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William A. Goddard, III, 316 Beckman Institute, x3093

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William A. Goddard, III, 316 Beckman Institute, x3093

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  1. Ch120a-Goddard-L01 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 <>

  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 ~ +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.

  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 = 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

  9. Trends: geometries of alkanes CH bond length = 1.095 ± 0.001A CC bond length = 1.526 ± 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 Snap 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 = 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

  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) = 95.0 + 2*7.3 = 109.6 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 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 = 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)

  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 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.

  27. CC double bond energies The bond energies for ethene are De=180.0, D0 = 169.9, D298K = 172.3 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 = 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

  28. 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

  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 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.

  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 2ndp bond of 89 kcal/mol comparable to the first 2P 4S-

  35. Bond energies

  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. 3 1 Infinite structure from tetrahedral bonding plus staggered bonds on adjacent centers 1 1 2 0 2 2 0 0 1 1 c 1 1 1 2nd layer 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

  38. The unit cell of diamond crystal c c f 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

  39. 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 Diamond Structure 5a 3a 1a 5 4b 6 3 5b 2b 7 4 1 3b 2 1c 1b 4a 2a

  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 Si2H6

  42. Comparisons of successive bond energies SiHn and CHn p lobe lobe lobe p 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 ≡ + like structure

  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 D6h symmetry. like structure

  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 orthogonal to both other bonds, leading to lower cost

  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