Chemical Bonding I:
Basic Concepts
Chapter 9
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Gilbert Lewis Atoms combine in order to achieve a more
stable electron configuration
=> Maximum stability is achieved when an
atom is isoelectric with a noble gas
When atoms interact to form a chemical bond,
only the outer regions are in contact
=> Valence electrons
Lewis dot symbols - symbol of element and
one dot for each valence electron
Valence electrons are the outer shell electrons of an
atom. The valence electrons are the electrons that
particpate in chemical bonding.
Group
e- configuration
# of valence e-
1A
ns1
1
2A
ns2
2
3A
ns2np1
3
4A
ns2np2
4
5A
ns2np3
5
6A
ns2np4
6
7A
ns2np5
7
9.1
Focus on main group elements
(representative elements)
(Groups 1A => 8A)
The Ionic Bond
1s2 => [He]
1s22s1
Li + F
1s22s22p5
[Li+] [ F -]
1s22s22p6 => [Ne]
Li
[Li+] + e-
e- +
F
[ F -]
Li+ +
F -
[Li+][ F -]
An ionic bond is the
electrostatic force that holds
together an ionic compound
=> large crystals, not individual
molecules
Ionic bonds form when a metal bonds
to a non-metal.
2 Ca (s) + O2 (g)
2 CaO (s)
Split into oxygen atoms
Ca + O
[Ca2+][
[Ar]
[Ar]4s2
O 2- ]
[Ne]
1s22s22p4
Transfer 2 electrons from Ca to O
Sometimes more than 2 atoms are involved
4 Li (s) + O2 (g)
1s22s1
2 Li + O
1s22s22p4
2 Li2O (s)
[He]
2 [Li+] [ O 2- ]
[Ne]
3 Mg (s) + 2 N2 (g)
Mg3N2 (s)
6 electrons
3
Mg + 2 N
3 [Mg2+] 2 [ N 3-]
transferred
[Ne]3s2
[Ne]
[Ne]
2
2
3
1s 2s 2p
We can predict which compounds will be ionic based on
ionization energies and electron affinities.
BUT
these are measured in the gas phase and ionic compounds
are generally solids => very different environment.
Electrostatic (Lattice) Energy
Lattice energy (E) is the energy required to completely separate
one mole of a solid ionic compound into gaseous ions.
Q+Q E=k
r
Coulomb’s
Law
Q+ is the charge on the cation
Q- is the charge on the anion
r is the distance between the ions
cmpd
Lattice energy (E) increases MgF2
MgO
as Q increases and/or
as r decreases.
Multiple
charge
Small ions
LiF
LiCl
lattice energy
2957 Q= +2,-1
3938 Q= +2,-2
More charge
1036
r F < r Cl
853
Shorter distance
Born-Haber Cycle for Determining Lattice Energy
Form ions
Think of this as a
variation of a
conservation of energy
problem.
Form crystalline
lattice
Form atoms in gas phase
o
DHoverall
= DHo1 + DHo2 + DHo3 + DHo4 + DHo5
Hess’ Law
In a series, as
lattice energy
increases, so
does the
melting point.
Lattice Energy is always positive!!
A covalent bond is a chemical bond in which two or more
electrons are shared by two atoms.
Covalent compounds contain only covalent bonds.
Why should two atoms share electrons?
F
+
7e-
F
F F
7e-
8e- 8e-
Share 2
electrons
Lewis structure of F2
single covalent bond
lone pairs
F
F
lone pairs
single covalent bond
lone pairs
F F
lone pairs
Valence electrons that are NOT
involved in bonding
(non-bonding)
Interactions between molecules in covalent
compounds are weaker than bonds in the molecule
and in ionic compounds.
Lewis structure of water
H
+
O +
H
single covalent bonds
H O H
or
H
O
H
2e- 2e
8eShare 2 e - Share 2 e Double bond – two atoms share two pairs of electrons
O C O
8e
8e
8e-
or
O
O
C
double bonds
Share 4 e Share 4 e -
double bonds
Triple bond – two atoms share three pairs of electrons
N N
8e- 8e
triple bond
or
N
N
triple bond
Share 6 e -
Lengths of Covalent Bonds
Bond
Type
Bond
Length
(pm)
C-C
154
CC
133
CC
120
C-N
143
CN
138
CN
116
Bond Lengths
Triple bond < Double Bond < Single Bond
Polar covalent bond or polar bond is a covalent
bond with greater electron density around one of the
two atoms
electron poor
region
H
electron rich
region
F
e- poor
H
d+
e- rich
F
d-
Electronegativity is the ability of an atom to attract
toward itself the electrons in a chemical bond.
Electron Affinity - measurable, Cl is highest
X (g) + e-
X-(g)
Gas phase
Electronegativity - relative, F is highest
Electronegativity is a relative measure of the
attraction an atom has for the shared electrons
in a bond
=> numerical scale
=> the higher the number, the stronger the
attraction
0.7
Willing to give
up electrons
1.0
4.0
Wants to fill its
orbitals
=> not measured directly, but calculated
using bond energies, atom size, ……..
9.5
9.5
Non-metal atom
electronegativities
Easier to fill
valence shell
e- +
F
[ F -]
>
Metal atom
electronegativities
Easier to empty
valence shell
Fr
[Fr+] + e-
F is the highest electronegativity because atom is
very small
=> valence shell is close to the nucleus
=> needs 1 electron to fill orbital
Fr is the lowest electronegativity because atom is
very large
=> valence shell does not have much interaction
with the nucleus
Classification of bonds by difference in electronegativity
Difference
Bond Type
0
Covalent
2
0 < and <2
Ionic
Polar Covalent
Increasing difference in electronegativity
Covalent
Polar Covalent
Ionic
share e-
partial transfer of e-
transfer eExample: FrF
F = 4.0 Fr = 0.7
Δ = 3.3
Example: CS2
S=C=S
C = 2.5 S = 2.5
Δ = 0.0
Classify the following bonds as ionic, polar covalent,
or covalent: The bond in CsCl; the bond in H2S; and
the NN and NH bonds in H2NNH2.
Cs – 0.7
Cl – 3.0
3.0 – 0.7 = 2.3
Ionic
H
H – 2.1
S – 2.5
2.5 – 2.1 = 0.4
S
Polar Covalent
N – 3.0
N – 3.0
3.0 – 3.0 = 0
Covalent
N – 3.0
H – 2.1
3.0 – 2.1 = 0.9
H
Polar Covalent
H
H
N=N
H
H
Bonds involving 2 atoms
of the same element
MUST be covalent.
Writing Lewis Structures
1. Draw skeletal structure of compound showing
what atoms are bonded to each other. Put least
electronegative element in the center. Element most
willing to share.
2. Count total number of valence e-. Add 1 for
each negative charge. Subtract 1 for each
positive charge.
3. Complete an octet for all atoms except
hydrogen Start with atoms bonded to the central atom.
Add lone pairs to central atom.
4. If structure contains too many electrons, form
double and triple bonds on central atom as
needed.
Write the Lewis structure of nitrogen trifluoride (NF3).
Step 1 – N is less electronegative than F, put N in center
Step 2 – Count valence electrons N - 5 (2s22p3) and F - 7 (2s22p5)
5 + (3 x 7) = 26 valence electrons
Step 3 – Draw single bonds between N and F atoms and complete
octets on N and F atoms.
Step 4 - Check, are # of e- in structure equal to number of valence e- ?
3 single bonds (3x2) + 10 lone pairs (10x2) = 26 valence electrons
6 electrons
F
N
F
F
20 electrons
N
F
F
F
Write the Lewis structure of the carbonate ion (CO32-).
Step 1 – C is less electronegative than O, put C in center
Step 2 – Count valence electrons C - 4 (2s22p2) and O - 6 (2s22p4)
-2 charge – 2e4 + (3 x 6) + 2 = 24 valence electrons
Step 3 – Draw single bonds between C and O atoms and complete
octet on C and O atoms.
Step 4 - Check, are # of e- in structure equal to number of valence e- ?
3 single bonds (3x2) + 10 lone pairs (10x2) = 26 valence electrons
Step 5 - Too many electrons, form double bond and re-check # of e-
-
Double
Bond
O
C
-2 Charge O
-
O
2 single bonds (2x2) = 4
1 double bond = 4
8 lone pairs (8x2) = 16
Total = 24
Two possible skeletal structures of formaldehyde (CH2O)
H
C
O
H
H
C
H
O
An atom’s formal charge is the difference between the
number of valence electrons in an isolated atom and the
number of electrons assigned to that atom in a Lewis
structure.
Shared between 2 atoms
formal charge
on an atom in
a Lewis
structure
=
total number
total number
of valence
of nonbonding
electrons in electrons
the free atom
-
1
2
(
total number
of bonding
electrons
The sum of the formal charges of the atoms in a molecule
or ion must equal the charge on the molecule or ion.
Extra bond => + formal charge
Remove bond => - formal charge
)
H
-1
+1
C
O
formal charge
on an atom in
a Lewis
structure
H
=
C – 4 eO – 6 e2H – 2x1 e12 e-
2 single bonds (2x2) = 4
1 double bond = 4
2 lone pairs (2x2) = 4
Total = 12
total number
total number
of valence
of nonbonding
electrons in electrons
the free atom
-
1
2
(
total number
of bonding
electrons
)
formal charge
= 4 -2 -½ x 6 = -1
on C
formal charge
= 6 -2 -½ x 6 = +1
on O
9.7
H
H
0
C
formal charge
on an atom in
a Lewis
structure
0
O
=
C – 4 eO – 6 e2H – 2x1 e12 e-
2 single bonds (2x2) = 4
1 double bond = 4
2 lone pairs (2x2) = 4
Total = 12
total number
total number
of valence
of nonbonding
electrons in electrons
the free atom
-
1
2
(
total number
of bonding
electrons
)
formal charge
= 4 - 0 -½ x 8 = 0
on C
formal charge
= 6 -4 -½ x 4 = 0
on O
9.7
Formal Charge and Lewis Structures
1. For neutral molecules, a Lewis structure in which there
are no formal charges is preferable to one in which
formal charges are present.
2. Lewis structures with large formal charges are less
plausible than those with small formal charges.
3. Among Lewis structures having similar distributions of
formal charges, the most plausible structure is the one in
which negative formal charges are placed on the more
electronegative atoms.
Which is the most likely Lewis structure for CH2O?
H
-1
+1
C
O
H
H
H
0
C
0
O
9.7
What are the formal charges on the atoms of the
carbonate ion (CO32-) ?
-
Double
Bond
O
C
O
-2 Charge O
O1 6 – 4 -1/2 (4) = 0
O1
O2 6 – 6 -1/2 (2) = -1
O3
-
O2
-
O3 6 – 6 -1/2 (2) = -1
C 4 – 0 -1/2 (8) = 0
Total of formal charges = 0 -1 -1 0 = -2 = ion charge
A resonance structure is one of two or more Lewis structures
for a single molecule that cannot be represented accurately by
only one Lewis structure.
Use double
headed arrow
+
+
O
O
O
O
O
O
= Equilibrium
What are the resonance structures of the
carbonate (CO32-) ion?
-
O
C
O
-
O
C
O
-
-
O
C
O
O
O
O
In reality, all bonds are determined to be equivalent.
Resonance structures – neither structure is correct
Animal analogy – Rhinoceros looks like a cross
between two mythical animals (griffin & unicorn)
=> it does not oscillate between the two forms
Benzene – C6H6
H
H
H
H
C
C
C
C
C
C
H
H
H
H
H
Kekulé structures
C
C
C
C
C
C
H
H
H
In benzene, all C – C bonds lengths are 140 pm
Normal C – C bond length is 154 pm
C = C bond length is 133 pm
and
Ring structure
Exceptions to the Octet Rule
The Incomplete Octet
BeH2
Be, B & Al are the most
common examples
Be – 2e2H – 2x1e4e-
H
Be
H
Cannot complete an octet
with only 4 electrons
BF3
B – 3e3F – 3x7e24e-
F
B
F
3 single bonds (3x2) = 6
9 lone pairs (9x2) = 18
Total = 24
F
9.9
Exceptions to the Octet Rule
Odd-Electron Molecules
NO
N – 5eO – 6e11e-
N
O
The Expanded Octet (central atom with principal quantum number n > 2)
Third row and below
SF6
S – 6e6F – 42e48e-
F
F
F
S
F
F
F
6 single bonds (6x2) = 12
18 lone pairs (18x2) = 36
Total = 48
9.9
Expanded Octet Rule Exceptions
 Larger atoms can have involvement with
unfilled d orbitals
 The larger the central atom, the more atoms
can surround it (especially if surrounding atoms
are small (ex: F)
What is we have a choice between an expanded
octet with fewer formal charges and one following
the octet rule with more formal charges?
For 3rd row and beyond, the preference is for the
structure that obeys the octet rule.
-
-
SO42-
O
O
-
O
S
O
O
-
Satisfies octet rule
Most likely structure
- OR -
O
S
O
O
-
Expanded octet
The enthalpy change required to break a particular bond in
one mole of gaseous molecules is the bond energy.
Bond Energy
DH0 = 436.4 kJ
H2 (g)
H (g) + H (g)
H
H
Cl2 (g)
Cl (g) + Cl (g) DH0 = 242.7 kJ Cl
Cl
HCl (g)
H (g) + Cl (g) DH0 = 431.9 kJ Cl
H
O2 (g)
O (g) + O (g) DH0 = 498.7 kJ O
O Double bond
N2 (g)
N (g) + N (g) DH0 = 941.4 kJ
N Triple bond
N
Bond Energies
Single bond < Double bond < Triple bond
Single
bonds
Average bond energy in polyatomic molecules
H2O (g)
OH (g)
H (g) + OH (g) DH0 = 502 kJ
H (g) + O (g)
DH0 = 427 kJ
502 + 427
= 464 kJ
Average OH bond energy =
2
Compare
strength of
single and
multiple
bonds.
Bond Energies (BE) and Enthalpy changes in reactions
Imagine reaction proceeding by breaking all bonds in the
reactants and then using the gaseous atoms to form all the
bonds in the products.
DH0 = total energy input – total energy released
= SBE(reactants) – SBE(products)
Note: We
are not
simply
taking
products reactants.
Use bond energies to calculate the enthalpy change for:
H2 (g) + F2 (g)
2HF (g)
H
H +F
F
H
F + H
F
DH0 = SBE(reactants) – SBE(products)
Type of
bonds broken
H
H
F
F
Type of
bonds formed
H
F
Number of
bonds broken
Bond energy
(kJ/mol)
Energy
change (kJ)
1
1
436.4
156.9
436.4
156.9
Number of
bonds formed
Bond energy
(kJ/mol)
Energy
change (kJ)
2
568.2
1136.4
DH0 = 436.4 + 156.9 – 2 x 568.2 = -543.1 kJ
H2 (g) + Cl2 (g)
2HCl (g)
DH0 = SBE(reactants) – SBE(products)
Type of
bonds broken
H
H
Cl
Cl
Type of
bonds formed
H
Cl
Number of
bonds broken
Bond energy
(kJ/mol)
1
1
436.4
242.7
Number of
bonds formed
Bond energy
(kJ/mol)
2
431.9
DH0 = 436.4 + 242.7 – 2 x 431.9 =
= -184.7 kJ
2H2 (g) + O2 (g)
2H2O (g)
DH0 = SBE(reactants) – SBE(products)
Type of
bonds broken
H
O
H
O
Type of
bonds formed
H
O
Number of
bonds broken
Bond energy
(kJ/mol)
2
1
436.4
498.7
Number of
bonds formed
Bond energy
(kJ/mol)
4
464.5
DH0 = (2 x 436.4) + 498.7 – (4 x 464.5) =
= -486.5 kJ