5 = nonpolar
between 0.5 and 2.0 = polar
Chemistry-140
Lecture 23
Bond Polarity
Examples:
F2
HF
LiF
4.0 - 4.0 = 0
4.0 - 2.1 = 1.9
4.0 - 1.0 = 3.0
nonpolar
polar
ionic
d+
d-
_
H F
d+ represents a partial positive charge
d- represents a partial negative charge
Chemistry-140
Lecture 23
Drawing Lewis Structures
 Sum the valence electrons from all atoms in the species.
 Write the atomic symbols for the atoms involved so as to
show which atoms are connected to which. Draw a single
bond between each pair of bonded atoms
 Complete the octets of the atoms bonded to the central
atom (i.e. the peripheral atoms)
Chemistry-140
Lecture 23
Drawing Lewis Structures
 Place leftover electrons on the central atom, even if it
results in the central atom having more than an octet
 If there are not enough electrons to give the central atom
an octet, form multiple bonds by pulling terminal electrons
from a peripheral atom and placing them into the bond
with the central atom
Chemistry-140
Lecture 23
Drawing Lewis Structures
Question:
Draw the Lewis structure for PCl3.
Chemistry-140
Lecture 23
Drawing Lewis Structures
Answer:
Step 1: Sum the valence electrons. P has 5 and each Cl
has 7 for a total of [5 + (3 x 7)] = 26 valence electrons
Chemistry-140
Lecture 23
Drawing Lewis Structures
Answer:
Step 2: Arrange atoms showing connectivity and draw a
single bond between atoms. NOTE: In a binary (two-
element) compound, the first element listed is usually the
central one with the others surrounding it
_
_ _
Cl P Cl
Cl
Chemistry-140
Lecture 23
Drawing Lewis Structures
Answer:
Step 3: Complete the octets on the atoms bonded to the
central atom. NOTE: This accounts for 24 of the 26
valence electrons
_
_ _
Cl P Cl
Cl
Chemistry-140
Lecture 23
Drawing Lewis Structures
Answer:
Step 4: Place the remaining electrons on the central atom to
complete the octet. Since this gives an octet to each atom
we are finished
_
_ _
Cl P Cl
Cl
Chemistry-140
Lecture 23
Drawing Lewis Structures
Question:
Draw the Lewis structure for HCN.
Chemistry-140
Lecture 23
Drawing Lewis Structures
Answer:
Step 1: Sum the valence electrons. H has 1, C has 4 and
N has 5 for a total of [1 + 4 + 5)] = 10 valence electrons
Chemistry-140
Lecture 23
Drawing Lewis Structures
Answer:
Step 2: Arrange atoms showing connectivity and draw a
single bond between atoms. NOTE: Since H can only
form one covalent bond it can never be the central atom.
The choices are HCN or HNC. Formula is written
HCN!!!
_ _
H C N
This accounts for 4 valence electrons
Chemistry-140
Lecture 23
Drawing Lewis Structures
Answer:
Step 3: Complete the octets on the atoms bonded to the
central atom.
_ _
H C N
BUT: There are only 6 valence electrons left.
If we put them on N we do not achieve an octet at C !!
Chemistry-140
Lecture 23
Drawing Lewis Structures
Answer:
Step 4: Try using multiple bonding to share the electrons
between C and N. A triple bond is required to give an
octet to each atom
_ =
H C =N
Chemistry-140
Lecture 24
Chapter 9:
Bonding & Molecular Structure:
Fundamental Concepts
 Chapter Highlights

bonding types

Lewis symbols & octets

ionic bonding & ionic lattices

covalent bonding & Lewis dot structures
 resonance structures
 breaking the octet rule & formal charge

bond order & bond energy

VSEPR molecular shape & polarity
Chemistry-140
Lecture 24
Formal Charge & Lewis Structures
 Formal charges: a way of assigning a relative charge to
each atom in the molecule
 When several different Lewis structures seem plausible, the
one in which the formal charges are minimized is generally
the preferred one.
Chemistry-140
Lecture 24
Assigning Formal Charge
 All bonding electrons are divided equally between the
atoms that form bonds
 All nonbonding electrons are assigned to the atom on which
they reside
Chemistry-140
Lecture 24
Assigning Formal Charge
 Formal charge: the number of valence electrons for the
element minus the number of electrons assigned by rules
1 and 2.
formal charge on an atom in a molecule =
{# valence electrons normally found for that atom [(# non-bonding electrons) + 1/2(# bonding electrons)]}
FC = VE - (NBE + 1/2BE)
Chemistry-140
Lecture 24
Applying Formal Charge to Lewis Structures
Question:
There are three possible structures for SCN-. Use formal
charge to decide the most likely structure.
Chemistry-140
Lecture 24
Drawing Lewis Structures
Answer:
Step 1: Sum the valence electrons. S has 6, C has 4 and N
has 5 and there is an extra electron represented by the
single negative charge of the ion.
Total of [6 + 4 + 5 + 1] = 16 valence electrons
Chemistry-140
Lecture 24
Drawing Lewis Structures
Answer:
Step 2: Arrange atoms showing connectivity and draw a
single bond between atoms.
_ _ [S C N]
_ _
[C S N]
_ _ [S N C]
Chemistry-140
Lecture 24
Drawing Lewis Structures
Answer:
Step 3: Complete the octets on the atoms bonded to the
central atom.
_ _ _
_
_
_
[ S C N ] [ C S N ] [ S N C ]BUT: Each of these leaves us with
only four electrons at the central atom!
Chemistry-140
Lecture 24
Drawing Lewis Structures
Answer:
Step 4: Use multiple bonding to share the electrons
between peripheral atoms and the central atom until
octets are achieved.
[S = C = N]-
[C = S = N]- [S = N = C]-
Chemistry-140
Lecture 24
Drawing Lewis Structures
Answer:
Step 5: Calculate formal charge for each atom.
[S = C = N]0
0
-1
[C = S = N]- [S = N = C]-2 +2 -1
0
+1
Note: the total formal charge on each
molecule is equal to the charge on the molecule
-2
Chemistry-140
Lecture 24
Drawing Lewis Structures
Answer:
Step 6: Decide on the most probable structure.
[S = C = N]0
0
-1
[C = S = N]- [S = N = C]-2 +2 -1
0
+1
The structure that results in the least amount of
formal charge separation throughout the molecule
-2
Chemistry-140
Lecture 24
Resonance Structures
 There are times when more than one Lewis structure
involving multiple bonds seems equally stable
_
O O =O
ozone
_
O=O O
 Resonance structures: structures that differ only in the
placement of electrons.
Chemistry-140
Lecture 24
Resonance Structures
 Resonance forms rapidly interconvert so that the structure
appears to be a blend of all the forms.
_
O O =O
_
O=O O
Chemistry-140
Lecture 24
Resonance Structures
 The molecule does not oscillate rapidly between two or
more different forms. There is only one form of the
molecule. Ozone has two equivalent O-O bonds whose
length is intermediate between single and double bonds
_
_
O O O
Chemistry-140
Lecture 24
Exceptions to the Octet Rule
 Most second-period elements (n = 2), notably C, N, O and F
are always observed with octets, but.…
 Molecules with an odd number of electrons.
 Molecules in which an atom has less than an octet.
 Molecules in which an atom has more than an octet.
Chemistry-140
Lecture 24
Odd Numbers of Electrons
 Some examples are ClO2, NO and NO2. They tend to be
very rare and very reactive. React to pair unpaired
electrons.
O
O
N
O
O
N
+
O
O
_
N
O
N
O
Chemistry-140
Lecture 24
Less Than an Octet
 Some lighter elements (H, Be, B) tend to be surrounded by
less than an octet of electrons.
 For example: Hydrogen has a valence-shell capacity of
only 2 electrons. Beryllium often is surrounded by four
electrons and boron is often surrounded by six electrons,
due mostly to their small size
Chemistry-140
Lecture 24
Less Than an Octet
 Formal charge assignments generally support these
deviations from the octet rule.
F
_
B F
F
Chemistry-140
Lecture 24
Consequences
For example, BF3 acts as a Lewis acid to form an adduct with
NH3 which is a Lewis base
B F
F
+
H
_
N H
H F
_
_
_
H N B F
H F
_
_
_
H
_
_
F
Chemistry-140
Lecture 24
More Than an Octet
 A larger group of compounds are those in which the central
atom is surrounded by more than an octet of electrons.
 Elements of the third period or lower are capable of
expanding their octets, due to the availability of d orbitals.
Chemistry-140
Lecture 24
More Than an Octet
 For example: After satisfying the octet rule for the
peripheral F-atoms in PF5, we have 10 valence electrons at
P-atom. The extra electrons can reside in a 3d-orbital.
_
F
F
_
P F
F F
_
Chemistry-140
Lecture 24
Lewis Structures
Question:
Draw the Lewis structure for IF4-.
Chemistry-140
Lecture 24
Drawing Lewis Structures
Answer:
Step 1: Sum the valence electrons. I has 7 and each F has
7 and we count 1 for the anion for a total of
[7 + (4 x 7)] +1 = 36 valence electrons
Chemistry-140
Lecture 24
Drawing Lewis Structures
Answer:
Step 2: Arrange atoms showing connectivity and draw a
single bond between atoms. Recall: In a binary (twoelement) compound, the first element listed and the
heavier is usually the central one with the others
surrounding it.
_
F
_
_
F I F
_
F
Chemistry-140
Lecture 24
Drawing Lewis Structures
Answer:
Step 3: Complete the octets on the atoms bonded to the
central atom. NOTE: This accounts for 32 of the 36
valence electrons
_
F
_
_
F I F
_
F
Chemistry-140
Lecture 24
Drawing Lewis Structures
Answer:
Step 4: We must place the remaining electrons on the
central atom even though this gives the I-atom more than
an octet.
_
F
_
_
F I F
_
F
Remember: No double bonds to F
Chemistry-140
Lecture 24
Term Test #2
Friday November 9th, 2001
6:30 P.M.
Chemistry-140
Lecture 24
Duration: 75 minutes
Contents: SIX “Problems”!!
Covers Material From Chapters 6-9
A Periodic Table & ALL Required
Constants will be Supplied
Chemistry-140
Lecture 25
Chapter 9:
Bonding & Molecular Structure:
Fundamental Concepts
 Chapter Highlights

bonding types

Lewis symbols & octets

ionic bonding & ionic lattices

covalent bonding & Lewis dot structures

resonance structures

breaking the octet rule & formal charge
 bond order & bond energy
 VSEPR molecular shape & polarity
Chemistry-140
Lecture 25
Molecular Shape
 The geometry of a molecule, along with its size determines
in large part its chemical behavior
_
Cl
_
_
Cl C Cl
_
Cl
Chemistry-140
Lecture 25
VSEPR Model
 The geometry of molecules, can be predicted using VSEPR
rules (valence-shell electron-pair repulsion)
(R. J. Gillespie & R. S. Nyholm)
 The best arrangement of electron pairs is the one that
maximizes the separation among them, since this
minimizes the repulsions among the electron pairs..
Chemistry-140
Lecture 25
Electron-pairs
 Electron pairs are differentiated as being bonding electron
pairs or nonbonding electron pairs (lone pairs).
 (For the purposes of VSEPR, a multiple bond counts as a
single bonding pair or as a single region of electrons.)
1 lone pair
_
_
_
H N H
H
3 bonding pairs
Chemistry-140
Lecture 25
Electron-pairs
 The electron-pair geometry is the geometry described by the
regions of electrons around the central atom.
_
_
_
H N H
NH3 (ammonia) has FOUR
electron pairs in total
H
tetrahedral geometry
Chemistry-140
Lecture 25
Electron-pairs
 The molecular geometry is the geometry described by the
atoms only!! (central atom and the peripheral atoms)
_
_
_
H N H
NH3 (ammonia) has THREE
bonding electron pairs & ONE lone pair
H
Triangular
pyramidal
N
Chemistry-140
Lecture 25
Building VSEPR Models
 Sketch the Lewis dot structure of the molecule or ion
 Count the total number of electron pairs around the central
atom (multiple bonds only count as one pair) and arrange
them in the way that minimizes electron-pair repulsions
 Describe the molecular geometry in terms of the angular
arrangement of bonding-pairs
Chemistry-140
Lecture 25
Electron-Pair Geometries (2)
AX2 Linear
180 o
Chemistry-140
Lecture 25
Electron-Pair Geometries (3)
AX3 Trigonal Planar
120 o
Chemistry-140
Lecture 25
Electron-Pair Geometries (4)
AX4 Tetrahedral
109.5 o
Chemistry-140
Lecture 25
Electron-Pair Geometries (5)
AX5 Trigonal
Bipyramidal
90 o
120 o
Equatorial
Axial
Chemistry-140
Lecture 25
Electron-Pair Geometries (6)
AX6 Octahedral
90 o
Chemistry-140
Lecture 25
Deviations From Electron-Pair Geometries
Chemistry-140
Lecture 25
Electron-Pair Repulsions
lone-pair//lone-pair
>
lone-pair//bond-pair
>
bond-pair//bond-pair
 Nonbonding electron pairs repel other electron pairs more
than do single-bonding electron pairs.
 Electrons in multiple bonds repel other electron pairs
more than do single-bonding electron pairs.
Chemistry-140
Lecture 25
Molecular Shapes
No lone pairs
Trigonal
Bipyramidal
Chemistry-140
Lecture 25
Molecular Shapes
One lone pair
Seesaw
Chemistry-140
Lecture 25
Molecular Shapes
Two lone pairs
T-Shaped
Chemistry-140
Lecture 25
Molecular Shapes
Three lone pairs
Linear
Chemistry-140
Lecture 25
VSEPR Models
Question:
Using the VSEPR model, predict the molecular geometry
of SnCl3-.
Chemistry-140
Lecture 25
VSEPR Models
Answer:
Step 1: The Lewis structure for the SnCl3- anion is:
_
Cl
_
_
Cl Sn
Cl
Chemistry-140
Lecture 25
VSEPR Models
Answer:
Step 2: The central atom is surrounded by one
nonbonding electron pair and three single bonds.
Therefore, FOUR electron pairs; tetrahedral electronpair geometry.
Cl
Cl
Sn
Cl
The molecular geometry is trigonal pyramidal
Chemistry-140
Lecture 25
VSEPR Models
Question:
Using the VSEPR model, predict the molecular geometry
of SF4.
Chemistry-140
Lecture 25
VSEPR Models
Answer:
Step 1: The Lewis structure for the SF4 molecule is:.
_
F
_
_
_
F S
F
F
Chemistry-140
Lecture 25
VSEPR Models
Answer:
Step 2: The central atom is surrounded by one
nonbonding electron pair and four single bonds.
Therefore, FIVE electron pairs; trigonal bipyramidal
electron -pair geometry, but there are two possiblities
F
F
F
S
F
F
F
S
F
F
Chemistry-140
Lecture 25
VSEPR Models
Answer:
Step 2: The central atom is surrounded by one
nonbonding electron pair and four single bonds.
Therefore, FIVE electron pairs; trigonal bipyramidal
electron -pair geometry, only one possibility is correct!
F
F
F
S
F
F
F
S
F
F
Chemistry-140
Lecture 25
Molecular Polarity
Chemistry-140
Lecture 25
Molecular Polarity
A polar molecule is one in which the centers of positive
and negative charge do not coincide.
CCl4
CHCl3
Chemistry-140
Lecture 25
Textbook Questions From Chapter # 9
Valence electrons, octet rule:
28, 32
Lewis structures:
38, 40
Bond properties:
46, 48, 50, 56
Bond polarity & formal charge:
62, 66
Molecular geometry:
76, 80, 82, 86, 96
Chemistry-140
Lecture 25
Term Test #2
Friday November 9th, 2001
6:30 P.M.
Chemistry-140
Lecture 25
Duration: 75 minutes
Contents: SIX “Problems”!!
Covers Material From Chapters 6-9
A Periodic Table & ALL Required
Constants will be Supplied