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

Teaching Assistants: Caitlin Scott <[email protected]>Hai Xiao [email protected]; Fan Liu <[email protected]>

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

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

mailto:[email protected]




Last time


Summary, bonding to form hydrides

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

Thus use 1A1 AH2 for SiH2 and CF2 get pyramidal AH3

Use 3B1 for CH2 get planar AH3.

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.


Now combine Carbon fragments to form larger molecules (old chapter 7)

Starting with the ground state of CH3 (planar), we bring two together to form ethane, H3C-CH3.

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

107.7º

111.2º

1.526A

1.095A1.086A120.0º


Bonding (GVB) orbitals of ethane (staggered)

Note nodal planes from

orthogonalization to CH bonds on

right C


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, SCH-CH Pauli requires that they be orthogonalized, which leads to a repulsion that increases exponentially with decreasing distance RCH-CH.

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


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.

qH ~ +0.15

qC ~ -0.45

This leads to qH ~ +0.15 and qC ~ -0.45.

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 CH3-C=C-CH3, 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 decrease by only 1/3. However overlap decreases exponentially.


Propane

Replacing an H of ethane with CH3, leads to propane

Keeping both CH3 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


Trends: geometries of alkanes

CH bond length = 1.095 ± 0.001A

CC bond length = 1.526 ± 0.001A

CCC bond angles

HCH bond angles


Bond energiesDe = EAB(R=∞) - EAB(Re) e for equilibrium)Get from QM calculations. Re is distance at minimum energy.


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 R0


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)


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 isD298(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)}.


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)


Bond energies for ethane

D0 = 87.5 kcal/mol

ZPE (CH3) = 18.2 kcal/mol,

ZPE (C2H6) = 43.9 kcal/mol,

De = D0 + 7.5 = 95.0 kcal/mol (this is calculated from QM)

D298 = 87.5 + 2.4 = 89.9 kcal/mol

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


The snap Bond energy

In breaking the CC bond of ethane the geometry changes from CC=1.526A, HCH=107.7º, CH=1.095A

To CC=∞, HCH=120º, CH=1.079A

Thus the net bond energy involves both breaking the CC bond and relaxing the CH3 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

De,snap (CH3-CH3) = 95.0 + 2*7.3 = 109.6 kcal/mol


Substituent effects on Bond energiesThe 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


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 C2H6 has 6 CH-CH interactions lost upon breaking the bond,

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.


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 explanation PUT IN ACTUAL RELAXATION ENERGIES


Bond energies: Compare to CF3-CF3

For CH3-CH3 we found a snap bond energy of

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

Because the relaxation of tetrahedral CH3 to planar gains 7.3 kcal/mol

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

Thus we might estimate that for CF3-CF3 the bond energy would be De = 109.6 kcal/mol, hence D298 ~ 110-5=105

Indeed the experimental value is D298=98.7±2.5 kcal/mol suggesting that the main effect in substituent effects is relaxation (the remaining effects might be induction and steric)


New material lecture 6, January 18, 2012


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)


Comparison of The GVB bonding orbitals of ethene

and methylene


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


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 pp orbitals.


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 od

Desnap = 180.0 + 4.7 = 184.7 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


Bond energies double bondsAlthough the ground state of CH2 is 3B1 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.


C=C bond energies


CC triple bonds

Starting with two CH radicals in the 4S- 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

De = 235.7, D0 = 227.7, D298K = 229.8 kcal/mol

Which can be compared to De of 180.0 for H2C=CH2 and 95.0 for H3C-CH3.


GVB orbitals of HCCH


GVB orbitals of CH 2P and 4S- state


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-


Bond energies


DiamondReplacing 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


Infinite structure from tetrahedral bonding plus staggered bonds on adjacent centers

Chair configuration

of cylco-hexane

Not shown: zero layer just like 2nd layer but above layer 13rd layer just like the 1st layer but below layer 2

2nd layer

1st layer

2nd layer

1st layer

2nd layer

1st layer

1

1

c 1

3 1

02

1

2

1

0

11

20


The unit cell of diamond crystalAn alternative view of the diamond structure is in terms of cubes of side a, that can be translated in the x, y, and z directions to fill all space.

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

• all 8 corners (but only 1/8 inside the cube): (0,0,0)• 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 this cube by multiples of a in the x,y,z directions

There are atoms at c c

c c

c c

c c

ff

f

f

f

f

ii

ii


Diamond Structure

12

1a

1c1b

32b

2a

Start with C1 and make 4 bonds to form

a tetrahedron.

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

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

Chair cyclohexane ring: 1-2-3-3b-7-1c

43b

3a

54b

4a

5b

5a

67


Properties of diamond crystals


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 Si2H6


Comparisons of successive bond energies SiHn and CHn

p lobe

p

lobep

p

lobe

lobe


Redo the next sections

Talk about heats formation first

Then group additivity

Then resonance etc


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


Allyl Radical


Allyl wavefunctions

It is about 12 kcal/mol


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


Heats of Formation


Bond energies


Bond energies

Both secondary


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


Resonance in thermochemical Calculations


Resonance energy butadiene


Allyl radical


Benzene resonance

william a. goddard, iii, [email protected] 316 beckman institute, x3093

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