Crystal Structure Determination I - Khwarizmi Science Society
Determination of Crystal Structure (From Chapter 10 of Textbook 2)
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Transcript of Determination of Crystal Structure (From Chapter 10 of Textbook 2)
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Determination of Crystal Structure(From Chapter 10 of Textbook 2)
Unit cell line positionsAtom position line intensity (known chemistry)
Three steps to determine an unknown structure: (1) Angular position of diffracted lines shape and size of the unit cell. (2) sizes of unit cell, chemical composition, density # atoms/unit cell (3) Relative intensities of the peaks positions of the atoms within the unit cell
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Preliminary treatment of data: Ensure true random orientation of the particles of the sample Remove extraneous lines from (1) K or other wavelength:
sin2;sin2 hklhkl dd
sinsin
In the analyzing step, the sin2 is used
22
2
sinsin
2.1
2
For mostradiations
filament contamination W KL radiation Diffraction by other substance
Calibration curve using known crystal
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Effect of sample height displacement
R
s cos2 radians) (in 2
s: sample height displacementR: diffractometer radius
+
s
Length of cossa larger error for low angle
peaks for the most accurate unit cell parameters it is generally better to use the high angle peaks for this calculation.
R
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Example for sample height displacement
Assume a crystal; cubic structure; a = 0.6 nm.Consider the error that can be introduced if the samplewas displaced by 100 microns (0.1 mm) for (100) and(400) diffraction peaks? Assume λ = 0.154 nm and R = 225 mm.
37.7154.0sin)6.0(2 100 d87.30154.0sin)15.0(2 400 d
The displacement cause these peaks to shift by 051.01081.8225/37.7cos1.022 4
044.01063.7225/87.30cos1.022 4
(100): = 7.395o
(400): = 30.892o
5982.0154.0395.7sin2 100 ad
5999.0154.0892.30sin2 400 ad
025.0022.0
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Pattern Indexing
→assign hkl values to each peak
Simplest example: indexing cubic pattern
sin2 hkld222 lkh
adhkl
2
2
222
2
222 4
sinsin2
alkhlkh
a
Constant for agiven crystal
222 lkhs Define
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Values for h2 + k2 + l2 for cubic system
SC: 1, 2, 3, 4, 5, 6, 8, … BCC: 2, 4, 6, 8, 10, 12, 14, ..
FCC: 3, 4, 8, 11, 12, 16, 19, 20, …Diamond: 3, 8, 11, 16, 19, …
SC BCC FCC Diamond S
100 X X X 1
110 X X 2
111 X 3
200 X 4
210 X X X 5
211 X X 6
220 8
221 X X X 9
300 X X X 9
310 X X 10
311 X 11
222 X 12
320 X X X 13
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SC: 1, 2, 3, 4, 5, 6, 8, … BCC: 2, 4, 6, 8, 10, 12, 14, ..
s is doubled in BCC,No s = 7 in SC
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Indexing Tetragonal system
2
2
2
22
2
1
c
l
a
kh
dhkl
hkld2/sin
2222
2
2
2222
2 )(42
sin ClkhAc
l
a
kh
dhkl
2222 4/;4/ cCaA
When l = 0 (hk0 lines), )(sin 222 khA
Possible values for l2 are 1, 4, 9, 16, …
2222 )(sin ClkhA
Possible values for h2 + k2: 1, 2, 4, 5, 8, 9, 10, …
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2
2
2
22
2 3
41
c
l
a
khkh
dhkl
2222
22
2
2222 )(
43sin ClkhkhA
c
l
a
khkh
2222 4/;3/ cCaA
Possible values for h2 + hk + k2 are 1, 3, 4, 7, 9, 12, … The rest of lines are Possible values for l2 are 1, 4, 9, 16, …
2222 )(sin ClkhkhA
Indexing Hexagonal system
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Example: sin2 sin2/3 sin2/4 sin2/70.097 0.0320.112 0.037 0.136 0.0450.209 0.0700.332 0.1110.390 0.1300.434 0.1450.472 0.1570.547 0.1820.668 0.2230.722 0.241
0.0240.0280.0340.0520.0830.0980.1090.1180.1370.1670.180
0.0140.0160.0190.0300.0470.0560.0620.0670.0780.0950.103
100
110
Let’s say A = 0.112
123456789
1011
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sin2 sin2-A sin2-3A0.0970.1120.1360.2090.3320.3900.4340.4720.5470.6680.7220.8060.879
00.0240.0970.2200.2780.3220.3600.4350.5560.6100.6940.767
0.0540.0980.1360.2110.3320.3860.4700.543
0.097 belongs to Cl2.What is the l?There are two linesbetween 100 and 110.Probably, 10l1 and10l2 0.024 and 0.097are different ls. 0.097/0.024 ~ 40.220/0.024 ~ 90.390/0.024 ~ 16 C = 0.02441
123456789
10111213
100
110
101102
002
112
,103004
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Indexing orthorhombic system:
2222
2
2
2
2
222
4sin ClBkAh
c
l
b
k
a
h
More difficult! Consider any two lines having indices hk0 and hkl Cl2 put it back get A and B. guess right (consistent) not right, try another guesses C
Indexing Monoclinic and Triclinic system Even more complex, 6 variables must have enoughdiffraction lines for the computer to indexing.
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Effect of Cell distortion on the powder Pattern
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http://www.ccp14.ac.uk/solution/indexing/index.html
Autoindexing
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Determination of the number of atoms in a unit cell
M
NVn C 0
VC: unit cell volume; : densityN0: Avogodro’s number;M: molecular weight;n: number of molecules in a unit cell
66054.1)10022.6()()10( 23
03338 C
C
VN
cm
gcmVnM
in Å3
in g/cm3
Indexing the power pattern shape and size of unit cell (volume)number of atoms in that unit cell.
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Determination of Atom positions
Relative intensities determine atomic positions(a trial and error process)
N
nnnnn lwkvhuifFpFI
12
22
)(2exp ; cossin
2cos1
(un vn wn): position of nth atom in a unit cell. Trial and error: known composition, known number of molecules, known structure eliminate some trial structure. Space groups and Patterson Function (selection of trial structures)
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Example: CdTe Chemical analysis which revealed:
49.8 atomic percent as Cd (46.6 weight percent) 50.2 atomic percent as Te (53.4 weight percent)
2.50:8.490009.1:16.127
4.53:
4.112
6.46Te:Cd
* Make powder diffraction and list sin2: index the pattern! Assume it is cubic
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0.04620.11940.16150.1790.2340.2750.3460.3910.4610.5040.5750.6160.6880.7290.799
1234568910111213141617
0.04620.05970.053830.044750.04680.045830.043250.043440.04610.045820.047920.047380.049140.045560.047
24681012141618202224263032
0.02310.029850.026920.022370.02340.022920.024710.024440.025610.02520.026140.025670.026460.02430.02497
348111216182024273235364043
0.01540.029850.020190.016270.01950.017190.019220.019550.019210.018670.017970.01760.019110.018220.01858
3811161924273235404349515659
0.01540.014930.014680.011190.012320.011460.012810.012220.013170.01260.013370.012570.013490.013020.01354
sin2 SCs
BCCs
FCCs
Diamonds
s
2sin
s
2sin
s
2sin
s
2sin
close
Veryweak
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0.04620.11940.16150.2340.2750.3460.3910.4610.5040.5750.6160.6880.7290.7990.84
3811161924273235404349515659
0.01540.014930.014680.014630.014470.014420.014480.014410.01440.014370.014330.014040.014290.014270.01424
veryclose
A 486.601413.02
542.1
01413.04 2
2
a
a
Diamond structure and Zinc blend structure: forbidden peaks! Which one has more peaks? removing line 4 first
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0.04620.11980.16150.1790.2340.2750.3460.3910.4610.5040.5750.6160.6880.7290.799
Alternative: Assume the first line is from s = 1, s = 2 and s = 3, …
2sin0462.0
sin2 0462.0
sin2 2 0462.0
sin3 2
1.0002.5933.4963.8745.0655.9527.4898.4639.97810.90912.44613.33314.89215.77917.294
2.0005.1896.9917.74910.13011.90514.97816.92619.95721.81824.89226.66729.78431.55834.589
3.0007.77910.48711.62315.19517.85722.46825.39029.93532.72737.33840.00044.67547.33851.883
1234568910111213141617
24681012141618202224263032
348111216182024273235364043
3811161924273235404349515659
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Larger error smaller , use the first three lines to fit a more correct A. 0.01444, Use this number to divide sin2 !
0.04620.11980.16150.1790.2340.2750.3460.3910.4610.5040.5750.6160.6880.7290.799
01444.0/sin2 3.1998.29611.18412.39616.20519.04423.96127.07831.92534.90339.82042.65947.64550.48555.332
3811161924273235404349515659
LargerDeviation
New A is 0.01428
01428.0
sin2
3.2358.38911.31012.53516.38719.25824.23027.38132.28335.29440.26643.13748.17951.05055.952
.
.
.
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0.1790/0.01413 = 12.66; 222 s = 12 (forbidden diffraction lines for diamond structure, OK for zinc-blend structure!) * = 5.82 g/cm3 then
67.95366054.1
82.5)486.6(
66054.1
3
CV
nM
M for CdTe is 112.4+127.6 = 240 n = 3.97 ~ 4 There are 4 Cd and 4 Te atoms in a unit cell.
* FCC based structure with 4 molecules in a unit cell – NaCl and ZnS; CdTe: NaCl or ZnS
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NaCl –Cd at 000 & Te at ½½½ + fcc translations
)2
1
2
1
2
1(2
)2
1
2
10(2)
2
10
2
1(2)0
2
1
2
1(2
1
lkhi
TeCd
lkhilkhilkhi
hkl
eff
eeeF
FCC
when h, k, l all even 22)(16 TeCdhkl ffF
when h, k, l all odd 22)(16 TeCdhkl ffF
ZnS –Cd at 000 & Te at ¼ ¼ ¼ + fcc translations
)4
1
4
1
4
1(2 lkhi
TeCdFCChkl effFF
when h + k + l is odd )(16 222
TeCdhkl ffF
when h + k + l = odd multiple of 2 = even multiple of 2
22)(16 TeCdhkl ffF
FCCF
Unmixed hkl only
22)(16 TeCdhkl ffF
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fCd + fTe = 100 and |fCd fTe| = 4 at sin / = 0 tofCd + fTe = 30.3 and |fCd fTe| = 1.7 at sin / ~ 1.0
sin
0.0 0.2 0.4 0.6 0.8 1.0 1.2
CdTe
48 37.7 27.5 21.8 17.6 14.3 12.052 41.3 30.3 24.0 19.5 16.0 13.3
TeCd
TeCd
TeCd Rff
ff
2
2
sin
0.0 0.2 0.4 0.6 0.8 1.0 1.2
TeCdR 625 482 426 415 381 318 379
Several hundreds
NaCl –Cd at 000 & Te at ½½½ + fcc translations
h, k, l all even: strong diffracted linesh, k, l all odd: week diffracted lines
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ZnS fit better!
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Order-Disorder Determination
temperaturehighlow TC
disorderorder
Example – Cu-Au system (AuCu3), TC = 390 oC
disordered ordered
Au
Cu
Cu-Au average
Substitutional solid solution A, B elements AB atoms’ arrangement
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Complete Disordered structure:
the probability of each site beingoccupied: ¼ Au, ¾ Cu simple FCC with fav
)2
1
2
10(2)
2
10
2
1(2)0
2
1
2
1(2
1lkhilkhilkhi
avhkl eeefF
4/)3( CuAuav fff
For mixed h, k, l Fhkl = 0For unmixed h, k, l Fhkl = (fAu + 3fCu)
disordered
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For mixed h, k, l Fhkl = (fAu fCu)For unmixed h, k, l Fhkl = (fAu + 3fCu)
Peaksshow up
Complete Ordered structure: 1 Au atom, at 000, three Cu atoms at ½ ½ 0, ½ 0 ½, 0 ½ ½.
)2
1
2
10(2)
2
10
2
1(2)0
2
1
2
1(2)000(2 lkhi
Cu
lkhi
Cu
lkhi
Culkhi
Auhkl efefefefF
)()()( lkilhikhiCuAuhkl eeeffF
ordered
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Define a long range order parameter S:
A
AA
F
FrS
1
rA: fraction of A sites occupied by the rightatoms; FA: fraction of A atoms in the alloy
complete order: rA = 1 S = 1;complete disorder: rA = FA S = 0
0 S 1
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AuCu3 : order parameter S
25.075.0)1( SFFSr AuAuAu
A-site is the 000 equipointrAu :fraction of Au atoms in 000 site the average atomic form factor
Average atomic factor for A-site
CuCuAuAu
CuAuAuAuavA
SfffSf
frfrf
75.075.025.075.0
)1(
CuAuCuAuavA ffffSf 75.025.0)(75.0
= fav
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(1 rA): is the fraction of Au occupying the B site in B site (1 rA)/3 Au and 1 (1rA)/3 Cu = (2 + rA)/3 Cu
3
)2(
3
)1( CuAuAuAuavB
frfrf
CuCuAuAuavB fSfSfff 75.025.025.025.0
CuAuAuCuavB ffffSf 75.025.0)(25.0
Average atomic factor for B-site
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The structure factor is
)2
1
2
10(2)
2
10
2
1(2)0
2
1
2
1(2)000(2 lkhilkhilkhi
avBlkhi
avAhkl eeefefF
For mixed h, k, l avBavAhkl ffF
)()(25.0)(75.0 CuAuavAuCuavCuAuhkl ffSfffSfffSF
For unmixed h, k, l avBavAhkl ffF 3
avavAuCuavCuAuhkl ffffSfffSF 43)(25.03)(75.0
CuAuCuAuavhkl fffffF 3)75.025.04( 4
Intensity |F|2 superlattice lines S2
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B
BB
F
FrS
1
Using different S definition
What would you get?Homework!
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Intensity weak diffuse background
If atoms A and B completely random in a solid solution diffuse scattering
2)( BAD ffkI k: a constant for any one composition
f decreases as sin/ increases ID as sin/
Weak signal, very difficult to measure
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fav
fZn
fCu
Example – Cu-Zn system (CuZn), TC = 460 oC
disordered ordered
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)1( )()2
1
2
1
2
1(2)000(2 lkhi
av
lkhi
avlkhi
avhkl efefefF
* Completely random: a BCC structure
ZnCuav fff 5.05.0
* Completely order: a CsCl
For h + k + l even Fhkl = fCu + fZn
For h + k + l odd Fhkl = 0
)()2
1
2
1
2
1(2)000(2 lkhi
ZnCu
lkhi
Znlkhi
Cuhkl effefefF
For h + k + l even Fhkl = fCu + fZn
For h + k + l odd Fhkl = fCu fZn
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A
AA
F
FrS
1
Define a long range order parameter S:
For h + k + l even Fhkl = fCu + fZn
For h + k + l odd Fhkl = S(fCu fZn)
(practice yourself)
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1
0
TCT
Different system
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Relative intensity from the superlattice line and the fundamental line:
* Case AuCu3: ignoring the multiplication factor and Lorentz-polarization factor, just look at the |F|2.
2
2
|3|
||
line) al(FundamentIntensity
line) ice(superlattIntensity
CuAu
CuAu
ff
ff
Assume sin/ = 0 f = z
09.0~|29379|
|2979|
line) al(FundamentIntensity
line) ice(superlattIntensity 2
2
About 10%, can be measured without difficulty.
Assume sin/ = 0.2
11.0~|6.21365|
|6.2165|
line) al(FundamentIntensity
line) ice(superlattIntensity 2
2
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* Case CuZn: the atomic number Z of Cu and Zn is close atomic form factor is close!!
2
2
||
||
line) al(FundamentIntensity
line) ice(superlattIntensity
ZnCu
ZnCu
F
S
ff
ff
I
I
Assume sin/ = 0 f = z
Assume sin/ = 0.2
0003.0|3029|
|3029|
||
||2
2
2
2
ZnCu
ZnCu
F
S
ff
ff
I
I
0003.0|4.226.21|
|4.226.21|
||
||2
2
2
2
ZnCu
ZnCu
F
S
ff
ff
I
I
About 0.03%, very difficult to measure
choose a proper wavelength to resolve the case!
![Page 42: Determination of Crystal Structure (From Chapter 10 of Textbook 2)](https://reader035.fdocuments.us/reader035/viewer/2022081506/56814f21550346895dbcb2bc/html5/thumbnails/42.jpg)
Resonance between the radiation and theK shell electrons larger absorption f
Produce extradifference in
fCu fZn
0013.0|)7.230()6.329(|
|)7.230()6.329(|
||
||2
2
2
2
ZnCu
ZnCu
F
S
ff
ff
I
I
Using Zn K radiation. f for Cu is -3.6 and for Zn is -2.7
About 0.13%, possible to be detected.