Concepts of Chemical Bonding
Brown, LeMay Ch 8
AP Chemistry
8.1: Types of “Inter-Atomic” Bonds
Increasing
Diff. of EN
1. Ionic: electrostatic attraction between oppositely
charged ions
2. Covalent: sharing of e- between two atoms
(typically between nonmetals)
3. Metallic: “sea of e-”; bonding e- are relatively free
to move throughout the 3D structure
Ionic
Covalent
Metallic
2
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•
•
3
The Octet Rule


Atoms tend to gain, lose, or share e- until they are
surrounded by 8 valence e- (have filled s and p
subshells) and are thus energetically stable.
Exceptions do occur (and will be discussed later.)
4
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.
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 sodium fluoride.
••
[ Na ]1+[: F : ]1••
5
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.
Ex: Which has a greater lattice energy?
NaCl or KCl
NaCl or MgS
6

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.
Most lose e- to end up with a filled or a half-filled
subshell.
7
8.4 - 8.5: 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)
8
8.6: Drawing Lewis Diagrams
1. Add up valence e- from all atoms in formula.
 If there is a charge, add e- (if an anion) or subtract e(if a cation).
2. Draw the “molecular skeleton”:
 Place the least EN atom(s) in the center.
 Array the remaining elements around the center and
connect them with a single bond. (When in doubt, put
the element written first in the formula in the center of
the molecule.)
3. Complete the octets of the outer (more EN) atoms first.
4. Place leftover e- on the central atom, even if it violates the
octet rule.
5. If the central atom does not have an octet, create multiple
bonds by sharing e- with the outer atoms.
9
Ex: Draw the Lewis structure, and name the
molecule.
SO42-
HCN
H2O2
CNS1-
10
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)
11

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
12
8.6: 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
13

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.
14
Examples: Draw at least 2 Lewis structures for each, then
calculate the FC of each atom.
SCN1-
N2O
BF3
15
8.7: Resonance Structures



Equivalent Lewis structures that describe a molecule
with more than one likely arrangement of eNotation: 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.
16
Bond Order

An indication of bond strength and bond length
 Single bond: 1 pair of e- shared
Ex: F2
•• ••
Longest,
:••F-F••:

Double bond: 2 pairs of e- shared
Ex: O2

weakest
O=O
Triple bond: 3 pairs of e- shared
Ex: N2
:N ≡ N:
Shortest,
strongest
17
Bond Order & Resonance Structures

To determine bond order with resonance
structures:
 Pick one bond and add up the integer bond
order in one resonance structure to the same
bond position in all other resonance structures.
 Divide the sum by the number of resonance
structures to find bond order.
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Examples
SO3
C6 H6
19
8.9: Bond enthalpy:
DH/mol to break a particular bond of substance (g)
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
20
Ex: CH4 (g) + Cl2 (g) → CH3Cl (g) + HCl (g) DHrxn = ?
Bond
C-H
H-Cl
C-C
Ave DH/mol
413
431
348
Bond
Cl-Cl
C-Cl
C=C
Ave DH/mol
242
328
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
21
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
CH4(g) + Cl2(g)
DHrxn
Release E,
form 1 C-Cl
and 1 H-Cl
bond
CH3Cl (g) + HCl (g)
DHrxn =  (Hbonds broken) +  (- Hbonds formed)
DHrxn =  (Hbonds broken) -  (Hbonds formed)
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.
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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.