1. 3D Symmetry _2
(Two weeks)

2. 3D lattice: Reading crystal7.pdf
Building the 3D lattices by adding another translation vector to existing 2D lattices
Oblique (symmetry 1) +
triclinic
General
Triclinic Primitive

3. Oblique (symmetry 2) +
projection
4 choices:

4. Double cell
side centered
Double cell
side centered

5. Double cell
body centered
Some people use
based centered,
some use body centered.
monoclinic

6. Rectangular (symmetry m) +
90o
90o
already exist!
Rectangular (symmetry g) +
: the same.
cm + ?

7. Rectangular (symmetry 2mm) +
P2mm
P2mg
p2gg
Orthorhombic
primitive

8. Double cell
side centered
Orthorhombic
base-centered
Orthorhombic
base-centered
Double cell
side centered

9. Orthorhombic
body-centered
rectangular

10. Centered Rectangular (symmetry 2mm) +
C2mm
the same

11. Face centered
orthorhombic

12. Square (symmetry 4, 4mm) + P4
P4mm
p4gm
Tetragonal
primitive

13. Tetragonal
Body centered
Tetragonal

14. p3 p3m1 p31m Hexagonal (symmetry 3, 3m) + not in this category
Hexagonal primitive
not in this category
Why?
Rhombohedral

15. Hexagonal primitive
Rhombohedral
triple cell

16. p6 p6mm p31m Hexagonal (symmetry 3m, 6, 6mm) +
can only located at positions:
Hexagonal primitive
p31m
Hexagonal & 6 related can only fit 3P!

17. 11 lattice types already
cubic (isometric)
Special case of orthorhombic (222) with a = b = c
Primitive (P)
Body centered (I)
Face centered (F)
Base center (C)
Tetragonal (I)?
Cubic
a = b  c
[100]/[010]/[001]
[111]
Tetragonal (P)

18. Another way to look as cubic:
Consider an orthorhombic and requesting the diagonal
direction to be 3 fold rotation symmetry
P222  P23
Primitive
I222  I23
Body centered
F222  F23
Face centered
C222  I23

19. Bingo!
14 Bravais lattices!

20. Lattice type - compatibility with - point group
reading crystal9.pdf.
Crystal Class
Bravais Lattices
Point Groups
Triclinic
P (1P)
1, 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, 3 , 32, 3m, 3 2/m
Hexagonal
P (3P)
6, 6 , 6/m, 622, 6mm, 6 m2,
6/m 2/m 2/m
Tetragonal
P (4P), I (4I)
4, 4 , 4/m, 422, 4mm, 4 2m,
4/m 2/m 2/m
Isometric (Cubic)
P (23P), F(23F), I (23I)
23, 2/m 3 , 432, 4 3m, 4/m 3 2/m

21. http://www.theory.nipne.ro/~dragos/Solid/Bravais_table.jpg  = P = I
= T P

22. = P
= I
= B
= T P

23. 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!

24. 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)

25. Monoclinic a  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).
Orthorhombic The standard convention is that c < a < b. 
Once you define the cell following the convention 
A, B, C centered

26. Hexagonal/ Rhombohedral
Crystal System
Symmetry Direction
Primary
Secondary
Tertiary
Triclinic
None
Monoclinic
[010]
Orthorhombic
[100]
[001]
Tetragonal
[100]/[010]
[110]
Hexagonal/ Rhombohedral
[120]/[1 0]
Cubic
[100]/[010]/ [001]
[111]

27. Consider 2P Monoclinic + 2
/2 P2
/2

28. How about 2I Monoclinic + 2
There is a lattice point
in the cell centered!

29. z
(1)
(3)
(2) z
z +1/2
(3)
(1)
(2)
New type of operation
In general
Screw axis 21 2

30. Specifying

31. For a 3-fold screw axis: 3 31 32
4-fold screw axis: 43 41 41 42 43

32. 42 n1 n2
……...
nm-2
nm-1
No chirality

33. 3 31 32 2 21 4 41 42 43

34. 6 61 62 63 64 65

35. 62

36. Example to combine lattice with screw symmetryD A
A: 2-fold + translation
(to arise at B, C, or D) B C
Rotation symmetry of B, C, and D is the same as A.
A: 2
P + 2 = P2

37. A: 21 21
P + 21 = P21 21 21
I + 2 = I2 or I + 21 = I21 A
A: 2  E: 21
Same, only
shifted E
A: 21  E: 2
I2 = I21

38. Hexagonal lattice (P and R) with 3, 31, 32. Case P first!
All translations in P have
component on c of 0 or unity! A
B C B
C
B and C: same point; B and C: equivalent point;
Having

39. P3
P31
P32

40. All translations of R has component on c of 1/3 or 2/3!
Case R! A E
D
2/3
All translations of R has
component on c of 1/3 or 2/3!
E D
1/3
Screw at
Designation of
Space group A 
D’ E’ 3 31 32
c/3
2c/3
2/3
c/3
2c/3 c
2/3
2c/3 c
4c/3 31 32 3 32 3 31 R3
R31
R32 R3 = =
Hexagonal lattice (P, R) + 3, 31, 32  P3, P31, P32, R3.

42. Square lattice P with 4, 41, 42, 43.
The translation of P have
component on c of 0 or unity! A C B
B
B
C
C A 4 41 42 43 
c/4
c/2
3c/4 B
/2 0
/2 c/4
/2 c/2
/2 3c/4
B
 0
 c/2
 c
 3c/2 B 4 41 42 43
B 2 21 P4
P41
P42
P43

43. P4
P41
P42
P43

44. Homework:
Discuss the cases of I4, I41, I42, I43.

45. 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.

46. Symmetry planes normal to the plane of projection
Graphical symbol
Translation
Symbol
Reflection plane
None
  m
Glide plane
1/2 along line
  a, b, or c
1/2 normal to plane
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

47. Symmetry planes parallel to plane of projection
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
  d
3/8
1/8
The presence of a d-glide plane automatically implies a centered lattice!

48. Symmetry Element
Graphical Symbol
Translation
Symbol
Identity
None
  1
2-fold ⊥ page
  2
2-fold in page
2 sub 1 ⊥ page
1/2
  21
2 sub 1 in page
3-fold
  3
3 sub 1
1/3
  31
3 sub 2
2/3
  32
4-fold
  4
4 sub 1
1/4
  41
4 sub 2
  42
4 sub 3
3/4
  43
6-fold
  6
6 sub 1
1/6
  61
6 sub 2
  62
6 sub 3
  63

49. Symmetry Element
Graphical Symbol
Translation
Symbol
6 sub 4
2/3
  64
6 sub 5
5/6
  65
Inversion
None
  1
3 bar
  3
4 bar
  4
6 bar
  6 = 3/m
2-fold and inversion
  2/m
2 sub 1 and inversion
  21/m
4-fold and inversion
  4/m
4 sub 2 and inversion
  42/m
6-fold and inversion
  6/m
6 sub 3 and inversion
  63/m

50. c-glide
 b
n-glide
|| c 21
 c
|| a n 2 2 21
b-glide m m
 c
|| b
 a

51. From the point group mmm  orthorhombic
For orthorhombic: primary direction is (100), secondary direction is (010), and tertiary is (001).
lattice
for orthorhombic: C
Short symbol
No. 17 orthorhombic
that can be derived

52. Principles for judging crystal system by space group
Cubic – The secondary symmetry symbol will always be either 3 or –3 (i.e. Ia3, Pm3m, Fd3m)
Tetragonal – The primary symmetry symbol will always be either 4, (-4), 41, 42 or 43 (i.e. P41212, I4/m, P4/mcc)
Hexagonal – The primary symmetry symbol will always be a 6, (-6), 61, 62, 63, 64 or 65 (i.e. P6mm, P63/mcm)
Trigonal – The primary symmetry symbol will always be a 3, (-3) 31 or 32 (i.e P31m, R3, R3c, P312)

53. 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).

54. What can we do with the space group information
contained in the International Tables?
1. Generating a Crystal Structure from its Crystallographic Description
2. Determining a Crystal Structure from Symmetry & Composition

55. Example: Generating a Crystal Structure
Description of crystal structure of Sr2AlTaO6
Space Group = Fm 3 m; a= 7.80 Å Atomic Positions
Atom x y z Sr
0.25 Al
0.0 Ta
0.5 O

56. From the space group tables 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 d
mmm
0 ¼ ¼, 0 ¾ ¼, ¼ 0 ¼,
¼ 0 ¾, ¼ ¼ 0, ¾ ¼ 0 8 c
4 3m
¼ ¼ ¼ , ¼ ¼ ¾ 4 b
m 3 m
½ ½ ½ a
000

57. Sr 8c; Al 4a; Ta 4b; O 24e
40 atoms in the unit cell
stoichiometry Sr8Al4Ta4O24  Sr2AlTaO6
F: face centered
 (000) (½ ½ 0) (½ 0 ½) (0 ½ ½) Sr
(000) (½½0) (½0½) (0½½)
8c: ¼ ¼ ¼  (¼¼¼) (¾¾¼) (¾¼¾) (¼¾¾)
¼ ¼ ¾  (¼¼¾) (¾¾¾) (¾¼¼) (¼¾¼) Al
¾ + ½ = 5/4 =¼
4a:  (000) (½ ½ 0) (½ 0 ½) (0 ½ ½)

58. Ta
(000) (½½0) (½0½) (0½½)
4b: ½ ½ ½  (½½½) (00½) (0½0) (½00)
(000) (½½0) (½0½) (0½½) O
x00
24e: ¼ 0 0  (¼00) (¾½0) (¾0½) (¼½½)
-x00
¾ 0 0  (¾00) (¼½0) (¼0½) (¾½½)
0x0
0 ¼ 0  (0¼0) (½¾0) (½¼½) (½¾½)
0-x0
0 ¾ 0  (0¾0) (½¼0) (½¾½) (0¼½)
00x
0 0 ¼  (00¼) (½½¼) (½0¾) (0½¾)
00-x
0 0 ¾  (00¾) (½½¾) (½0¼) (0½0¼)

59. Bond distances:
Al ion is octahedrally coordinated by six O
Al-O distance
d = 7.80 Å  − − − = 1.95 Å
Ta ion is octahedrally coordinated by six O
Ta-O distance
d = 7.80 Å  − − − = 1.95 Å
Sr ion is surrounded by 12 O
Sr-O distance: d = 2.76 Å

60. Determining a Crystal Structure from
Symmetry & Composition
Example:
Consider the following information:
Stoichiometry = SrTiO3 Space Group = Pm 3 m a = 3.90 Å Density = 5.1 g/cm3

61. First step:
calculate the number of formula units per unit cell :
Formula Weight SrTiO3 =  (16.00) = g/mol (M)
Unit Cell Volume = (3.9010-8 cm)3 = 5.93  cm3 (V)
(5.1 g/cm3)(5.93  cm3) : weight in a
unit cell
( g/mole) / (6.022 1023/mol) : weight
of one molecule of SrTiO3

62.  number of molecules per unit cell : 1 SrTiO3.
 (5.1 g/cm3)(5.93  cm3)/
( g/mole/6.022 1023/mol) = 0.99
 number of molecules per unit cell : 1 SrTiO3.
From the space group tables (only part of it) 6 e
4mm
x00, -x00, 0x0,
0-x0,00x, 00-x 3 d
4/mmm
½ 0 0, 0 ½ 0, 0 0 ½ c
0 ½ ½ , ½ 0 ½ , ½ ½ 0 1 b
m 3 m
½ ½ ½ a
000

63. Calculate the Ti-O bond distances:
Sr: 1a or 1b; Ti: 1a or 1b
 Sr 1a Ti 1b or vice verse
O: 3c or 3d
Evaluation of 3c or 3d:
Calculate the Ti-O bond distances:
d 3c) = 2.76 Å (0 ½ ½) d 3d) = 1.95 Å (½ 0 0, Better)
Atom x y z Sr
0.5 Ti O