Lecture 9 January 26, 2011
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
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>
Caitlin Scott <cescott@caltech.edu>
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
1
Course schedule
Friday January 14: L3 and L4
Monday January 17: Caltech holiday (MLKing)
Wednesday January 19: wag L5 and L6
Friday January 21: wag L7 and L8, caught up
Monday January 24: wag L7 and L8
Wag rotator
cuff operation
Wednesday January 26: wag L9 and L10
Friday January 28: wag participates in a retreat for our
nanotechnology project with UCLA
Friday January 28: wag L11 Back on schedule
Monday January 31: wag L12
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
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Last time
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
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Bond energies
De = EAB(R=∞) - EAB(Re)
e for equilibrium)
Get from QM calculations. Re is
distance at minimum energy.
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
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Bond energies
De = EAB(R=∞) - EAB(Re)
Get from QM calculations. Re is
distance at minimum energy
D0 = H0AB(R=∞) - H0AB(Re)
H0=Ee + ZPE is enthalpy at T=0K
ZPE = S(½Ћw)
This is spectroscopic bond energy from
ground vibrational state (0K)
Including ZPE changes bond distance
slightly to R0
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
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Bond energies
De = EAB(R=∞) - EAB(Re)
Get from QM calculations. Re is
distance at minimum energy
D0 = H0AB(R=∞) - H0AB(Re)
H0=Ee + ZPE is enthalpy at T=0K
ZPE = S(½Ћw)
This is spectroscopic bond energy from
ground vibrational state (0K)
Including ZPE changes bond distance
slightly to R0
Experimental bond enthalpies at 298K and atmospheric pressure
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)}.
(H =
E + pV assuming© an
ideal gas)
6
Ch120a-Goddard-L09
copyright 2011 William A. Goddard III, all rights reserved
Snap Bond Energy: Break bond without relaxing the fragments
Snap
DErelax = 2*7.3 kcal/mol
Adiabatic
D
Desnap (109.6snap
kcal/mol) De (95.0kcal/mol)
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
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Details 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 can be calculated from QM)
D298 = D0 + 2.4 = 87.5 + 2.4 = 89.9 kcal/mol
This is the quantity we will quote in discussing bond breaking
processes
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© copyright 2011 William A. Goddard III, all rights reserved
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Bond energies: Compare to CF3-CF3
CH3-CH3 De-snap = De + DErelax = 95.0 + 2*7.3 = 109.6 kcal/mol
(relaxation of tetrahedral CH3 to planar gains 7.3 kcal/mol)
For CF3-CF3, there is no little relaxation since CF3 wants to be
pyramidal, FCF~111º, estimate DErelax ~ 2 kcal/mol
Assume that De-snap = 109.6 kcal/mol for CF3-CF3 (as CH3-CH3)
Predict De(CF3-CF3) ~ 110 – 4 = 106
Assume DZPE (C2F6) ~ DZPE (C2H6)* sqrt(MH/MF) ~ 2.1
D0 (C2F6) = De –DZPE ~ 106 – 2.1 = 102
Thus D298 (C2F6) ~ D0 + 2.4 = 104
Experimental is D298=98.7±2.5 kcal/mol
Additional weakening of CC bond may be due to induction (Csp3
bond weaker because of charge transfer to F) and steric (F—F
nonbonded interactions)
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© copyright 2011 William A. Goddard III, all rights reserved
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Substituent effects on CC Bond energies (kcal/mol @ 298K)
The strength
of a CC
bond
changes
from 89.9 to
70 kcal/mol
as replace
Hs with
CH3s.
Goddard
believes is
mostly due
to fragment
relaxation
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
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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)
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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 rule 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 1.2 kcal/mol, due to Intraatomic Exchange (residual triplet coupling
12
of s,p
on same center) © copyright 2011 William A. Goddard III, all rights reserved
Ch120a-Goddard-L09
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
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© copyright 2011 William A. Goddard III, all rights reserved
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Geometries ethene
N state (planar) RCC = 1.339A (double bond
(twisted 90°) RCC = 1.47A (single s bond).
Ethane: RCC = 1.526 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 1.47A
T state (twisted 90°) RCC = 1.47A (single s bond).
(planar 0 °) RCC = 1.57A (Orthogonalization of the triple
coupled pp orbitals)
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© copyright 2011 William A. Goddard III, all rights reserved
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CC double bond energies
The bond energies for ethene are
De=180.0, D0 = 169.9, D298K = 172.3 kcal/mol
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 CH2 so that
Desnap = 180.0 + 4.7 = 184.7 kcal/mol
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© copyright 2011 William A. Goddard III, all rights reserved
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Energies H2C=CH2
Snap
DErelax = 2*2.35 kcal/mol
HCH=117.6°133°
Adiabatic
D
Desnap (184.7snap
kcal/mol) De (180.0kcal/mol)
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
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CC double bond energies
The bond energies for ethene are
De=180.0, D0 = 169.9, D298K = 172.3 kcal/mol
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 CH2 so that
Desnap = 180.0 + 4.7 = 184.7 kcal/mol
Since the Desnap = 109.6 kcal/mol, for H3C-CH3,
The p bond adds 75.1 kcal/mol to the bonding. (compare to
65kcal/mol rotational barrier)
Twisted ethylene, De = 180 – 65 = 115; Desnap = 115 + 5 =120.
This is 10 kcal/mol larger than for ethane.
May be due to sp2 vs sp3 of the effect of CH repulsions
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
17
Predict F2C=CF2 bond energy
F2C (3B1) + CF2 (3B1)
Relax. E. = 2*57 kcal/mol
F2C=CF2
F2C (1A1) + CF2 (1A1)
De-smap = 184.7 kcal/mol (from CH2-CH2)
De = 185 – 2* 57 = 71 kcal/mol, D0 = 71 – Dzpe = 71 – 2.3=69
D298
= D0 + 2.4 = 71,©Exper=
75
Ch120a-Goddard-L09
copyright 2011 William A. Goddard III, all rights reserved
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bond energy of F2C=CF2
Ethene Desnap = 180.0 + 4.7 = 184.7 kcal/mol
Consider the bond energy of F2C=CF2,
Assume Desnap(F2C=CF2) = 180.0 + 4.7 = 184.7 kcal/mol
But the snap electronic state is 3B1 which is 57 kcal/higher
than 1A1
Thus for CF2 the fragment relaxation is 2*57 = 114 kcal/mol
(there is also a correction from the FCF angle of C2F4 and
that of CF2 (3B1)
Predict adiabatic De= 185-114 = 71 kcal/mol.
D0 = De – Dzpe ~ 71 – 2.3 = 69
Thus D298 = D0 + 2.4 = 71 kcal/mol
The experimental value is D298 ~ 75 kcal/mol, close to the
prediction
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
19
Bond energies double bonds
ground state of CH2 is 3B1 by 9.3
kcal/mol, but substitution of one or
both H with CH3 leads to 1A1 ground
states.
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
20
CC triple bonds
Starting with two CH radicals in the 4S- state we can form
ethyne (acetylene) with two p bonds and a s 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
De = 235.7, D0 = 227.7, D298K = 229.8 kcal/mol
Which can be compared to De of 180.0 for H2C=CH2 and
95.0 for H3C-CH3.
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
21
HC=CH bond energy
HC (4S-) + CH (4S-)
Desnap = 270
HC=CH
Ch120a-Goddard-L09
Relax. E. = 2*17 kcal/mol
HC (2P) + CH (2P)
De = 235.7
© copyright 2011 William A. Goddard III, all rights reserved
22
Comparison of CC bond energies
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
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Triple Bond energies
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
24
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 Ch120a-Goddard-L09
on the next slide © copyright 2011 William A. Goddard III, all rights reserved
25
Infinite structure from tetrahedral bonding plus staggered bonds
on adjacent centers
2nd layer
3
1
1
1st layer
1
2
02
2
2nd layer
0
0
1c
1st layer
1
1
1
1
2nd layer
1st layer
Chair
configuration
of cylcohexane
Not shown: zero layer just like 2nd layer but above layer 1
st layer but below layer 2
3rd
layer just like the©1copyright
Ch120a-Goddard-L09
2011 William A. Goddard III, all rights reserved
26
c
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.
c
f
c
i
c
i f
f
f
f
i c
i
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 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
thisCh120a-Goddard-L09
cube by multiples©of
a in 2011
theWilliam
x,y,zA. Goddard
directions
copyright
III, all rights reserved
c
27
Diamond Structure
Now bond one of these
atoms, C2, to 3 new C
so that the bond are
staggered with respect
to those of C1.
5a
3a
1a
4b
2b
5
6
3
4
2
1b
4a
2a
1
Start with C1 and
make 4 bonds to form
a tetrahedron.
5b
3b
1c
7
Continue this process.
Get unique structure:
diamond
Note: Zig-zag chain
1b-1-2-3-4-5-6
Chair cyclohexane
ring: 1-2-3-3b-7-1c
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
28
Properties of diamond crystals
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
29
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.
Ch120a-Goddard-L09
© copyright 2011
William
Goddard III,
all rights reserved
Note
for Si, that the average
bond
isA.much
different
than for Si H30
Comparisons of successive bond energies SiHn and CHn
p
lobe
lobe
lobe
p
lobe
p
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
p
31
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
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
32
Examples of special
planes
c
c/l
b/k
b
a a/h
To denote all equivalent
planes we use {h,k,l} so
that
From Wikipedia
{1,0,0} for cubic
includes the 3 cases in
the first row)
A number with a bar
Ch120a-Goddard-L09
indicates
negative
© copyright 2011 William A. Goddard III, all rights reserved
33
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
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
34
The Si Crystal viewed
from the [001] direction
[001]
[010]
[110]
[010
[100]
(001) Surface
[100]
[1,-1,0]
Ch120a-Goddard-L09
1st Layer 
2nd Layer 
3rd Layer 
4th Layer 
RED
GREEN
ORANGE
WHITE
not show bonds
35
© copyright 2011 William A. Goddard III, all rights reserved
to 5th layer
The Si Crystal (100) surface, unreconstructed
Projection of
bulk cubic cell
Surface zig-zag row
Every red atom was
bonded to two Si that
are now removed,
thus two dangling
bond orbitals (like
1A1 state) sticking
out of plane
(100) VIEW
1st Layer 
2nd Layer 
3rd Layer 
4th Layer 
Surface unit©cell
P(1x1)
copyright
2011 William A. Goddard III, all rights reserved
Ch120a-Goddard-L09
RED
GREEN
ORANGE
WHITE36
Si(100) surface (unreconstructed)
viewed (nearly) along the [110] direction
Ch120a-Goddard-L09
Each surface atom has
two dangling bond orbitals
pointing to left and right,
along
[1,-1,0]
direction
© copyright
2011 William
A. Goddard III, all rights reserved
37
The (100) Surface Reconstruction
viewed (nearly) along the [110] direction
Spin pair dangling bond orbitals of adjacent atoms in
[1,-1,0] direction (originally 2nd near neighbors
Get one strong s bond but leave two dangling bond orbitals on
adjacent now bonded atoms (form weak p bond in plane)
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
38
Si(100) 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
Surface
orginal cell New cell
length
Lateral
bond
length
7.6A
displacements
2.4A
3.8A
0.7A 0.7A
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
39
Si(100) surface reconstructed (top view)
Rows of dimer
pairs are parallel
New unit cell
reconstructed
surface
P(2x1)
original unit cell
unreconstructed
surface
P(1x1)
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
40
Get 2x2 unit cell but atom at
center is equivalent to atom at
corner, therform c(2x2)
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
41
Two simple patterns for (100) Surface
Reconstruction
Dimer rows alternate
C(2x2), high energy
Ch120a-Goddard-L09
Dimer rows parallel
P(2x1), low energy
© copyright 2011 William A. Goddard III, all rights reserved
42
P(2x1) more stable than c(2x2) by ~ 1kcal/mol
The Sisurf-Si2nd-Sisurf bond for c(2x2) opens up to 120º
because the Sisurf move opposite directions
110º
120º
120º
Ch120a-Goddard-L09
110º
For P(2x1) the Sisurf move
the same directions and
Sisurf-Si2nd-Sisurf bond
43
© copyright 2011 William A.
Goddard III, all
reserved
remains
atrights
110º
New material
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
44
Construct (111) surface
using cubic unit cell
4
2
2
4
Atom #1 bonded to 3 atoms #2
3
4
4
2
1
Each #2 is bonded straight down
to an atom#3
2
4
1
4
1
4
1
2
2
Ch120a-Goddard-L09
2
2
3c
Go straight down to atom #1
Each #2 is bonded to 3 atoms #1 4
in top layer. Get hexagonal double
layer
0
1 4
3
Start at diagonal atom #0
Each atom #3 is bonded to 3
atom#4.
2
2
2
1
1
2
2
© copyright 2011 William A. Goddard III, all rights reserved
45
Si(111) surface (alternate construction)
Start with red atom on top, bond to 3 green atoms in 2nd layer
Each green atom is
bonded to 2 other
1st layer atoms plus
a 3rd atom straight
down (not shown)
The 3rd layer atoms
bond to 3 4th layer
atoms in orange
(now white)
Surface unit cell
P(1x1)
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
46
Reconstruction of Si(111) 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 2x1 unit
cell.
Ch120a-Goddard-L09
Indeed freshly
cleaved Si(111) at
low temperature
does show 2x1
Surface unit cell
47
P(1x1)
© copyright 2011 William A. Goddard III, all rights reserved
LEED experiments (Schlier and Farnsworth,
1959) observed 7th Order Spots  7x7 unit
cell (49 1x1 cells)
From 1959 to 1981 many models proposed to fit various
experiments or calculations.
Binnig et al., 1981 did first STM image of Si (7x7) and saw 12
bright spots in 7x7 cell, showed that every previous model
was incorrect
Takayanagi et al.,
1985, proposed the
DAS Model that
explained the
experiments
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
48
two 7x7 cells
What kind of interactions can go over a
7x7 region, with cell size 26.6 by 26.6 A?
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
49
Origin of complex reconstruction of Si(111)
In 49 surface unit cells have 49 dangling bonds. Since
cohesive energy of Si crystal is 108 kcal/mol expect
average bond energy must be 108/2 = 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 = 1.1 eV per surface atom
Calculated value = 1.224 eV snap and 1.200 ev relaxed.
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
50
Consider bonding an atom on top of 3 dangling bonds
T4
H3
T4
T4
H3
T4
Get 3x2 unit
cell
By adding a cap of one adatom Si per 3 top layer Si, can tie
offCh120a-Goddard-L09
all original dangling
bonds.
ThusA. 4816
© copyright
2011 William
Goddard III, all rights reserved
51
Consider bonding an atom on top of 3 dangling bonds
Two ways to do this. T4 and H3
T4 (observed)
Stabilize by 0.1
eV per site
Ch120a-Goddard-L09
H3 (not observed)
Destabilize by
0.15 eV per site
© copyright 2011 William A. Goddard III, all rights reserved
52
T4 versus H3 site bonding to dangling bonds
Energy increases
by 0.15 eV per
original surface
atom or 0.45 eV
per new adatom
Energy decreases
by 0.10 eV per
original surface
atom or 0.30 eV per
new adatom
Angle between
bond A and bond
B is 180º bad
overlap  orthog
Angle between bond
A and bond B is
100º ok overlap
no orthog
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
53
H3 reconstruction
1
1
0
1
2
1
1
0
1
Top layer labeled 1
2nd layer green
Addon layer 0, blue
Need just 1/3
Monolayer to tie up
bonds.
Surface energy
increases by 0.13 eV
Because 0-1-2 is linear
Unit cell
1
1
0
Ch120a-Goddard-L09
1
0
© copyright 2011 William A. Goddard III, all rights reserved
54
H3 reconstruction, √3 x √3
Top layer labeled 1
2nd layer green
Addon layer 0, blue
Need just 1/3
Monolayer to tie up
bonds.
Surface energy
increases by 0.13 eV
1
1
√3
0
1
1
1
0
1
1
1
1
0
Ch120a-Goddard-L09
1
0
© copyright 2011 William A. Goddard III, all rights reserved
55
T4 reconstruction √3 x √3
1
1
1
2
1
1
1
Top layer labeled 1
2nd layer green
Addon layer 0, blue
Need just 1/3
Monolayer to tie up
bonds.
Surface energy
decreases by 0.10 eV
Because 0-1-2 ~ 100º
Unit cell
1
1
Ch120a-Goddard-L09
1
© copyright 2011 William A. Goddard III, all rights reserved
56
T4 reconstruction 2x2
1
1
1
2
1
1
1
Top layer labeled 1
2nd layer green
Addon layer 0, blue
Need just 1/3
Monolayer to tie up
bonds, leave dangling
bond orbital
Surface energy
decreases by 0.08 eV
Per 2x2 cell
Unit cell
1
1
Ch120a-Goddard-L09
1
© copyright 2011 William A. Goddard III, all rights reserved
57
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
58
The (111) 7x7 DAS Surface
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
59
The (111) 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
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
60
The (111) 7x7 DAS Surface Layers
(purple, brown and blue atoms have one dangling bond)
1st
2nd
18 + 18 red
atoms, all
bonded to
1st layer
3rd
4th
First unreconstructed layer
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
61
The (111) 7x7 DAS Surface
Ch120a-Goddard-L09
12-membered
ring at corner of
cell
© copyright 2011 William A. Goddard
III, all rights reserved
62
The (111) 7x7 DAS Surface
Side view
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
63
The (111) 7x7 DAS Surface Cornerhole
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
64
Si(111) 7x7
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
65
The (111) 7x7 DAS Layer Positions
1
2
3
3
4
5
6
7
7
8
9
REF
Ch120a-Goddard-L09
REF
REF
© copyright 2011 William A. Goddard III, all rights reserved
69
The (111) 3x3
DAS Surface Unit Cell
Side view
Ch120a-Goddard-L09
Top view
12-membered rings
© copyright 2011 William A. Goddard III, all rights reserved
70
The (111) 5x5
DAS Surface Unit Cell
Side view
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
71
The (111) 5x5
DAS Surface Unit Cell
Top view
12- and 8-membered rings
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
72
The (111) 9x9
DAS Surface Unit Cell
Side view
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
73
The (111) 9x9
DAS Surface Unit Cell
Top view
12- and 8-membered rings
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
74
DAS Surface Energies (PBE DFT)
Energy, eV/1x1 Cell
1.09
Regression
Ab Initio
1.08
1.078
1.070
1.07
1.068
1.06
1.055
1.048
1.05
1.044
1.04
3
5
7
9
11
13
DAS Cell Size
Unreconstructed relaxed surface: 1.200 eV/1x1 cell
Infinite© DAS
model: 1.107 eV/1x1 cell
Ch120a-Goddard-L09
copyright 2011 William A. Goddard III, all rights reserved
75
DAS Reconstruction Driving Force
• 49 unpaired electrons (1/2 Si-Si bond) per 7x7 cell @ 1.2 eV
= 58.8 eV/cell
• DAS 7x7 Surface energy = 51.2 eV/cell (19 unpaired
electrons)
• Energy reduction due to reconstruction = 7.6 eV
• Difference is due to strain
• Bond length range = 2.31 – 2.50 Å (equilibrium 2.35 Å)
• Bond angle range = 91 – 117º (Equilibrium 109.4°)
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
76
DAS Surface Energy Contributions
Energy, eV/1x1 Cell
1.2
0.6
0.0
0.0
0.1
0.2
0.3
-0.6
(DAS Model Cell Size) -1
1x1
Ch120a-Goddard-L09
T4
8R
12R
F
D
TOTAL
© copyright 2011 William A. Goddard III, all rights reserved
77
DAS Surface Energies:
Sequential Size Change Model
5
Energy, eV/16x16 Cell
0
1
3
5
7
9
11
13
-5
-10
-15
SSC Irregular-odd and even
SSC regular-odd
-20
SSC Cell Size
Real-time STM by Shimada & Tochihara, 2003
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
78
DAS Surface Energies:
Origin of a finite cell size
Energy, eV/1x1 Cell
1.4
SSC Irregular-odd and even
SSC regular-odd
DFT
1.3
1.2
1.1
1.0
1
3
5
7
9
11
13
Cell Size
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
79
The (110) plane (outlined in green, layer 1)
[001]
3
1
[-1,1,0]
1
1
2
02
2
0
0
1c
1
1
[010]
1
1
[100]
Ch120a-Goddard-L09
[110]
© copyright 2011 William A. Goddard III, all rights reserved
80
Si(110) surface (top view)
Cut through
cubic unit cell
surface unit
cell P(1x1)
Surface
atoms red
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
81
Si(110) surface (viewed nearly along [-1,1,0] direction)
One dangling bond
electron per surface atom
Surface atoms red
bulk atoms orange
[1,1,0]
[001]
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
82
Reconstruction of (110) surface, surface atoms only
side view
(along [-1,1,0])
Showing just 2 dangling
bond orbitals
54.7º
54.7º
Top view
(from [-1,-1,0])
[001]
[-1,1,0]
Ch120a-Goddard-L09
[1,1,0]
[001]
© copyright 2011 William A. Goddard III, all rights reserved
83
Reconstruction of (110) surface, surface atoms only
We have a chain of dangling bond
orbitals along the [-1,1,0] direction,
each tilted by 35.3º from the [110]
(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
[-1,1,0] direction
Ch120a-Goddard-L09
side view
(along [-1,1,0])
Showing just 2 dangling
bond orbitals
54.7º
54.7º
54.7º
[110]
[001]
© copyright 2011 William A. Goddard III, all rights reserved
84
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.
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
85
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 sp3 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 acidLewis 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
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
86
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 13-15
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
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
87
Lecture 10 January 26, 2011
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
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>
Caitlin Scott <cescott@caltech.edu>
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
88
Last time
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
89
T4 versus H3 site bonding to dangling bonds
Energy increases
by 0.15 eV per
original surface
atom or 0.45 eV
per new adatom
Energy decreases
by 0.10 eV per
original surface
atom or 0.30 eV per
new adatom
Angle between
bond A and bond
B is 180º bad
overlap  orthog
Angle between bond
A and bond B is
100º ok overlap
no orthog
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
90
H3 reconstruction
1
1
0
1
2
1
1
0
1
Top layer labeled 1
2nd layer green
Addon layer 0, blue
Need just 1/3
Monolayer to tie up
bonds.
Surface energy
increases by 0.13 eV
Because 0-1-2 is linear
Unit cell
1
1
0
Ch120a-Goddard-L09
1
0
© copyright 2011 William A. Goddard III, all rights reserved
91
H3 reconstruction, √3 x √3
Top layer labeled 1
2nd layer green
Addon layer 0, blue
Need just 1/3
Monolayer to tie up
bonds.
Surface energy
increases by 0.13 eV
1
1
√3
0
1
1
1
0
1
1
1
1
0
Ch120a-Goddard-L09
1
0
© copyright 2011 William A. Goddard III, all rights reserved
92
T4 reconstruction √3 x √3
1
1
1
2
1
1
1
Top layer labeled 1
2nd layer green
Addon layer 0, blue
Need just 1/3
Monolayer to tie up
bonds.
Surface energy
decreases by 0.10 eV
Because 0-1-2 ~ 100º
Unit cell
1
1
Ch120a-Goddard-L09
1
© copyright 2011 William A. Goddard III, all rights reserved
93
T4 reconstruction 2x2
1
1
1
2
1
1
1
Top layer labeled 1
2nd layer green
Addon layer 0, blue
Need just 1/3
Monolayer to tie up
bonds, leave dangling
bond orbital
Surface energy
decreases by 0.08 eV
Per 2x2 cell
Unit cell
1
1
Ch120a-Goddard-L09
1
© copyright 2011 William A. Goddard III, all rights reserved
94
The (111) 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
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
95
The (111) 7x7 DAS Surface Layers
(purple, brown and blue atoms have one dangling bond)
1st
2nd
18 + 18 red
atoms, all
bonded to
1st layer
3rd
4th
First unreconstructed layer
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
96
The (111) 7x7 DAS Surface
Ch120a-Goddard-L09
12-membered
ring at corner of
cell
© copyright 2011 William A. Goddard
III, all rights reserved
97
The (111) 7x7 DAS Surface Cornerhole
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
98
Si(111) 7x7
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
99
The (110) plane (outlined in green, layer 1)
[001]
3
1
[-1,1,0]
1
1
2
02
2
0
0
1c
1
1
[010]
1
1
[100]
Ch120a-Goddard-L09
[110]
© copyright 2011 William A. Goddard III, all rights reserved
103
Si(110) surface (viewed nearly along [-1,1,0] direction)
One dangling bond
electron per surface atom
Surface atoms red
bulk atoms orange
[1,1,0]
[001]
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
104
Reconstruction of (110) surface, surface atoms only
We have a chain of dangling bond
orbitals along the [-1,1,0] direction,
each tilted by 35.3º from the [110]
(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
[-1,1,0] direction
Ch120a-Goddard-L09
side view
(along [-1,1,0])
Showing just 2 dangling
bond orbitals
54.7º
54.7º
54.7º
[110]
[001]
© copyright 2011 William A. Goddard III, all rights reserved
105
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.
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
106
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 sp3 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 acidLewis 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
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
107
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 13-15
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
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
108
New material
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
109
The
(110) plane (outlined in green, layer 1)
[001]
Cut through cubic unit cell
3
1
[-1,1,0]
1
1
2
02
2
0
0
1c
1
1
surface unit
cell P(1x1)
[010]
1
1
[100]
As atoms
top layer
Ch120a-Goddard-L09
[110]
Ga atoms
top layer
[001]
[-1,1,0]
© copyright 2011 William A. Goddard III, all rights reserved
110
Reconstruction of (110) surface, side view along [-1,1,0]
Si has
dangling
bond
electron at
each surface
atom
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
54.7º
54.7º
As
Ga
54.7º
[110]
Si (110)
[001]
Ch120a-Goddard-L09
GaAs (110)
© copyright 2011 William A. Goddard III, all rights reserved
111
Reconstruction of GaAs(110) surface
As has 3 covalent bonds,
leaving 2 electrons in 3s
lone pair, Ga has 3
covalent bonds leaving 0
eletrons in 4th orbital
Ga
As 54.7º
54.7º
Top view
(from [-1,-1,0])
[001]
[-1,1,0]
Ch120a-Goddard-L09
[1,1,0]
side view
[001] (along [-1,1,0])
112
© copyright 2011 William A. Goddard III, all rights reserved
Reconstruction of (110) GaAs
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
113
III-V reconstruction
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
114
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
115
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
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
118
4
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+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
0
Ga
3/4
1
5.25e
2.75e
2
As
0
Ga
1
1
3/4
5/4
3/4
3/4
3/4
5/4
a
1
5/4
3/4
5/4
g
5/4
Ch120a-Goddard-L09
1
3/4
5/4
5/4
a
1
Net Q =0
2
As
1
1
5/4
3/4
g
3/4
5/4
0
Ga
1
3/4
5/4
a
1
3/4 III, all rights reserved
© copyright 5/4
2011 William A. Goddard
5/4
3/4
5/4
5/4
2
As
3/4
g
3/4
3/4119
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)
Ch120a-Goddard-L09
1st Layer 
2nd Layer 
3rd Layer 
th Layer 
© copyright 2011 William A. Goddard4III, all
rights reserved
RED
GREEN
ORANGE
WHITE 120
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
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
121
Charges for 2x1 GaAs(100)
2nd layer ga
has 3 e
1
2
1
5/4
2
2nd layer, ga
5/4
3/4
3/4
3/4
3/4
3/4
3/4
3/4
3/4
5/4
2e As-ga
bond
1
2
2e As LP
5/4
3/4
3/4
1st layer As
has 5.5 e
Top layer, As
3/4
Ch120a-Goddard-L09
1
5/4
3rd layer, as
2
5/4
Each surface
As has extra
3/4
0.5 e  dimer
3/4
3/4 has extra 1e
2e As-As
3/4
Not stable 122
© copyright bond
2011 William A. Goddard III, all rights reserved
3/4
Now consider a missing row of As for GaAs(100)
1
5/4
1
Top layer, As
2nd layer, ga
5/4
3/4
3/4
3/4
3/4
3/4
0
ga empty
LP
3/4
0
3rd layer, as
2nd layer ga
has 2.25e
0
3/4
1st layer As
has 5.5 e
3/4
Ch120a-Goddard-L09
Each 2nd layer ga
next to missing
0
As is deficient by
3/4
0.75e extra 0.5 e
3/4
3/4
 4 ga are
3/4
123
missing
© copyright 2011 William A. Goddard III, all rights
reserved 3e
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
Ch120a-Goddard-L09
Extra 1e
Extra 1e
missing 3e
© copyright 2011 William A. Goddard III, all rights reserved
124
Different views of GaAs(100)4x2 reconstruction
-1.0e
Previous page, 3
As dimer rows
then one missing
Ch120a-Goddard-L09
+1.5e
Two missing As row
plus missing Ga row
Exposes 3rd row As
Agrees with experiment
Hashizume
PhysIII,Rev
B 51,
4200
© copyright
2011 Williamet
A.al
Goddard
all rights
reserved
(1995)
125
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
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
126
Intrinsic semiconductors
+
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
127
Excitation energy
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
128
To be added – band states
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
129
To be added – band states
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
130
Semiconducting properties
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
131
Semiconducting properties
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
132
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
133
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
134
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
135
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
136
Ch120a-Goddard-L09
© copyright 2011 William A. Goddard III, all rights reserved
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Ch120a-Goddard-L09
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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
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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
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stop
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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+
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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.
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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
thisCh120a-Goddard-L09
out in the 1920’s without
theWilliam
aid A.ofGoddard
computers.
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