1

Chemical Bonding II

Molecular Geometry

Valence Bond Theory

Phys./Chem. Properties

Quantum Mechanics

Sigma & Pi bonds

Hybridization

MO theory

2

Molecular Geometry

3-D arrangement of atoms

3

VSEPR

Valence-shell electron-pair

repulsion theory.

All valence shell e- pairs

(or e- “domains”) repel each other.

(Coulomb’s law as applied to the

repulsion of valence electrons.)

Determines the geometry of

e- domains around central atom.

4

VSEPR Theory: Rules

1.Single, double, triple bonds and

lone pairs are treated as one e-

domain (approximation)

2.Apply VSEPR to any one

resonance structure

3.Geometry– move e- domains as

far apart as possible in 3-D space.

5

AB2 (with no lone pairs)

Beryllium chloride BeCl2

— Be—

Cl—Be—Cl

180o

linear geometry

e-

atoms

6

AB3 (with no lone pairs)

Boron trifluoride BF3

B

F

FF

Trigonal planar

120o

7

C

AB4 (with no lone pairs)

Methane CH4

tetrahedron

109.5o

8

AB5 (with no lone pairs)

Phosphorus Pentachloride PCl5

trigonal bipyramid

equatorial

axial90o

120o

9

SF

F

F

F

F

F

AB6 (with no lone pairs)

Sulfur hexafluoride SF6

octahedral

All angles

90o & 180o

10

Quiz: Name That Shape!

AB2

AB3

AB4

AB5

AB6

11

VSEPR: Lone Pairs

Molecules in which the

central atom has lone pair(s)

ABxEy

central atom

surrounding atoms

lone pairs on A

12

ABxEy

Geometry is similar to ABx, but

nonbonding pairs are treated like

bonding pairs to determine

geometry of the e- domains.

VSEPR

13

Bonding e- take up less space

than nonbonding e-

Example: water

Bonding e- are

“focused” between

the nuclei.

14

VSEPR

lone pair lone pair bonding pair

vs vs vs

lone pair bonding pair bonding pair

repulsion repulsion repulsion

Decreasing e- pair repulsion

15

AB2E

Sulfur dioxide SO2

O S—O

OSO angle < 120o

Draw Lewis structure

S

O O3 e- domains

16

Shape: e- Pairs vs. Atoms

S

O O

e- domains are trigonal planar

S

O O

molecule is bent linear

(always state molecular shape)

17e- tetrahedral trigonal pyramidal

AB3E

Ammonia H—N—H

H

18

AB2E2

Water H—O—H

e- tetrahedral molecule: bent linear

19

Comparison

AB4 AB3E AB2E2

Methane ammonia water

109.5 107.3 104.5

CH

H

H

H

NH

H

HO

H

H

20

AB4E

Sulfur tetrafluoride SF4

S

F F

F FPredict: trigonal bipyramid

for e- domains.

Draw Lewis structure

21

AB4E: SF4

trigonal bipyramid: 2 choices

F

F

F

F

F

F

F

F

distorted tetrahedron or seesaw

22

AB3E2

ClF3 F

F

F

molecule: T-shaped

A little weird.

23

AB2E3Three possibilities for I3

-

Want nonbonding e-

domains farthest apart.

24

AB2E3

I3-

molecule: linear

I

I

-

Lone pairs always go on

equatorial position(s)

25

AB5E

BrF5

molecule: square pyramidal

FF

FF

F

26

AB4E2: 2 Possibilities

Want

nonbonding

e- pairs

farthest

apart.

1

2

XeF4

27

AB4E2

XeF4

molecule: square planar

F

FF

F

28

You Try It

Predict geometry and

approximate bond angles for:

AlCl4-

XeF2

XeOF2

Text: Tables 10.1 and 10.2.

29

Bond Polarity

FH

Shift of e- density toward F,

Thus HF is polar.

d+ d-

F is more

electronegative

than H.

30

Overall Molecular Polarity

O C O

Even though CO2 has polar bonds,

it is nonpolar since the individual

bond polarities add to zero.“Dipoles” are a vector quantities.

Symmetrical = nonpolar

“pull test”

(Recall the “Regents” rule about

lone pairs on the central atom.)

31

Molecular Polarity

Molecule GeometryDipole

Strength

HF Linear 1.92

HBr Linear 1.08

Water Bent 1.87

SO2 Bent 1.60

32

Molecular Polarity

NH

H

H

Which has larger polarity?

NH3 NF3

NF

F

F

m =0.24 Dm = 1.46 D

33

Dipole Moments

Predict whether the following

molecules are polar.

IBr

CH2Cl2AlCl3

Remember to distinguish between

bond polarity and molecular polarity.

34

Both H2 & F2 have single bonds,

but…

Bond

length

Bond

energy

H2 74 pm 436 kJ

F2 142 pm 151 kJ

Shortfall of VSEPR

35

Quantum Mechanics…

Valence Bond Theory

e- in molecule occupy

blended atomic orbitals

…to the rescue

Molecular Orbital Theory

molecule has

“molecular orbitals”

36

Valence Bond Theory

Atomic orbitals (s, p, d …)

of the valence electrons

“hybridize” or mix to form new

orbitals for the molecule.

37

Valence Bond Theory

2s 2pcarbon

atomic

orbitalsC

How can carbon form four equal

bonds with four hydrogen atoms

using its atomic ‘s’ and ‘p’ orbitals?

Consider tetrahedral CH4

38

CH4: sp3 Hybridization

2s 2p

carbon

hybridized

orbitals

carbon

atomic

orbitalsC

Csp3 (all the same)

39

Hybridization Analogy

s p p p

4 sp3 hybrid orbitals

40

sp3 Hybridization

It takes energy to form hybrid

orbitals, but this energy is

more than compensated by

bond formation.

41

sp3 Orbital Shapes

Each

orbital

can hold

2 e-

tetrahedral

One s & three p orbitals change

to four sp3 hybrid orbitals.

42

CH4 Bonding

The 4 sp3 hybrid

orbitals of C bond

(overlap) with the

4 1s atomic orbitals

of the H atoms.

H

HHH

C

43

NH3 sp3 Bonding

nitrogen

hybridized

orbitals

nitrogen

atomic

orbitals

2s 2pN

sp3 (all the same)

N

44

sp3

bonding orbitalslone pair

NH3 sp3 Bonding

45

Hybridization

First use VSEPR to predict the

arrangement of e- pairs, then

use hybridization to predict the

type of bonding.

46

sp Hybridization

Be atomic

orbitals

2s 2pBe

Be hybrid

orbitals

sp 2p

Be

47

sp Hybridization: BeCl2

BeCl Cl

one atomic p

orbital of Cl

two sp hybrid

orbitals of Be

48

sp2 Hybridization

B atomic

orbitals

2s 2p

B

B hybrid

orbitals

sp2 2p

B

49

sp2 Hybridization: BF3

B

F F

F

one p orbital

of F

three sp2

hybrid orbitals

of B

50

Hybridization

Hybridization for 2nd Period

elements (C, N, O, F) explains the

“octet” rule, since there are 4 hybrid

orbitals formed from one s and three

p atomic orbitals.

s p sp3

51

Hybridization: Review

1.Not applied to isolated atoms

2.First determine VSEPR geometry

3.Mix nonequivalent atomic orbitals

of central atom to form hybrid

orbitals

52

Hybridization: Review

4.Requires energy, but energy is

more than returned by bond

formation

5.Covalent bonds formed by

overlap of hybrid-hybrid and/or

hybrid-unhybridized orbital

53

Hybridization: Try It

Determine hybridization in:

AlBr3PF3

HgCl2

54

Let’s Not Forget d Orbitals

For elements in the 3rd Period and

higher, hybridization can also

include ‘d’ orbitals.

d-orbital hybridization is still

being debated!

55

SF

F

F

F

F

F

d Hybridization

SF6 VSEPR predicts

octahedral geometry

56

Hybridization: SF6

S atomic orbitals

S hybridized sp3d2 orbitals

3s 3p 3d

sp3d2 3d

57

Hybridization

You Try It .

What is the VSEPR geometry

and the hybridization in PBr5?

58

Summary: Hybrid Orbitals

2 e- pairs (sp) linear

3 e- pairs (sp2)

trigonal planar

4 e- pairs (sp3)

tetrahedral

e- domains (bonding & nonbonding pairs)

59

Summary: Hybrid Orbitals

5 e- pairs (sp3d)

trigonal bipyramid

6 e- pairs (sp3d2)

octahedral

60

Multiple Bonds

In VSEPR, no distinction was made

among single, double, triple bond or lone

pair. All were counted as an “e- domain”.

Each carbon is trigonal planar

C2H4 C CH

H

H

H

61

Double Bond: C2H4

2s 2pC atomic

orbitals

C hybrid

orbitals

sp2 pz

+

unhybridized

p orbital

62

Double Bond: C2H4

C hybrid

orbitals

sp2 pz

+

trigonal planar dumb bell

63

Double Bonds: C2H4

C C

H

HH

H

64

Double Bond: C2H4

Sigma (s) and Pi (p) Bonds

(model)

65

Pi & Sigma Bonds

Pi (p) bond: covalent bond formed

by sideways e- overlap above and

below the plane connecting atoms

(weaker than s)

Sigma (s) bond: covalent bond

formed by e- overlap along the

axis connecting atoms

66

Double Bond: C2H4

The double bond is one sigma and one

pi bond between the carbon atoms

C CH

H

H

H

67

Triple Bond: H-C=C-H

2s 2pC atomic

orbitals

C hybrid

orbitals

sp py pz

+unhybridized

p orbitals

68

Triple Bond: C2H2

Hybridize s bonds only

hybridized

sp orbitals

unhybridized

p orbitals

69

Hybridizing Shortcuts

1.Single bonds: sigma2.Double bonds: one sigma, one pi3.Triple bonds: one sigma, two pi4.Hybridize: add the number of

sigma bonds plus lone pairs

s p d

70

You Try It

Describe the bonding and

hybridization for each atom in:

•formaldehyde, CH2O

•hydrogen cyanide, HCN

(assign formal charges)

71

Delocalized Molecular Orbitals

An explanation of resonance.

O OOO O O

ozone

Ozone is a blend of the two

resonance structures.

Delocalized Molecular Orbitals

72

The p component of the double bond

is delocalized over the molecule.

sp2

sp2

sp2

73

Delocalized “MOs”

e- in s bonds– localized

e- in p bonds-- delocalized

H

H

H

H

H

H

H

H

H

H

H

H

benzene

74

Benzene

sp2

sp2

C

C

C C

C

C

Sigma bonds

(localized)

75

Benzene

p bonds

(delocalized)

sp2 pz

+

76

Benzene

Bond order 1.5

H

H

H

H

H H

“Conjugated” double bonds are

more stable (less reactive)

77

Carbonate Ion

+ resonance

structuresO—CO

O

2-

C: sp2 fors bonds (localized)

plus a 2pz for p bond (delocalized)

O: each has a 2pz orbital (delocalized)

78

Carbonate Ion

o

o

oC

Additional stability due to

delocalization of the pi bond.o

o

oC

79

You Try It

Would you predict the NO3- anion

to have additional stability due to

delocalization of the p electrons?

80

Polyatomic Ions

Many common polyatomic ions

have delocalized pi bonds,

partially accounting for their

additional stability in chemical

reactions.

CO3-2, NO3

- , ClO3- etc.

81

Ionic & Covalent

CaCO3

C

O

OOCa

2-

2+

Trigonal planar, sp2, 120o

Ionic compounds containing polyatomic

ions have both ionic and covalent bonds!