Brown, LeMay Ch 8
AP Chemistry
Monta Vista High School
8.1: Types of “Inter-Atomic” or Intra molecular Bonding a.k.a
bonding
Increasing
Diff. of EN
1. Ionic: electrostatic attraction between oppositely charged
ions. Ex. NaCl, K2SO4. Generally solids.
2. Covalent: sharing of e- between two atoms (typically
between nonmetals). Ex. CO2, SO2. Generally gases and
liquids
3. Metallic: “sea of e-”; bonding e- are relatively free to move
throughout the 3D structure
Ionic
Ex. Fe, Al, generally solids
Covalent
Metallic
4. Covalent Network: large number of atoms/molecules
bonded in a network through covalent bonding. Ex. SiO2,
Si, Ge, Diamond, Graphite
2
8.2: Ionic Bonding
Results as atoms lose or gain e- to achieve a noble gas
e- configuration; is typically exothermic.
 The bonded state is lower in energy (and therefore
more stable).
 Electrostatic attraction results from the opposite
charges. Electrostatic attraction (force) determines
the strength of ionic bond, just like electro negativity
difference determines the strength of covalent bond.
 Occurs when diff. of EN of atoms is > 1.7 (maximum is
3.3: CsF)
 Can lead to interesting crystal structures (Ch. 11).
 Use brackets when writing Lewis symbols of ions.
Ex: Draw the Lewis symbol of fluoride.

Animation of Ionic/Covalent Bonding
Animation of LiCl Crystal

••
[ : F :]1••
3
Lattice Energy

Measurement of the energy of stabilization present
in ionic solids
DHlattice = energy required to completely separate 1
mole of solid ionic compound into its gaseous ions
Q Q
DH lattice 
r  r

Electrostatic attraction (and thus lattice energy)
increases as ionic charges increase and as ionic
radii decrease. Animation showing formation of a lattice
4
 The
lattice energy of NaCl is the
energy given off when Na+ and Clions in the gas phase come together
to form the lattice of alternating Na+
and Cl- ions in the NaCl crystal (in this
case lattice energy is negative) or it
may be defined as the amount of
energy required to break 1 mole of
solid NaCl into its ions in gaseous
state (in this case lattice energy will
be positive).

Na+(g) + Cl-(g)  NaCl(s) ΔHo = -787.3 kJ/mol

The lattice energies for the alkali metal
halides is therefore largest for LiF
(smallest size, smallest d) and smallest
for CsI ( largest size, largest d), as
shown below.
 Values
of Lattice Energies of Alkali
Metals Halides (kJ/mol)
F-
Cl-
Br-
I-
Li+ 1036 853 807 757
Na+ 923 787 747 704
K+ 821 715 682 649
Rb+ 785 689 660 630
Cs+ 740 659 631 604
Data taken from Purdue website
Understanding Check
Ex: Which has a greater lattice energy?
Why?
NaCl or KCl
NaCl or MgS
Covalent Bonding

Atoms share e- to achieve noble gas configuration
that is lower in energy (and therefore more stable).

Occurs when diff. of EN of atoms is ≤ 1.7
 Polar covalent:
0.3 < diff. of EN ≤ 1.7 (e- pulled closer to more EN
atom)
 Nonpolar covalent:
0 ≤ diff. of EN ≤ 0.3 (e- shared equally)
(ionic vs. covalent bonding in youtube video)
 Coordinate Covalent: Shared pair contributed by only
one of the two sharing species. Ex. Lewis acids and
bases
8
Cl-Cl Bond in Cl2 molecule
Animation of bonding in Chlorine Molecule
9
23.5: Metallic bonding


Metallic elements have low I.E.; this means valence eare held “loosely”.
A metallic bond forms between metal atoms because of
the movement of valence e- from atom to atom to atom
in a “sea of electrons”. The metal thus consists of
cations held together by negatively-charged e- "glue.“

This results in excellent thermal &
electrical conductivity, ductility, and
malleability.

A combination of 2 metals is called an
alloy.
10
Free e- move rapidly in
response to electric fields,
thus metals are excellent
conductors of electricity.
Free e- transmit kinetic energy
rapidly, thus metals are
excellent conductors of
heat.
Layers of metal atoms are
difficult to pull apart
because of the movement
of valence e-, so metals
are durable.
However, individual atoms are held loosely to other
atoms, so atoms slip easily past one another, so metals
are ductile.
Covalent
(a.k.a.
covalent
network)
Graphite
Atoms
bonded
in a
covalent
network
Very hard
Very high
MP
Covalent
bonds Generally
insoluble
Variable
conductivity
Diamond
C (diamond
& graphite)
SiO2
(quartz)
Ge, Si, SiC,
BN
SiO2
The Octet Rule
Atoms tend to gain, lose, or share euntil they are surrounded by 8 e- in their
outermost energy level (have filled s and
p sub shells) and are thus energetically
stable.
 Exceptions do occur (and will be
discussed later.)

Visualizing atomic orbitals
13
Lewis symbols
Valence e-:
 e- in highest energy level and involved
in bonding; all elements within a group
on P.T. have same # of valence e-
Gilbert N.
Lewis
(1875 – 1946)
Lewis symbol (or electron-dot symbol):
Shows a dot only for valence e- of an atom or ion.
 Place dots at top, bottom, right, and left sides and
in pairs only when necessary (Hund’s rule).
 Primarily used for representative elements only
(Groups 1A – 8A)
Ex: Draw the Lewis •symbols of C and N.

•C•
•
•
:N•
•
14

Transition metals typically form +1, +2,
and +3 ions.
 It is observed that transition metal
atoms first lose both “s” e-, even though
it is a higher energy subshell. Cr2+,
Cr3+
 Most lose e- to end up with a filled or a
half-filled subshell. Ex. Cu+ ion
15
Lewis Structures
Lewis structures are used to depict
bonding pairs and lone pairs of electron
in the molecule.
 Step 1
 Total number of valence electrons in the
system: Sum the number of valence
electrons on all the atoms . Add the total
negative charge if you have an anion.
Subtract the charge if you have a cation.
 Example: CO32
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Step 2
Number of electrons if each atom is to be happy:
Atoms in our example will need 8 e (octet rule) or
2 e ( hydrogen). So, for the ex.
 Step 3
 Calculate number of bonds in the system:
Covalent bonds are made by sharing of e. You
need 32 and you have 24. You are 8 e deficient. If
you make 4 bonds ( with 2 e per bond) , you will
make up the deficiency. Therefore,



# of bonds= ( e in step 2- e in step 1)/2 =(3224)/2= 4 bonds
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Step 4
 Draw the structure: The central atom is C (
usually the atom with least electro negativity
will be in the center). The oxygens surround it .
Because there are four bonds and only three
atoms, there will be one double bond.
 Step 5
 Double check your answer by counting total
number of electrons.
Drawing Lewis Structures (youtube video)
18
Level 1 Practice for Lewis Structurs
Draw the Lewis structures for the
following using above steps. Show
work!
 A.Cl2
 B. CH2Cl2
 C. NH3
 D. NaCl
19
Level 2 practice on Lewis Structures
SO4
2-
H2O2
HCN
CNS1-
20
8.8: Exceptions to the Octet Rule

Odd-electron molecules:
Ex: NO or NO2 (involved in breaking down
ozone in the upper atmosphere)

Incomplete octet:
H2
He
BeF2
BF3
NH3 + BF3 → NH3BF3 (Lewis acid/base rxn)
21

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)
PCl5
SF6
22
8.6: 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
23

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.
24
Examples: Draw at least 2 Lewis structures for each, then
calculate the FC of each atom in each structure.
1SCN
N2 O
BF3
25
8.7: Resonance Structures
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: O3


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” e- structure is an “average” of all
resonance structures.
26
Bond Order

An indication of bond strength and bond length
 Single bond: 1 pair of e- shared
Ex: F2

:F-F
:
•• ••
Longest,
weakest
Double bond: 2 pairs of e- shared
Ex: O2

•• ••
O=O
Triple bond: 3 pairs of e- shared
Ex: N2
:N ≡ N:
Shortest,
strongest
27
Bond Order & Resonance
Structures

To determine bond order with resonance
structures:
 Determine the bond order at one position
in one resonance structure, and add it to
the bond orders at the same bond
position in all other resonance structures.
 Divide the sum by the number of
resonance structures to find bond order.
28
Examples: draw the Lewis structure and determine the bond S-O,
C-C, and C-H bond orders
SO3
C6H6
29
8.9: Bond enthalpy:

Amount 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 DHrxn .
Ex: CH4 (g) + Cl2 (g) → CH3Cl (g) + HCl (g) DHrxn = ?
 1 C-H & 1 Cl-Cl bond are broken (per mole)
 1 C-Cl & 1 H-Cl bond are formed (per mole)
DHrxn ≈  (Hbonds broken) -  (Hbonds formed)
Note: this is the “opposite” of Hess’ Law where
DHrxn = DHproducts – Dhreactants
Bond Enthalpy link
30
Ex: CH4 (g) + Cl2 (g) → CH3Cl (g) + HCl (g) DHrxn =
?
Bond
C-H
H-Cl
C-C
Ave DH/mol
413
431
348
Bond
Ave DH/mol
Cl-Cl
242
C-Cl
328
C=C
614
DHrxn ≈  (Hbonds broken) -  (Hbonds formed)
DHrxn ≈ [(1(413) + 1(242)] – [1(328) + 1(431)]
DHrxn ≈ -104 kJ/mol
DHrxn = -99.8 kJ/mol (actual)
Note:
2 C-C
≠ 1 C=C
2(348) = 696 kJ ≠ 614 kJ
31
Ex: CH4(g) + Cl2(g) → CH3Cl(g) + HCl(g) DHrxn=?
*CH3(g) + H(g) + 2 Cl(g)
H
Absorb E,
break 1 C-H
and 1 Cl-Cl
bond
Release E,
form 1 C-Cl
and 1 H-Cl
bond
CH4(g) + Cl2(g)
DHrxn
CH3Cl (g) + HCl (g)
DHrxn =  (Hbonds broken) +  (- Hbonds formed)
DHrxn =  (Hbonds broken) -  (Hbonds formed)