monoclinic + 2
DESCRIPTION
Monoclinic + 2. 2D. P 2. p 2. c. b. a. z. (1). (2). z. z +1/2. (3). New type of operation. 2. Screw axis. Specifying. For a 3-fold screw axis:. 3. 3 1. 3 2. P 3 1. P 3 2. 4-fold screw axis:. 4 3. 4 1. 4 1. 4 2. 4 3. 4 2. n 1. n 2. ……. n m-2. n m-1. No chirality. - PowerPoint PPT PresentationTRANSCRIPT
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3D Symmetry _2(Two weeks)
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3D lattice: Reading crystal7.pdf
Oblique (symmetry 1) + 3T
2T
1T
2T
1T
3T
321 TTT
133221 TTTTTT
General
P1
Triclinic Primitive
Building the 3D lattices by adding another translation vector to existing 2D lattices
triclinic
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Oblique (symmetry 2) + 3T
2T
1T
3T
projection
4 choices in order to maintain2 fold rotation symmetry:
)2
1
2
1(),0
2
1(),
2
10(),00(
3T
Is this position OK? X
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2T
1T
3T
oTTTT 901323
)00(
P2
)2
10(
2T
1T
3T
oTTTT 901'
32'
3
Double cellside centered
A2321 TTT
'321 TTT
)02
1(
23'
3 2 TTT
'
3T
2T
1T
3T Double cell
side centered
13'
3 2 TTT
'
3T
B2
zTTT 213 00
oTTTT 901'
32'
3
'321 TTT
zTTT 213 02
1
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'3T
)2
1
2
1(
2T
1T 3T
Double cellbody centered
123'
3 2 TTTT
I2
oTTTT 901'
32'
3
'321 TTT
IBA 222
zTTT 213 2
1
2
1
general 21 TT
Some people usebased centered,some use body centered.
IP 2 ,2 monoclinic
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Rectangular (symmetry m) + 3T
90o
90o oTTTT 901321 321 TTT
2T
1T
3T
zTyTT 213 2
1
zTyTT 222 213
A2
Rectangular (symmetry g) + 3T
: the same.
cm + ?3T
1T
2T
already exist!
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Rectangular (symmetry 2mm) + 3T
P2mmP2mgp2gg
3T
)2
1
2
1(),0
2
1(),
2
10(),00(
)2
1
2
1(),0
2
1(),
2
10(),00( zzzz
2T
1T
3T
oTTTTTT 90132321
)00(
P222
321 TTT
Orthorhombicprimitive
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)2
10(
A222
Orthorhombicbase-centered
)02
1(
2T
1T
3T
'3T
2T
1T
3T Double cell
side centered
13'
3 2 TTT
'
3T
B222
Orthorhombicbase-centered
Double cellside centered
23'
3 2 TTT
oTTTTTT 901'
32'
321
'321 TTT
oTTTTTT 901'
32'
321
'321 TTT
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'3T
)2
1
2
1(
2T
1T 3T
123'
3 2 TTTT
I222'321 TTT
IA 222222
oTTTTTT 90132321
rectangular
AB 222 ,222
C222
Orthorhombicbody-centered
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Centered Rectangular (symmetry 2mm) + 3T
C2mm )2
1
2
1(),0
2
1(),
2
10(),00( zzzz
)00( z
2T
1T
3T
oTTTTTT 90132321
C222
321 TTT
)2
1
2
1( z
2T
1T
'3T
oTTTTTT 901'
32'
321
C222
'321 TTT
the same
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)02
1( z
2T
1T
3T Face centered
'3T
F222oTTTTTT 901
'32
'321
'
321 TTT
FCIP 222 ,222 ,222 ,222 orthorhombic
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Square (symmetry 4, 4mm) + 3T
P4P4mmp4gm
)2
1
2
1(),00( zz
2T
1T
3T
oTTTTTT 90132321
321 TTT
)00( z
Tetragonalprimitive
P4
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)2
1
2
1( z
TetragonalBody centered
'3T
2T
1T 3T
I4
'321 TTT
oTTTTTT 901'
32'
321
zTTT 213 2
1
2
1
IP 4 ,4 Tetragonal
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Hexagonal (symmetry 3, 3m) + 3T
1T
2T
p31mp3m1
)3
1
3
2(),00( zz
)00( z Hexagonal primitive P3
)3
1
3
2( z Rhombohedral R3
p3
not in this categoryWhy?
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)00( z Hexagonal primitive P3
1T 2T
3T
o
o
TTTT
TT
90
120
1323
21
321 TTT
)3
1
3
2( z Rhombohedral R3
1T
2T
triple cell
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ab
c
1/32/3
1/3
2/3
a = b = c; = = 90o
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Hexagonal (symmetry 3m, 6, 6mm) + 3T
2T
1T
can only located at positions:3T
)00( z
p6p6mm
)00( z Hexagonal primitive P3p31m
RP 3 ,3 Hexagonal & 6 related can only fit 3P!
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cubic (isometric)
Primitive (P)Body centered (I)Face centered (F)Base center (C)
Special case of orthorhombic (222) with a = b = cCheck the 3 fold rotation symmetry in 111 direction
a = b cTetragonal (P)
Tetragonal (I)?
11 lattice types already
FIP 23 ,32 ,23 Cubic
[100]/[010]/[001] [111]
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Another way to look as cubic:Consider an orthorhombic and requesting the diagonaldirection to be 3 fold rotation symmetry
222P 23P
222F 23F
222I 23I
222C 23C3 fold rotation symmetry doesnot exist in base centered cell
Primitive
Body centered
Face centered
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Bingo!
14 Bravais lattices!
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Crystal Class Bravais Lattices Point Groups
Triclinic P (1P) 1,
Monoclinic P (2P), C(2I) 2, m, 2/m
Orthorhombic P(222P), C(222C) F(222F), I(222I) 222, mm2, 2/m 2/m 2/m
Rhombohedral P (3P), 3R 3, , 32, 3m, 2/m
Hexagonal P (3P) 6, , 6/m, 622, 6mm, m2,6/m 2/m 2/m
Tetragonal P (4P), I (4I) 4, , 4/m, 422, 4mm, 2m,4/m 2/m 2/m
Isometric (Cubic)
P (23P), F(23F), I (23I) 23, 2/m, 432, 3m, 4/m2/m
Lattice type - compatibility with - point groupreading crystal9.pdf.
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http://www.theory.nipne.ro/~dragos/Solid/Bravais_table.jpg
= P = I
= T P
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= P = I
= B
= T P
http://users.aber.ac.uk/ruw/teach/334/bravais.php
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Next, we can put the point groups to the compatible lattices, just like the cases in 2D space group.
3D Lattices (14) + 3D point groups 3D Space group
There are also new type of symmetry shows up in 3D space group, like glide appears in 2D space (plane) group!
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The naming (Herman-Mauguin space group symbol) is the same as previously mentioned in 2D plane group!
The first letter identifies the type of lattice:•P: Primitive; I: Body centered; F: Face centered•C: C-centered; B: B-centered, A: A-centered
The next three symbols denote symmetry elements in certain directions depending on the crystal system. (See next page)
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Monoclinica b = 90o; c b = 90o. b axis is chosen to correspond to a 2-fold axis of rotational symmetry axis or to be perpendicular to a mirror symmetry plane. Convention for assigning the other axes is c < a. a c is obtuse (between 90º and 180º).
OrthorhombicThe standard convention is that c < a < b.
Once you define the cell following the convention A, B, C centered
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Crystal SystemSymmetry Direction
Primary Secondary Tertiary
Triclinic None
Monoclinic [010]
Orthorhombic [100] [010] [001]
Tetragonal [001] [100]/[010] [110]
Hexagonal/Rhombohedral [001] [100]/[010] [120]/[1 0]
Cubic[100]/[010]/
[001] [111] [110]
1
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Monoclinic + 2
P2
p2
2D
TBAT2
1@
Consider 2P Monoclinic + 2
/2
P2/2
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ab
cA ? AT
cba2
1
2
1
2
1T
T ||T
How about 2I Monoclinic + 2
There is a lattice pointin the cell centered!
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A
T
T
||T
ba2
1
2
1
zz
z +1/2
New type of operation
,BAT
||TT
2
1@
2
Screw axis
(1)(2)
(3)
T
(1)(2)
(3)
21
1,, BAT
||1 TT
2
1@
In general
)2/cot()2/( Tx
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TBAT2
1@
TBATTAT T 2
1@ )(
||,||
TBATTAT T 2
1@ )(
||,,||,
A
T
B
T||T
T
A
(1)(2)
(3)||,TB
T||T
T
,A
(1)
(2)
(3)||, TB
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TBAT2
1@ T
2
1@
)2/cot()2/( Tx
TBATTAT T 2
1@ )(
||,||
)2/cot()2/( Tx
TBATTAT T 2
1@ )(
||,,||,
)2/cot()2/( Tx
Crystal diffraction 2.ppt page 31
Tx
Tx
||,TB
A
B
A
Tx
||, TB
,A
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,2/3A
T
3
1 T
Specifying ,T
Tmn
Tn
m
,....3
4,
3
3
,3
2,
3
1,0
TT
TTT
T
TTTT
3
1
3
4 ,0
3
3
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For a 3-fold screw axis:
TTT
3
2,
3
1,0
3T
3
2T
231 32
4-fold screw axis: TTTT
4
3,
4
2,
4
1,0
T
4
3 T
43
41
T
4
1 T
41 42 43
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T
42
T
4
2
n1 n2 ……...
Tn
1 T
n
2
nm-2 nm-1
Tn
m 2 T
n
m 1
No chirality
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41 42 434
212 31 323
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61 62 63 64 656
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62
T
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Example to combine lattice with screw symmetry
P + 2 = P2
A
BC
D A: 2-fold + translation(to arise at B, C, or D)
ccTT
or 0 : all ||
Rotation symmetry of B, C, and D is the same as A.
A: 2
||,, TBAT
P
BAT
,, BAT
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P + 21 = P21
A: 21
I + 2 = I2 or I + 21 = I21
cbaT
2
1
2
1
2
1
2/,, cBAT
E
A
A: 2 E: 21
A: 21 E: 2Same, onlyshifted
I2 = I21
BBAT ccc 2/2/,2/,
2/,2/, cc BAT
21
21
21
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Hexagonal lattice (P and R) with 3, 31, 32. Case P first!
A
'T ''T
} 1{ 3/43/2 AA All translations in P havecomponent on c of 0 or unity!
ccT
1or 0||
',3/2,3/2
' BAT
''
2,3/42,3/4'
BAT
',3/2,3/2
'' CAT
''
2,3/42,3/4''
CAT
B
B
C
C
B and C: same point; B and C: equivalent point;
2,3/4,3/2 , BBHaving
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P3
P31
P32
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All translations of R hascomponent on c of 1/3 or 2/3!
A
'T ''T
} 1{ 3/43/2 AA
1/3
2/3
'3/,3/2,3/2
'cDAT
''
3/2,3/42,3/4'
cDAT
'3/2,3/2,3/2
''cEAT
''
3/22,3/42,3/4''
cEAT
DD E
E
A '3/,3/2 cD
'3/2,3/2 cE
Screw atD’ E’
Designation ofSpace group
331
32
0c/3
2c/3
2/32/32/3
c/32c/3
c
2c/3c
4c/3
2/32/32/3
31
32
3
32
331
R3R31
R32
R3R3
==
Hexagonal lattice (P, R) + 3, 31, 32 P3, P31, P32, R3.
Case R!
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The translation of P havecomponent on c of 0 or unity!
A
A441
42
43
'T
''T
} 1{ 2/32/ AAA
Square lattice P with 4, 41, 42, 43.
',2/,2/
' BAT
B''
2,2,'
BAT
B '''3,2/33,2/3
' BAT
B
',2/,2/
'' CAT
C
''2,2,
'' CAT
C
'''3,2/33,2/3
'' CAT
C
0
c/4c/2
3c/4
B/2 0/2 c/4/2 c/2/2 3c/4
B 0 c/2 c
3c/2
B441
42
43
B221
221
P4P41
P42
P43
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P4 P41
P42P43
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How to obtain Herman-Mauguin space group symbol by reading the diagram of symmetry elements?
First, know the Graphical symbols used for symmetry elements in one, two and three dimensions!
International Tables for Crystallography (2006). Vol. A, Chapter 1.4, pp. 7–11.
http://www.kristall.uni-frankfurt.de/media/exercises/Symbols-for-symmetryelements-ITC-Vol.A2.pdf
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Symmetry planes normal to the plane of projection
Symmetry plane Graphical symbol Translation Symbol
Reflection plane None m
Glide plane 1/2 along line a, b, or c
Glide plane1/2 normal to plane
a, b, or c
Double glide plane
1/2 along line &1/2 normal to plane (2 glide vectors)
e
Diagonal glide plane
1/2 along line, 1/2 normal to plane (1 glide vector)
n
Diamond glide plane
1/4 along line &1/4 normal to plane
d
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1/2 along line
1/2 normal toplane
a, b, c
a, b, c
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n-glide
d-glide
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e-glide
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1/8
Symmetry plane Graphical symbol Translation Symbol
Reflection plane None m
Glide plane 1/2 along arrow a, b, or c
Double glide plane
1/2 along either arrow
e
Diagonal glide plane
1/2 along the arrow
n
Diamond glide plane
1/8 or 3/8 along the arrows
d3/8
Symmetry planes parallel to plane of projection
The presence of a d-glide plane automatically implies a centered lattice!
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Symmetry Element
Graphical Symbol Translation Symbol
Identity None None 12-fold page⊥ None 22-fold in page None 2
2 sub 1 page⊥ 1/2 21
2 sub 1 in page 1/2 21
3-fold None 33 sub 1 1/3 31
3 sub 2 2/3 32
4-fold None 44 sub 1 1/4 41
4 sub 2 1/2 42
4 sub 3 3/4 43
6-fold None 66 sub 1 1/6 61
6 sub 2 1/3 62
6 sub 3 1/2 63
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Symmetry Element
Graphical Symbol Translation Symbol
6 sub 4 2/3 64
6 sub 5 5/6 65
Inversion None 13 bar None 34 bar None 46 bar None 6 = 3/m
2-fold and inversion
None 2/m
2 sub 1 and inversion
None 21/m
4-fold and inversion
None 4/m
4 sub 2 and inversion
None 42/m
6-fold and inversion
None 6/m
6 sub 3 and inversion
None 63/m
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mb-glide
2
21
2 21
n-glide
c-glide
|| b a
ba2
1
2
1
m
n|| a
b
c
c
|| c
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latticec
c
b
b
a
a
plane
|| axis
plane
|| axis
plane
|| axisfor orthorhombic:
n
2
2 2
2 2 111
mncbmC
)( 22 1 mnn dncbaem
mcmC 12 2 2
Cmcm Short symbol172hD
No. 17 orthorhombicthat can be derived
For orthorhombic: primary direction is (100), secondary direction is (010), and tertiary is (001).
From the point group mmm orthorhombic
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Principles for judging crystal system by space group
•Cubic – The secondary symmetry symbol will always be either 3 or (i.e. Ia3, Pm3m, Fd3m)•Tetragonal – The primary symmetry symbol will always be either 4, , 41, 42 or 43 (i.e. P41212, I4/m, P4/mcc)•Hexagonal – The primary symmetry symbol will always be a 6, , 61, 62, 63, 64 or 65 (i.e. P6mm, P63/mcm)•Trigonal – The primary symmetry symbol will always be a 3, , 31 or 32 (i.e P31m, R3, R3c, P312)
3
4
6
3
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•Orthorhombic – All three symbols following the lattice descriptor will be either mirror planes, glide planes, 2-fold rotation or screw axes (i.e. Pnma, Cmc21, Pnc2)•Monoclinic – The lattice descriptor will be followed by either a single mirror plane, glide plane, 2-fold rotation or screw axis or an axis/plane symbol (i.e. Cc, P2, P21/n)•Triclinic – The lattice descriptor will be followed by either a 1 or a (-1).
http://chemistry.osu.edu/~woodward/ch754/sym_itc.htm
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1. Generating a Crystal Structure from its Crystallographic Description
What can we do with the space group informationcontained in the International Tables?
2. Determining a Crystal Structure from Symmetry & Composition
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Example: Generating a Crystal Structure
http://chemistry.osu.edu/~woodward/ch754/sym_itc.htm
Description of crystal structure of Sr2AlTaO6
Space Group = Fmm; a= 7.80 ÅAtomic Positions
Atom x y z
Sr 0.25 0.25 0.25
Al 0.0 0.0 0.0
Ta 0.5 0.5 0.5
O 0.25 0.0 0.0
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From the space group tables
http://www.cryst.ehu.es/cgi-bin/cryst/programs/nph-wp-list?gnum=225
32 f 3m xxx, -x-xx, -xx-x, x-x-x,xx-x, -x-x-x, x-xx, -xxx
24 e 4mm x00, -x00, 0x0, 0-x0,00x, 00-x
24 d mmm 0 ¼ ¼, 0 ¾ ¼, ¼ 0 ¼,¼ 0 ¾, ¼ ¼ 0, ¾ ¼ 0
8 c 3m ¼ ¼ ¼ , ¼ ¼ ¾
4 b mm ½ ½ ½
4 a mm 000
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Sr 8c; Al 4a; Ta 4b; O 24e
40 atoms in the unit cellstoichiometry Sr8Al4Ta4O24 Sr2AlTaO6
F: face centered (000) (½ ½ 0) (½ 0 ½) (0 ½ ½)
8c: ¼ ¼ ¼ (¼¼¼) (¾¾¼) (¾¼¾) (¼¾¾) ¼ ¼ ¾ (¼¼¾) (¾¾¾) (¾¼¼) (¼¾¼)
¾ + ½ = 5/4 =¼
Sr
Al
4a: 0 0 0 (000) (½ ½ 0) (½ 0 ½) (0 ½ ½)
(000) (½½0) (½0½) (0½½)
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Ta
4b: ½ ½ ½ (½½½) (00½) (0½0) (½00)
O
24e: ¼ 0 0 (¼00) (¾½0) (¾0½) (¼½½)
(000) (½½0) (½0½) (0½½)
(000) (½½0) (½0½) (0½½)
¾ 0 0 (¾00) (¼½0) (¼0½) (¾½½)
x00
-x00
0 ¼ 0 (0¼0) (½¾0) (½¼½) (½¾½)0x0
0-x0 0 ¾ 0 (0¾0) (½¼0) (½¾½) (0¼½)
0 0 ¼ (00¼) (½½¼) (½0¾) (0½¾)00x
00-x 0 0 ¾ (00¾) (½½¾) (½0¼) (0½0¼)
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Bond distances:Al ion is octahedrally coordinated by six OAl-O distanced = 7.80 Å = 1.95 Å
Ta ion is octahedrally coordinated by six OTa-O distanced = 7.80 Å = 1.95 Å
Sr ion is surrounded by 12 OSr-O distance: d = 2.76 Å
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Determining a Crystal Structure fromSymmetry & Composition
Example:Consider the following information:Stoichiometry = SrTiO3
Space Group = Pmma = 3.90 ÅDensity = 5.1 g/cm3
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First step:calculate the number of formula units per unit cell :Formula Weight SrTiO3 = 87.62 + 47.87 + 3 (16.00) = 183.49 g/mol (M)
Unit Cell Volume = (3.9010-8 cm)3 = 5.93 10-23 cm3 (V)
(5.1 g/cm3)(5.93 10-23 cm3) : weight in aunit cell
(183.49 g/mole) / (6.022 1023/mol) : weightof one molecule of SrTiO3
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number of molecules per unit cell : 1 SrTiO3.
(5.1 g/cm3)(5.93 10-23 cm3)/(183.49 g/mole/6.022 1023/mol) = 0.99
6 e 4mm x00, -x00, 0x0,0-x0,00x, 00-x
3 d 4/mmm ½ 0 0, 0 ½ 0, 0 0 ½
3 c 4/mmm 0 ½ ½ , ½ 0 ½ , ½ ½ 0
1 b mm ½ ½ ½
1 a mm 000
From the space group tables (only part of it)
http://www.cryst.ehu.es/cgi-bin/cryst/programs/nph-wp-list?gnum=221
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Sr: 1a or 1b; Ti: 1a or 1b Sr 1a Ti 1b or vice verseO: 3c or 3d
Evaluation of 3c or 3d: Calculate the Ti-O bond distances:d (O @ 3c) = 2.76 Å (0 ½ ½) d (O @ 3d) = 1.95 Å (½ 0 0, Better)
Atom x y z
Sr 0.5 0.5 0.5
Ti 0 0 0
O 0.5 0 0