Lecture 6 January 18, 2012 ΔHf, Group additivity CC Bonds diamond,

Lecture 6 January 18, 2012

CC Bonds diamond, ΔHf, Group additivity

Nature of the Chemical Bond with applications to catalysis, materials science, nanotechnology, surface science, bioinorganic chemistry, and energy

Course number: Ch120a

Hours: 2-3pm Monday, Wednesday, Friday

William A. Goddard, III, wag@wag.caltech.edu

316 Beckman Institute, x3093

Charles and Mary Ferkel Professor of Chemistry, Materials

Science, and Applied Physics,

California Institute of Technology

Teaching Assistants: Caitlin Scott < cescott@caltech.edu

>

Hai Xiao xiao@caltech.edu

; Fan Liu <fliu@wag.caltech.edu>

Ch120a-Goddard-L07,08 © copyright 2011 William A. Goddard III, all rights reserved

Goddard-

Last time


2

Summary, bonding to form hydrides

General principle: start with ground state of AH n form the ground state of AH n+1 and add H to

Thus use 1 A 1 AH

2 for SiH

2 and CF

2 get pyramidal AH

3

Use 3 B

1 for CH

2 get planar AH

3

.

For less than half filled p shell, the presence of empty p orbitals allows the atom to reduce electron correlation of the

(ns) pair by hybridizing into this empty orbital.

This has remarkable consequences on the states of the Be,

B, and C columns.


3

Now combine Carbon fragments to form larger molecules

(old chapter 7)

Starting with the ground state of CH

3

(planar), we bring two together to form ethane, H

3

C-CH

3

.

As they come together to bond, the CH bonds bend back from the CC bond to reduce overlap

(Pauli repulsion or steric interactions between the CH bonds on opposite C).

At the same time the 2pp radical orbital on each

C mixes with 2s character, pooching it toward the corresponding hybrid orbital on the other C

120.0

º 1.086A

107.7

º

1.095A

1.526A

Ch120a-Goddard-L07,08

111.2

º

© copyright 2011 William A. Goddard III, all rights reserved

4

Bonding (GVB) orbitals of ethane (staggered)

Note nodal planes from orthogonalization to CH bonds on right C


5

Staggered vs. Eclipsed

There are two extreme cases for the orientation about the

CC axis of the two methyl groups

The salient difference between these is the overlap of the CH bonding orbitals on opposite carbons.

To whatever extent they overlap, S

CH-CH

Pauli requires that they be orthogonalized, which leads to a repulsion that increases exponentially with decreasing distance R

CH-CH

.

The result is that the staggered conformation is favored over eclipsed by 3.0 kcal/mol


6

Alternative interpretation

The bonding electrons are distributed over the molecule, but it is useful to decompose the wavefunction to obtain the net charge on each atom.

This leads to q

H

~ +0.15 and q

C

~ -0.45. q

C q

H

~ -0.45

~ +0.15

These charges do NOT indicate the electrostatic energies within the molecule, but rather the electrostatic energy for interacting with an external field.

Even so, one could expect that electrostatics would favor staggered.

The counter example is CH

3

-C=C-CH

3

, which has a rotational barrier of 0.03 kcal/mol (favoring eclipsed). Here the CH bonds are ~ 3 times that in CH3-CH3 so that electrostatic effects would


7

Propane

Replacing an H of ethane with CH

3

, leads to propane

Keeping both CH

3 groups staggered leads to the unique structure

Details are as shown. Thus the bond angles are

HCH = 108.1 and 107.3 on the CH3

HCH =106.1 on the secondary C

CCH=110.6 and 111.8

CCC=112.4,

Reflecting the steric effects


8

Trends: geometries of alkanes

CH bond length = 1.095 ± 0.001A

CC bond length = 1.526 ± 0.001A

CCC bond angles

HCH bond angles


9

Bond energies

D e

= E

AB

(R= ∞) - E e for equilibrium)

AB

(R e

)

Get from QM calculations. Re is distance at minimum energy.


10

Bond energies

D e

= E

AB

(R= ∞) - E

AB

(R e

)

Get from QM calculations. Re is distance at minimum energy

D

0

= H

0AB

(R= ∞) - H

0AB

(R e

)

H

0

=Ee + ZPE is enthalpy at T=0K

ZPE =

S(

½Ћ w

)

This is spectroscopic bond energy from ground vibrational state (0K)

Including ZPE changes bond distance slightly to R

0


11

Bond energies

D e

= E

AB

(R= ∞) - E

AB

(R e

)

Get from QM calculations. Re is distance at minimum energy

D

0

= H

0AB

(R= ∞) - H

0AB

(R e

)

H

0

=Ee + ZPE is enthalpy at T=0K

ZPE =

S(

½Ћ w

)

This is spectroscopic bond energy from ground vibrational state (0K)

Including ZPE changes bond distance slightly to R

0

Experimental bond enthalpies at 298K and atmospheric pressure

D

298

D

298

(A-B) = H

298

– D

0

=

0

∫

298

(A) – H

298

(B) – H

298

(A-B)

[C p

(A) +C p

(B) – C p

(A-B)] dT = 2.4 kcal/mol if A and

B are nonlinear molecules (C p

(A) = 4R).

{If A and B are atoms D

298

– D

0

= 0.9 kcal/mol (C p

(A) = 5R/2)}.

12


Bond energies, temperature corrections

Experimental measurements of bond energies, say at 298K, require an additional correction from QM or from spectroscopy.

The experiments measure the energy changes at constant pressure and hence they measure the enthalpy,

H = E + pV (assuming an ideal gas)

Thus at 298K, the bond energy is

D

298

(A-B) = H

298

(A) – H

298

(B) – H

298

(A-B)

D

298

– D

0

=

0

∫

298 [C p

(A) +C p

(B) – C p

(A-B)] dT =2.4 kcal/mol if A and B are nonlinear molecules (C p

(A) = 4R).

{If A and B are atoms D

298

– D

0

= 0.9 kcal/mol (C p

(A) = 5R/2)}.


13

Snap Bond Energy: Break bond without relaxing the fragments

De snap

D

Snap

D

E relax

= 2*7.3 kcal/mol

Adiabatic

D e

(95.0kcal/mol)


14

Bond energies for ethane

D

0

= 87.5 kcal/mol

ZPE (CH

3

) = 18.2 kcal/mol,

ZPE (C

2

H

6

) = 43.9 kcal/mol,

D e

= D

0

+ 7.5 = 95.0 kcal/mol (this is calculated from QM)

D

298

= 87.5 + 2.4 = 89.9 kcal/mol

This is the quantity we will quote in discussing bond breaking processes


15

The snap Bond energy

In breaking the CC bond of ethane the geometry changes from

CC=1.526A, HCH=107.7

º, CH=1.095A

To CC= ∞, HCH=120º, CH=1.079A

Thus the net bond energy involves both breaking the CC bond and relaxing the CH

3 fragments.

We find it useful to separate the bond energy into

The snap bond energy (only the CC bond changes, all other bonds and angles of the fragments are kept fixed)

The fragment relaxation energy.

This is useful in considering systems with differing substituents.

For CH3 this relation energy is 7.3 kcal/mol so that

D e,snap

(CH

3

-CH

3


) = 95.0 + 2*7.3 = 109.6 kcal/mol


16

Substituent effects on Bond energies

The strength of a CC bond changes from 89.9 to 70 kcal/mol as the various H are replace with methyls.Explanations given include:

•Ligand CC pair-pair repulsions

•Fragment relaxation

•Inductive effects


17

Ligand CC pair-pair repulsions:

Each H to Me substitution leads to 2 new CH bonds gauche to the original CC bond, which would weaken the CC bond.

Thus C

2

H

6 bond, has 6 CH-CH interactions lost upon breaking the

But breaking a CC bond of propane loses also two addition

CC-CH interactions.

This would lead to linear changes in the bond energies in the table, which is approximately true.

However it would suggest that the snap bond energies would decrease, which is not correct.


18

Fragment relaxation

Because of the larger size of Me compared to H, there will be larger ligand-ligand interaction energies and hence a bigger relaxation energy in the fragment upon relaxing form tetrahedral to planar geometries.

In this model the snap bond enegies are all the same.

All the differences lie in the relaxation of the fragments.

This is observed to be approximately correct

Inductive effect

A change in the character of the CC bond orbital due to replacement of an H by the Me.

Goddard believes that fragment relaxation is the correct


19

Bond energies: Compare to CF

3

-CF

3

For CH

3

-CH

3 we found a snap bond energy of

D e

= 95.0 + 2*7.3 = 109.6 kcal/mol

Because the relaxation of tetrahedral CH

3

7.3 kcal/mol to planar gains

For CF

3

-CF

3

, there is no such relaxation since CF3 wants to be pyramidal, FCF~111 º

Thus we might estimate that for CF

3

-CF

3 would be D e

= 109.6 kcal/mol, hence D

298 the bond energy

~ 110-5=105

Indeed the experimental value is D

298

=98.7

±2.5 kcal/mol suggesting that the main effect in substituent effects is relaxation (the remaining effects might be induction and steric)


20

New material lecture 6, January 18, 2012


21

CH2 +CH2  ethene

Starting with two methylene radicals (CH

2

) in the ground state ( 3 B

1

) we can form ethene

(H2C=CH2) with both a s bond and a p bond.

The HCH angle in CH2 was 132.3

º, but Pauli Repulsion with the new s bond, decreases this angle to 117.6

º (cf with 120º for CH

3

)


22

Comparison of The GVB bonding orbitals of ethene and methylene


23

Twisted ethene

Consider now the case where the plane of one CH

2

90 º with respect to the other (about the CC axis)

This leads only to a s bond. The nonbonding p l and p r orbitals can be combined into singlet and triplet states is rotated by

Here the singlet state is referred to as N (for Normal) and the triplet state as T.

Since these orbitals are orthogonal, Hund’s rule suggests that T is lower than N (for 90º). The K lr

~ 0.7 kcal/mol so that the splitting should be ~1.4 kcal/mol.

Voter, Goodgame, and Goddard [Chem. Phys. 98 , 7 (1985)] showed that N is below T by 1.2 kcal/mol, due to Intraatomic Exchange ( s,p on same center) 24

Twisting potential surface for ethene

The twisting potential surface for ethene is shown below. The

N state prefers θ=0º to obtain the highest overlap while the T state prefers θ=90º to obtain the lowest overlap


25

geometries

For the N state (planar) the CC bond distance is 1.339A, but this increases to

1.47A for the twisted form with just a single s bond.

This compares with 1.526 for the CC bond of ethane.

Probably the main effect is that twisted ethene has very little CH

Pauli Repulsion between CH bonds on opposite C, whereas ethane has substantial interactions.

This suggests that the intrinsic CC single bond may be closer to

1.47A

For the T state the CC bond for twisted is also 1.47A, but increases to 1.57 for planar due to Orthogonalization of the triple coupled p p orbitals.


26

CC double bond energies

The bond energies for ethene are

D e

=180.0, D

0

= 169.9, D

298K

= 172.3 kcal/mol

Breaking the double bond of ethene, the HCH bond angle changes from 117.6

º to 132.xº, leading to an increase of 2.35 kcal/mol in the energy of each CH

2 so that

D esnap

= 180.0 + 4.7 = 184.7 kcal/mol

Since the D esnap

= 109.6 kcal/mol, for H3C-CH3,

The p bond adds 75.1 kcal/mol to the bonding.

Indeed this is close to the 65kcal/mol rotational barrier.

For the twisted ethylene, the CC bond is De = 180-65=115

Desnap = 115 + 5 =120. This increase of 10 kcal/mol compared to ethane might indicate the effect of CH repulsions


27

bond energy of F

2

C=CF

2

The snap bond energy for the double bond of ethene od

D esnap

= 180.0 + 4.7 = 184.7 kcal/mol

As an example of how to use this consider the bond energy of F

2

C=CF

2

,

Here the 3 B

1 state is 57 kcal/higher than 1 A

1 so that the fragment relaxation is 2*57 = 114 kcal/mol, suggesting that the F

2

C=CF

2 bond energy is D snap

~184-114 = 70 kcal/mol.

The experimental value is D298 ~ 75 kcal/mol, close to the prediction


28

Bond energies double bonds

Although the ground state of CH2 is 3 B

1 by 9.3 kcal/mol, substitution of one or both H with CH3 leads to singlet ground states. Thus the CC bonds of these systems are weakened because of this promotion energy.


29

C=C bond energies


30

CC triple bonds

Starting with two CH radicals in the 4

S

state we can form ethyne (acetylene) with two p bonds and a s bond.

This leads to a CC bond length of 1.208A compared to 1.339 for ethene and 1.526 for ethane.

The bond energy is

D e

= 235.7, D

0

= 227.7, D

298K

= 229.8 kcal/mol

Which can be compared to De of 180.0 for H2C=CH2 and

95.0 for H3C-CH3.


31

GVB orbitals of HCCH


32

GVB orbitals of CH 2

P and 4

S

- state


33

CC triple bonds

Since the first CC s bond is D e

=95 kcal/mol and the first CC p bond adds 85 to get a total of 180, one might wonder why the

CC triple bond is only 236, just 55 stronger.

The reason is that forming the triple bond requires promoting the CH from 2

P to 4

S

, which costs 17 kcal each, weakening the bond by 34 kcal/mol. Adding this to the 55 would lead to a total 2 nd p bond of 89 kcal/mol comparable to the first

2

P

4

S

-


34

Bond energies


35


36

Diamond

Replacing all H atoms of ethane and with methyls, leads to with a staggered conformation

Continuing to replace H with methyl groups forever, leads to the diamond crystal structure, where all C are bonded tetrahedrally to four C and all bonds on adjacent C are staggered

A side view is

This leads to the diamond crystal structure. An expanded view is on the next slide


37

Infinite structure from tetrahedral bonding plus staggered bonds on adjacent centers

2 nd layer

3 1

1

1st layer 1

2

0

2

2 nd layer

2 0

0 1

1 c 1

1st layer

1

1

2 nd layer

1st layer

Chair configuration of cylcohexane

Not shown: zero layer just like 2 nd layer but above layer 1

3 rd layer just like the 1 st layer but below layer 2

38

The unit cell of diamond crystal c f

An alternative view of the c i c diamond structure is in terms of i f cubes of side a, that can be f f translated in the x, y, and z f i directions to fill all space.

i c

Note the zig-zag chains c-i-f-i-c f and cyclohexane rings (f-i-f)-(i-f-i) c

There are atoms at

•all 8 corners (but only 1/8 inside the cube): (0,0,0) c

•all 6 faces (each with ½ in the cube): (a/2,a/2,0), (a/2,0,a/2),

(0,a/2,a/2)

•plus 4 internal to the cube: (a/4,a/4,a/4), (3a/4,3a/4,a/4),

(a/4,3a/4,3a/4), (3a/4,a/4,3a/4),

Thus each cube represents 8 atoms.

All other atoms of the infinite crystal are obtained by translating c c

39


Diamond Structure

Start with C1 and make 4 bonds to form a tetrahedron.

4 b

2 b

6

4

2

1 b

5

3

1

4 a

2 a

5 a

3 a

1 a


5 b

3 b

1 c

7

Now bond one of these atoms, C2, to 3 new C so that the bond are staggered with respect to those of C1.

Continue this process.

Get unique structure: diamond

Note: Zig-zag chain

1 b

-1-2-3-4-5-6

Chair cyclohexane ring: 1-2-3-3 b

-7-1 c


40

Properties of diamond crystals


41

Properties of group IV molecules (IUPAC group 14)

1.526

There are 4 bonds to each atom, but each bond connects two atoms.

Thus to obtain the energy per bond we take the total heat of vaporization and divide by two.

Note for Si, that the average bond is much different than for Si

2

H

42

6

Comparisons of successive bond energies SiH n and CH n p lobe lobe p

Ch120a-Goddard-L07,08 © copyright 2011 William A. Goddard III, all rights reserved lobe lobe p p

43

Redo the next sections

Talk about heats formation first

Then group additivity

Then resonance etc


44

Benzene and Resonance referred to as Kekule or VB structures


45

Resonance


46

Benzene wavefunction

≡ is a superposition of the VB structures in (2) benzene as

+


47

More on resonance

That benzene would have a regular 6-fold symmetry is not obvious. Each VB spin coupling would prefer to have the double bonds at ~1.34A and the single bond at ~1.47 A (as the central bond in butadiene)

Thus there is a cost to distorting the structure to have equal bond distances of 1.40A.

However for the equal bond distances, there is a resonance stabilization that exceeds the cost of distorting the structure, leading to D

6h symmetry.


48

Cyclobutadiene

For cyclobutadiene, we have the same situation, but here the rectangular structure is more stable than the square.

That is, the resonance energy does not balance the cost of making the bond distances equal.

1.34 A

1.5x A

The reason is that the pi bonds must be orthogonalized, forcing a nodal plane through the adjacent C atoms, causing the energy to increase dramatically as the 1.54 distance is reduced to 1.40A.

For benzene only one nodal plane makes the pi bond


49

graphene

Graphene: CC=1.4210A

Bond order = 4/3

Benzene: CC=1.40 BO=3/2

Ethylene: CC=1.34 BO = 2

CCC=120 °

Unit cell has 2 carbon atoms

1x1 Unit cell

This is referred to as graphene


50

1x1 Unit cell

Graphene band structure

Unit cell has 2 carbon atoms

Bands: 2p p orbitals per cell

 2 bands of states each with N states where N is the number of unit cells

2 p electrons per cell  2N electrons for

N unit cells

The lowest N MOs are doubly occupied, leaving N empty orbitals.

The filled 1 st band touches the empty 2 nd band at the

Fermi energy

Get semi metal


2 nd band

1 st band


51

Graphite

Stack graphene layers as ABABAB

Can also get ABCABC Rhombohedral

AAAA stacking much higher in energy

Distance between layers = 3.3545A

CC bond = 1.421

Only weak London dispersion attraction between layers

D e

= 1.0 kcal/mol C

Easy to slide layers, good lubricant

Graphite: D

0K

=169.6 kcal/mol, in plane bond = 168.6

Thus average in-plane bond = (2/3)168.6 = 112.4 kcal/mol

112.4 = sp 2 s

+ 1/3 p

Diamond: average CCs = 85 kcal/mol  p

= 3*27=81 kcal/mol


52

energetics


53

Allyl Radical


54

Allyl wavefunctions

It is about 12 kcal/mol


55

Cn

What is the structure of C

3

?


56

Cn


57

Energetics Cn

Note extra stability of odd C n by 33 kcal/mol, this is because odd

C n has an empty p x orbital at one terminus and an empty p other, allowing stabilization of both p systems y on the


58

Stability of odd Cn


59


60

Bond energies and thermochemical calculations


61

Bond energies and thermochemical calculations


62

Heats of Formation


63

Heats of Formation


64

Heats of Formation


65

Heats of Formation


66

Bond energies


67

Bond energies


68

Bond energies

Both secondary


69


70

Average bond energies


71

Average bond energies


72

Real bond energies

Average bond energies of little use in predicting mechanism


73

Group values


74

Group functions of propane


75

Examples of using group values


76

Group values


77

Strain


78

Strain energy cyclopropane from Group values


79

Strain energy c-C3H6 using real bond energies


80

Stained GVB orbitals of cyclopropane


81

Benson Strain energies


82

Resonance in thermochemical Calculations


83

Resonance in thermochemical Calculations


84

Resonance energy butadiene


85

Allyl radical


86

Benzene resonance


87

Benzene resonance


88

Benzene resonance


89

Benzene resonance


90

Benzene resonance


91

Lecture 6 January 18, 2012 ΔHf, Group additivity CC Bonds diamond,

Lecture 6 January 18, 2012

CC Bonds diamond, ΔHf, Group additivity

Properties of diamond crystals

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Lecture 6 January 18, 2012 ΔHf, Group additivity CC Bonds diamond,