William A. Goddard, III, [email protected] 316 Beckman Institute, x3093

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L07,08 1

Nature of the Chemical Bond with applications to catalysis, materials

science, nanotechnology, surface science, bioinorganic chemistry, and energy

William A. Goddard, III, [email protected] Beckman Institute, x3093

Charles and Mary Ferkel Professor of Chemistry, Materials Science, and Applied Physics,

California Institute of Technology

Lecture 11 January 31, 2014Graphite, graphene, bucky balls, bucky tubes

Course number: Ch120aHours: 2-3pm Monday, Wednesday, Friday

Teaching Assistants:Sijia Dong <[email protected]>Samantha Johnson <[email protected]>

mailto:[email protected]

mailto:[email protected]


From lecture 6


Bond energiesDe = EAB(R=∞) - EAB(Re) Get from QM calculations. Re is distance at minimum energyD0 = H0AB(R=∞) - H0AB(Re) H0=Ee + ZPE is enthalpy at T=0KZPE = S(½Ћw) This is spectroscopic bond energy from ground vibrational state (0K)Including ZPE changes bond distance slightly to R0Experimental bond enthalpies at 298K and atmospheric pressure D298(A-B) = H298(A) – H298(B) – H298(A-B)

D298 – D0 = 0∫298 [Cp(A) +Cp(B) – Cp(A-B)] dT =2.4 kcal/mol if A and

B are nonlinear molecules (Cp(A) = 4R). {If A and B are atoms D298 – D0 = 0.9 kcal/mol (Cp(A) = 5R/2)}.(H = E + pV assuming an ideal gas)


Snap Bond Energy: Break bond without relaxing the fragments

Snap

Adiabatic

DErelax = 2*7.3 kcal/mol

DsnapDesnap (109.6 kcal/mol) De (95.0kcal/mol)


CH2 +CH2 ethene

Starting with two methylene radicals (CH2) in the ground state (3B1) we can form ethene (H2C=CH2) with both a s bond and a p bond.

The HCH angle in CH2 was 132.3º, but Pauli Repulsion with the new s bond, decreases this angle to 117.6º (cf with 120º for CH3)

3B13B1

3B1


Twisted etheneConsider now the case where the plane of one CH2 is rotated by 90º with respect to the other (about the CC axis)This leads only to a s bond. The nonbonding pl and pr orbitals can be combined into singlet and triplet states

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 Klr ~ 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)


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


CC double bond energies

Breaking the double bond of ethene, the HCH bond angle changes from 117.6º to 132.xº, leading to an increase of 2.35 kcal/mol in the energy of each CH2 so that

Desnap = 180.0 + 4.7 = 184.7 kcal/mol

Since the Desnap = 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

The bond energies for ethene are

De=180.0, D0 = 169.9, D298K = 172.3 kcal/mol


bond energy of F2C=CF2

The snap bond energy for the double bond of ethene of

Desnap = 180.0 + 4.7 = 184.7 kcal/mol

9

1A1

3B1

57 kcal/mol

As an example of how to use this consider the bond energy of F2C=CF2,

Here the 3B1 state is 57 kcal/higher than 1A1 so that the fragment relaxation is 2*57 = 114 kcal/mol, suggesting that the F2C=CF2 bond energy is Dsnap~184-114 = 70 kcal/mol.

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


CC triple bonds

Since the first CCs bond is De=95 kcal/mol and the first CCp 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 2P to 4S-, which costs 17 kcal each, weakening the bond by 34 kcal/mol. Adding this to the 55 would lead to a total 2nd p bond of 89 kcal/mol comparable to the first

2P

4S-



Cn

What is the structure of C3?


Cn


Energetics Cn

Note extra stability of odd Cn by 33 kcal/mol, this is because odd Cn has an empty px orbital at one terminus and an empty py on the other, allowing stabilization of both p systems


Stability of odd Cn



Bond energies and thermochemical calculations


Bond energies and thermochemical calculations


Heats of Formation


Heats of Formation


Heats of Formation


Heats of Formation


Bond energies


Bond energies


Bond energies

Both secondary



Average bond energies


Average bond energies


Real bond energies

Average bond energies of little use in predicting

mechanism


Group values


Group functions of propane


Examples of using group values


Group values


Strain


Strain energy cyclopropane from Group values


Strain energy c-C3H6

using real bond

energies


Stained GVB orbitals of cyclopropane


Benson Strain energies


Allyl radical


Allyl Radical


Allyl wavefunctions

It is about 12 kcal/mol


Resonance in thermochemical Calculations


Resonance in thermochemical Calculations


Resonance energy butadiene


Benzene resonance


Benzene resonance


Benzene resonance


Benzene resonance


Benzene resonance


Benzene and Resonance

referred to as Kekule or VB structures


Resonance


Benzene wavefunction

≡ +

benzene as

is a superposition of the VB structures in (2)


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 D6h symmetry.


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.

1.5x A

1.34 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 orthogonal to both other bonds, leading to lower cost


graphene

This is referred to as graphene

Graphene: CC=1.4210ABond order = 4/3Benzene: CC=1.40 BO=3/2Ethylene: CC=1.34 BO = 2CCC=120°Unit cell has 2 carbon atoms

1x1 Unit cell


Graphene band structureUnit cell has 2 carbon atomsBands: 2pp orbitals per cell2 bands of states each with N states where N is the number of unit cells2 p electrons per cell 2N electrons for N unit cellsThe lowest N MOs are doubly occupied, leaving N empty orbitals.

1x1 Unit cell

1st band2nd band

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

Get semi metal


Graphite

Stack graphene layers as ABABABCan also get ABCABC RhombohedralAAAA stacking much higher in energy

Distance between layers = 3.3545ACC bond = 1.421Only weak London dispersion attraction between layersDe = 1.0 kcal/mol CEasy to slide layers, good lubricant

Graphite: D0K=169.6 kcal/mol, in plane bond = 168.6Thus average in-plane bond = (2/3)168.6 = 112.4 kcal/mol112.4 = sp2 s + 1/3 pDiamond: average CCs = 85 kcal/mol p = 3*27=81 kcal/mol


energetics


Stopped Feb. 4, 2013


Graphene: generalize benzene in all directions


Have to terminate graphene: two simple cases

Armchair edge

For each edge atom break two

sp2 sigma bonds but form bent pi bond in plane

111.7 – 20 = 92 kcal/mol

Length = 3*1.4=4.2A

22 kcal/molA

Zig-zag edge

For each edge atom break sp2

sigma bond, maybe not break

pi bond?

111.7/2 = 56 kcal/mol per

dangling bond

Length = 1.4*sqrt(3)=

2.42A

23 kcal/mol/AThus both graphene

ribbon surfaces (edges) have

similar energies


C60 flat sheet

Cut from graphene

6 arm chair pairs @92

5 zig-zag atoms @56

Total cost 832 kcal/mol!


C60 fullerene

No broken bonds

Just ~11.3 kcal/mol strain at each atom

678 kcal/mol

Compare with 832 kcal/mol for flat sheet

Lower in energy than flat sheet by 154 kcal/mol!


First observation

Heath, Smalley, Krotos

Laser evaporation of carbon + supersonic nozzle

Observe various sized clusters in mass spect

Change various conditions found peak at C60!

Smalley and Krotos each independently postulated futball (soccer ball structure) ~1986

^ H. W. Kroto, J. R. Heath, S. C. O'Brien, R. F. Curl and R. E. Smalley (1985). "C60: Buckminsterfullerene". Nature

318: 162–163. doi:10.1038/318162a0.

http://en.wikipedia.org/wiki/Fullerene

http://en.wikipedia.org/wiki/Digital_object_identifier

http://dx.doi.org/10.1038/318162a0


Nature 1985: discovery of C60


Evidence for C60, Nature 1985

760 torr He

10 torr He

maximize cluster-cluster reactions in integration cup


1985-1990 Many papers on C60, no definitive proof that it had fullerene structure, lots of skepticism


1985-1990 Many papers on C60, no definitive proof that it had fullerene structure, lots of skepticism

Then, Nature 347, 354 - 358 (27 September 1990)W. Krätschmer, Lowell D. Lamb, K. Fostiropoulos & Donald

R. Huffman; Solid C60: a new form of carbon

In 1990 physicists W. Krätschmer and D.R. Huffman for the first time produced isolable quantities of C60 by causing an arc

between two graphite rods to burn in a helium atmosphere and extracting the carbon condensate so formed using an organic

solvent.

A new form of pure, solid carbon has been synthesized consisting of a somewhat disordered hexagonal close

packing of soccer-ball-shaped C60 molecules. Infrared spectra and X-ray diffraction studies of the molecular

packing confirm that the molecules have the anticipated 'fullerene' structure.

Mass spectroscopy shows that the C70 molecule is present at levels of a few per cent.


Nature 1990, Krätschmer, Lamb, Fostiropoulos, Huffman

Sears arc welder with flowing He, get soot of C60. grams per hour


NMR the key experiment

Carbon 13 NMR spectrum of C705 peaks, definitive proof of fullerene structure

Carbon 13 NMR spectrum of C601 peak

Definitive proof that C60 is fullerene


Polyyne chain

precursors fullerenes, all even



C540

All fullerens have 12 pentagonal rings

Documents

William A. Goddard, III, [email protected] 316 Beckman Institute, x3093