3D Symmetry _2(Two weeks)

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

Oblique (symmetry 2) + 3T

2T

1T

3T

projection

4 choices:

)2

1

2

1(),0

2

1(),

2

10(),00(

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

'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

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!

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

)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

'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

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

)02

1( z

2T

1T

3T Face centered

'3T

F222oTTTTTT 901

'32

'321

'

321 TTT

FCIP 222 ,222 ,222 ,222 orthorhombic

Square (symmetry 4, 4mm) + 3T

P4P4mmp4gm

)2

1

2

1(),00( zz

2T

1T

3T

oTTTTTT 90132321

321 TTT

)00( z

Tetragonalprimitive

P4

)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

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?

)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

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!

cubic (isometric)

Primitive (P)Body centered (I)Face centered (F)Base center (C)

Special case of orthorhombic (222) with a = b = c

a = b cTetragonal (P)

Tetragonal (I)?

11 lattice types already

FIP 23 ,32 ,23 Cubic

[100]/[010]/[001] [111]

Another way to look as cubic:Consider an orthorhombic and requesting the diagonaldirection to be 3 fold rotation symmetry

P222 P23

F222 F23

I222 I23

C222 I23

Primitive

Body centered

Face centered

Bingo!

14 Bravais lattices!

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.

http://www.theory.nipne.ro/~dragos/Solid/Bravais_table.jpg

= P = I

= T P

= P = I

= B

= T P

http://users.aber.ac.uk/ruw/teach/334/bravais.php

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!

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)

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

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

Monoclinic + 2

P2

p2

2D

TBAT2

1@

Consider 2P Monoclinic + 2

/2

P2/2

ab

cA ? AT

cba2

1

2

1

2

1T

T ||T

How about 2I Monoclinic + 2

There is a lattice pointin the cell centered!

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

,2/3A

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

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

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

41 42 434

212 31 323

61 62 63 64 656

62

T

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

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

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

P3

P31

P32

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!

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

P4 P41

P42P43

Homework:

Discuss the cases of I4, I41, I42, I43.

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

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

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!

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

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

mb-glide

2

21

2 21

n-glide

c-glide

|| b a

ba2

1

2

1

m

n|| a

b

c

c

|| c

latticec

c

b

b

a

a

plane

|| axis

plane

|| axis

plane

|| axisfor orthorhombic:

n

2

2 2

2 2 111

mncbmC

122 dncbam

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

•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)

Principles for judging crystal system by space group

•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

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

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

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

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)

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)

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

Determining a Crystal Structure fromSymmetry & Composition

Example:Consider the following information:Stoichiometry = SrTiO3

Space Group = Pmma = 3.90 ÅDensity = 5.1 g/cm3

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

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

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