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CHEMICAL
BONDING
Cocaine

1
2

For more help

Gulfam Hussain
e_gulfam@yahoo.com
+923336275670
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
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

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
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
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
8

Electron
Electron
Distribution in
Distribution in
Molecules
Molecules

• Electron distribution
is depicted with

Lewis electron
dot structures

• Valence electrons
are distributed as
shared or BOND
PAIRS and
unshared or LONE

PAIRS.

G. N. Lewis
1875 - 1946
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.
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
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
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.
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
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
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

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
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.
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

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.
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
••
••

••
••
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
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.
23

Urea, (NH2)2CO
24

Urea, (NH2)2CO
1. Number of valence electrons = 24 e2. Draw sigma bonds.

O
O
H N
H N
H
H

C
C

N H
N H
H
H
25

Urea, (NH2)2CO
3. Place remaining electron pairs in the
molecule.
••
••
••
••

H N
H N
H
H

••
••

O
O

••
••

C
C

••
••

N H
N H
H
H
26

Urea, (NH2)2CO
4. Complete C atom octet with double
bond.
••

O
••

H N
H

C

•
•
••

N
H

H
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
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
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
Formal Atomic Charges
Definition of Formal Charge:

• Formal charge=
Group no. – 1/2 BEs - LPEs

30
Carbon Dioxide, CO2

6 - (1 / 2)(4) - 4
••
•
•

O

••

C

4 - (1 / 2)(8) - 0

O

•
•

=

0

=

0

31
32

Calculated Partial Charges in CO2

Yellow = negative & red = positive
Relative size = relative charge
Thiocyanate Ion, SCN

-

6 - (1/2)(2) - 6 = -1

5 - (1/2)(6) - 2 = 0

••
•
•

S
••

C

N

•
•

4 - (1/2)(8) - 0 = 0

33
34

Thiocyanate Ion, SCN

-

••

••
•
•

S
••

C

N

•
•

•
•

S

C

••

N

••
•
•

S

C

N

•
•

••

Which is the most important resonance form?

•
•
Calculated Partial Charges
in SCN-

All atoms negative, but
All atoms negative, but
most on the S
most on the S

35

••
•
•

S
••

C

N

•
•
36

Boron Trifluoride, BF3
••

F

•
•
••

•
•

F
••

•
•

•
•
••

B

•
•

B -1

F

••
•
•

F
••

•
•

F +1
•
•

•
•

F

•
•

••

What if we form a B—F double
bond to satisfy the B atom octet?
MOLECULAR
GEOMETRY

37
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.
39

Electron Pair Geometries
Figure 9.12
40

No. of e- Pairs
Around Central
Example
Atom
2

F—Be—F

Geometry
linear

180Þ
F
3

planar
trigonal

B
F

F
120Þ
H

109Þ

4
C
H

tetrahedral
H
H
41

Electron Pair Geometries
Figure 9.12
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
Bond Properties

43

bond order, bond length, bond energy, bond polarity

Buckyball in HIV-protease
Bond Order

# of bonds between a pair of atoms

Double bond

Single bond

Acrylonitrile
Triple
bond

44
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

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

Bond Length
• Bond length is the distance between the nuclei of
two bonded atoms.
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

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

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
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?
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 )
+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

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
Linus Pauling, 1901-1994

The only person to receive two unshared
Nobel prizes (for Peace and Chemistry).
Chemistry areas: bonding, electronegativity,
protein structure

55
Electronegativity
Figure 9.9

56
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).
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
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
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.
61

Dipole Moments

Why are some molecules polar but others
are not?
Molecular Polarity
Molecules will be polar if
a)
bonds are polar
AND
b)
the molecule is NOT “symmetric”

All above are NOT polar

62
Polar or Nonpolar?
Compare CO2 and H2O. Which one is polar?

63
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
65

Microwave
oven

Consequences of H2O
Polarity
66

Polar or Nonpolar?
• Consider AB3 molecules: BF3, Cl2CO, and NH3.
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.
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.
Is Methane, CH4, Polar?
H
C
H

H
H

Methane is symmetrical and is NOT polar.

69
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
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
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.

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Bonding

  • 2. 2 For more help Gulfam Hussain e_gulfam@yahoo.com +923336275670
  • 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
  • 8. 8 Electron Electron Distribution in Distribution in Molecules Molecules • Electron distribution is depicted with Lewis electron dot structures • Valence electrons are distributed as shared or BOND PAIRS and unshared or LONE PAIRS. G. N. Lewis 1875 - 1946
  • 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.
  • 24. 24 Urea, (NH2)2CO 1. Number of valence electrons = 24 e2. Draw sigma bonds. O O H N H N H H C C N H N H H H
  • 25. 25 Urea, (NH2)2CO 3. Place remaining electron pairs in the molecule. •• •• •• •• H N H N H H •• •• O O •• •• C C •• •• N H N H H H
  • 26. 26 Urea, (NH2)2CO 4. Complete C atom octet with double bond. •• O •• H N H C • • •• N H H
  • 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
  • 30. Formal Atomic Charges Definition of Formal Charge: • Formal charge= Group no. – 1/2 BEs - LPEs 30
  • 31. Carbon Dioxide, CO2 6 - (1 / 2)(4) - 4 •• • • O •• C 4 - (1 / 2)(8) - 0 O • • = 0 = 0 31
  • 32. 32 Calculated Partial Charges in CO2 Yellow = negative & red = positive Relative size = relative charge
  • 33. Thiocyanate Ion, SCN - 6 - (1/2)(2) - 6 = -1 5 - (1/2)(6) - 2 = 0 •• • • S •• C N • • 4 - (1/2)(8) - 0 = 0 33
  • 35. Calculated Partial Charges in SCN- All atoms negative, but All atoms negative, but most on the S most on the S 35 •• • • S •• C N • •
  • 36. 36 Boron Trifluoride, BF3 •• F • • •• • • F •• • • • • •• B • • B -1 F •• • • F •• • • F +1 • • • • F • • •• What if we form a B—F double bond to satisfy the B atom octet?
  • 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.
  • 40. 40 No. of e- Pairs Around Central Example Atom 2 F—Be—F Geometry linear 180Þ F 3 planar trigonal B F F 120Þ H 109Þ 4 C H tetrahedral H H
  • 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
  • 43. Bond Properties 43 bond order, bond length, bond energy, bond polarity Buckyball in HIV-protease
  • 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.
  • 61. 61 Dipole Moments Why are some molecules polar but others are not?
  • 62. Molecular Polarity Molecules will be polar if a) bonds are polar AND b) the molecule is NOT “symmetric” All above are NOT polar 62
  • 63. Polar or Nonpolar? Compare CO2 and H2O. Which one is polar? 63
  • 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
  • 66. 66 Polar or Nonpolar? • Consider AB3 molecules: BF3, Cl2CO, and NH3.
  • 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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