1. 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,
316 Beckman Institute, x3093
Charles and Mary Ferkel Professor of Chemistry, Materials Science, and Applied Physics, California Institute of Technology
Teaching Assistants: Caitlin Scott
Hai Xiao Fan Liu
Ch120a-Goddard-L01

2. Last time

3. 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.

4. 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
120.0º
1.086A
1.095A
107.7º
1.526A
111.2º

5. Bonding (GVB) orbitals of ethane (staggered)
Note nodal planes from orthogonalization to CH bonds on right C

6. 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

7. 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 ~ and qC ~
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.

8. 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 = and on the CH3
HCH =106.1 on the secondary C
CCH=110.6 and 111.8
CCC=112.4,
Reflecting the steric effects

9. Trends: geometries of alkanes
CH bond length = ± 0.001A
CC bond length = ± 0.001A
CCC bond angles
HCH bond angles

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

11. Bond energies De = EAB(R=∞) - EAB(Re)
Get from QM calculations. Re is distance at minimum energy
D0 = H0AB(R=∞) - H0AB(Re)
H0=Ee + ZPE is enthalpy at T=0K
ZPE = S(½Ћw)
This is spectroscopic bond energy from ground vibrational state (0K)
Including ZPE changes bond distance slightly to R0

12. Bond energies De = EAB(R=∞) - EAB(Re)
Get from QM calculations. Re is distance at minimum energy
D0 = H0AB(R=∞) - H0AB(Re)
H0=Ee + ZPE is enthalpy at T=0K
ZPE = S(½Ћw)
This is spectroscopic bond energy from ground vibrational state (0K)
Including ZPE changes bond distance slightly to R0
Experimental 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)

13. 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
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)}.

14. Snap Bond Energy: Break bond without relaxing the fragments
DErelax = 2*7.3 kcal/mol
Adiabatic
Dsnap
De (95.0kcal/mol)
Desnap (109.6 kcal/mol)

15. Bond energies for ethane
D0 = 87.5 kcal/mol
ZPE (CH3) = 18.2 kcal/mol,
ZPE (C2H6) = 43.9 kcal/mol,
De = D = 95.0 kcal/mol (this is calculated from QM)
D298 = = 89.9 kcal/mol
This is the quantity we will quote in discussing bond breaking processes

16. 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) = *7.3 = kcal/mol

17. 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

18. 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.

19. Fragment relaxation Inductive effect
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

20. Bond energies: Compare to CF3-CF3
For CH3-CH3 we found a snap bond energy of
De = *7.3 = 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 = 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)

21. New material lecture 6, January 18, 2012

22. 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)

23. Comparison of The GVB bonding orbitals of ethene and methylene

24. Twisted ethene
Consider 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)

25. 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

26. 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 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.

27. CC double bond energies
The bond energies for ethene are
De=180.0, D0 = 169.9, D298K = kcal/mol
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 = = kcal/mol
Since the Desnap = 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 = =115 Desnap = =120. This increase of 10 kcal/mol compared to ethane might indicate the effect of CH repulsions

28. bond energy of F2C=CF2
The snap bond energy for the double bond of ethene od
Desnap = = 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~ = 70 kcal/mol.
The experimental value is D298 ~ 75 kcal/mol, close to the prediction

29. Bond energies double bonds
Although 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.

30. C=C bond energies

31. 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 for ethene and for ethane.
The bond energy is
De = 235.7, D0 = 227.7, D298K = kcal/mol
Which can be compared to De of for H2C=CH2 and 95.0 for H3C-CH3.

32. GVB orbitals of HCCH

33. GVB orbitals of CH 2P and 4S- state

34. 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-

35. Bond energies

37. 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

38. Chair configuration of cylco-hexane
Infinite structure from tetrahedral bonding plus staggered bonds on adjacent centers
2nd layer 1 c 3 2
1st layer
2nd layer
1st layer
2nd layer
Chair configuration of cylco-hexane
1st layer
Not shown: zero layer just like 2nd layer but above layer 1
3rd layer just like the 1st layer but below layer 2

39. The unit cell of diamond crystal
An 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) c c i i f f f f i i c c f c c
There are atoms at
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

40. 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
Chair cyclohexane ring: b-7-1c
Diamond Structure 5a 3a 1a 5 4b 6 3 5b 2b 7 4 1 3b 2 1c 1b 4a 2a

41. Properties of diamond crystals

42. 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

43. Comparisons of successive bond energies SiHn and CHn
lobe
lobe
lobe p
lobe p p

44. Redo the next sections Talk about heats formation first
Then group additivity
Then resonance etc

45. Benzene and Resonance
referred to as Kekule or VB structures

46. Resonance

47. Benzene wavefunction benzene as ≡ +
is a superposition of the VB structures in (2) ≡ +
like structure

48. 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.
like structure

49. 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 orthogonal to both other bonds, leading to lower cost

50. graphene Graphene: CC=1.4210A Bond order = 4/3 Benzene: CC=1.40 BO=3/2
This is referred to as 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

51. Graphene band structure
1x1 Unit cell
Unit cell has 2 carbon atoms
Bands: 2pp 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 1st band touches the empty 2nd band at the Fermi energy
2nd band
Get semi metal
1st band

52. Graphite Stack graphene layers as ABABAB
Can also get ABCABC Rhombohedral
AAAA stacking much higher in energy
Distance between layers = A
CC bond = 1.421
Only weak London dispersion attraction between layers
De = 1.0 kcal/mol C
Easy to slide layers, good lubricant
Graphite: D0K=169.6 kcal/mol, in plane bond = 168.6
Thus average in-plane bond = (2/3)168.6 = kcal/mol
112.4 = sp2 s + 1/3 p
Diamond: average CCs = 85 kcal/mol  p = 3*27=81 kcal/mol

53. energetics

54. Allyl Radical

55. Allyl wavefunctions
It is about 12 kcal/mol

56. Cn
What is the structure of C3?

57. Cn

58. 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

59. Stability of odd Cn

61. Bond energies and thermochemical calculations

62. Bond energies and thermochemical calculations

63. Heats of Formation

64. Heats of Formation

65. Heats of Formation

66. Heats of Formation

67. Bond energies

68. Bond energies

69. Bond energies
Both secondary

71. Average bond energies

72. Average bond energies

73. Average bond energies of little use in predicting mechanism
Real bond energies
Average bond energies of little use in predicting mechanism

74. Group values

75. Group functions of propane

76. Examples of using group values

77. Group values

78. Strain

79. Strain energy cyclopropane from Group values

80. Strain energy c-C3H6 using real bond energies

81. Stained GVB orbitals of cyclopropane

82. Benson Strain energies

83. Resonance in thermochemical Calculations

84. Resonance in thermochemical Calculations

85. Resonance energy butadiene

86. Allyl radical

87. Benzene resonance

88. Benzene resonance

89. Benzene resonance

90. Benzene resonance

91. Benzene resonance