Bonding in solids

Types of solids

• metallic

• ionic

• covalent

• molecular

Types of Crystals

Metallic Crystals

• Lattice points occupied by metal atoms

• Held together by metallic bonds

• Soft to hard, low to high melting point

• Good conductors of heat and electricity

11.6

Cross Section of a Metallic Crystal

nucleus &

inner shell e-

mobile “sea”

of e-

Types of Crystals

Ionic Crystals

• Lattice points occupied by cations and anions

• Held together by electrostatic attraction

• Hard, brittle, high melting point

• Poor conductor of heat and electricity

CsCl ZnS CaF2

11.6

Types of Crystals

Covalent Crystals

• Lattice points occupied by atoms

• Held together by covalent bonds

• Hard, high melting point

• Poor conductor of heat and electricity

11.6diamond graphite

carbon

atoms

Types of Crystals

Molecular Crystals

• Lattice points occupied by molecules

• Held together by intermolecular forces

• Soft, low melting point

• Poor conductor of heat and electricity

11.6

!"# $%:&'( 3)* +"

7

electronic conductors

8

conductor

semiconductor

insulator

metals

electrical conductivity decreases

with increasing temperature,

whereas in semiconductors,

electrical conductivity increases

with increasing temperature.

band theory of solids

9

formation of a band

10

overlap of orbitals = band

11

occupation of bands

12

13

Fermi level

•N electrons for N MO’s

•The hightest occupied MO

PHYS 624: Electronic Band Structure of Solids 14

Is there a Fermi energy of intrinsic Semiconductors?

•If is defined as the energy separating the

highest occupied from the lowest unoccupied level,

then it is not uniquely specified in a solid with an

energy gap, since any energy in the gap meets this

test.

•People nevertheless speak of “the Fermi energy”

on an intrinsic semiconductor. What they mean is

the chemical potential, which is well defined at any

non-zero temperature. As , the chemical

potential of a solid with an energy gap approaches

the energy of the middle of the gap and one

sometimes finds it asserted that this is the “Fermi

energy”. With either the correct of colloquial

definition, does not have a solution in a

solid with a gap, which therefore has no Fermi

surface!

PHYS 624: Electronic Band Structure of Solids 15

p-type and n-type semiconductors

!PURE

DOPPED"

thermal excitation

• when 2N e- are present, the band is full and it is an insulator at T = 0

• When T>0, there is thermal excitation to form a semiconductor

16

P-type, N-type semiconductor

17

dopants can trap electrons

dopants can carry excess electrons

Light Emitting Diode

LED vs. Laser Diode

• standard enthalpy change accompanying the separation of the species that compose the solid

Lattice enthalpy

The Solution Process for NaCl

!Hsoln = Step 1 + Step 2 = 788 – 784 = 4 kJ/mol6.7

Born-Haber cycle

crystal structure

A crystalline solid possesses rigid and long-range order. In a

crystalline solid, atoms, molecules or ions occupy specific

(predictable) positions.

An amorphous solid does not possess a well-defined

arrangement and long-range molecular order.

A unit cell is the basic repeating structural unit of a crystalline

solid.

Unit Cell

lattice

point

Unit cells in 3 dimensions 11.4

At lattice points:

• Atoms

• Molecules

• Ions

11.4

The identification of crystal

planes - intersection distances

(1a, 1b, infinity c)

(3a, 2b, infinity c)

(-1a, 1b, infinity c)

(infinity a, 1b, infinity c)

Miller indices (hkl)

26

,-. /01 2345-. 678

11.5

Formation of X-ray

• bremsstrahlung (brake ray)

28

Extra distance = BC + CD = 2d sin" = n# (Bragg Equation)11.5

The Bragg Law

glancing angle

X rays of wavelength 0.154 nm are diffracted from a

crystal at an angle of 14.170. Assuming that n = 1,

what is the distance (in pm) between layers in the

crystal?

n# = 2d sin " n = 1 " = 14.170 # = 0.154 nm = 154 pm

d =n

2si

n"

=1 x 154 pm

2 x sin14.17= 77.0 pm

11.5

William Henry

Bragg

William Lawrence

Bragg

31

Typical X-ray powder diffraction

pattern

32

Typical Structures

11.4

Fig. 11.29

11.4

11.4

Shared by 8

unit cells

Shared by 2

unit cells

11.4

11.4

1 atom/unit cell

(8 x 1/8 = 1)

2 atoms/unit cell

(8 x 1/8 + 1 = 2)

4 atoms/unit cell

(8 x 1/8 + 6 x 1/2 = 4)

11.4

When silver crystallizes, it forms face-centered cubic

cells. The unit cell edge length is 409 pm. Calculate

the density of silver.

d = m

VV = a3 = (409 pm)3 = 6.83 x 10-23 cm3

4 atoms/unit cell in a face-centered cubic cell

m = 4 Ag atoms107.9 g

mole Agx

1 mole Ag

6.022 x 1023 atomsx = 7.17 x 10-22 g

d = m

V

7.17 x 10-22 g

6.83 x 10-23 cm3= = 10.5 g/cm3

11.4

ABA vs. ABC packing

42

hcp vs. ccp

• hexagonal close-packed structure

– ABA

– coordination number 12

• cubic close-packed structure

– ABC

– coordination number 12

43

Structures of ionic crystals

44

6-coordinated rock-salt structure8-coordinated CsCl structure

Fig. 11.25

Fig. 11.27

radius-ratio rule

• radius ratio = r(smaller)/r(larger)

• ratio = 0.732

– CsCl structure

• 0.414 < ratio < 0.732

– rock salt structure

– 6 coordination

• ratio < 0.414

– 4 coordination

– ZnS (sphalerite)47

Fig. 11.p460top

High temperatre

superconductor :

HTSC

YBa2Cu3O7 :

123 compound

amorphous solid

• 9$%: 1; <#

– =;> ?@A BCDE 1;( FG "HI

*"J !"> FGDK L7 MN

– long range OPQR FGS TU

– (V) WH

49

Fig. 11.31

quartz crystal glass

2D structures of SiO2