© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L07,08 Ch120a- Goddard- L01 1 Nature of the Chemical Bond with applications.

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L07,08 Ch120a- Goddard- L01 1 Nature of the Chemical Bond with applications to catalysis, materials science, nanotechnology, surface science, bioinorganic chemistry, and energy William A. Goddard, III, wag@wag.caltech.eduwag@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 xiao@caltech.edu Lecture 6 January 18, 2012 CC Bonds diamond, ΔHf, Group additivity Course number: Ch120a Hours: 2-3pm Monday, Wednesday, Friday

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L07,08 3 Summary, bonding to form hydrides General principle: start with ground state of AH n and add H to form the ground state of AH n+1 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.

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L07,08 4 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 107.7º 111.2º 1.526A 1.095A 1.086A120.0º

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

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L07,08 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. q H ~ +0.15 q C ~ -0.45 This leads to q H ~ +0.15 and q C ~ -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 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 decrease by only 1/3. However overlap decreases exponentially.

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L07,08 8 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

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

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L07,08 10 Bond energies D e = E AB (R=∞) - E AB (R e ) e for equilibrium) Get from QM calculations. Re is distance at minimum energy.

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L07,08 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 =  ½Ћ  ) This is spectroscopic bond energy from ground vibrational state (0K) Including ZPE changes bond distance slightly to R 0

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L07,08 12 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 =  ½Ћ  ) 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 (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)}. (H = E + pV assuming an ideal gas)

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

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L07,08 14 Snap Bond Energy: Break bond without relaxing the fragments Snap Adiabatic  E relax = 2*7.3 kcal/mol D snap De snap (109.6 kcal/mol) D e (95.0kcal/mol)

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L07,08 15 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

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

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

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L07,08 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 C 2 H 6 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.

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

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L07,08 20 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 to planar gains 7.3 kcal/mol For CF 3 -CF 3, there is no such relaxation since CF3 wants to be pyramidal, FCF~111º Thus we might estimate that for CF 3 -CF 3 the bond energy would be D e = 109.6 kcal/mol, hence D 298 ~ 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)

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L07,08 22 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  bond and a  bond. The HCH angle in CH2 was 132.3º, but Pauli Repulsion with the new  bond, decreases this angle to 117.6º (cf with 120º for CH 3 )

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L07,08 24 Twisted ethene Consider now the case where the plane of one CH 2 is rotated by 90º with respect to the other (about the CC axis) This leads only to a  bond. The nonbonding  l and  r 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 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 (  on same center)

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

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L07,08 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  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  orbitals.

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L07,08 27 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 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  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 D e =180.0, D 0 = 169.9, D 298K = 172.3 kcal/mol

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L07,08 28 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

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L07,08 29 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.

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L07,08 31 CC triple bonds Starting with two CH radicals in the 4  - state we can form ethyne (acetylene) with two  bonds and a  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.

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L07,08 34 CC triple bonds Since the first CC  bond is D e =95 kcal/mol and the first CC  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  to 4  -, which costs 17 kcal each, weakening the bond by 34 kcal/mol. Adding this to the 55 would lead to a total 2 nd  bond of 89 kcal/mol comparable to the first 24-24-

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

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L07,08 38 Infinite structure from tetrahedral bonding plus staggered bonds on adjacent centers Chair configuration of cylco- hexane 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 2 nd layer 1st layer 2 nd layer 1st layer 2 nd layer 1st layer 1 1 c1 31 0 2 1 2 1 0 1 1 2 0

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L07,08 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) 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 cc cc cc cc f f f f f f i i i i

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L07,08 40 Diamond Structure 1 2 1a1a 1c1c 1b1b 3 2b2b 2a2a 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 1 b -1-2-3-4-5-6 Chair cyclohexane ring: 1-2-3-3 b -7-1 c 4 3b3b 3a3a 5 4b4b 4a4a 5b5b 5a5a 6 7

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L07,08 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 Si 2 H 6

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L07,08 48 More on resonance like structure 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.

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L07,08 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. 1.5x 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

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L07,08 50 graphene 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

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L07,08 51 Graphene band structure Unit cell has 2 carbon atoms Bands: 2p  orbitals per cell  2 bands of states each with N states where N is the number of unit cells 2  electrons per cell  2N electrons for N unit cells The lowest N MOs are doubly occupied, leaving N empty orbitals. 1x1 Unit cell 1 st band 2 nd band The filled 1 st band touches the empty 2 nd band at the Fermi energy Get semi metal

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L07,08 52 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  + 1/3  Diamond: average CCs = 85 kcal/mol   = 3*27=81 kcal/mol

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L07,08 58 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 y on the other, allowing stabilization of both  systems

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L07,08 Ch120a- Goddard- L01 1 Nature of the Chemical Bond with applications.

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© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L07,08 Ch120a- Goddard- L01 1 Nature of the Chemical Bond with applications.

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