crystal structure & bonding: lec-09

Kapil AdhikariMt. Annapurna CampusPokhara, Nepal

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Date: March 23, 2018Contact: [email protected]

Crystal structure &bonding: Lec-09

KAPIL ADHIKARI

B.Sc. CSIT

Reference Book:Garcia Narciso, Damask Arthur, Physics for Computer Sci-ence Students, Springer-Verlag

Crystal Structure

A crystalline solid is a three-dimensional, periodic array of atoms or molecules called a crystal structure.All crystal structures can be included in one of fourteen lattice arrangements.

A crystal structure can be specified by a periodic space lattice and an atom or group of atoms placedat or around each lattice point. The atom or group of atoms constitutes the basis. The space lattice isa regular periodic arrangement of points in space and is purely a mathematical abstraction. To obtaina crystal structure, we must place at or around each lattice point a basis of atoms. This group of atomsmust be identical in composition, arrangement, and orientation.

NaCl Structure

Figure 1: Crystal Structure of NaCl

This is a face-centered cubic (fcc) Bravais lattice. The basisconsists of a Na atom and a Cl atom separated by one-half the body diagonal. Other materials having the samestructure include KBr, KCl, AgBr, MgO, MnO, and PbS.

CsCl StructureThis is a simple cubic Bravais lattice. The basis consistsof a Cl atom at the corner and a Cs atom separated byone-half the body diagonal. Other materials having thisstructure include AgMg, AINi, CuZn (brass), and BeCu.

Figure 2: Crystal Structure of CsCl


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KAPIL ADHIKARI

B.Sc. CSIT



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KAPIL ADHIKARI

B.Sc. CSIT


Diamond Structure

Figure 3: Crystal Structure of Diamond

It is a face-centered cubic. The basis consists oftwo identical atoms, one is situated at the corner ofthe cube and the other is displaced by one-quarterthe body diagonal along that diagonal. The dia-mond structure can be thought of as being made oftwo interlocked fcc structures, each with one atomper lattice point. These two fcc structures are dis-placed from each other by 1

4 d along the body diago-nal.

The structure of diamond gives rise to a tetrahedralbond arrangement where each atom can be considered tobe at the center of a tetrahedron forming one bond witheach of its four nearest neighbors. These appear to belocated at the four corners of the tetrahedron. Materialshaving this type of structure include C (carbon in the di-amond structure), Si (silicon), Ge (germanium), and Sn(tin).

Figure 4: Tetrahedral arrangement of atomsin diamond resulting from the crystal struc-ture. Each carbon atom in the diamond struc-ture is at the center of a tetrahedron with itsfour nearest neighbors located at the cornersof the tetrahedron.

Crystal Bonding

Ionic Crystals (Ionic Bonding) Ionic crystals consist ofpositive and negative ions in a periodic array. This is a re-sult of an atom losing one or more electrons and anotheratom capturing them. For this to occur, the exchange mustbe energetically favorable. This means that if one startswith two neutral atoms, say Na and Cl, and goes throughthe process of ionizing one of them, then giving the re-sulting electron or electrons to the other and bringing thetwo ions together, the overall energy of the bound pairmust be less than the energy of the initial state of two neu-tral atoms.This situation occurs (although not exclusively)with the alkali elements that have a weakly bound s elec-tron and the halogen elements that need one electron inorder to have an inert configuration. To ionize Na, an en-ergy of 5.14 eV must be provided, which may be writtenas

Na + 5.14eV → Na+ + e−

On the other hand, when Cl captures one electron, this electron will become bound, and the energyreleased will be 3.62eV. We may write the reaction as

Cl + e− → Cl− + 3.62eV

If now we bring Na+ and Cl− together until the separation between them is 2.51Å (this is the equilibriumseparation between the centers of the two ions in the NaCl molecule), the Coulomb attraction potentialenergy Ep will be

Ep = −1

4πε0

e2

r= −9×109Nm2/C2 (1.6 × 10−19)2

2.51 × 10−10m= −9.18×10−19 J = −5.73eV


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KAPIL ADHIKARI

B.Sc. CSIT


This represents a decrease of energy from the situation where the two ions are infinitely far apart;that is, this represents a release of energy. The net energy released is

E = 3.62eV + 5.73eV − 5.14eV = 4.22eV

If E = 0 is the reference energy state when the Na and Cl atoms are far apart, when they are ions atr = 2.51Å their energy is −4.22eV. In other words, they fall into a potential well of depth −4.22eV. Thisis the energy needed to break apart the NaCl molecule and restore each ion to electrical neutrality. In anionic crystal, the binding energy is the sum of all the attractive forces (between all ions of opposite sign)and all the repulsive forces (between ions of like sign). For an ionic crystal this energy can be written as

E = −α1

4πε0

e2

r(1)

where α is known as the Madelung’s constant and r is the distance between two nearest ions. TheMadelung constant is different for different crystal structures.

For the fcc NaCl structure, α = 1.7476. The Na+ ion has six nearest neighbors at a distance r (six Cl−

ions). Therefore the energy contribution is

E6 = −61

4πε0

e2

r(2)

The next nearest neighbors are 12 Na+ at a distance (2)1/2r and contribute a repulsion energy of

E12 = +121

4πε0

e2

21/2r(3)

At the next distance there are eight Cl− at (3)1/2r

E8 = −81

4πε0

e2

31/2r(4)

Then there are six Na+ at 2r

E6 = +61

4πε0

e2

2r(5)

and so on. The total binding energy obtained by combining all these contribution is given by

ETotal = −1

4πε0

e2

r

[6 −

12√

2+

8√

3−

62+ ...

]and the summation of the series was evaluated by Madelung for fcc crystals as 1.7476. The total bindingenergy is then

ETotal = −1.74761

4πε0

e2

r(6)

Ionic crystals are extremely hard and have a high melting point. We should also note that all the electronsare bound to the ions, which, with the electron exchange, now have completely filled electron subshells.Because there are no free or loosely bound electrons to transport charge when an electric field is appliedacross the crystal, simple ionic crystals do not conduct electricity and belong to the class of insulators.

Covalent Bond Another important and very strong type of bond is the covalent bond. A good exampleof this is the H2 molecule. The bond comes about because both electrons shared by both nuclei.

The covalent bond is very strong, for example, 7.4 eV for diamond, 12.3 eV for SiC. Moreover, be-cause all available valence electrons pair up and orbit around pairs of atoms, it is difficult to dislodgethem to conduct electricity and therefore covalently bonded solids belong to the class of insulators orsemiconductors.


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KAPIL ADHIKARI

B.Sc. CSIT


Metallic Bond The metallic bond can be thought of as the limiting case of the covalent bond in whichelectrons are shared by all the ions in the crystal. When a crystal is formed of atoms that have a fewweakly bound electrons in the outer subshells, these electrons become free from the individual atoms,which thereupon acquire stable closed subshell configurations. The energy needed to liberate theseloosely bound electrons is more than compensated for by the decrease in energy resulting from the bind-ing.

The metallic bond is in many ways similar to the ionic bond in the sense that the main role is playedby the electrostatic attraction between unlike charges; however, there is a big difference. Whereas inthe ionic crystal the position of the positive and negative charges and therefore the directionality of theforces is fixed, in the metallic bond they are not. The electrostatic attraction comes from all directions.This is important, because it explains why a small deformation in a nearly perfect metal crystal does notcause a fracture. Whether we compress, twist, or pull a piece of metal, the cohesive forces are still thereand coming from all directions. Pure metals are ductile and malleable.

Molecular or van der Waals Bond Van der Waals bond is the bond due to a weak, short-range attractiveforce between atoms or molecules caused by their dipole moments, often arising in otherwise nonpolaratoms or molecules from a temporary shift of orbital electrons to one side of one atom or molecule.

Problems1. Copper has a face-centered cubic structure with a one-atom basis. The density of copper is 8.96g/cm3

and its atomic weight is 63.5g/mole. What is the length of the unit cube of the structure?2. Assuming that atoms in a crystal structure are arranged as close-packed spheres, what is the ratio of

the volume of the atoms to the volume available for the simple cubic structure? Assume a one-atombasis.

3. Repeat the above Problem for the body-centered cubic structure.4. Repeat the above Problem for the face-centered cubic structure.5. The dissociation energy of the KF molecule is 5.12 eV . The ionization energy for K is 4.34 eV,

and the electron affinity of F is 4.07 eV. What is the equilibrium separation constant for the KFmolecule?

crystal structure & bonding: lec-09

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