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Introduction

Neighbour’s envy, Owner’s pride,
Silicon Valley’s delight, a girl’s best friend,

Miner’s blackjack ! Is that you?

Learning Objectives

On completion of this topic you will be able to:
1. Identify the crystal system of diamond, NaCl, ZnS and CsCl, graphite
2. Derive packing factor for diamond structures
3. Differentiate between diamond and zinc blende
4. Give minimum 2 polymorphic forms of carbon

Introduction

Most of the high density materials have close packed structures, the hcp, the fcc or modifications of
the basic structures studied so far. We shall now look at some of the important modified cubic
structures. Materials that are made up of tetravalent elements, i.e the semiconductors , insulators etc
are found to possess a diamond structure which is a modified FCC structure whereas ionic crystals like
NaCl, KBr etc also possess a modified FCC structure, but of a different type.

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The Diamond Structure

The diamond cubic structure is a characteristic of materials that have its constituent materials
bound in tetrahedral bonds. Insulators such as carbon, semiconductors like Germanium, Silicon etc
exhibit this crystal structure. The diamond structure is an FCC with a two atom basis. If one atom of the
basis is at the FCC lattice point, say (0,0,0) the other atom of the basis is at (¼ a, ¼ a, ¼ a ) where a is
the lattice constant of the FCC lattice. Therefore the diamond structure can be arrived at by attaching
one atom to each FCC lattice point (x,y,z) and the other at a point ( (x+a/4), (y+1/4), (z+a/4) ).

An alternative way of describing the diamond structure is to as one with 2 interpenetrating FCC sub-
lattices with one FCC penetrating upto ¼th the body diagonal of the other. The figure given below
illustrates the diamond structure. The second penetrating FCC structure can be seen by connecting the
second atoms of the basis that is attached to the original FCC lattice.(shown in greenshade)

Fig 1: The diamond structure

Note: The 2nd atom of the basis is indicated in a different colour (green) only for the sake of easy
identification. Both atoms are of the same radius and element.

The unit cell of this new structure is taken as the unit cell of the FCC with the additional 4 atoms
which are enclosed within it i.e the second atoms attached to the point A on the base, the base-center,
the face-centre on the left side and the face-centre of the front face of the original FCC lattice. The
figure 2 shows a the unit cell of the diamond structure with co-ordinates of each point and the

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tetrahedral bonds of the nearest neighbours. The diamond unit cell has 4 terahedral bonds as shown in
the figure.

Fig 2 : The diamond unit cell

2D representation of the diamond unit cell

In a 2D projection of the diamond unit cell, the base plane of the unit cell ABCD is represented
by a square and the atoms in the base plane are represented by their z co-ordinates. The face-centered
atoms of all 4 side faces are projected down and represented by numbers indicating their height from the
base in terms of the lattice constant a. Figure 3 gives how the 3D unit cell is depicted in an equivalent 2D
representation. All atoms whose z-co-ordinates differ by ¼ are nearest neighbour’s involved in the
tetrahedral bond. Therefore the nearest neighbour distance would be that between say A and atom at J.

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Fig 3: 2D projection of diamond unit cell

Determination of packing fraction of diamond

The co-ordination number of an atom in diamond structure is obviously 4 as can be seen from the unit
cell of diamond and is loosely packed.

The unit cell has 8 corner atoms, 6 face-centered atoms and 4 atoms inside the cube. Since each corner
atom is shared by 8 adjacent unit cells and each face-centered atom is shared by 2 unit cells,

the number of atoms /unit cell = (⅛) x 8 + (½) x 6 + 4 = 8

(corner atoms) (fc atoms)

To determine the distance between nearest neighbour’s one needs to estimate the distance AJ as shown
in the figure below.

O

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The atom at J is displaced from that at A by a/4 along x-axis and a/4 along Y axis , therefore

AO2 =  a 2 +  a 2 2a 2 = a2 ;
=

 4   4  16 8

Or

AJ 2 = a 2 + a 2 = 3a 2
8 16 16

3
AJ = a = 2r ; where r is the radius of the atoms.

4

Therefore the radius of the atom,r = 3
a

8

Volume occupied by the atoms in the unit cell,v = no. of atoms in the unit cell x volume of an atom

=8 x 4 πr 3 =8 x 4 3 3 a3 = π 3 a3
π
3 83
3 16

Volume of the unit cell, V = a3
Therefore.

Packing fraction of diamond unit cell = v = π 3
= 0.34

V 16

Thus the diamond structure is found to be the most loosely packed structure. Diamond (made of
carbon atoms) which is the hardest material also has the smallest packing fraction.

The Zinc Blende or Sphalerite Structure.

Zinc blende is similar to diamond cubic structure, but the two FCC sub-lattices are occupied by
atoms of two different elements. Some of the important compounds which possess this structure are
semiconductors like InSb, GaAs and also ZnS and CuCl. Zinc sufide in diamond structure is generally
called Zinc Blende . 2 samples of zinc blende ores are shown below. Its color is usually yellow, brown, or
gray to gray-black, and it may be shiny or dull. Iron content in the ore makes it opaque/black. It is also
called the miner’s blackjack or mock lead.

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Zinc sulfide has a lattice constant of 0.596nm . The sulfide ions are quite large (atomic radius
184 pm) relative to the size of the zinc ions (74 pm). In the diamond structure sulfide ion (shown in grey)
occupies the FCC lattice points with the Zinc ions(shown in red) is the other atom of the basis. When
compared to diamond saphalerite structure is more soft. Thus we can see that the strength of the
tetrahedral bond determines the hardness of the material.

2 unit cells of ZnS structure

ZnS has an allotropic form which is a hcp structure with 2 atom basis. The hexagonal form is known
wurtzite. Both sphalerite and wurtzite are intrinsic, wide-bandgap semiconductors. The cubic form has a
band gap of 3.54 eV at 300 K whereas the hexagonal form has a band gap of 3.91 eV. A transition from
the sphalerite form to the wurtzite form occurs at around 1293.15 K.

Zinc sulfide, with addition of few ppm of suitable activator, is used as phosphor in many applications,
from cathode ray tubes through x-ray screens to glow in the dark products. Copper provides long glow
time and the familiar glow-in-the-dark greenish color. Zinc sulfide is also used as an infrared optical
material, transmitting from visible wavelengths to over 12 micrometres. It can be used planar as an
optical window or shaped into a lens.

Graphite Structure

Graphite is an allotropic form of carbon . In case of graphite the carbon atoms are arranged in regular
hexagons in flat parallel layers and each carbon atom covalently bonds with three atoms in a this
hexagonal layer. The remaining electron is delocalized between the planes forming a weak bond between
the parallel layers. Since the bonding between these parallel layers are weak, the layers are easily
separable from each other and is the cause for softness and lubricating action of graphite. The

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delocalized electron results in the moderate conductivity of graphite in comparison to carbon in diamond
structure. The figure given below shows the graphite structure.

The spacing between the carbon atoms in the hexagonal layer is about 1.42A° and the spacing between
the parallel layers is 3.4°.

The Sodium chloride structure

NaCl is an ionic crystal and in this crystal the sodium and chlorine ions are situated side-by side
in the FCC lattice. Like diamond it is also an FC lattice with a two atom basis. It can also be thought of as
having two FCC sub-lattices . Cl ion occupies the FCC lattice and the second atom of the basis is the Na
ion which is at a distance (a/2, 0, 0) from the Cl ion. Each ion in the NaCl crystal has six nearest
neighbour ions at a distance of a/2. The ionic radius of sodium is about 0.98A° wheras that of chlorine
ion is 1.81 A°.There are 4 molecules of NaCl in each unit cell.

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The NaCl structure

The Caesium Chloride Structure

CsCl structure is essentially a simple cubic lattice with 2 atom basis, where one atom is attached to the
lattice point and the other at ( a/2 , a/2, a/2) from it. Both caesium and chlorine have approximately
the same atomic radii. Chlorine ion occupies the SC lattice point with the Cs ion occupying the body
centre of the cube since (a/2, a/2, a/2 ) represents the body centre of a cube of side a. Thus each
caesium ion has eight nearest chlorine atoms and the structure resembles that of a body centre of the
atom.

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Check your understanding

Choose the right answer from the options given below:

1. The crystal structure of germanium is

a) HCP b) FCC c) BCC d)diamond cube e) none of the above

2. Packing factor of daimond structure is

a)π/6 b)√3π/16 c) )√3π/8 d) π/3√2

3. Fill in the blank with the right answer.
In spite of having the smallest packing fraction, carbon in diamond structure is hard because of
____________________________________________.

4. The crystal structure of graphite is

a) HCP b) FCC c) BCC d)diamond cube e) none of the above

5. Allotropic forms of carbon are d)graphite e)Fullrene
a)Saphalerite b) diamond c)Wurtzite

Check the correct answers on page 10.

Summary

On completion of this topic you have learned that:

1. Diamond structure is an FCC with 2 atom basis.One atom at FCC lattice point and other
(a/4,a/4,a/4) from it. It can also be thought of as having 2 FCC sub-lattices. Example of
materials with diamond structure: Germanium, silicon, carbon, etc

2. Saphalerite or Zinc blende has diamond structure but has a basis of 2 different elements
Example ZnS, CuCl, GaAs,InSb etc. bigger atoms occupy FCC lattice points with the samller atom
of the basis at the a/4,a/4,a/4 positions away from its pair.

3. NaCl is an FCC cryatal with 2 atom basis. One atom at FCC lattice and other at (a/2,0,0) from it.

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4. CsCl is an SC with 2 atom basis.One atom(Cl) in the SC lattice and other(Cs) at a distance of
a/2,a/2,a/2) from it.

5. Graphite ia an allotropic form of carbon with carbon atoms arranged in parallel hexagonal layers
with each carbon atom forming covalent bonds with 3 nearest neighbours and one electron
delocalized and shared between parallel layers.

Activity

Tabulate the hardness of the materials and the bond strengths for the crystals starting from that of C
down to Pb in the periodic table and their crystal structures. What do you infer from the study.

Suggested Reading

1. Solid State Physics by Kittel
2. Engineering Physics-I by P.K.Palaniswamy

Answers to CYU. 4.e 5. b,d&e

1.d 2.b 3.strength of its tetrahedral bond

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