1

Chapter Three Crystal binding

Why do atoms form crystals or solids?

Answer : Interatomic forces that bind atoms.

Atoms bind due to the Coulomb attractive forces between electrons and neighboring atomic ions.

Contents:

☻ Types and strengths of binding forces

☻ Reason for crystal structure formation

☻ Mechanical properties of crystals

[ stress ] = [ elastic constant ] x [ strain ]

2

Cohesive energy U

≡ the energy that must be added to the crystal to separate itscomponents into neutral free atoms at rest

≡ –Energy of free atoms Crystal energy

Hence, U>0 to form a stable solid

• Magnitude ~ 1-10eV except for the inert gas crystals (0.02-0.2eV)

• U ≤ Eion (Ionization energy = Binding energy of valence electrons)

• U controls the melting temperature and bulk modulus

3

Types of bonds

(a) Van der Waals(Molecular)

(b) Covalent

(c) Metallic

(d) Ionic

Electrons localized among atoms

Electrons shared by the neighboring atoms

Electrons free to move through sample

Electrons transferred to adjacent atoms

(a)

+---

-+

---

-+---

-

+---

-+

---

-+---

-

+---

-

(b)

+

+ +

++

(c)

+ +

+

+

+

(d)

+

+

+

+

- -

--

-

4

(a) Molecular bonding Inert gas crystals : He, Ne, Ar, Kr, Xe, Rn

Transparent Insulators – completely filled outer electron shells

Weakly bonding – van der Waals bonding

FCC structures except for He3 and He4high ionization energies

low melting temperatures

12.1314.0015.7621.56Ionization energy (eV)

161.4115.883.8124.56Melting temperature (K)

0.160.120.080.02Cohesive energy (eV/atom)XenonKryptonArgonNeon

What holds an inert gas crystal ?

5

Phase diagrams of (a) 4He and (b) 3He.

6

Van der Waals –London InteractionConsider two identical inert gas atoms

R

Neutral: positive nucleus + spherically symmetric distribution of electron charge

No interaction between atoms → No cohesion (NO solid) ?Attractive interactionFluctuating dipole-dipole interaction

between the atoms

Reviews:

r )rn(r dP 3 rrr∫=r

n(r)

d -q+q←= qdP

r

P

r

as r>>dElectric fields Attractive force Repulsive force

421

rPP~F

321

rPP~U3r

P~E

7

Inert gas solids

• On average spherically symmetric distribution of electron charge with thepositive nucleus in the center 0P =

r

• But thermal fluctuations (finite T) cause instantaneous electric dipole moment

0P 0(t)P 2 ≠→≠r

--

--+

--

--+ 0P≠

v0P=

vfluctuations

• On adjacent atoms if the dipoles are random there could be no net force(time average)

• But dipole induces a dipole in neighboring atoms that always gives an attractive force

8

Model for inert gas solid – two identical linear harmonic oscillators

R

x2

-e+ex1

-e+e

p1 and p2 are the momenta of these two oscillators

C is the force constant

Hamiltonian for the unperturbed system – no Coulomb interaction

22

222

1

21

o Cx21

2mpCx

21

2mpH +++=

Hamiltonian for Coulomb interaction energy of the system

321

2

Rx,x Rxx2e

21− → <<

2

2

1

2

21

22

1 xRe

xRe

xxRe

ReH

−−

+−

−++=

9

Normal mode transformation -- symmetric (s) and anti-symmetric (a)

2ppp ;

2ppp

2xx x;

2xxx

21a

21s

21a

21s

−≡

+≡

−≡

+≡

Total Hamiltonian after the transformation

+++

−+= 2

a3

22a2

s3

22s x

R2eC

21

2mpx

R2eC

21

2mpH

22

222

1

21

o Cx21

2mpCx

21

2mpH +++=

mCω ,H oo =

mR/2eC

mCω

mR/2eC

mCω , H

32a

a

32s

s

+==

−==Two frequencies of the coupled oscillators

symmetric (s) and anti-symmetric (a)

10

+

×

−+

+=

+=

+=

+

×

−+

−=

−==

K

K

2

3

2

3

2

o

2/1

3

232

a

2

3

2

3

2

o

2/1

3

232

s

CR2e

21

)21(

21

CR2e

211ω

CR2e1

mC

mR/2eCω

CR2e

21

)21(

21

CR2e

211ω

CR2e1

mC

mR/2e-Cω

Therefore, the zero point energy of the coupled oscillators islowered from the uncoupled oscillators by

oo 21

21 ωω hh +

as 21

21 ωω hh +

The zero point energyThe uncoupled oscillators

The coupled oscillators

6

2

3

2

o RA2

CR2e

81ω

21U −=

×

−=∆ h Attractive

interaction

11

The van der Waals interaction, the London interaction, the induced dipole-dipole interaction

litypolarizabi electronic theis re whe

ω2CeωA 2

o2

4

o

α

αhh ≡=0 RAU 6 <−=∆

P2induced

P1 fluctuation“Polarizability” of the atom E P α=

rr

31

12 RPE P αα ==

rr

6

21

321

RP~

RPP~U α−−

R

< 0

12

What limits attraction ? -- Repulsive force (Pauli exclusion principle)

Two electrons cannot have all their quantum number s the same.

A B A B

Charge distributions overlap

• When charge distributions of two atoms overlap, there is a tendency for electronsfrom atom B to occupy in part states of atom A occupied by electrons of atom A,and vice versus.

• Pauli exclusion principle prevents multiple occupancy, and electron distribution of atoms with closed shells can overlap only if accompanies by the partial promotion of electrons to unoccupied high energy state of the atom.

The electron overlap increases the total energy of the system and gives a repulsive contribution to the interaction.

13

0 RBU 12 >=∆Empirical formula for such repulsive potential

RR

ε4

RA

RB

612

612

−

=

−=

σσ

The total potential for inert gas system

the Lennard-Jones potential

where empirical parameters A=4εσ6 and B=4εσ12 are determined from independent measurements made in the gas phase.

Values of ε[energy] and σ[length] are shown in Table 4.

0.85 0.90 0.95 1.00 1.05 1.10 1.15 1.20

-2

0

2

4

6

8

UPauli

UvdW

U

U(R

)/4ε

R/σ

U(r) = UPauli + UvdW

14

N atoms in the crystal

( ) ∑

−

=

≠ ji

6

nnij

12

nnijtotal RpRp

4εN21U σσ

where Rnn is nearest neighbor distance andpijRnn is the distance between atom i and atom j

( )

∑−

∑=

≠≠

6

nnji6ij

12

nnji12ij

total Rp1

Rp1 4εN

21U σσ

dimensionlessBoth lattice sums can be done for any structure.

Sum of 1/pn converges rapidly for large n.

More distant neighbors have more influence on the latter term than the former term.

15

FCC structure, 45392 .14p1 13188 .12

p1

ji6ijji

12ij

== ∑∑≠≠

and

HCP structure, 45489 .14p1 13229 .12

p1

ji6ijji

12ij

=∑=∑≠≠

and

Both structures have 12 nearest neighbors.

25330.12p1 11418.9

p1

ji6ijji

12ij

=∑=∑≠≠

andBCC structure,

BCC structure has 8 nearest neighbors.

16

Cohesive energy of inert gas crystals at 0K-- minimum Utotal (Equilibrium)

−

=

−−−=

)6)(45.14()12)(13.12(

σR

Rσ (14.45)(6) 2Nε

Rσ)6)(45.14(

Rσ12)(12.13)( 2Nε

dRdU

6

13

12

7

6

13

12total

minimum a is )4N)(15.2()R( U,09.1Rat ototalo εσ −==

=0

1.091.101.111.14Ro/σ

3.983.653.402.74σ (Å)

4.354.013.763.13Ro (Å)XenonKryptonArgonNeon FCC structure

DeviationQuantum corrections

17

Expect structure to form crystals which have lowest energy,largest cohesive energy

Gibbs free energy : G=U – TS + PV

Assuming T=0, P=0, and no kinetic energy of atomic motion

-8.62-8.61-8.24-5.69

12.1312.139.116.2

14.4514.4512.258.4

FCCHCPBCCSC

∑≡≠ ji

12ijp1α

ε/NUtot

∑≡≠ ji

6ijp

1β

FCC is favored.

18

(d) Ionic bonding Alkali halides

Electron transfers between atoms to form two oppositely charged ions.Strong electrostatic forces dominate.

Electron configuration : closed electronic shellsFor examples, LiF : Li+ (1S2) instead of Li (1S22S)

F- (1S22S22p6) instead of F (1S22S22p5)Like inert gas atoms w/. some distortion of charge distribution near

the region of contact with neighboring atomsbutCharge distribution is spherically symmetric.

Electron density distributionin the base plane of NaCl

relative electron concentration

19

Need to consider the ionization energies and electron affinities of atoms

energy that must be supplied in order to remove an electron from a neutral atom

Ionization energy I

energy that is gained when an additional electron is added to a neutral atom

Electron affinity A

Ionic bonding is produced whenever an element w/. a relatively low ionization energy is combined with an element w/. a high electron affinity.

20

e.g. NaCl crystal What is its cohesive energy ?Valence electron loosely bound to ionIonization energy =5.14eV(energy to remove electron from Na)

Na+

e-Na :

Cl

e-Cl : Seven valence electrons tightly bounddesire a filled outer shell Electron affinity energy = 3.61eV

+ 3.61eVe- + Cl-Cl

Na Na++ 5.14eV + e-

NaCl :

Na+ Cl-

r=2.81Å

Ion bind by electrostatic attraction U~-e2/4πεr ~ -7.9eVNa+ + Cl- = NaCl +7.9eV

Ionic bound Na + Cl = NaCl + 6.37eVcohesive energy

21

N ions in the crystal and Uij is the interaction energy between ions i and j

( i ≠ j )

rqq

)r

exp( λUij

jiijij +−=

ρlong range electrostatic

short range Pauli repulsive

CGS

Rpqq

)Rexp( λUij

jiij +−=

ρ where R = nearest neighbor distance

−−=

+−== ∑∑≠≠

Rαq)Rexp( zλ N

Rpqq

)Rexp( λ NzUU

2

ji ij

ji

jiijtot

ρ

ρ

p

αj ij

'∑ ±≡where z = number of nearest Madelung constantneighbors of any ion

22

minimum Utotal (Equilibrium)

zλραq)

ρR(expR

0R

q Nα)ρRexp(

ρNzλ

dRdU

2o2

o

2

2tot

=−

=+−−=

At equilibrium Ro,

−−=

oo

2

tot Rρ1

Rq NαU

Madelung energy

Short range repulsiveρ = 0.1Ro

Madelung constant α : geometric sum

depends on relative distance, number, and sign of neighboring atoms----- crystal structures and basis

23

One dimension : line of ions of alternating signs

+ - + +rR

+ +- - - -

p

αij ij

'∑ ±=

( )

•••+−+−=+

•••+−+−=⇒

•••+−+−=

±=∑

4x

3x

2xxx1n and

41

31

2112

4R1

3R1

2R1

R12

rRα

432

ij ij

'

l

α

α = 2 ln(2) =1.386 for one dimensional chain

24

In three dimensions, it is more complicated to calculate α.

very slowly convergent

very long range electrostatic forces

Special mathematical tricks are used to calculate Madelung constant.

1.6414ZnS (Wurtzite)

1.63814GaAs (Zinc blende)1.76278CsCl (BCC)1.74766NaCl (FCC)

αCoordinate Nostructure

Higher coordination number gives larger Madelung constant.It depends on the structure of the crystal but not unit cell dimensions.

25

Madelung energy

o

2

M Rq Nα

41Eπε

−= in a binary ionic crystalw/. 2N ions in the crystal

Sometimes, the core-core repulsive energy in other general form is considered.

noo

2

total RNA

Rq Nα

41E +−=πε

Equilibrium

TdSPdVdE +−= The first law of thermodynamics

0dVdE

=At T=0, the equilibrium sample volume is determined by

0dRdE

o

=

26

The equilibrium nearest-neighbor distance

( )1n1

2eqo q α

nA 4R−

−=

πε

−−=

n11

Rq Nα

41E eq

o

2

total πεThen, the total energy

3.4×10-120

10.65

3.57CsCl

1.0×10-100

8.55

3.17KCl

1.8×10-995.0×10-882.3×10-892.6×10-79A(J mn)

8.386.417.306.20n

2.823.322.572.01(Å)NaClNaFLiClLiFcrystal

eqoR

TdVdP

V1

−≡κ V=Na3=NCRo3The isothermal compressibility

( )κα

πε2

4eqo

q RC 361n +=

( )( )4eq

o

2

RC 361nq 1

πεα

κ−

=eqT

2

2

dVEdV1

=κ or

27

(b) Covalent bonding C, Si, GeTetrahedral bond

C C Organic chemistry / diamond

Si Si

Ge Ge

4 atoms in the valence band bond to 4 neighboring atoms

Semiconductor}7.3eV/atom

4.6eV/atom

3.9eV/atom

diamond

Tetrahedral bonding

Nature of chemical bonds in a diamond or zinc blende structure

28

(1/4,1/4,1/4)

(0,0,0)

(1/2,1/2,0)

(0,1/2,1/2)

(1/2,0,1/2)

High electron concentration

Tetrahedral sp3 bond

Four lobes emanate from an atom at the center of a cube. Other atoms are at the ends of the dotted lines and lobes point from them toward the cube center.

The bond is usually formed from two electrons, one from each atom participating in the bond.

Electron forming the bond tend to be partially localized in theregion between two atoms joined by the bond.

The spins of two electrons in the bond are antiparallel.

distortion of electron cloud around atoms

spin dependent coulomb energy

29

Calculated valence electron concentration in Ge.

30

Consider simple covalent bond : H - H

Both hydrogen atoms would like to form a filled outer shell-- share electrons

Two cases : ↑↑ (same spins on electrons)↑↓ (opposite spins on electrons)

ψ2

r

ψ

↑↓

↑↑r

↑↓

↑↑

Pauli exclusion principle forbids two electrons with the same states.

↑↑ same spins: electrons must stay apart↑↓ opposite spins: electrons can occupy the same place

31

Pauli exclusion principle modifies the distribution of charge according to spin orientation.

Energy is lower when electrons spend time between nuclei --attractive Coulomb interaction from both

Spin-dependent Coulomb energy

Exchange interaction

32

Neutral H has only one electron → covalent bonding with one other atom

But, there would be a hydrogen bond between them under certain conditions

~ 0.1eVbeing formed only between the most electronegative atoms,

such as F, O, and N.

.

F- F-

H+HF2

- is stabilized by a hydrogen bond.

In the extreme ionic form of the hydrogen bond, the hydrogen atom loses its electron to another atom in the molecule; the bare proton forms the hydrogen bond.

The hydrogen bond connects only two atoms.

33

(c) metallic bonding most metalsHigh electrical conductivity : a large number of electrons in a

metal are free to move .conduction electrons

Outer electrons of atoms that form metals are loosely bound.

The potential energy barrier between atoms is reduced, the electron energy may be well above the potential energy maximum and their wave functions are then nearly plane waves in regions between atoms.

Weak binding, 1~5eV/atom Metals tend to crystallize in relatively closed packed structures :

hcp, fcc, bcc, …

34

Mechanical properties of solid

Bond --- harmonic oscillation

( )orrkFrrr

−−=Crystal --- A collection of harmonic oscillatorsa homogeneous continuous medium rather than a periodic array of atoms

Apply forces displacements of atoms

stress strain ε

l

l uAk

Aku

AF

==dimensionless

elastic constant [Nt/m2]

σ = C ε

1 D Elastic regime

stress [Nt/m2]

Elastic behavior is the fundamental distinction between solids and fluids.Elasticity describes the dimensional change under external stresses.