Lecture 13 February 3, 2010

1. Lecture 13 February 3, 2010 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.edu 316 Beckman Institute, x3093 Charles and Mary Ferkel Professor of Chemistry, Materials Science, and Applied Physics, California Institute of Technology Teaching Assistants: Wei-Guang Liu <wgliu@wag.caltech.edu> Ted Yu <tedhyu@wag.caltech.edu>

2. Course schedule Wednesday Feb. 3, 2pm L13 TODAY(caught up) Friday Feb. 5, 2pm L14 Midterm given out on Friday. Feb. 5, due on Wed. Feb. 10 It will be four hour take home, open notes

3. Last time

4. Reconstruction of (110) surface, side view along [-1,1,0] Surface As has 3 covalent bonds to Ga, with 2 e in 3s lone pair, relaxes upward until average bond angle is 95º Surface Ga has 3 covalent bonds leaving 0 e in 4th orbital, relaxes downward until average bond angle is 119º. GaAs angle 0º 26º Si has dangling bond electron at each surface atom 54.7º 54.7º As Ga 54.7º [110] Si (110) GaAs (110) [001]

5. Reconstruction of (110) GaAs

6. Reconstruction of GaAs(110) surface, discussion We consider that bulk GaAs has an average of 3 covalent bonds and one donor acceptor (DA) bond. But at the surface can only make 3 bonds so the weaker DA bond is the one broken to form the surface. The result is that GaAs cleaves very easily compared to Si. No covalent bonds to break. As has 3 covalent bonds, leaving 2 electrons in 3s lone pair. AsH3 has average bond angle of 92º. At the GaAs surface As relaxes upward until has average bond angle of 95º Ga has 3 covalent bonds leaving 0 eletrons in 4th orbital. GaH3 has average bond angle of 120º. At the GaAs surface Ga relaxes downward until has average bond angle of 119º. This changes the surface Ga-As bond from 0º (parallel to surface to 26º. Observed in LEED experiments and QM calculations

7. Analysis of charges Ga As Ga As Ga As Bulk structure: each As has 3 covalent bonds and one Donor-accepter bond(Lewis base – Lewis acid). This requires 3+2=5 electrons from As and 3+0=3 electrons from Ga. We consider that each bulk GaAs bond has 5/4 e from As and ¾ e form Ga. Each surface As has 5/4+1+1+2 = 5.25e for a net charge of -0.25 each surface Ga has ¾+1+1+0= 2.75 e for a net charge of +0.25 Thus considering both surface Ga and As, the (110) is neutral 5.25e 2.75e Net Q =0 0 2 0 2 0 2 1 1 1 1 1 1 1 1 1 1 5/4 3/4 3/4 3/4 5/4 5/4 5/4 3/4 3/4 5/4 5/4 3/4 5/4 3/4 3/4 5/4 5/4 3/4 a g a g a g 5/4 3/4 3/4 5/4 5/4 3/4 5/4 3/4 3/4 5/4 5/4 3/4

8. The GaAs (100) surface, unreconstructed Every red surface atom is As bonded to two green 2nd layer Ga atoms, but the other two bonds were to two Ga that are now removed. This leaves three non bonding electrons to distribute among the two dangling bond orbitals sticking out of plane (like AsH2) 1st Layer  RED 2nd Layer  GREEN 3rd Layer  ORANGE 4th Layer  WHITE

9. GaAs(100) surface reconstructed (side view) For the perfect surface, As in top layer, Ga in 2nd layer, As in 3rd layer, Ga in 4th layer etc. For the unreconstructed surface each As has two bonds and hence three electrons in two nonbonding orbitals. Expect As atoms to dimerize to form a 3rd bond leaving 2 electrons in nonbonding orbitals. Surface As-As bonds As Ga As Ga As Ga As Ga

10. Charges for 2x1 GaAs(100) 2 2 2 2 2nd layer ga has 3 e 1 1 Top layer, As 2nd layer, ga 5/4 5/4 3/4 3/4 3/4 3/4 3/4 3/4 3/4 3/4 5/4 2e As-ga bond 5/4 3rd layer, as 1 1 2e As LP 5/4 5/4 Each surface As has extra 0.5 e  dimer has extra 1e Not stable 3/4 3/4 3/4 3/4 3/4 1st layer As has 5.5 e 3/4 3/4 2e As-As bond 3/4

11. Now consider a missing row of As for GaAs(100) 0 0 0 0 1 1 Top layer, As 2nd layer, ga 5/4 5/4 3/4 3/4 3/4 3/4 3/4 3/4 ga empty LP 3rd layer, as 2nd layer ga has 2.25e Each 2nd layer ga next to missing As is deficient by 0.75e extra 0.5 e  4 ga are missing 3e 3/4 3/4 3/4 1st layer As has 5.5 e 3/4 3/4 3/4

12. Consider 1 missing As row out of 4 Extra 1e missing 3e -1-1-1+3=0 net charge Extra 1e Thus based on electron counting expect simplest surface reconstruction to be 4x2. This is observed Extra 1e Extra 1e missing 3e

13. Different views of GaAs(100)4x2 reconstruction -1.0e +1.5e Two missing As row plus missing Ga row Exposes 3rd row As Agrees with experiment Previous page, 3 As dimer rows then one missing Hashizume et al Phys Rev B 51, 4200 (1995)

14. summary • Postulate of surface electro-neutrality • Terminating the bulk charges onto the surface layer and considering the lone pairs and broken bonds on the surface should lead to: • the atomic valence configuration on each surface atom. For example As with 3 covalent bonds and a lone pair and Ga with 3 covalent bonds and an empty fourth orbital • A neutral surface • This leads to the permissible surface reconstructions

15. Excitation energies semiconductors

16. To be added – band states

17. To be added – band states

18. Semiconducting properties

19. Semiconducting properties

22. To be added – band states IP(P)=4.05 eV 0.054 eV Remove e from P, add to conduction band = 4.045-4.0 = 0.045 eV Thus P leads to donor state just 0.045eV below LUMO or CBM

24. To be added – band states EA(Al)=5.033 eV 0.045 eV Add e to Al, from valence band = 5.1 -5.033 = 0.067 eV Al leads to acceptor state just 0.067eV above HOMO or VBM

30. New material

31. Homonuclear Diatomics

32. Homonuclear Diatomics Molecules – the valence bond view Consider bonding two Ne atoms together Clearly there will be repulsive interactions as the doubly occupied orbitals on the left and right overlap, leading to repulsive interactions and no bonding. In fact as we will consider later, there is a weak attractive interaction scaling as -C/R6, that leads to a bond of 0.05 kcal/mol, but we ignore such weak interactions here The symmetry of this state is 1Sg+

33. Halogen dimers Next consider bonding of two F atoms. Each F has 3 possible configurations (It is a 2P state) leading to 9 possible configurations for F2. Of these only one leads to strong chemical binding This also leads to a 1Sg+ state. Spectroscopic properties are listed below . Note that the bond energy decreases for Cl2 to Br2 to I2, but increases from F2 to Cl2. we will get back to this later.

34. Di-oxygen or O2 molecule Next consider bonding of two O atoms. Each O has 3 possible configurations (It is a 3P state) leading to 9 possible configurations for O2. Of these one leads to directly to a double bond This suggests that the ground state of O2 is a singlet state. At first this seemed plausible, but by the late 1920’s Mulliken established experimentally that the ground state of O2 is actually a triplet state, which he had predicted on the basis of molecular orbitial (MO) theory. This was a fatal blow to VB theory, bringing MO theory to the fore, so we will consider next how Mulliken was able to figure this out in the 1920’s without the aid of computers.

35. The homonuclear diatomic correlation diagram Mulliken knew the ordering of the atomic orbitals and considered how combinations of the atomic orbitals would change as the nuclei were pushed togtether to eventually form a united atom. First consider the separate atoms limit where there is a large but finite distance R separating the atoms. The next slide shows the combinations formed from 1s, 2s, and 2p orbitals.

36. Separated atoms limit Note that in each case we get one bonding combination (no new nodal plane) and one antibonding combination (new nodal plane, red lines)

37. Splitting of levels General nodal arguments allow us to predict that But which is lower of and which is lower of Here the nodal plane arguments do not help

38. At large R 2ps better bonding than 2pp In earlier lectures we considered the strength of one-electron bonds where we found that Since the overlap of ps orbitals is obviously higher than pp We expect that bonding antibonding

39. Separated atom limit MO notation Separated atoms notation

40. United atom limit Next consider the limit in which the two nuclei are fused together to form a united atom For N2 this would lead to a Si atom. Here we get just the normal atomic aufbau states 1s < 2s < 2p < 3s < 3p < 4s,3d < 4p etc But now we consider an itty bity elongation of the Si nucleus toward two N nuclei and how the atomic states get perturbed For the 1s orbital all that happens is that the energy goes up (less electron density on the nuclei) and the symmetry becomes sg

41. 2s and 2p united atom orbitals Similarly 2s just goes to 2sg (and a lower binding) But the 2p case is more interesting For the 2ps state the splitting of the nuclei lead to increased density on the nuclei and hence increased binding while for 2pp there is no change in density Thus 2psu < 2ppu

42. Summarizing united atom limit Note for 3d, the splitting is 3ds < 3dp < 3dd Same argument as for 2p

43. Summary more united atom levels

44. Correlation diagram for Carbon row homonuclear diatomics C2 N2 O2 F2 United atom limit O2+ separated atom limit N2+

45. Using the correleation diagram Choices for N2 In order to use the correlation diagram to predict the states of diatomic molecules, we need to have some idea of what effective R to use (actually it is the effective overlap with large R small S and small R large S). Mulliken’s original analysis [Rev. Mod. Phys. 4, 48 (1932)] was roughly as follows. 1. N2 was known to be nondegenerate and very strongly bound with no low-lying excited states 2 4 2 2 4 2 4 2 2 2 2

46. N2 MO configurations This is compatible with several orderings of the MOs Largest R 2 4 2 4 2 2 2 4 2 2 Smallest R 2

47. N2+ But the 13 electron molecules BeF, BO, CO+, CN, N2+ Have a ground state with 2S symmetry and a low lying 2S sate. In between these two 2S states is a 2P state with spin orbital splitting that implies a p3 configuration This implies that Is the ground configuration for N2 and that the low lying states of N2+ are This agrees with the observed spectra

48. Correlation diagram for Carbon row homonuclear diatomics C2 N2 O2 F2 United atom limit O2+ separated atom limit N2+

49. 1s and 2s cases A A B B

50. Bond Anti BO 1 2 2.5 3 2.5 2 1 0