Chapter 8 Chemical Bonding Basic Concepts

When atoms interact to form chemical bonds, only the outer VALENCE electrons take part.
Therefore, we need a tool for keeping track of valence electrons, e.g., the Lewis Dot symbol.
e.g.,
Na
+
7 v.e’s
l v.e.

When these two elements combine to form a compound
2 Na (s) + Cl2 (g)  NaCl (s)
e.g.
Na
+
[Ne]3s1

Cl
Cl
 NaCl (s)
[Ne]3s23p7
What’s happening?
Na (g)  Na+ (g) + e- (ionizes, loses e-) an electron configuration of [Ne]
Cl

(g) + e –  Cl – (g)
an electron configuration of [Ar]
In the crystal lattice, we have Na+ and Cl- ions; strong electrostatic attractions between the
positive ions and negative cations keep the NaCl unit together.
ionic bonding - electrostatic attractions that hold ions together in an ionic compound.
e.g. remember Eionic = k q1q2/r

The strength of interaction depends on charge magnitude and distance between them.

Other examples of ionic compounds
Na+Br-, Li+Br-, NH4Cl- etc.

we also have
2 Na  +

O
 Na2O (s)
i.e. each Na atom gives its electron to the O central atom  note again stable noble gas
configurations
Lattice Energies
 The stability of ionic compounds depends on two main factors
1. The electron affinity of one of the elements;
2. The ionization energy of the other
2

Note that both the electron affinities and ionization potentials are gas-phase reactions. How are
they related to the stability of solid materials?

Lattice energy (latH) - the energy needed to separate one mole of a solid ionic compound into its
gaseous ions, e.g., for NaCl (s)  Na+ (g) + Cl- (g) latH lattice energy.

Determining lattice energies  we can’t measure directly the lattice energy of a compound.
Therefore, we must use the Born-Haber Cycle  this is a very good example of the utility of
Hess’s Law to estimate a thermodynamic property that can’t be measured easily.

An example of the Born-Haber Cycle. Determination of the lattice energy of KCl (s)
K (s) + ½ Cl2 (g)  KCl (s)
fH = -436 kJ/mol

Note that KCl (s) is a very stable compound; this stability results primarily from the lattice
energy.

We need to set up a Hess’s Law process for KCl (s)
K (s)  K (g)
K (g)  K+ (g) + e½ Cl2 (g)  C l (g)
Cl (g) + e-  Cl- (g)

H = Hsub = 90 kJ
H = I1 = 419 kJ
H = ½ (bond dissociation energy of Cl) = 122 kJ
H = EA1 = -349 kJ
Add up the reactions
K (s) + ½ Cl2 (g)  K+ (g) + Cl- (g)

but we know
KCl (s)  K (s) + ½ Cl2 (g)
 total reaction KCl (s)  K+ (g) + Cl- (g)

H = 282 kJ
H = -fH = 436 kJ/mole
H = 718 kJ/mol KCl
Note: this means we have to expend 718 kJ/mol of KCl (s) to break apart the ions in the crystal
lattice. This is a very large amount of energy. Hence, the greater magnitude of the lattice
energy, the greater the stability of the ionic compound.
Example
 Note: Lattice energy of MgCl2, MgO are very large
MgCl2  2326 kJ/mol why  doubly charged cation; increased energy of attraction
MgO  3795 kJ/mol  Coulombic interactions are great.
3
Lattice energies vs. formulae for ionic compounds.
 A good example; NaCl2 would be an unstable ionic compound. Assuming that the lattice energy
of NaCl2 is similar to that of MgCl2 (i.e., near 2300 kJ/mol), there is not enough energy in the
crystal lattice to compensate for the very high I1 + I2 values of Na (I1 + I2 for Na+ = 5056 kJ/mol).
Covalent Bonding

Covalent bonds - a bond in which the electrons are shared by two atoms.
e.g.
H2  H-H, F2  F- F, Cl2  Cl-Cl

For many electron atoms (like F and Cl), we again to worry only about the outermost (valence)
electrons.

Let’s look at the Cl2 example.

Each Cl has 3 lone pairs

What about the Cl2 molecule?
lone pairs (non bonding)
Lone pair
Cl
Cl
bonding electrons

The structures we have just drawn are called Lewis structures.

let’s redraw F2
the dash here represents the bonding electrons
Note both Cl2 and F2 satisfy the filled valence shell rule (otherwise known as the ‘octet rule’ for
lighter elements).

What about O2? How can we satisfy the octet rule for 2 O atoms?

By sharing two e-s between each
O =O
O
atom.



4
The oxygen atoms in the O2 molecule satisfy the filled valence shell rule by the formation of a
double bond.
Filled Valence Shell rule  Atoms participate in the formation of bonds (either ionic or
covalent) in order to satisfy their valence shell requirements.
Note that for the first two rows in the periodic table, atoms other than H tend to form bonds until
they end up being surrounded by 8 valence electrons (the noble gas configuration). This is
known as the octet rule.
check out N2 
:NN: (triple bond)

Note that the octet rule works mainly for the second row elements. Filled valence shells
can have more than 8 electrons after Z=14 (Si). This is generally termed octet expansion.

for hydrocarbons
CH4

H
C
H
H
H
Compounds that contain only covalent bonds are called covalent compounds. There are two
main types of covalent compounds, i.e., molecular covalent compounds (CO2, C2H4) and
network covalent compounds (SiO2, BeCl2). The network covalent compound are characterized
by an extensive “3-D” network bonding
Comparison between Ionic and Covalent Compounds
Ionic Compounds
Covalent Compounds
usually solids with very high melting points
usually low melting solids, gases or liquids
conduct electricity when molten (melted)
don’t conduct electricity when molten
usually quite water soluble and they are aren’t very soluble in water and are non
electrolytes in aqueous solution
electrolytes
e.g. NaCl
CCl4
Electronegativity
 Electronegativity is defined as the ability of an atom to attract electrons towards itself in a
molecule. We usually denote the electronegativity using the symbol  (pronounced ‘chi’)

As an example, if we examine the H-F molecule, experimental evidence suggests that in the H-F
covalent bond, there is an unequal sharing of the bonding electrons.

H-F +
5


The symbolism  denotes a partial “+” charge on the hydrogen atom, while the  denotes a
partial “-“ charge on F atom.

Why does this happen? The ability of the atom to attract electron density (the electronegativity),
is related to the electron affinity and the ionization energy. It is found that atoms having with a
high E. A. and a high I1 (significant ability to attract an electron and very little tendency to lose
an electron) have high  values.

Compare the following elements.
Na
Low I1
High E.A.
low 

-
F
High I1
Low E.A.
high 
The electronegativity scale was devised by Linus Pauling based on available ionization energies
and electron affinities. This is another example of the use of Hess’s law to estimate a
thermodynamic property that we can’t directly measure.
Trends in the  Values
1. Across a row  The  values generally increase as we proceed from left to right in the periodic
table. This is a reflection of both the decreasing electron affinity and the increasing ionization
potential as we move across the row.
2. Down a group  The  values generally decrease as we descend the group. This is due to the
increase in the electron affinity and the decrease in the I1 values as we move from top to bottom in
the group.
3. Transition metals  Essentially constant  values

Can we use the electronegativity values to help us deduce the type of bonding in compounds?
Note: as a rough guide
 values
0.0 <  < 0.5
0.5    1.9
2.0    3.3

bond type
non-polar covalent
polar covalent
Ionic bond
Look at electronegativity differences ( values) in the following compounds.
Three different compounds with
NaCl Na,  = 0.9; Cl,  = 3.0
 = 3.0 - 0.9 = 2.1  ionic bond.
(note: no units!)
6
Cl2
Cl,  = 3.0
H Cl H,  = 2.1

 = 3.0-3.0 = 0.0 non-polar covalent bond
 = 3.0 –2.1 = 0.9  polar covalent
Note   values between 2.0 - 3.3 do not mean that the electron is completely transferred.
Even CsF (highest electronegativity difference) has some covalent character (a partial sharing of
the electron does occur between the two atoms).
Writing Lewis Structures
7
An Outline for Drawing Lewis Structures
1. Predict arrangement of atoms (i.e., predict the skeletal
arrangement of the molecule or ion).
 The H always is always a terminal atom, bonded to ONE
OTHER ATOM ONLY. A halogen atom is usually a
terminal atom. Note that the central atom usually has the
least negative electron affinity.
2. Count total number of valence shell electrons (include
ionic charges).
3. Place 1 pair electrons (sigma bond, ) between each pair
of bonded atoms (i.e., the central atom and each one of
the terminal atoms).
4. Place remaining electrons around the terminal atoms to
satisfy the filled valence shell rule. (lone pairs).
5. All remaining electrons are assigned to the central atom.
Atoms in the 3rd or higher row can have more than eight
electrons around them.
 If a central atom does not have a filled valence shell (i.e.,
surrounded by 8 electrons for elements that obey the octet
rule), use a lone pair of electrons from a terminal atom to
make a pi () bond.
8
Example #1
NH3
H
Step 1:
skeletal structure
N
H
Step 2:
H
Count the total number of electrons
Number of electrons = 1 x 5 + 3 x 1 = 8 electrons
(note: an N atom has 5 valance electrons since it is in group 5, an O atom would
have 6 valence electrons, a Cl atom would have 7 valence electrons, etc.)
H
Step 3:
N
H
Draw a single covalent bond between the N central atom
and the H terminal atoms.
H
Note: accounts for 8 - (3 x 2) = 2. All but 2e-‘s.
Step 4. These remaining electrons cannot be placed on the H terminal atoms (they already have
a filled valence shell).
Step 5. The remaining must be placed on N to satisfy the filled valence shell rule for the N
atom.
H
H
N
H
Example #2
NO3Step 1.
Skeletal structure
O
O
N
O
Step 2.
Total number of electrons = 1 x 5 + 3 x 6 +1 = 24 (note: as the nitrate anion has
a –1 charge, we must add in that electron).
Step 3.
Draw the single bonds between the N and O atoms.


O
O
N
O
9



Number of electrons remaining = 24- (3x2) = 18 electrons
Step 4.
Place the remaining electrons on the oxygen terminal atoms.
O
O
N
O
10
Note: as there are only 6 electrons surrounding the N atom, we must apply Step 6, i.e. we
will move a lone pair from one of the oxygen atoms to make a double bond between a
nitrogen atoms and a terminal oxygen atoms.
O
N
O
O
Formal Charges
 Definition: formal charge on atom = number of valences electrons – number of non-bonding - ½
the number of bonding electrons. The formal charge in a Lewis Structure is a bookkeeping
“device” that helps us keep track of the electrons “associated” with certain atoms in a molecule
vs. the valence e-‘s in the isolated atom! How does it work?
Example #1
H
H
NH3

N
H
Lewis Structure
formal charge on N = 5 - 2 - ½ (6) = O
(start with central atom)
formal charge on H, f.c.(H) = 1 - 0 - ½ (2) = 0

Compare with NH4+
H
H
N
H
H
f.c.(N) = 5 - 0 - ½ (8) = +1
f.c.(H) = 1 - 0 - ½ (2) = 0
a “1+” charge is assigned to N in NH4+
11






Another example CO32Ob
Oa
assign formal charges
on the C central atom f.c.(C) = 4 - 0 - ½ (8) = 0
for the terminal O atoms
f.c.(Oa) = 6 - 4 - ½ (4) = 0
f.c. (Ob) = 6 - 6 - ½ (2) = -1
2O
O
C
O
Rules for formal charges
1. Neutral molecules   formal charges = 0 (the NH3 example)
2. Anions; cations   formal charges = charge of ion (e.g. CO32-, NH4+, NO3-)
3. For molecules where the possibility of multiple Lewis Structures with different formal charges
exist
 Neutral molecule - choose the structure with the fewest formal charges.
 Structures with large formal charges are less likely than ones with small formal charges
(i.e., +2, +3 less likely than +1, -1)
 Two Lewis Structures with similar formal charge distribution  negative formal charges on
more electronegative atom
Resonance Structures
 Examine the NO3- anion.
f.c.(N) = 5 - 0 - ½ (8) = +1
f.c.(Oa) = 6 - 4 - ½ (4) = 0
 f.c.(Ob) = 6 - 6 - ½ (2) = -1
12

How do these structures differ? They differ in the location of the N=O double bond. They are
said to be resonance structures. The actual structure of the molecule is a combination of all three
resonance structures (the resonance hybrid).
O
O
O
O
N
O
O
N
O
O
O
O
N
O
O
O
O
O

Another example e.g., ozone

The actual structure is in between these two.

Experimental evidence for resonance.

The resonance hybrid for benzene is diagrammed below.

Experimentally, we would expect the C=C and C-C bonds in benzene to be different in bond
length, if the actual structure of benzene is that of one of the resonance structures only (i.e., the
average distance between the nuclei of two bonded atoms in a molecule).
C=C 
C-C 
Look at an example, benzene C6H6
bond length = 133 pm = 0.133 nm
bond length = 0.154 nm
13

However, for the benzene carbon-carbon bond length is 0.140 nm  halfway between the C=C
and C-C bonds.
Exceptions to the Octet Rule
 e.g. Be
1s2 2s2
BeH2

Group 3A elements boron and Al compounds
BF3, AlCl3, BCl3.
F
B
F

Cl
F
Cl
B
Cl
Cl
Cl
Al
Cl
BF3 is stable  B has a tendency to pick up an unshared e- pair from another compound
e.g. BF3 + NH3  BF3NH3 the B-N bond is an example of a coordinate covalent bond, or a
“dative” bond  i.e. a bond in which one of the atoms donates both bonding electrons.
F
F
Dative or coordinate
covalent bond
B
F
N
F
F
F
Odd e- molecules
 These molecules have uneven numbers of electrons  no way that they can form octets.
Examples
N
O
O
and
N
O
Valence Shells having more than 8 Electrons (Expanded Octets) - atoms having more than 8 valance
shell electrons is possible with atomic number 14 and above.
Cl
Example
PCl5
Cl
Cl
P
Cl
Cl
14

Reason - elements in this category can use the energetically low-lying d orbitals to accommodate
extra electrons (for the P atom and the Cl atom in the above examples, the 3d orbitals).

Look at HClO3
O


H O Cl O
Due to the high formal charge on the electronegative
2 Cl atom (f.c.(Cl) = 7-2-1/2 (6) = +2), this

resonance structure would make only a very small contribution to the overall resonance hybrid.
Note that with the possibility of using the low lying d-orbitals on the Cl atom to accommodate
extra electron pairs, we may write other Lewis structures.
O
O
H O Cl O

H O Cl O
O
H O Cl O
Note: these structures reduce the formal charges.
Bond energies and Thermochemistry
 Look at the energy required to break 1 mole of gaseous diatomic molecular into their constituent
gaseous atoms.
H2 (g)  H (g) + H (g)
H = 436.4 kJ
Cl2 (g)  Cl (g) + Cl (g)
H = 242 kJ

The enthalpy changes are called bond dissociation energies. In the above examples, the enthgalpy
changes are designated D (H-H) an D (Cl-Cl). We can also apply this to molecules.
e.g.
CO (g)  C (g) + O (g)
H = 745 kJ

Denote the H of this reaction D(C=O)

What about dissociating methane into C + 4 H’s?
CH4 (g)  C(g) + 4 H (g)

H = 1650 kJ/mol CH4
Note 4 C-H bonds in CH4  D (C-H) = 412 kJ/mol
H2O (g)  2 H (g) + O (g)
H = 929 kJ/mol H2O
15

but it takes more energy to break the first O-H bond.
e.g.,
H2O (g)  H (g) + OH (g)
HO (g)  H (g) + O (g)
H = 502 kJ/mol H2O
H = 427 kJ/mol H2O

Note: we realize that all chemical reactions involve the breaking and reforming of chemical
bonds. For bonds to break, we must add energy. When new bonds are formed, energy is
released.

rxnH   D(bonds broken) -  D(bonds formed)
Example
Calculate the enthalpy associated with the combustion of C2H6 to form CO2 and H2O using
(1) the bond energy method; and
(2) the fH’s for the products and reactants.
Start with balanced equation
C2H6 (g) + 7/2 O2 (g)  2 CO2 (g) + 3 H2O (g)
rxnH/mol C2H6 =  D(bonds broken) -  D(bonds formed)
Bonds Broken
Bonds Formed
1 C-C
1 x 348 kJ = 348 kJ
4 C=O 4 x 799 kJ = 3196 kJ
6 C-H
6 x 413 kJ = 2478 kJ
6 H-O 6 x 463 kJ =
7/2 O=O
7/2 x 495 kJ = 1732 kJ
2778 kJ
rxnH = -1.51 * 103 kJ/mol C2H6
= 348 kJ + 2478 kJ + 1732 kJ - 3196 kJ – 2778 kJ = -1416 kJ = -1.42 x 103 kJ
rxnH = 3 fH [H2O (g)] + 2 fH [CO2 (g)] - fH [C2H6 (g)]
fH [CO2 (g)] = -394 kJ/mol
fH [C2H6 (g)] = -85 kJ/mol
fH [H2O (g)] = -244 kJ/mol
 rxnH
= {2 * -394 + -244 * 3 - (-85)} kJ
= 1.44 x 103 kJ/mol C2H6
16


These are close but not quite exact. Why?
The bond energies we use are averaged bond energies, i.e., the energy required to break a C-H
bond in CH4 may not be the same as the energy needed to break a C-H bond in C2H6, C3H8, etc.

This is a good approximate for equations involving diatomic species.

We can only use the above equations for GAS PHASE REACTIONS ONLY.