1

structure prediction, chemical

bonding

2

Lewis structures

• Atoms listed in order of increasing EN, no

connectivity implied

• CSPF

• (PNF2)4

3

after the Lewis

Structure

• determine the steric number

number of bonds (single and multiple)

number of unshared pairs

or sometimes single unpaired electrons such as in

NO2

• then assign geometry using

4

the following guidelines

SN geometry of electrongroups

2 linear (180°)3 trigonal planar (120°)4 tetrahedral (109.5°)5 trigonal pyramidal6 octahedral (90°)7 pentagonal bipyramid

5

6

Cartoon representations

7

8will need to refine

bond angles

considering

• the different sizes of

single bonds vs. multiple bonds

lone pairs vs. single bonds

9

today in dj

• One chemical fad is/was the isolobal analogy.

Draw a chemical species or fragment that is

isolobal with the methylene group, CH2,

derived from methane by loss of two (2) H

atoms.

10

• Repulsions between electron groups are in the order

LP-LP > LP-BP > BP-BP

• when lone pairs are present the bond angles are smaller than those

predicted for the idealized geometry in rule 1.

• lone pairs choose the site with the most volume available to them, or the

least sterically hindered site.

• if all sites are equivalent, the lone pairs will be trans to one another.

• multiple bonds occupy more space than single bonds.

• bonding pairs to electronegative substituents occupy less space than those

to less electronegative substituents

11

• If the central atom is in third row or below in the periodic chart there are

two possibilities:

if the substituents are oxygen atoms or halogens, the above rules hold.

if the substituents are less electronegative than halogens, the lone pairs will

occupy a non-bonding s orbital and the bonding to the substituents will be

primarily through p orbintals with resulting bond angles of about 90°.

• Describe Structures by the arrangement of atoms. Unshared electrons

(lone pairs) influence the structure (arrangement of atoms about central

atom) but are not considered when describing the molecular structure.

We “see” atoms’ positions, but not the position of the unshared electrons.

12

some examples

• SeBr5-

• ICl3

• CoF63-

• OF2

• COCl2

13

Today’s stuff

• Next time, 63 thru 72

• Look at 76-82 as well…

• Exam 1, 20 October.. Up to Matrix

representations.

14

Today’s q

• What would be the biggest bond angle in

BClBrI ?

15

some reality

Molecule Angle(°) Molecule Angle(°) Molecule Angle(°)

H-X-H H-X-H H-X-H

H2O 104.5 NH3 107.3 CH4 109.5

H2S 92.2 PH3 93.3 SiH4 109.5

H2Se 91.0 AsH3 91.8 GeH4 109.5

H2Te 89.5 SbH3 91.3 SnH4 109.5

16

more reality

Molecule X-O-Y H-C-H

H2O 104.5 CH4 109.5

F2O 103.2 CH3Cl 110.5

Cl2O 111 CH2Cl2 112.0

(CH3)2O 111 CH3Br 111.2

CH3OH 109 CH3I 111.4

CH3OH 109.3

H3C-CH3 109.3

17

any refinement?

• SeBr5-

• ICl3

• CoF63-

• OF2

• COCl2

18 experimental

confirmation of the

postulated

hybridizations?

19for s-p hybrids, bond

angle can give % s

character

• the relationship to right

can give relative s and p

character for equivalent

spx hybridscos =

s

s 1=

p 1

p

20

consider O=CF2

cos108° =s

s 1= 0.309

s = 0.24• note that the orbitals

have more “p”

character due to either

steric effects or

electronegativity of F

substituents

• Remember that p

orbitals have greater

spacial extension!

More e- density to F

21

hybridization and bond

angles

22

hybridization

energetics

• orbital mixing in

hybridization usually

requires energy

• consider C hybridizing to

sp3

• this promotion costs

400 kJ/mole

• Payback? Stronger

bonds, more stable

product!

2s

2p

sp3 hybrids

23

factors influencing

hybridization

• energetics of mixing

how much energy needed for promotion?

• bond energies

depends on the system

will BE’s return energy required for hybridization

• electron pair repulsions

less important in larger atoms

24consider

rehybridization and

repulsions

• for PH3 vs NH3 similar rehybrization energies

are required

• electron pair in ns orbital must be promoted to

sp3 orbital energy (for P, E 600 kJ/mole)

• sizes of atoms are different

75 vs 110 pm

• Therefore orbital mixing is not advantageous

25

Sterics vs. Electronics

What is more important?

Steric effects or electronic effects?

26

Lone Pairs can be

insignificantSn(C5Ph5)2

Has two electrons in valence shell..

27 or stereochemically

active (Janiak, et al.,

Chem. Ber.(1988) 121

p1745)

28

consider reality

Molecule Angle(°) Molecule Angle(°) Molecule Angle(°)

H-X-H H-X-H H-X-H

H2O 104.5 NH3 107.3 CH4 109.5

H2S 92.2 PH3 93.3 SiH4 109.5

H2Se 91.0 AsH3 91.8 GeH4 109.5

H2Te 89.5 SbH3 91.3 SnH4 109.5

29

results

• smaller bond angles in PH3 and lower

congeners indicate low degrees of hybridization

• lone pair is in “s”orbital

• Much lower basicity!!

30

more reality

Molecule X-O-Y H-C-H

H2O 104.5 CH4 109.5

F2O 103.2 CH3Cl 110.5

Cl2O 111 CH2Cl2 112.0

(CH3)2O 111 CH3Br 111.2

CH3OH 109 CH3I 111.4

CH3OH 109.3

H3C-CH3 109.3

31

bond energies,

electronegativities

• this factor can be difficult to deconvolute since

electronegativity difference leads to stronger

bonding because of ionic resonance forms

A-B A+-B-

32

Bent’s Rule

• More electronegative substituents prefer hybrid orbitals

of less “s” character and the less elctronegative

substituents prefer orbitals having more “s” character.

• the “p” character tends to concentrate in orbitals of

weaker covalency and “s” character concentrates in

orbitals of stronger covalency

covalency depends on electronegativity difference and orbital

overlap

33

for example

• PMeCl4

• PMe2Cl3

34

VSEPR is a lie that

works

• we think of H2O as

having this structure:

H

O

H

35

the lone pairs in H2O

look like this:

36

Valence Bond Theory

• developed by Pauling

• invokes overlap

• molecular orbitals are formed from products of

1 electron atomic orbitals

= (1) (2)

• hybridization is required to obtain the correct

geometry

37

today

• Introduction to the chemical bond- yet another

time…

38

Valence Bond Model

and refinements

39

(a) = 1sA• 1sB

(b) delocalization = 1sA(1) 1sB(2) +1sA(2) 1sB(1)

(c) shielding by electrons

(d) ionic resonance forms H-H H+- H- H-- H+ = 1sA(1) 1sB(2) +

1sA(2) 1sB(1)+ ( 1sA(1)1sB(1) + 1sA(2) 1sB(2))

40

aspects of VB theory

• hybridization

• delocalization is not inherent

• ionic forms are not inherent

41

Linear Combinations

of Atomic Orbitals

• a molecular orbital theory

• can use symmetry to make building MO’s

easier

• start with diatomics, O2 and NO

• finish with several triatomics, BeH2 H2O, CO2

and NO2-

some these latter will be “solved” using group theory

42

LCAO assumptions

• same size/energy orbitals equal distribution of electrons, therefore each atomic

orbital makes an equal contribution to the molecular orbital. This condition does

not hold in heterodiatomic molecules.

• use the orbital approximation

the wave function of the N electrons in an atom can be written as the product of the 1

electron wavefunctions, each a fcn of n, l , ml and ms.

for electrons in low-lying orbitals (near the nucleus) in a molecule the wavefunction

resembles that of the lowlying corresponding orbital in the atom.

a reasonable first approximation to the molecular wave functioncan be obtained from

linear combinations of the atomic orbitals, and specifically, the valence orbitals of the

atoms involved.

43

For H2 one can form:

(1) = (1) + (2)

(2) = (1) (2)

44

Consider H2: by

overlap of 1s orbitals

• See the spreadsheets:

just overlapping wavefcns

VB formalism

And LCAO method

45

an important

foundation

• orbitals of different symmetry cannot mix to

form molecular orbitals

• for diatomic molecules, the z axis contains the

line joining the nuclei.

• the px and py orbitals have the same symmetry

as do the pz and s orbitals.

46

combinations

• “s” orbitals

47

bonding and

antibonding orbitals

• antibonding

• bonding

+

+

+

+

48

And still more

• Other combos

• Which are bonding?

• Antibonding?

49

Start with O2

• What are valence orbitals on O atoms?

2s and 2p orbitals

• What orbitals can overlap and form

bonding/antibonding combinations?

2s and 2pz can form bonds

2px and 2py form bonds

• What of relative energies?

50

Cache ab initio MO’s for

O2

• s orbitals give these sigma

bonding and antibonding

orbitals in O2

• (s- )= (2s) + (2s)

• (s- )= (2s) - (2s)

51

pz orbitals give sigma

bonding and antibonding

orbitals

52px and py orbitals give

degenerate and *

orbitals

53

MO diagram for O2

• s orbitals give bonding

and antibonding orbitals

• p orbitals give and

bonding and antibonding

orbitals

*not completely true!

np

p

p

ns

np

ns

54

differences between homo-

and hetero- binuclear

molecules

• expect that electron density

should favor proximity to

more electronegative atom

(bonding MO’s are located on

more en atom)

• antibonding MO’s should

have more character of less

en atom

np

ns

np

ns

N O

55

compare orbitals for

O2 and NO

56compare p orbital

and * orbitals on NO

and O2

57

compare p orbital

and * orbitals

58

differences between homo-

and hetero- binuclear

molecules

• note that in most cases, the

bonding orbitals that are

occupied have more electron

density on the more

electronegative atom

• antibonding MO’s have more

character on the less en atom

np

ns

np

ns

N O

59

VSEPR is a lie that

works

• we think of H2O as

having this structure:

H

O

H

60

the lone pairs in H2O

look like this:

61

This is the orbital

mixing!

bonding

Non-

bonding

62

What is % s

• Hard to tell

• Not exactly25% like sp3, eh.

63

Isolobal analogy

• Orbital similarities between molecular

fragments have predictive power of the types

of compounds formed.

64

What is meant by

isolobal?

To determine whether two molecules are isolobal one must

consider the frontier orbitals (the valence orbitals including the

HOMO and the LUMO).

65

two molecular fragments are

considered isolobal if:

1) the number

2) the symmetry

3) the electron occupancy and

4) the approximate energy

of the frontier orbitals are the same.

66

What’s the use?

• Isolobal analogy:

compounds which have frontier orbitals which have the same

symmetry

and electron occupation

• will tend to form analogous complexes.

67

Making fragments

H

H HH

-H•

H HH

-e-

H HH

•CH3 CH3+

68

What is isolobal with

•CH3 and CH3+?

H

H HH

-H•

H HH

-e-

H HH

•CH3 CH3+

H•

Cl•

H+

HB

HH

69

Simple predictions 1.

• Since •CH3 forms a compound with the •:NH2

radical, so will the isolobal H• and Cl•

fragments.

70

Pictorially,

H

HH

•CH3 NH

H

+ H3C NH2

therefore:

H•

Cl•N

H

H

+

Cl NH2

H NH2

71

however

• these fragments are isolobal to each other.

72

Unlike fragments, CH3+

H

HH H

NH

H+

exists, therefore

H

NH

H

H

NH

H

H

and probably existH

B

HH

+

+