Lecture 9 January 26, 2011 Si, GaAs surfaces

Transcription

1 Lecture 9 January 26, 20 Si, GaAs surfaces 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 36 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> Caitlin Scott <cescott@caltech.edu>

2 Course schedule Friday January 4: L3 and L4 Monday January 7: Caltech holiday (MLKing) Wednesday January 9: wag L5 and L6 Friday January 2: wag L7 and L8, caught up Monday January 24: wag L7 and L8 Wednesday January 26: wag L9 and L0 Friday January 28: wag participates in a retreat for our nanotechnology project with UCLA Friday January 28: wag L Back on schedule Monday January 3: wag L2 Wag rotator cuff operation 2

3 Last time 3

4 Bond energies D e = E AB (R= ) - E AB (R e ) e for equilibrium) Get from QM calculations. Re is distance at minimum energy. 4

5 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 = Σ(½Ћω) This is spectroscopic bond energy from ground vibrational state (0K) Including ZPE changes bond distance slightly to R 0 5

6 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 = Σ(½Ћω) 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 = [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) 6

7 Snap Bond Energy: Break bond without relaxing the fragments Snap ΔE relax = 2*7.3 kcal/mol Adiabatic D snap De snap (09.6 kcal/mol) D e (95.0kcal/mol) 7

8 Details Bond energies for ethane D 0 = 87.5 kcal/mol ZPE (CH 3 ) = 8.2 kcal/mol, ZPE (C 2 H 6 ) = 43.9 kcal/mol, D e = D = 95.0 kcal/mol (this can be calculated from QM) D 298 = D = = 89.9 kcal/mol This is the quantity we will quote in discussing bond breaking processes 8

9 Bond energies: Compare to CF 3 -CF 3 CH 3 -CH 3 D e-snap = De + ΔErelax = *7.3 = 09.6 kcal/mol (relaxation of tetrahedral CH 3 to planar gains 7.3 kcal/mol) For CF 3 -CF 3, there is no little relaxation since CF 3 wants to be pyramidal, FCF~º, estimate ΔE relax ~ 2 kcal/mol Assume that D e-snap = 09.6 kcal/mol for CF 3 -CF 3 (as CH 3 -CH 3 ) Predict D e (CF 3 -CF 3 ) ~ 0 4 = 06 Assume Δ ZPE (C 2 F 6 ) ~ Δ ZPE (C 2 H 6 )* sqrt(m H /M F ) ~ 2. D 0 (C 2 F 6 ) = D e Δ ZPE ~ = 02 Thus D 298 (C 2 F 6 ) ~ D = 04 Experimental is D 298 =98.7±2.5 kcal/mol Additional weakening of CC bond may be due to induction (Csp 3 bond weaker because of charge transfer to F) and steric (F F nonbonded interactions) 9

10 Substituent effects on CC Bond energies 298K) The strength of a CC bond changes from 89.9 to 70 kcal/mol as replace Hs with CH 3 s. Goddard believes is mostly due to fragment relaxation 0

11 CH 2 +CH 2 ethene Starting with two methylene radicals (CH 2 ) in the ground state ( 3 B ) we can form ethene (H 2 C=CH 2 ) with both a σ bond and a π bond. The HCH angle in CH 2 was 32.3º, but Pauli Repulsion with the new σ bond, decreases this angle to 7.6º (cf with 20º for CH 3 )

12 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 σ bond. The nonbonding π l and π 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 ~.4 kcal/mol. Voter, Goodgame, and Goddard [Chem. Phys. 98, 7 (985)] showed that N is below T by.2 kcal/mol, due to Intraatomic Exchange (residual triplet coupling of σ,π on same center) 2

13 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 Rotational barrier 3

14 Geometries ethene N state (planar) R CC =.339A (double bond (twisted 90 ) R CC =.47A (single σ bond). Ethane: R CC =.526 A Main effects: twisted ethene little CH Pauli Repulsion between CH bonds on opposite C, ethane has substantial interactions. the intrinsic CC single bond may be closer to.47a T state (twisted 90 ) R CC =.47A (single σ bond). (planar 0 ) R CC =.57A (Orthogonalization of the triple coupled pπ orbitals) 4

15 CC double bond energies The bond energies for ethene are D e =80.0, D 0 = 69.9, D 298K = 72.3 kcal/mol Breaking the double bond of ethene, the HCH bond angle changes from 7.6º to 32.xº, leading to an increase of 2.35 kcal/mol in the energy of each CH 2 so that D esnap = = 84.7 kcal/mol 5

16 Energies H 2 C=CH 2 Snap Adiabatic ΔE relax = 2*2.35 kcal/mol HCH= D snap De snap (84.7 kcal/mol) D e (80.0kcal/mol) 6

17 CC double bond energies The bond energies for ethene are D e =80.0, D 0 = 69.9, D 298K = 72.3 kcal/mol Breaking the double bond of ethene, the HCH bond angle changes from 7.6º to 32.xº, leading to an increase of 2.35 kcal/mol in the energy of each CH 2 so that D esnap = = 84.7 kcal/mol Since the D esnap = 09.6 kcal/mol, for H 3 C-CH 3, The π bond adds 75. kcal/mol to the bonding. (compare to 65kcal/mol rotational barrier) Twisted ethylene, De = = 5; D esnap = =20. This is 0 kcal/mol larger than for ethane. May be due to sp 2 vs sp 3 of the effect of CH repulsions 7

18 Predict F 2 C=CF 2 bond energy F 2 C ( 3 B ) + CF 2 ( 3 B ) F 2 C=CF 2 Relax. E. = 2*57 kcal/mol F 2 C ( A ) + CF 2 ( A ) D e-smap = 84.7 kcal/mol (from CH2-CH2) D e = 85 2* 57 = 7 kcal/mol, D0 = 7 Δ zpe = 7 2.3=69 D 298 = D = 7, Exper= 75 8

19 bond energy of F 2 C=CF 2 Ethene D esnap = = 84.7 kcal/mol Consider the bond energy of F 2 C=CF 2, Assume D esnap (F 2 C=CF 2 ) = = 84.7 kcal/mol But the snap electronic state is 3 B which is 57 kcal/higher than A Thus for CF 2 the fragment relaxation is 2*57 = 4 kcal/mol (there is also a correction from the FCF angle of C2F4 and that of CF 2 ( 3 B ) Predict adiabatic D e = 85-4 = 7 kcal/mol. D 0 = D e Δ zpe ~ = 69 Thus D 298 = D = 7 kcal/mol The experimental value is D 298 ~ 75 kcal/mol, close to the prediction 9

20 Bond energies double bonds ground state of CH 2 is 3 B by 9.3 kcal/mol, but substitution of one or both H with CH 3 leads to A ground states. 20

21 CC triple bonds Starting with two CH radicals in the 4 Σ - state we can form ethyne (acetylene) with two π bonds and a σ bond. This leads to a CC bond length of.208a compared to.339 for ethene and.526 for ethane. The bond energy is D e = 235.7, D 0 = 227.7, D 298K = kcal/mol Which can be compared to De of 80.0 for H2C=CH2 and 95.0 for H3C-CH3. 2

22 HC=CH bond energy HC=CH D esnap = 270 HC ( 4 Σ - ) + CH ( 4 Σ - ) Relax. E. = 2*7 kcal/mol HC ( 2 Π) + CH ( 2 Π) D e =

23 Comparison of CC bond energies 23

24 Triple Bond energies 24

25 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 25

26 Infinite structure from tetrahedral bonding plus staggered bonds on adjacent centers 2 nd layer 3 st layer nd layer c st layer 2 nd layer Chair configuration st layer of cylcohexane Not shown: zero layer just like 2 nd layer but above layer 3 rd layer just like the st layer but below layer 2 26

27 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 f and cyclohexane rings (f-i-f)-(i-f-i) c c There are atoms at all 8 corners (but only /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 c f i c f c i f i c f i f c c 27

28 4 b 2 b Diamond Structure 5 a 3 a a b 4 a 2 a 5 b 3 b c 7 Start with C 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 C. Continue this process. Get unique structure: diamond Note: Zig-zag chain b Chair cyclohexane ring: b -7- c 28

29 Properties of diamond crystals 29

30 Properties of group IV molecules (IUPAC group 4).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. 30 Note for Si, that the average bond is much different than for Si H

31 Comparisons of successive bond energies SiH n and CH n p lobe lobe p lobe lobe p p 3

32 Miller indices A 3D crystal is characterized by a unit cell with axes, a, b, c that can be translated by integer transations along a, b, c to fill all space. The corresponding points in the translated cells are all equivalent. Passing a plane through any 3 such equivalent points defines a plane denoted as (h,k,l). An equally spaced set of planes parallel to (h,k,l) pass through all equivalent points. Put the origin on a point in one of these parallel planes. The closest one will intersect the unit vectors at a/h, b/k, and c/l. c These are called Miller indices c/l b/k b a a/h 32

33 Examples of special planes c c/l a a/h To denote all equivalent planes we use {h,k,l} so that indicates negative b/k b {,0,0} for cubic includes the 3 cases in the first row) A number with a bar From Wikipedia 33

34 Crystallographic directions A lattice vector can be written as Rmnp = m a + n b + p c where m,n,p are integers. This is denoted as [m,n,p] The set of equivalent vectors is denoed as <m,n,p> Examples are shown here. From Wikipedia 34

35 The Si Crystal viewed from the [00] direction [00] [00] [0] [00] [00 [00] (00) Surface st Layer RED 2 nd Layer GREEN 3 rd Layer ORANGE 4 th Layer WHITE [,-,0] not show bonds to 5 th layer 35

36 The Si Crystal (00) surface, unreconstructed Projection of bulk cubic cell Surface unit cell P(x) Surface zig-zag row Every red atom was bonded to two Si that are now removed, thus two dangling bond orbitals (like A state) sticking out of plane (00) VIEW st Layer RED 2 nd Layer GREEN 3 rd Layer ORANGE 4 th Layer WHITE 36

37 Si(00) surface (unreconstructed) viewed (nearly) along the [0] direction Each surface atom has two dangling bond orbitals pointing to left and right, along [,-,0] direction 37

38 The (00) Surface Reconstruction viewed (nearly) along the [0] direction Spin pair dangling bond orbitals of adjacent atoms in [,-,0] direction (originally 2 nd near neighbors Get one strong σ bond but leave two dangling bond orbitals on adjacent now bonded atoms (form weak π bond in plane) 38

39 Si(00) surface reconstructed (side view) Surface atoms now bond to form dimers (move from 3.8 to 2.4A) Get row of dimes with doubled surface unit cell One strong σ bond, plus weak π bond in plane orginal cell New cell Surface length length bond Lateral 7.6A 3.8A 2.4A displacements 0.7A 0.7A 39

40 Si(00) surface reconstructed (top view) New unit cell reconstructed surface P(2x) Rows of dimer pairs are parallel original unit cell unreconstructed surface P(x) 40

41 Get 2x2 unit cell but atom at center is equivalent to atom at corner, therform c(2x2) 4

42 Two simple patterns for (00) Surface Reconstruction Dimer rows alternate C(2x2), high energy Dimer rows parallel P(2x), low energy 42

43 P(2x) more stable than c(2x2) by ~ kcal/mol The Sisurf-Si2nd-Sisurf bond for c(2x2) opens up to 20º because the Sisurf move opposite directions 20º 0º 20º 0º For P(2x) the Sisurf move the same directions and Sisurf-Si2nd-Sisurf bond remains at 0º 43

44 New material 44

45 Construct () surface using cubic unit cell Start at diagonal atom #0 Go straight down to atom # Atom # bonded to 3 atoms #2 Each #2 is bonded to 3 atoms # in top layer. Get hexagonal double layer Each #2 is bonded straight down to an atom#3 Each atom #3 is bonded to 3 atom# c

46 Si() surface (alternate construction) Start with red atom on top, bond to 3 green atoms in 2 nd layer Each green atom is bonded to 2 other st layer atoms plus a 3 rd atom straight down (not shown) The 3 rd layer atoms bond to 3 4 th layer atoms in orange (now white) Surface unit cell P(x) 46

47 Reconstruction of Si() surface Each surface atom has a single dangling bond electron, might guess that there would be some pairing of this with an adjacent atom to form a 2x unit cell. Indeed freshly cleaved Si() at low temperature does show 2x Surface unit cell P(x) 47

48 LEED experiments (Schlier and Farnsworth, 959) observed 7th Order Spots 7x7 unit cell (49 x cells) From 959 to 98 many models proposed to fit various experiments or calculations. Binnig et al., 98 did first STM image of Si (7x7) and saw 2 bright spots in 7x7 cell, showed that every previous model was incorrect Takayanagi et al., 985, proposed the DAS Model that explained the experiments 48

49 two 7x7 cells What kind of interactions can go over a 7x7 region, with cell size 26.6 by 26.6 A? 49

50 Origin of complex reconstruction of Si() In 49 surface unit cells have 49 dangling bonds. Since cohesive energy of Si crystal is 08 kcal/mol expect average bond energy must be 08/2 = 54 kcal/mol (each atom has 4 bonds, but double count the bonds) (H3Si-SiH3 bond energy is 74 kcal/mol) Thus each dangling bond represents ~ 27 kcal/mol of surface energy =. ev per surface atom Calculated value =.224 ev snap and.200 ev relaxed. 50

51 Consider bonding an atom on top of 3 dangling bonds T 4 H 3 T 4 T 4 H 3 T 4 Get 3x2 unit cell By adding a cap of one adatom Si per 3 top layer Si, can tie off all original dangling bonds. Thus

52 Consider bonding an atom on top of 3 dangling bonds Two ways to do this. T 4 and H 3 T 4 (observed) H 3 (not observed) (a) T 4 (b) H 3 SIDE TOP SIDE TOP Stabilize by 0. ev per site Destabilize by 0.5 ev per site 52

53 T4 versus H3 site bonding to dangling bonds Energy increases by 0.5 ev per original surface atom or 0.45 ev per new adatom Angle between bond A and bond B is 80º bad overlap orthog H 3 T 4 A B B A Energy decreases by 0.0 ev per original surface atom or 0.30 ev per new adatom Angle between bond A and bond B is 00º ok overlap no orthog HOMO H3 HOMO T4 53

54 0 2 H3 reconstruction 0 Top layer labeled 2 nd layer green Addon layer 0, blue Need just /3 Monolayer to tie up bonds. Surface energy increases by 0.3 ev Because 0--2 is linear Unit cell 0 0 SIDE 54

55 H3 reconstruction, 3 x Top layer labeled 2 nd layer green Addon layer 0, blue Need just /3 Monolayer to tie up bonds. Surface energy increases by 0.3 ev 0 0 SIDE 55

56 T4 reconstruction 3 x 3 02 Top layer labeled 2 nd layer green Addon layer 0, blue Need just /3 Monolayer to tie up bonds. Surface energy decreases by 0.0 ev Because 0--2 ~ 00º Unit cell

57 T4 reconstruction 2x Top layer labeled 2 nd layer green Addon layer 0, blue Need just /3 Monolayer to tie up bonds, leave dangling bond orbital Surface energy decreases by 0.08 ev Per 2x2 cell Unit cell 0 57

58 58

59 The () 7x7 DAS Surface 59

60 The () 7x7 DAS Surface Layers (purple, brown and blue atoms have one dangling bond) Adatoms on Top layer These adatoms protrude from the surface so that they show up prominently in STM 60

61 The () 7x7 DAS Surface Layers (purple, brown and blue atoms have one dangling bond) st 2 nd red atoms, all bonded to st layer 3 rd 4 th First unreconstructed layer 6

62 The () 7x7 DAS Surface 2-membered ring at corner of cell 62

63 The () 7x7 DAS Surface Side view 63

64 The () 7x7 DAS Surface Cornerhole 64

65 Si() 7x7 65

66 The () 7x7 DAS Layer Positions REF REF REF 66

67 The () 3x3 DAS Surface Unit Cell Side view Top view 2-membered rings 67

68 The () 5x5 DAS Surface Unit Cell Side view 68

69 The () 5x5 DAS Surface Unit Cell Top view 2- and 8-membered rings 69

70 The () 9x9 DAS Surface Unit Cell Side view 70

71 The () 9x9 DAS Surface Unit Cell Top view 2- and 8-membered rings 7

72 DAS Surface Energies (PBE DFT).09 Energy, ev/x Cell Regression Ab Initio DAS Cell Size Unreconstructed relaxed surface:.200 ev/x cell Infinite DAS model:.07 ev/x cell 72

73 DAS Reconstruction Driving Force 49 unpaired electrons (/2 Si-Si bond) per 7x7 ev = 58.8 ev/cell DAS 7x7 Surface energy = 5.2 ev/cell (9 unpaired electrons) Energy reduction due to reconstruction = 7.6 ev Difference is due to strain Bond length range = Å (equilibrium 2.35 Å) Bond angle range = 9 7º (Equilibrium 09.4 ) 73

74 DAS Surface Energy Contributions.2 Energy, ev/x Cell (DAS Model Cell Size) - x T4 8R 2R F D TOTAL 74

75 DAS Surface Energies: Sequential Size Change Model 5 Energy, ev/6x6 Cell SSC Irregular-odd and even SSC regular-odd -20 SSC Cell Size Real-time STM by Shimada & Tochihara,

76 DAS Surface Energies: Origin of a finite cell size Energy, ev/x Cell SSC Irregular-odd and even SSC regular-odd DFT Cell Size 76

77 The (0) plane (outlined in green, layer ) [00] c [-,,0] [00] [00] [0] 77

78 Si(0) surface (top view) Cut through cubic unit cell surface unit cell P(x) Surface atoms red 78

79 Si(0) surface (viewed nearly along [-,,0] direction) One dangling bond electron per surface atom Surface atoms red bulk atoms orange [,,0] [00] 79

80 Reconstruction of (0) surface, surface atoms only side view (along [-,,0]) Showing just 2 dangling bond orbitals 54.7º 54.7º Top view (from [-,-,0]) [00] [,,0] [-,,0] [00] 80

81 Reconstruction of (0) surface, surface atoms only We have a chain of dangling bond orbitals along the [-,,0] direction, each tilted by 35.3º from the [0] (vertical) axis They will want to tilt toward the vertical axis, reducing their angle from 35.3º). This leads to moving the surface atoms toward the bulk. There could be 2 by 2 pairing to double the surface unit cell in the [-,,0] direction [0] side view (along [-,,0]) Showing just 2 dangling bond orbitals 54.7º 54.7º 54.7º [00] 8

82 The zincblende or sphalerite structure Replacing each C atom of the diamond structure alternately with Ga and As so that each Ga is bonded to four As and each As is bonded to four Ga leads to the zincblende or sphalerite structure (actually zincblende is the cubic form of ZnS and the mineral sphalerite is cubic ZnS with some Fe) As at corners: (0,0,0) As at face centers: (a/2,a/2,0), (a/2,0,a/2), (0,a/2,a/2) Ga 4 internal sites: (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 has 4 As and 4 Ga. 82

83 Bonding in GaAs Making a covalent bond between to each atoms, one might have expected tetrahedral As to make 3 bonds with a left over lone pair pointing away from the 3 bonds, while Ga might be expected to make 3 covalent bonds, with an empty sp 3 orbital point away from the 3 bonds, as indicated here, where the 3 covalent bonds are shown with lines, and the donor acceptor (DA) or Lewis acid- Lewis base bond as an As lone pair coordinated with and empty orbital on Ga Of course the four bonds to each atom will adjust to be equivalent, but we can still think of the bond as an average of ¾ covalent and ¼ DA 83

84 Other compounds Similar zincblende or sphalerite compounds can be formed with Ga replaced by B, Al,In and /or As replaced by N, P, Sb, or Bi. They are call III-V compounds from the older names of the columns of the periodic table (new UIPAC name 3-5 compounds). In addition a hexagonal crystal, called Wurtzite, also with tetrahedral bonding (but with some eclipsed bonds) is exhibited by most of these compounds. In addition there are a variety of similar II-VI systems, ZnS, ZnSe, CdTe, HgTe, etc 84

85 Lecture 0 January 26, 20 Si, GaAs surfaces 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 36 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> Caitlin Scott <cescott@caltech.edu> 85

86 Last time 86

87 T4 versus H3 site bonding to dangling bonds Energy increases by 0.5 ev per original surface atom or 0.45 ev per new adatom Angle between bond A and bond B is 80º bad overlap orthog H 3 T 4 A B B A Energy decreases by 0.0 ev per original surface atom or 0.30 ev per new adatom Angle between bond A and bond B is 00º ok overlap no orthog HOMO H3 HOMO T4 87

88 0 2 H3 reconstruction 0 Top layer labeled 2 nd layer green Addon layer 0, blue Need just /3 Monolayer to tie up bonds. Surface energy increases by 0.3 ev Because 0--2 is linear Unit cell 0 0 SIDE 88

89 H3 reconstruction, 3 x Top layer labeled 2 nd layer green Addon layer 0, blue Need just /3 Monolayer to tie up bonds. Surface energy increases by 0.3 ev 0 0 SIDE 89

90 T4 reconstruction 3 x 3 02 Top layer labeled 2 nd layer green Addon layer 0, blue Need just /3 Monolayer to tie up bonds. Surface energy decreases by 0.0 ev Because 0--2 ~ 00º Unit cell

91 T4 reconstruction 2x Top layer labeled 2 nd layer green Addon layer 0, blue Need just /3 Monolayer to tie up bonds, leave dangling bond orbital Surface energy decreases by 0.08 ev Per 2x2 cell Unit cell 0 9

92 The () 7x7 DAS Surface Layers (purple, brown and blue atoms have one dangling bond) Adatoms on Top layer These adatoms protrude from the surface so that they show up prominently in STM 92

93 The () 7x7 DAS Surface Layers (purple, brown and blue atoms have one dangling bond) st 2 nd red atoms, all bonded to st layer 3 rd 4 th First unreconstructed layer 93

94 The () 7x7 DAS Surface 2-membered ring at corner of cell 94

95 The () 7x7 DAS Surface Cornerhole 95

96 Si() 7x7 96

97 The (0) plane (outlined in green, layer ) [00] c [-,,0] [00] [00] [0] 97

98 Si(0) surface (viewed nearly along [-,,0] direction) One dangling bond electron per surface atom Surface atoms red bulk atoms orange [,,0] [00] 98

99 Reconstruction of (0) surface, surface atoms only We have a chain of dangling bond orbitals along the [-,,0] direction, each tilted by 35.3º from the [0] (vertical) axis They will want to tilt toward the vertical axis, reducing their angle from 35.3º). This leads to moving the surface atoms toward the bulk. There could be 2 by 2 pairing to double the surface unit cell in the [-,,0] direction [0] side view (along [-,,0]) Showing just 2 dangling bond orbitals 54.7º 54.7º 54.7º [00] 99

100 The zincblende or sphalerite structure Replacing each C atom of the diamond structure alternately with Ga and As so that each Ga is bonded to four As and each As is bonded to four Ga leads to the zincblende or sphalerite structure (actually zincblende is the cubic form of ZnS and the mineral sphalerite is cubic ZnS with some Fe) As at corners: (0,0,0) As at face centers: (a/2,a/2,0), (a/2,0,a/2), (0,a/2,a/2) Ga 4 internal sites: (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 has 4 As and 4 Ga. 00

101 Bonding in GaAs Making a covalent bond between to each atoms, one might have expected tetrahedral As to make 3 bonds with a left over lone pair pointing away from the 3 bonds, while Ga might be expected to make 3 covalent bonds, with an empty sp 3 orbital point away from the 3 bonds, as indicated here, where the 3 covalent bonds are shown with lines, and the donor acceptor (DA) or Lewis acid- Lewis base bond as an As lone pair coordinated with and empty orbital on Ga Of course the four bonds to each atom will adjust to be equivalent, but we can still think of the bond as an average of ¾ covalent and ¼ DA 0

102 Other compounds Similar zincblende or sphalerite compounds can be formed with Ga replaced by B, Al,In and /or As replaced by N, P, Sb, or Bi. They are call III-V compounds from the older names of the columns of the periodic table (new UIPAC name 3-5 compounds). In addition a hexagonal crystal, called Wurtzite, also with tetrahedral bonding (but with some eclipsed bonds) is exhibited by most of these compounds. In addition there are a variety of similar II-VI systems, ZnS, ZnSe, CdTe, HgTe, etc 02

103 New material 03

104 [00] 2 The (0) plane (outlined in green, layer ) [00] c [0] [-,,0] [00] Cut through cubic unit cell surface unit cell P(x) As atoms top layer Ga atoms top layer [00] [-,,0] 04

105 Reconstruction of (0) surface, side view along [-,,0] Si has dangling bond electron at each surface atom 54.7º 54.7º 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 9º. GaAs angle 0º 26º 54.7º Ga As [0] Si (0) [00] GaAs (0) 05

106 Top view (from [-,-,0]) Reconstruction of GaAs(0) surface As has 3 covalent bonds, leaving 2 electrons in 3s lone pair, Ga has 3 covalent bonds leaving 0 eletrons in 4 th orbital Ga As 54.7º 54.7º [00] [,,0] [-,,0] [00] side view (along [-,,0]) 06

107 Reconstruction of (0) GaAs 07

108 III-V reconstruction 08

109 09

110 Reconstruction of GaAs(0) 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 20º. At the GaAs surface Ga relaxes downward until has average bond angle of 9º. This changes the surface Ga-As bond from 0º (parallel to surface to 26º. Observed in LEED experiments and QM calculations 0

111 Analysis of charges Bulk structure: each As has 3 covalent bonds and one Donoraccepter 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+++2 = 5.25e for a net charge of each surface Ga has ¾+++0= 2.75 e for a net charge of Thus considering both surface Ga and As, the (0) is neutral 5.25e 2.75e Net Q = Ga As Ga As Ga As 3/4 5/4 3/4 5/4 3/4 5/4 5/4 3/4 5/4 3/4 5/4 3/4 5/4 3/4 5/4 3/4 5/4 3/4 a g a g a g 3/4 5/4 3/4 5/4 3/4 5/4 3/4 5/4 3/4 5/4 3/4

112 The GaAs (00) surface, unreconstructed Every red surface atom is As bonded to two green 2 nd 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 AsH 2 ) st Layer RED 2 nd Layer GREEN 3 rd Layer ORANGE 4 th Layer WHITE 2

113 GaAs(00) surface reconstructed (side view) For the perfect surface, As in top layer, Ga in 2 nd layer, As in 3 rd layer, Ga in 4 th 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 3 rd bond leaving 2 electrons in nonbonding orbitals. Surface As-As bonds As Ga As As Ga Ga As 3

114 Charges for 2x GaAs(00) 2 nd layer ga has 3 e 2e As-ga bond 2e As LP st layer As has 5.5 e 3/4 3/ /4 5/4 3/4 3/4 3/4 5/4 5/ /4 3/4 5/4 3/4 5/4 3/4 5/4 2e As-As bond 3/4 3/4 3/4 3/4 3/4 3/4 Top layer, As 2 nd layer, ga 3 rd layer, as Each surface As has extra 0.5 e dimer has extra e Not stable 4

115 Now consider a missing row of As for GaAs(00) Top layer, As ga empty LP st layer As has 5.5 e 3/4 3/4 3/4 5/4 3/ /4 2 nd layer ga has 2.25e 3/4 5/4 3/4 3/4 3/4 3/4 3/4 3/4 0 2 nd layer, ga 3 rd layer, as Each 2 nd layer ga next to missing As is deficient by 0.75e extra 0.5 e 4 ga are missing 3e 5

116 Consider missing As row out of =0 net charge Extra e missing 3e Extra e Thus based on electron counting expect simplest surface reconstruction to be 4x2. This is observed Extra e Extra e missing 3e 6

117 Different views of GaAs(00)4x2 reconstruction -.0e Previous page, 3 As dimer rows then one missing +.5e Two missing As row plus missing Ga row Exposes 3 rd row As Agrees with experiment Hashizume et al Phys Rev B 5, 4200 (995) 7

118 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 8

119 Intrinsic semiconductors + - 9

120 Excitation energy 20

121 To be added band states 2

122 To be added band states 22

123 Semiconducting properties 23

124 Semiconducting properties 24

125 25

126 26

127 27

128 28

129 29

130 30

131 To be added band states IP(P)=4.05 ev ev Remove e from P, add to conduction band = = ev Thus P leads to donor state just 0.045eV below LUMO or CBM 3

132 32

133 33

134 To be added band states EA(Al)=5.033 ev ev Add e to Al, from valence band = = ev Al leads to acceptor state just 0.067eV above HOMO or VBM 34

135 35

136 36

137 37

138 stop 38

139 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/R 6, that leads to a bond of 0.05 kcal/mol, but we ignore such weak interactions here The symmetry of this state is Σ g + 39

140 Halogen dimers Next consider bonding of two F atoms. Each F has 3 possible configurations (It is a 2 P state) leading to 9 possible configurations for F 2. Of these only one leads to strong chemical binding This also leads to a Σ g + state. Spectroscopic properties are listed below. Note that the bond energy decreases for Cl 2 to Br 2 to I 2, but increases from F 2 to Cl 2. we will get back to this later. 40

141 Di-oxygen or O 2 molecule Next consider bonding of two O atoms. Each O has 3 possible configurations (It is a 3 P state) leading to 9 possible configurations for O 2. Of these one leads to directly to a double bond This suggests that the ground state of O 2 is a singlet state. At first this seemed plausible, but by the late 920 s Mulliken established experimentally that the ground state of O 2 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 920 s without the aid of computers. 4