CHEMICAL
BONDING
Cocaine
1
Chemical Bonding
Problems and questions —
How is a molecule or
polyatomic ion held
together?
Why are atoms distributed at
strange angles?
Why are molecules not flat?
Can we predict the structure?
How is structure related to
chemical and physical
properties?
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Chemical Bonding:
Objectives
Objectives are to understand:
1. valence e- distribution
in molecules and ions.
2. molecular structures
3. bond properties and their effect
on molecular properties.
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Forms of Chemical Bonds
4
• There are 2 extreme forms of
connecting or bonding atoms:
• Ionic—complete transfer of 1
or more electrons from one
atom to another
• Covalent—some valence
electrons shared between atoms
• Most bonds are
somewhere in between.
Is electron transfer from an
element of low IE (metal) to an
element of high affinity for
electrons (nonmetal)
2 Na(s) + Cl2(g) --->
2 Na+ + 2 ClIonic compounds are
primarily between metals
(1A and 2A and transition)
and nonmetals
(O and halogens).
Ionic
Bonds
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Electrostatic Forces
COULOMB’S LAW
(charge on +)(charge on -)
Force of attraction =
(distance between ions)2
As ion charge increases, the attractive force
increases
_______________.
As the distance between ions increases, the
decreases
attractive force ________________.
This idea is important and will come up
many times in future discussions!
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Importance of Coulomb’s Law
NaCl, Na+ and Cl-,
m.p. 804 oC
MgO, Mg2+ and O2m.p. 2800 oC
Lattice Energy
1 Na(g) + ½ Cl2(g) ---> 1 Na+ Cl• Energy released when one mole of solid crystal
formed when ions in the gas phase combine
• Energy related to melting point and solubility
• Only calculated not measure directly
• Net energy is independent of the path so use
“Born – Haber cycle” to calculate Energy released
(exothermic)
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Covalent Bonding
The bond arises from the mutual attraction
of 2 nuclei for the same electrons.
Electron sharing results.
HA + HB
HA
HB
Bond is a balance of attractive and
repulsive forces.
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Electron
Distribution in
Molecules
G. N. Lewis
1875 - 1946
• Electron
distribution is
depicted with
Lewis electron
dot structures
• Valence electrons
are distributed as
shared or BOND
PAIRS and
unshared or LONE
PAIRS.
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Bond and Lone Pairs
• Valence electrons are distributed
as shared or BOND PAIRS and
unshared or LONE PAIRS.
••
H
Cl
•
•
••
shared or
bond pair
lone pair (LP)
This is called a LEWIS ELECTRON
DOT structure.
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Bond Formation
A bond can result from a “head-to-head”
overlap of atomic orbitals on
neighboring atoms.
••
H
+
Cl
••
••
•
•
H
Cl
•
•
••
Overlap of H (1s) and Cl (2p)
Note that each atom has a single,
unpaired electron.
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Valence Electrons
Electrons are divided between
core and valence electrons
B 1s2 2s2 2p1
Core = [He] , valence = 2s2 2p1
Br [Ar] 3d10 4s2 4p5
Core = [Ar] 3d10 ,
valence = 4s2 4p5
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Rules of the Game
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# of valence electrons = Group number
•For Groups 3A - 4A # of bond pairs = group
number.
• For
Groups 5A -7A ,
BP’s = 8 - Group #.
Rules of the Game
•Except
for H (and sometimes
atoms of 3rd and higher
periods),
BP’s + LP’s = 4
This observation is called the
OCTET RULE
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Building a Dot Structure
Ammonia, NH3
1. Decide on the central atom; never H.
Central atom is atom of lowest affinity
for electrons.
Therefore, N is central
2. Count valence electrons
H = 1 and N = 5
Total = (3 x 1) + 5
= 8 electrons / 4 pairs
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Building a Dot Structure
3.
Form a single bond
between the central atom and
each surrounding atom
4.
Remaining electrons form
LONE PAIRS to complete octet as
needed.
3 BOND PAIRS and 1 LONE PAIR.
H N H
H
••
H N H
H
Note that N has a share in 4 pairs (8 electrons),
while H shares 1 pair.
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Sulfite ion, SO32Step 1. Central atom = S
Step 2. Count valence electrons
S= 6
3 x O = 3 x 6 = 18
Negative charge = 2
TOTAL = 26 e- or 13 pairs
Step 3. Form bonds
O
10 pairs of electrons are
now left.
O
S
O
Sulfite ion, SO32Remaining pairs become lone pairs, first
on outside atoms and then on central
atom.
••
•
•
O
••
•
•
O
••
•
•
••
S
••
O
••
•
•
Each atom is surrounded by an octet
of electrons.
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Carbon Dioxide, CO2
C
1. Central atom = _______
16 or __
8 pairs
2. Valence electrons = __
3. Form bonds.
This leaves 6 pairs.
4. Place lone pairs on outer atoms.
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Carbon Dioxide, CO2
4. Place lone pairs on outer atoms.
5. So that C has an octet, we shall form
DOUBLE BONDS between C and O.
The second bonding pair forms a pi
(π) bond.
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Double and
even triple
bonds are
commonly
observed for C,
N, P, O, and S
H2CO
SO3
C2F4
Sulfur Dioxide, SO2
1. Central atom = S
2. Valence electrons = 18 or 9 pairs
3. Form double bond so that S has an octet
— but note that there are two ways of doing
this.
bring in
left pair
••
•
•
O
••
••
S
OR bring in
right pair
••
O
••
•
•
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RESONANCE STRUCTURES
This leads to the following structures.
These equivalent structures are called
RESONANCE STRUCTURES.
The true structure is a HYBRID of the two.
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Urea, (NH2)2CO
1. Number of valence electrons = 24 e2. Draw sigma bonds.
O
H N
H
C
N
H
H
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Urea, (NH2)2CO
3. Place remaining electron pairs in the
molecule.
••
•
•
O
••
H N
H
•
•
••
C
N
H
H
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Urea, (NH2)2CO
4. Complete C atom octet with double
bond.
••
O
••
H N
H
•
•
••
C
N
H
H
Violations of the Octet Rule
Usually occurs with B and elements of
higher periods.
BF3
SF4
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Boron Trifluoride
B
• Central atom = _____________
3 + 3(7) or
• Valence electrons = __________
12
electron pairs = __________
• Assemble dot structure
The B atom has a share in
only 6 e (or 3 pairs).
Boron in many molecules is
electron deficient.
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Sulfur Tetrafluoride, SF4
• Central atom = S
34 or ___
17 pairs.
• Valence electrons = ___
• Form sigma bonds and distribute electron
pairs.
5 pairs around the S
atom. A common
occurrence outside the
2nd period.
Formal Charges
•Formal charge
= Group # – 1/2 (# of BE)
- (# LP electrons)
• The predominant resonance
structure of a molecule is the one
with charges as close to 0 as
possible.
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Carbon Dioxide
6
- (1 / 2 )(4 )
- 4
••
•
•
4
O
=
••
C
- (1/ 2)(8) - 0
O
=
•
•
0
FC = Group no. – 1/2 (no. of BE) - (no. of LP)
0
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Thiocyanate Ion, SCNWhich is the most important resonance form?
••
•
•
S
••
C
N
•
•
•
•
••
S
C
N
•
•
••
••
•
•
S
C
N
•
•
••
FC = Group no. – 1/2 (no. of BE) - (no. of LP)
Thiocyanate Ion,
6 - (1/2)(2) - 6 = -1
SCN-
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5 - (1/2)(6) - 2 = 0
••
•
•
S
C
N
•
•
••
4 - (1/2)(8) - 0 = 0
FC = Group no. – 1/2 (no. of BE) - (no. of LP)
Isoelectric, CS2, CO2
. These equivalent structures are called
ISOELECTRIC STRUCTURES.
Different atoms with the same structure
•
•
O
••
•
•
C O
••
••
•
•
S
C S
••
•
•
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