3. 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?
3
4. 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
5. 5
Essentially complete
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
ClTherefore, ionic compds.
exist primarily between
metals at left of periodic
table (Grps 1A and 2A
and transition metals) and
Ionic
Ionic
Bonds
Bonds
6. Covalent Bonding
The bond arises from the mutual
attraction of 2 nuclei for the same
electrons. Electron sharing results.
(Screen 9.5)
HA + HB
HA
HB
Bond is a balance of attractive and repulsive
forces.
6
7. 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.
7
9. 9
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.
10. 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.
10
11. Valence Electrons
Valence Electrons
Electrons are divided between
core and
valence electrons
B 1s 2 2s 2 2p 1
Core = [He] , valence = 2s 2 2p 1
Br [Ar] 3d 10 4s 2 4p5
Core = [Ar] 3d 10 , valence = 4s 2 4p 5
11
12. Rules of the Game
12
No. of valence electrons of a main group
atom = Group number
• For Groups 1A-4A (14), no. of bond pairs
= group number.
• For Groups 5A (15)-7A (17), BP’s = 8 - Grp. No.
13. Rules of the Game
• No. of valence electrons of an atom =
Group number
• For Groups 1A-4A (14), no. of bond pairs =
group number
• For Groups 5A (15)-7A (17), BP’s = 8 Grp. No.
• Except for H (and sometimes atoms of
3rd and higher periods),
BP’s + LP’s = 4
This observation is called the
OCTET RULE
13
14. Building a Dot Structure
Ammonia, NH 3
1. Decide on the central atom ( the
atom with lowest electron affinity );
never H.
Hydrogen atoms are always
terminal.
Therefore, N is central
2. Count valence electrons
H = 1 and N = 5
Total = (3 x 1) + 5
= 8 electrons; 4 pairs
14
15. Building a Dot Structure
3.
Form a single bond
between the central atom
and each surrounding
atom
H
4.
Remaining electrons form
LONE PAIRS to complete octet asH
needed.
3 BOND PAIRS and 1 LONE
PAIR.
Note that N has a share in 4 pairs (8
electrons), while H shares 1 pair.
N H
H
••
N H
H
15
16. 16
Sulfite ion, SO
Sulfite ion, SO
223
3
Step 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
O
10 pairs of electrons are
10 pairs of electrons are
now left.
now left.
O
O
S
S
O
O
17. Sulfite ion, SO
Sulfite ion, SO
223
3
17
Remaining 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.
18. Carbon Dioxide, CO2
Carbon Dioxide, CO2
1. Central atom = _______
2. Valence electrons = __ or __
pairs
3. Form bonds.
O
C
O This leaves 6 pairs.
4. Place lone pairs on outer atoms.
••
•
•
O
••
C
••
O
••
•
•
18
19. 19
Carbon Dioxide, CO2
Carbon Dioxide, CO2
4. Place lone pairs on outer atoms.
••
••
••
••
O
O
••
••
••
••
C
C
O
O
••
••
••
••
5. So that C has an octet, we shall form
DOUBLE BONDS between C and O.
••
••
••
••
O
O
••
••
C
C
••
••
O
O
••
••
••
••
••
••
O
O
••
••
The second bonding pair forms a
C
C
pi ( π )
O
O
••
••
••
••
bond.
20. Double and
even triple
bonds are
commonly
observed for
C, N, P, O,
and S
20
H2CO
SO3
C2F4
••
••
O
O
••
••
C
C
O
O
••
••
••
••
21. Sulfur Dioxide, SO2
Sulfur Dioxide, SO2
1.
2.
Central atom = S
Valence electrons = 18 or 9 pairs
••
O
•
•
••
••
S
••
O
•
•
••
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
••
•
•
21
22. Sulfur Dioxide, SO 22
Sulfur Dioxide, SO
22
OR bring in
OR bring in
right pair
right pair
bring in
bring in
left pair
left pair
••••
••••
O
O
••
••
S
S
••••
••••
O
O
••
••
•
•
•
•
This leads to the following structures.
••
••
••
••
••
••
O S O
O S O
••
••
••
••
••
••
••
••
••
••
O S O
O S O
••
••
••••
••••
••
••
These equivalent structures are called
RESONANCE STRUCTURES . The true
electronic structure is a HYBRID of the
two.
27. Violations of the Octet Rule
Boron Trifluoride
• Central atom = _____________
• Valence electrons = __________ or
electron pairs = __________
• Assemble dot structure
••
••
••
••
••
••
F
F
••
••
••
••
F
F
••
••
B
B
••
••
F
F
••
••
••
••
The B atom has a
The B atom has a
share in only 6 pairs of
share in only 6 pairs of
electrons (or 3 pairs).
electrons (or 3 pairs).
B atom in many
B atom in many
molecules is electron
molecules is electron
deficient.
deficient.
27
28. Violations of the Octet Rule
Sulfur Tetrafluoride, SF4
•
•
•
Central atom =
Valence electrons = ___ or ___ pairs.
Form sigma bonds and distribute electron
pairs.
••
••
••
••
F
F
••
••
••
••
••
••
F
F
••
••
••
••
S
S
••
••
F
F
••
••
••
••
••
••
F
F
••
••
••
••
5 pairs around the S
5 pairs around the S
atom. A common
atom. A common
occurrence outside the
occurrence outside the
2nd period.
2nd period.
28
29. Violations of the Octet Rule
Odd # of electrons, NO2
•
•
•
Central atom =
Valence electrons = ___ or ___ pairs.
Form sigma bonds and distribute electron
pairs.
•
N
•
••
O
••
N
•
•
••
••
O
••
•
•
O
••
••
O
••
29
38. MOLECULAR GEOMETRY
VSEPR
• V alence S hell E lectron
P air R epulsion theory.
• Most important factor in
determining geometry is
relative repulsion
between electron pairs.
38
Molecule adopts
Molecule adopts
the shape that
the shape that
minimizes the
minimizes the
electron pair
electron pair
repulsions.
repulsions.
42. Structure Determination by
VSEPR
Ammonia, NH 3
There are 4 electron pairs at the
corners of a tetrahedron.
H
••
N
H
H
lone pair of electrons
in tetrahedral position
N
H
H
H
The ELECTRON PAIR GEOMETRY is
tetrahedral.
42
44. Bond Order
# of bonds between a pair of atoms
Double bond
Single bond
Acrylonitrile
Triple
bond
44
45. 45
Bond Order
Fractional bond orders
in resonance structures.
Consider NO 2 -
••
N
••
N
••• •••
••
••
O
O• •O
O
••
••
••
••
The N—O bond order = 1.5
Total # of e- pairs used for a type of bond
Bond order =
Total # of bonds of that type
3 e - pairs in NO bonds
Bond order =
2 N — O bonds
46. 46
Bond Order
Bond order is proportional to two important bond
properties:
(a)
(b)
bond strength
bond length
414 kJ
123 pm
110 pm
745 kJ
47. 47
Bond Length
• Bond length is the distance between the nuclei of
two bonded atoms.
48. 48
Bond Length
Bond length depends on
bond order.
Bond distances measured
Bond distances measured
using CAChe software. In
using CAChe software. In
Angstrom units where 1 A =
Angstrom units where 1 A =
10-2 pm.
10-2 pm.
49. 49
Using Bond Energies
Estimate the energy of the reaction
H—H + Cl—Cl ----> 2 H—Cl
Net energy = ∆ H rxn =
= energy required to break bonds
- energy evolved when bonds are made
H—H = 436 kJ/mol
H—H = 436 kJ/mol
Cl—Cl = 242 kJ/mol
Cl—Cl = 242 kJ/mol
H—Cl = 432 kJ/mol
H—Cl = 432 kJ/mol
50. 50
Using Bond Energies
Estimate the energy of the reaction
H—H + Cl—Cl ----> 2 H—Cl
H—H = 436 kJ/mol
H—H = 436 kJ/mol
Cl—Cl = 242 kJ/mol
Cl—Cl = 242 kJ/mol
H—Cl = 432 kJ/mol
H—Cl = 432 kJ/mol
Sum of H-H + Cl-Cl bond energies = 436
kJ + 242 kJ = +678 kJ
2 mol H-Cl bond energies = 864 kJ
Net = ∆H = +678 kJ - 864 kJ = -186 kJ
51. Molecular Polarity
51
Boiling point = 100 ˚C
Boiling point = -161 ˚C
Why do water and methane
differ so much in their
boiling points?
Why do ionic compounds dissolve in
water?
52. Bond Polarity
52
HCl is POLAR because it
has a positive end and
a negative end.
+δ -δ
••
••
H Cl••
•
•
••
••
Cl has a greater share in
bonding electrons than
does H.
Cl has slight negative charge (-δ ) and H has
slight positive charge (+ δ )
53. +δ -δ
••
H Cl
••
•
•
Bond Polarity
Due to the bond polarity, the
H—Cl bond energy is
GREATER than expected
for a “pure” covalent bond.
BOND
BOND
“pure” bond
“pure” bond
real bond
real bond
ENERGY
ENERGY
339 kJ/mol calc’d
339 kJ/mol calc’d
432 kJ/mol measured
432 kJ/mol measured
Difference = 92 kJ. This difference is
Difference = 92 kJ. This difference is
proportional to the difference in
proportional to the difference in
ELECTRONEGATIVITY,, χ ..
ELECTRONEGATIVITY χ
53
54. 54
Electronegativity, χ
χ is a measure of the ability of an atom
in a molecule to attract electrons to
itself.
Concept proposed by
Linus Pauling
1901-1994
55. Linus Pauling, 1901-1994
The only person to receive two unshared
Nobel prizes (for Peace and Chemistry).
Chemistry areas: bonding, electronegativity,
protein structure
55
57. 57
4
3.5
Cl
N
3
2.5
F
O
C
S
H
Si
2
Electronegativity, χ
See Figure 9.9
P
1.5
1
0.5
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14 15
16
17
18
• F has maximum χ .
• Atom with lowest χ is the center atom
in most molecules.
• Relative values of χ determine BOND
POLARITY (and point of attack on a
molecule).
58. Bond Polarity
Which bond is more polar (or
DIPOLAR)?
O—H
O—F
χ
3.5 - 2.1
3.5 - 4.0
∆
1.4
0.5
OH is +δ
more polar than OF
-δ
-δ
O
+δ
H
+δ
O
-δ
F
and polarity is “reversed.”
58
59. Molecular Polarity
Molecules—such as HCl and H 2 O— can be POLAR
(or dipolar).
They have a DIPOLE MOMENT. The polar HCl
molecule will turn to align with an electric field.
Figure 9.15
59
60. Molecular Polarity
60
The magnitude of the
dipole is given in
Debye units. Named
for Peter Debye
(1884 - 1966). Rec’d
1936 Nobel prize for
work on x-ray
diffraction and dipole
moments.
64. 64
Carbon Dioxide
•
•
O
••
C
Positive C atom
Positive C atom
is reason CO22 +
is reason CO +
H22O gives H22CO33
H O gives H CO
O
•
•
••
-0.75
• CO2 is NOT polar even
though the CO bonds
are polar.
• CO2 is symmetrical.
+1.5
-0.75
67. Molecular Polarity, BF3
F
B
F
F
B—F bonds in BF3 are polar.
B—F bonds in BF3 are polar.
But molecule is symmetrical and
But molecule is symmetrical and
NOT polar
NOT polar
67
B atom is
positive and
F atoms are
negative.
68. Molecular Polarity, HBF2
H
B
F
F
B—F and B—H bonds in HBF2 are
B—F and B—H bonds in HBF2 are
polar. But molecule is NOT
polar. But molecule is NOT
symmetrical and is polar.
symmetrical and is polar.
68
B atom is
positive but H
& F atoms are
negative.
69. Is Methane, CH4, Polar?
H
C
H
H
H
Methane is symmetrical and is NOT polar.
69
70. Is CH3F Polar?
F
C
H
H
H
C—F bond is very polar.
C—F bond is very polar.
Molecule is not symmetrical and
Molecule is not symmetrical and
so is polar.
so is polar.
70
71. Substituted Ethylene
• C—F bonds are MUCH more polar than C—
H bonds.
• Because both C—F bonds are on same
side of molecule, molecule is POLAR.
71
72. Substituted Ethylene
72
• C—F bonds are MUCH more polar than C—H
bonds.
• Because both C—F bonds are on opposing
ends of molecule, molecule is NOT POLAR.
Hinweis der Redaktion
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