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Lecture 8 Lecture 8
Tu 12.08.2025 Allotropy
Thursday Group Graphene: Lattice and Motif
meets on sep 11 Fullerene

for Exp 3 on XRD Nanotube: Armchair, zigzag and chiral

1 2

1 2

1. Crystalline structure of Elements

Structure of Solids
2. Crystalline structure of
Compounds and ‘solid solutions’
Raghavan Ch 5

3 4

3 4

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4 th. Group: Carbon

Allotropy
The phenomenon of the same element in the
same state (solid/liquid/gas) occurring in different
structural forms.
Fe Monatomic BCC at Room Temperature (Ferrite)

Fe Cubic Close-Packed (CCP), Bravais lattice: FCC, Austenite

Above 910 C
5 6

5 6

Allotropes of C

Graphite
Diamond

Buckminster Fullerene Carbon Nanotubes Graphene

1985 1991 2004
7 8

7 8

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3. Motif

Relation between crystal

and lattice?
Structure of
Crystal = Lattice + Motif
Graphene
Motif or basis: an atom or a
group of atoms associated
9
with each lattice point 10

9 10

Graphene: A 2D crystal = Lattice + Motif?

Crystal = Lattice + Motif

How What
to to
repeat repeat

1 12
1

11 12

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Graphene: Are the centres of all atoms lattice points? Graphene: Only the centres of alternate atoms are lattice points

13 14

13 14

Lattice of graphene Unit cell of the lattice of Graphene

primitive hexagonal lattice

ℎ𝑝

𝑎 𝛾 𝑏=𝑎
= 120°

𝑥 𝑦

15
16

15 16

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How to build crystal of graphene from its lattice? How to build crystal of graphene from its lattice?

Single Primitive hexagonal

atom per Lattice: ℎ𝑝
lattice
point of an 𝑎 𝛾 𝑏=𝑎 For 2D lattice we 𝑎 𝛾 𝑏=𝑎
hp lattice = 120° use small p = 120°
is not ℎ𝑝: 2𝐷 𝑃𝑟𝑖𝑚𝑖𝑡𝑖𝑣𝑒
𝑥 𝑦 𝑥 𝑦
graphene. 𝐻𝑒𝑥𝑎𝑔𝑜𝑛𝑎𝑙

For 3D lattice we use

primitive hexagonal lattice Capital P
ℎ𝑝 ℎ𝑃: 3𝐷 Primitive
Hexagonal

17 18

17 18

How to build crystal of graphene from its lattice? How to build crystal of graphene from its lattice?

𝑎 𝛾 𝑏=𝑎 𝑎 𝛾 𝑏=𝑎
= 120° = 120°

𝑥 𝑦 𝑥 𝑦

primitive hexagonal lattice primitive hexagonal lattice

ℎ𝑝 ℎ𝑝

𝐶 00
𝐺𝑟𝑎𝑝ℎ𝑒𝑛𝑒 = ℎ𝑝 𝑙𝑎𝑡𝑡𝑖𝑐𝑒 + 2 𝑎𝑡𝑜𝑚 𝑚𝑜𝑡𝑖𝑓 3 2 1
𝐶
19
33 20

19 20

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Unit cell of graphene: primitive or nonprimitive? Pitfall

!
Two atoms per cell = ×4 + 1
" Primitive vs. Nonprimitive
𝑎 𝑏=𝑎
unit cell
𝛾
= 120°

𝑥 𝑦
Primitive unit cell contains ONE LATTICE POINT per cell

!
One lattice point per cell = ×4 ⇒ 𝑃𝑟𝑖𝑚𝑖𝑡𝑖𝑣𝑒
And NOT one atom per cell.
"

21 22

21 22

Every lattice has to be a Bravais lattice.

Bravais lattice is a classification of lattices.

23 24

23 24

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Description of Motif
An atom or a group of atoms associated
with each lattice point
1. No. of atoms

2. Chemical identity of atoms

Truncated Icosahedron
3. Location of atoms

Displacement coordinates with respect to Icosahedron: A Platonic solid (a regular solid)

the lattice point in the crystal coordinate Truncated Icosahedron: An Archimedean solid
system (fractional coordinates) 25

25 26

Other closed-shell fullerenes

C70 All closed-shell fullerenes have 12

pentagonal faces.

C76
C78
C84
27 28

27 28

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American architect,
author, designer, futurist,
inventor, and visionary.

He was expelled from Harvard twice:

1. first for spending all his money partying with a
Vaudeville troupe,
2. for his "irresponsibility and lack of interest".

what he, as an individual, could do to improve humanity's

condition, which large organizations, governments, and private
R. Buckmnster Fuller
enterprises inherently could not do.
29

29 30

C60 Crystal = FCC lattice + C60 motif

Montreal Biosphere in Montreal, Canada 32

31 32

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Carbon
Nanotube

33 34

33 34

Multiwalled nanotube

36

35 36

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Chiral or wrapping
vector Tube
Circumference

Tube axis

38

37 38

Lecture 9
Nanotube
Lecture 9 Wrapping vector
W 13.08.2025 Graphite: Lattice and Motif
Diamond: lattice and Motif
Close-Packing of spheres: 1D

39 40

39 40

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a1
zigzig (n,0)
Wrapping 𝜃=chiral angle
vector a2 w ra
p p in Electrical
description of g ve
c to r For a given (n,m) nanotube, if n = m, the nanotube is metallic;

nanotube: (n,m)=(6,3) if n − m is a multiple of 3, then the nanotube is semiconducting with a very small band
(𝒏, 𝒎) gap,

otherwise the nanotube is a moderate semiconductor.

a rm Thus all armchair (n=m) nanotubes are metallic,

cha
ir (
n ,n and nanotubes (5,0), (6,4), (9,1), etc. are semiconducting.
)
In theory, metallic nanotubes can carry an electrical current density of 4×10^9 A/cm2
which is more than 1,000 times greater than metals such as copper[23].

41 42

Diamond
Graphite a = 2.46 Å Sp3 hybridization Þ 4 covalent bonds
c = 6.70 Å Þ Tetrahedral bonding

c
x y

A
www.scifun.ed.ac.uk/
B Location of atoms:
Lattice: Simple Hexagonal 8 Corners
Motif: 4 carbon atoms Proof:LeftAas an exercise 6 face centres
000; 2/3 1/3 0; 2/3 1/3 1/2; 1/3 2/3 1/2
43 4 one on each of the 4 body diagonals 44

43 44

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Diamond Cubic Crystal: Lattice & motif?

Diamond Cubic Crystal: Lattice & motif?
y y 1
1
M 0,1 2 0,1
0,1 2 0,1
R y M
S D R C
P N
Q
D R C R y M 1 L
M S 3
D 3 1 L P N
Q 4 4
L 4 4 Q
K D T 1
C T Q 1 1
1 L 0,1
T 0,1 K 2 S 2
S 2 C
A
B
x 2
T 1 K 3 N
1 K 3 N
4 4
A
B
x 4 4
A P B
A P B
x x
1 0,1
1 0,1 0,1
0,1 2
2
Diamond Cubic Crystal Projection of the unit cell on
Projection of the unit cell on the bottom face of the cube
Diamond Cubic Crystal = FCC lattice + motif:
the bottom face of the cube
000; ¼¼¼ 46
= FCC lattice + motif: 000; ¼¼¼ 45

45 46

Pitfall 1D: Diamond Cubic Lattice

E
Diamond Cubic TIC
T
Lattice LA
C
UBI
Diamond Cubic FCC 2 atom ì 000 C
Crystal Structure = Lattice + Motif í1 1 1 D
î4 4 4 N
O
AM
DI
There are only three Bravais Lattices: SC, BCC, FCC.

O
N
47 48
48

47 48

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Metals and Alloys

1. Metallic bond: Nondrectional (Fact)
ÞAs many bonds as geometrically possible (to
lower the energy)
Þ Close packing

Diamond Diamond 2. Atoms as hard sphere (Assumption)

Cubic Lattice Cubic Crystal
3. Elements (identical atoms)

1, 2 & 3 Þ Elemental metal crystals:

close packing of equal hard spheres
49 50

49 50

Close packing of equal hard spheres Close packing of equal hard spheres
Arrangement of equal nonoverlapping spheres
to fill space as densely as possible 1-D packing

Sphere packing problem: A chain of spheres

What is the densest packing of spheres in 3D?

Kepler’s conjecture, 1611

p
P.E = = 0.74
3 2
Kissing Number Problem P.E.=
occupied length
Kissing Number= 2
=1
What is the maximum number of spheres that total length
can touch a given sphere?
Close-packed direction of atoms
Coding Theory
Internet data transmission 51 52

51 52

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Lecture 10
Th 13.08.2025
(as Friday)
53 54

53 54

Close packing of equal hard spheres

Square Packing

1-D packing

A chain of spheres

Hexagonal Packing

occupied length
P.E.= =1 Kissing Number= 2
total length

Close-packed direction of atoms

55 56

55 56

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𝐴𝑟𝑒𝑎 𝑜𝑐𝑐𝑢𝑝𝑖𝑒𝑑
2 𝐷 𝑃𝑎𝑐𝑘𝑖𝑛𝑔 𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 𝑃𝐸 =
𝑇𝑜𝑡𝑎𝑙 𝐴𝑟𝑒𝑎
Square Packing

𝜋
𝑃𝐸 = ≈ 0.78
4

Hexagonal Packing 𝜋
𝑃𝐸 = ≈ 0.78
4

57 58

57 58

Close packing of equal hard spheres Close packing of equal hard spheres
2-D packing 3-D packing
First layer A
A hexagonal layer of atoms Close-packed plane of atoms
Second layer B
A A A
A
C C
C

B B B

Third layer A or C
A A A A

C C C

B B B

A A A A

C C C

B B B

A A A A

Close-packed directions? 3

occupied area p
P.E.= = = .907 Kissing Number=6
total area 2 3
1940 L. Fejes Toth : Densest packing of circles in plane
59 60

59 60

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Two important close packing realized in nature

All Possible stacking sequences giving rise to
closest packing

Close packed crystals:

…ABABAB… Hexagonal close packed (HCP)
61 …ABCABC… Cubic close packed (CCP) 62

61 62

Geometrical properties of ABAB stacking

Hexagonal close-packed (HCP) crystal
z
A A
b = a
g= 12 0 ° A A B
C C C

A
a
B B B

A A A A
½ ½
C

B
C

B
C

B
c B
A A A A
A
C C C
y
B B B
½ ½
A A A A

A and B do not have identical neighbours

Corner and inside atoms do not have
x the same neighbourhood
Either A or B as lattice points, not both
Lattice: Simple hexagonal Motif: Two atoms:
Unit cell: a rhombus based prism with a=b¹c; a=b=90°, g=120°
000; 2/3 1/3 1/2
The unit cell contains only one lattice point (simple) but two atoms (motif)
ABAB stacking = HCP crystal = Hexagonal P lattice + 2 atom motif 000 hcp lattice hcp crystal
63
2/3 1/3 1/2

63 64

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Hexagonal Close-Packed Crystal Structures

65 66

65 66

c/a ratio of an ideal HCP crystal

A A A B
Some HCP crystals
A
C C C

B B B A Element c/a
A A A A
Be 1.567
C C C
c B
B B B

A A A A
A Ti 1.587
C C C

B B B Mg 1.623
A A A A

Idea HCP 1.633

A single B atom sitting on a base of three A atoms forms a regular tetrahedron Zn 1.856
with edge length a = 2R
Cd 1.886
The same B atom also forms an inverted tetrahedron with three A atoms sitting
above it
c 2 2
c = 2 × height of a tetrahedron of edge length a = ≅ 1.633..
a 3
67 68

67 68

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Geometrical properties of ABCABC stacking

Geometrical properties of ABCABC stacking B
C

All atoms are equivalent

A
and their centres form a
C
lattice
Ö3 a
B
A A A A
C
A
Motif: single atom 000
C C

B B B

A A A A What is the Bravais lattice?

C C C

B B B

A A A A

C C C

B B B

A A A A

ABCABC stacking
= CCP crystal
= FCC lattice + single atom motif 000

69 70

69 70

Close packed planes in the FCC unit cell of cubic close packed crystal
Stacking sequence?

Body
diagonal

Close packed planes: {1 1 1}

71 72

71 72

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Find the mistake in the following figure from a website:

Stacking sequence?

http://www.tiem.utk.edu/~gross/bioed/webmodules/spherefig1.gif

73 74

73 74

Table 5.1

Coordination Number and Packing Efficiency

CW HW

Crystal Coordination Packing

Structure number efficiency

Diamond cubic (DC) 4

0.32
Simple cubic (SC) 6
0.52
Body-centred cubic 8

Face-centred cubic 12 0.68

0.74

Empty spaces are distributed in various voids

75

75

19