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William A. Goddard, III, wag@kaist.ac.kr

Lecture 11, October 13, 2009. Nature of the Chemical Bond with applications to catalysis, materials science, nanotechnology, surface science, bioinorganic chemistry, and energy. Course number: KAIST EEWS 80.502 Room E11-101 Hours: 0900-1030 Tuesday and Thursday.

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William A. Goddard, III, wag@kaist.ac.kr

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  1. Lecture 11, October 13, 2009 Nature of the Chemical Bond with applications to catalysis, materials science, nanotechnology, surface science, bioinorganic chemistry, and energy Course number: KAIST EEWS 80.502 Room E11-101 Hours: 0900-1030 Tuesday and Thursday William A. Goddard, III, wag@kaist.ac.kr WCU Professor at EEWS-KAIST and Charles and Mary Ferkel Professor of Chemistry, Materials Science, and Applied Physics, California Institute of Technology Senior Assistant: Dr. Hyungjun Kim: linus16@kaist.ac.kr Manager of Center for Materials Simulation and Design (CMSD) Teaching Assistant: Ms. Ga In Lee: leeandgain@kaist.ac.kr Special assistant: Tod Pascal:tpascal@wag.caltech.edu EEWS-90.502-Goddard-L10

  2. Schedule changes, reminder L11: Oct. 13, as scheduled, 9am L12: Oct. 14, new lecture 2pm, replaces Oct 8 L13: Oct. 15, as scheduled, 9am EEWS-90.502-Goddard-L10

  3. Last time EEWS-90.502-Goddard-L10

  4. The bonding states of BH and AlH 3S+ The ground state of BH and AlH is obtained by bonding to the p orbital (leading to the 1S+ state. However the bond to the lobe orbital (leading to the 3P state) is also quite strong. 1P 2P+ H 3P 1S+ The bond to the lobe orbital is weaker than the p, but it is certainly significant EEWS-90.502-Goddard-L10

  5. BH2 and AlH2 128º Starting from the ground state of BH or AlH, the second bond is to a lobe orbital, to form the 2A1 state. Just as for the 3B1 state of CH2 and SiH2 the bond for BH2 opens by several degrees to 131º while the bond to the AlH2 closes down by ~9º. θe Re EEWS-90.502-Goddard-L10

  6. BH3 and AlH3 Bonding the 3rd H to the 2A1 state of BH2, leads to planar BH3 or AlH3. But there is no 4th bond since there remain no additional unpaired orbitals to bond to. EEWS-90.502-Goddard-L10

  7. Bonding to halides (AXn, for X=F,Cl,… The ground state of F has just one singly-occuped orbital and hence bonding to C is in many ways similar to H, leading to CF, CF2, CF3, and CF4 species. However there are significant differences. Thus CF leads to two type so bonds, p and lobe just like CH Covalent bond expect (CH) actual bond (CF) 2P 80 kcal/mol 120 kcal/mol p bond 63 kcal/mol 63 kcal/mol 4S- lobe bond EEWS-90.502-Goddard-L10

  8. How can CF lead to such a strong bond, 120 vs 80 kcal/mol? Consider the possible role of ionic character in the bonding In the extreme limit + C F C+ F- E (R=∞) = 0 E (R=∞) =IP (C) – EA (F) =11.3 – 3.4 =7.9 eV IP (C) = 11.3 eV = 260 kcal/mol EA (F) = 3.40 eV = 78.4 kcal/mol Can Coulomb attraction make up this difference? EEWS-90.502-Goddard-L10

  9. Estimate energy of pure ionic bond for CF Ionic limit (C+ + F-) 7.9 eV 14.4/R 14.4/1.27 = 11.3 eV Covalent limit (C + F) Net bond = 11.3-7.9 = 3.4 eV= 78 kcal/mol Re=1.27A Ionic estimate ignores shielding and pauli repulsion for small R. Thus too large Units for electrostatic interactions E=Q1*Q2/(e0*R) where e0 converts units (called permittivity of free space) E(eV) = 14.4 Q1(e units)*Q2(e units)/R(angstrom) E(kcal/mol) = 332.06 Q1(e units)*Q2(e units)/R(angstrom) EEWS-90.502-Goddard-L10

  10. CF has strong mixture of covalent and ionic character + Pure covalent bond ~ 80 kcal/mol (based on CH) Pure ionic bond ~ 78 kcal/mol (ignore Pauli and shielding) Net bond = 120 kcal/mol is plausible for 2P state But why is the bond for the 4S- state only 63, same as for covalent bond? EEWS-90.502-Goddard-L10

  11. Consider ionic contribution to 4S-state + C F C+ F- E (R=∞) = 0 E (R=∞) =IP (C) – EA (F) =16.6 – 3.4 =13.2 eV To mix ionic character into the 4S- state the electron must be pulled from the sz lobe orbital. This leads to the (2s)1(2p)2 state of C+ rather than the ground state (2s)2(2p)1 which is 123 kcal/mol = 5.3 eV higher Thus ionic bond is NEGATIVE (78-123=-45 kcal/mol) EEWS-90.502-Goddard-L10

  12. Bonding the 2nd F to CF With the 4S- state at 57 kcal/mol higher than 2P, we need only consider bonding to 2P, leading to the 1A1 state. Bad Pauli repulsion increases FCF angle to 105º 3B1 57 kcal/mol 1A1 EEWS-90.502-Goddard-L10

  13. Now bond 3rd F to form CF3 The 3rd CF bond should be much weaker than 1st two. This strong preference for CF to use p character makes conjugated flourocarbons much less stabe than corresponding saturated compounds. Thus C4H6 prefers butadiene but C4F6 prefers cylcobutene Get pyramidal CF3 (FCF angle ~ 112º) In sharp contrast to planar CH3 Of course the 4th bond to form CF4 leads to a tetrahedral structure EEWS-90.502-Goddard-L10

  14. 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. EEWS-90.502-Goddard-L10

  15. Some old and some new EEWS-90.502-Goddard-L10

  16. 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 EEWS-90.502-Goddard-L10

  17. Bonding (GVB) orbitals of ethane (staggered) Note nodal planes from orthogonalization to CH bonds on right C EEWS-90.502-Goddard-L10

  18. 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 EEWS-90.502-Goddard-L10

  19. 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. EEWS-90.502-Goddard-L10

  20. 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 EEWS-90.502-Goddard-L10

  21. Trends: geometries of alkanes CH bond length = 1.095 ± 0.001A CC bond length = 1.526 ± 0.001A CCC bond angles HCH bond angles EEWS-90.502-Goddard-L10

  22. Bond energies The figure at the left shows how the energy changes as the bond distance between A and B is increased to infinity. The difference between the energy of the AB molecule at equilibrium (Re) and the energy at infinity is denoted as De. This is the quantity to which QM calculations lead directly. Experimentally, even at T=0K, each vibrational mode of the molecule has a zero-point energy, ZPE = ½ Ћw. Summing over all modes leads to the ZPE of the molecule. Including this ZPE changes the bond distance slightly to R0 and the bond energy to D0. This is the value obtained spectroscopically EEWS-90.502-Goddard-L10

  23. 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)}. EEWS-90.502-Goddard-L10

  24. 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 EEWS-90.502-Goddard-L10

  25. 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 EEWS-90.502-Goddard-L10

  26. 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 EEWS-90.502-Goddard-L10

  27. 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. EEWS-90.502-Goddard-L10

  28. 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 EEWS-90.502-Goddard-L10

  29. Bond energies: Compare to CF3-CF3 For CH3-CH3 we found a snap bond energy of D298K = 89.9 + 2*7.3 = 104.5 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 D298K = 104.5 kcal/mol, indeed the experimental value is ~ 109 kcal/mol suggesting that the main effect is relaxation (the remaining effects might be induction and steric) EEWS-90.502-Goddard-L10

  30. 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) EEWS-90.502-Goddard-L10

  31. Comparison of The GVB bonding orbitals of ethene and methylene EEWS-90.502-Goddard-L10

  32. 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 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 kcal/mol, because of Intraatomic Exchange EEWS-90.502-Goddard-L10

  33. 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 EEWS-90.502-Goddard-L10

  34. EEWS-90.502-Goddard-L10

  35. Bond energies double bonds EEWS-90.502-Goddard-L10

  36. Bond energies EEWS-90.502-Goddard-L10

  37. EEWS-90.502-Goddard-L10

  38. 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 EEWS-90.502-Goddard-L10

  39. Salient structural features of Diamond Staggered infinite zig-zag chains Six-membered rings These rings correspond to the chair form of cyclohexane (c-C6H12), which is more stable than boat by 5.6 kcal/mol (boat has 2 eclipsed CC bonds) EEWS-90.502-Goddard-L10

  40. 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 and cyclohexane rings 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 EEWS-90.502-Goddard-L10

  41. CH + CH Snap Compare to CF-CF Compare single, double, triple bonds EEWS-90.502-Goddard-L10

  42. L11 Butane, cyclohexane, hexane EEWS-90.502-Goddard-L10

  43. L11,L12 Diamond structure Band gap, conduction band, valence band. Impurity levels Si, p and n type (111), (100), (110) surfaces Reconstruction Ohmic contacts EEWS-90.502-Goddard-L10

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