Lecture 8 January 28, Silicon crystal surfaces

Lecture 8 January 28, 203 Nature of the Chemical Bond with applications to catalysis, materials science, nanotechnology, surface science, bioinorganic chemistry, and energy Course number: Ch20a Hours: 2-3pm Monday, Wednesday, Friday 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: Caitlin Scott <cescott@caltech.edu> Hai Xiao xiao@caltech.edu; Fan Liu <fliu@wag.caltech.edu> Silicon crystal surfaces Ch20a-

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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 3

Infinite structure from tetrahedral bonding plus staggered bonds on adjacent centers 2 nd layer 3 st layer 2 0 2 2 nd layer 2 0 0 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 4

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) 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 copyright a in 20 the William x,y,z A. Goddard directions III, all rights reserved c c f i c f c i f f i c f c i f c c 5

4 b 2 b Diamond Structure 5 a 3 a a 6 5 3 4 2 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 --2-3-4-5-6 Chair cyclohexane ring: -2-3-3 b -7- c 6

Properties of diamond crystals 7

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. Note for Si, that the average copyright 20 bond William is A. much Goddard III, different all rights reserved than for Si H 8

Comparisons of successive bond energies SiH n and CH n p lobe lobe p lobe lobe p p 9

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 a a/h b 0

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

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 2

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 3 to 5 th layer

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 4

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 5

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 s bond but leave two dangling bond orbitals on adjacent now bonded atoms (form weak p bond in plane) 6

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 s bond, plus weak p bond in plane orginal cell New cell Surface length length bond Lateral 7.6A 3.8A 2.4A displacements 0.7A 0.7A 7

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) 8

Get 2x2 unit cell but atom at center is equivalent to atom at corner, therform c(2x2) 9

Two simple patterns for (00) Surface Reconstruction Dimer rows alternate C(2x2), high energy Dimer rows parallel P(2x), low energy 20

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 2

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#4. 4 2 2 2 3 0 4 4 2 2 3 3 c 4 4 4 2 4 2 2 2 4 2 2 4 2 2 22

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) 23

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) 24

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 25

two 7x7 cells What kind of interactions can go over a 7x7 region, with cell size 26.6 by 26.6 A? 26

New material 27

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. 28

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 48 6 29

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) Stabilize by 0. ev per site Destabilize by 0.5 ev per site 30

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 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 3

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 32

H3 reconstruction, 3 x 3 0 3 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 0 0 33

T4 reconstruction 3 x 3 2 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 34

T4 reconstruction 2x2 2 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 35

The () 7x7 DAS Surface 37

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 38

The () 7x7 DAS Surface Layers (purple, brown and blue atoms have one dangling bond) st 2 nd 8 + 8 red atoms, all bonded to st layer 3 rd 4 th First unreconstructed layer 39

The () 7x7 DAS Surface 2-membered ring at corner of cell 40

The () 7x7 DAS Surface Side view 4

The () 7x7 DAS Surface Cornerhole 42

Si() 7x7 43

The () 7x7 DAS Layer Positions 2 3 3 4 5 6 7 7 8 9 REF REF REF 47

The () 3x3 DAS Surface Unit Cell Side view Top view 2-membered rings 48

The () 5x5 DAS Surface Unit Cell Side view 49

The () 5x5 DAS Surface Unit Cell Top view 2- and 8-membered rings 50

The () 9x9 DAS Surface Unit Cell Side view 5

The () 9x9 DAS Surface Unit Cell Top view 2- and 8-membered rings 52

Energy, ev/x Cell DAS Surface Energies (PBE DFT).09.08.07.06.05.04.070 Regression Ab Initio.048.044 3 5 7 9 3 DAS Cell Size.055.068.078 Unreconstructed relaxed surface:.200 ev/x cell Infinite DAS model:.07 ev/x cell 53

DAS Reconstruction Driving Force 49 unpaired electrons (/2 Si-Si bond) per 7x7 cell @.2 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 = 2.3 2.50 Å (equilibrium 2.35 Å) Bond angle range = 9 7º (Equilibrium 09.4 ) 54

Energy, ev/x Cell DAS Surface Energy Contributions.2 0.6 0.0 0.0 0. 0.2 0.3-0.6 (DAS Model Cell Size) - x T4 8R 2R F D TOTAL 55

Energy, ev/6x6 Cell DAS Surface Energies: Sequential Size Change Model 5 0 3 5 7 9 3-5 -0-5 SSC Irregular-odd and even SSC regular-odd -20 SSC Cell Size Real-time STM by Shimada & Tochihara, 2003 56

Energy, ev/x Cell DAS Surface Energies: Origin of a finite cell size.4.3 SSC Irregular-odd and even SSC regular-odd DFT.2..0 3 5 7 9 3 Cell Size 57

The (0) plane (outlined in green, layer ) [00] 2 3 2 0 2 0 0 c [-,,0] [00] [00] [0] 58

Si(0) surface (top view) Cut through cubic unit cell surface unit cell P(x) Surface atoms red 59

Si(0) surface (viewed nearly along [-,,0] direction) One dangling bond electron per surface atom Surface atoms red bulk atoms orange [,,0] [00] 60

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] 6

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] 62

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. 63

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 64

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 65

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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 68

H3 reconstruction, 3 x 3 0 3 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 0 0 69

T4 reconstruction 3 x 3 2 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 70

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 72

The () 7x7 DAS Surface Layers (purple, brown and blue atoms have one dangling bond) st 2 nd 8 + 8 red atoms, all bonded to st layer 3 rd 4 th First unreconstructed layer 73

The () 7x7 DAS Surface 2-membered ring at corner of cell 74

The () 7x7 DAS Surface Cornerhole 75

Si() 7x7 76

The (0) plane (outlined in green, layer ) [00] 2 3 2 0 2 0 0 c [-,,0] [00] [00] [0] 80

Si(0) surface (viewed nearly along [-,,0] direction) One dangling bond electron per surface atom Surface atoms red bulk atoms orange [,,0] [00] 8

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[00] 2 The (0) plane (outlined in green, layer ) [00] 3 2 0 2 0 0 c [0] [-,,0] [00] Cut through cubic unit cell surface unit cell P(x) As atoms top layer Ga atoms top layer [00] [-,,0] 87

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) GaAs (0) [00] 88

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]) 89

Reconstruction of (0) GaAs 90

III-V reconstruction 9

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 95

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 -0.25 each surface Ga has ¾+++0= 2.75 e for a net charge of +0.25 Thus considering both surface Ga and As, the (0) is neutral 5.25e 2.75e Net Q =0 0 2 0 2 0 2 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 96

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 97

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 As Ga As Ga Ga As 98

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 2 2 3/4 5/4 3/4 3/4 3/4 5/4 5/4 2 2 3/4 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 99

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 0 0 0 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 00

Consider missing As row out of 4 ---+3=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 0

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) 02

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 03

Intrinsic semiconductors + - 04

Excitation energy 05

To be added band states 06

To be added band states 07

Semiconducting properties 08

Semiconducting properties 09

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 6

To be added band states EA(Al)=5.033 ev 0.045 ev Add e to Al, from valence band = 5. -5.033 = 0.067 ev Al leads to acceptor state just 0.067eV above HOMO or VBM 9

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