disorder in crystals

Disorder in crystals

Disorder in crystals

All lattice points are not always

the same.

Apatite Ca3(PO4)2

Apatite Ca3(PO4)2

Ca2+

Apatite Ca3(PO4)2

Ca2+ 0.98Å

Apatite Ca3(PO4)2

Ca2+ 0.98Å

Sr2+ 1.12Å

Apatite Ca3(PO4)2

Ca2+ 0.98Å

Sr2+ 1.12Å

Group II

Be Mg Ca Sr Ba Ra

Group II

Be Mg Ca Sr Ba Ra

2+ in ionic compounds

Group II

Be Mg Ca Sr Ba Ra

2+ in ionic compounds

88Sr – 86% of naturally occuring 38

Group II

Be Mg Ca Sr Ba Ra

2+ in ionic compounds

88Sr – 86% of naturally occuring 38

90Sr – radioactive isotope product of nuclear weapons testing

38

Apatite Ca3(PO4)2

Ca2+ 0.98Å

Sr2+ 1.12Å

If Sr2+ replacesCa2+ consistently,the structure changes.

Apatite Ca3(PO4)2

Ca2+ 0.98Å

Sr2+ 1.12Å

If Sr2+ replacesCa2+ consistently,the structure changes.This is not disorder.

Apatite Ca3(PO4)2

Ca2+ 0.98Å

Sr2+ 1.12Å

If Sr2+ replacesSome Ca2+ randomly,the structure is disordered.

If a crystal contains 90% Ca and 10% Sr, each

M2+ site will appear to be Ca/Sr 90/10% based

on diffraction data.

Defects in Crystals

Defects in Crystals

Disorder implies that all

positions are occupied, but the

occupation of some sites may

not be consistent.

Defects in Crystals

A defect is a break in the

infinite lattice.

Defects in Crystals

A defect is a break in the

infinite lattice. Some sites

that would normally be occupied

in a perfect lattice, are open.

Color center defect

-

h + Cl- Cl + e-

Color center defect

-

h + Cl- Cl + e-

Cl 0.99 ÅCl- 1.81 Å

The uncharged Cl is not affected by

the + charges and is considerably

smaller than the Cl-.

The uncharged Cl is not affected by

the + charges and is considerably

smaller than the Cl-. The Cl can move

through, and leave, the lattice.

The uncharged Cl is not affected by

the + charges and is considerably

smaller than the Cl-. The Cl can move

through, and leave, the lattice. The

electron can be trapped in the octahedral

vacancy left by the Cl-.

Anion missing; replaced by e-.

Anion missing; replaced by e-.

The overall lattice is not disturbed.

Anion missing; replaced by e-.

This does not have to be the same site vacatedBy the Cl-.

Color center defect

Anion missing; replaced by e-.

Color center defect

The presence of e- in a void leads to an electronic transition in the visible range.

In a real (as opposed to a ‘perfect’)

Crystal, a small portion of the sites

will be unoccupied.

In a real (as opposed to a ‘perfect’)

Crystal, a small portion of the sites

will be unoccupied.

This is called a Shottky defect.

+-

Perfect

+-

Perfect Real

+-

Perfect Real

In ionic crystals, charges still

must balance.

Shottky Defect

Shottky Defect: a void that doesnot disturb the structure.

Shottky Defect in metal.

Other defects may alter the lattice.

+-

Interstitial site:

+-

Interstitial site: position between ions or atoms which can be occupied by anotherion or atom.

+-

Interstitial site: position between ions or atoms which can be occupied by anotherion or atom.

+-

Move ion from normal site tointerstitial site.

Frenkel defect: lattice is distorted whenan ion is moved to an interstitial site.

Defects tend to be dynamic.

Nonstoichiometric Compounds

Wüstite

Wüstite

FeO

Wüstite

FeO

= O

= Fe

Wüstite

FeO

= O

= Fe

+2 -2

Wüstite

FeO

= O

= Fe

+2 -2

Fe0.85-0.95O

If there is less than 1 Fe per O,Fe must be in more than 1 ox. State.

Wüstite

FeO

= O

= Fe2+, Fe3+

+2 -2Fe0.85-0.95O

Fe0.85-0.95O

Fe0.85O

Fe0.85-0.95O

Fe0.85O

Fe2+x ; Fe3+

0.85-x

Fe0.85-0.95O

Fe0.85O

2x + 3(0.85-x) = 2

Fe2+x ; Fe3+

0.85-x

Fe0.85-0.95O

Fe0.85O

2x + 3(0.85-x) = 2

Fe2+x ; Fe3+

0.85-x

2x + 2.55 –3x = 2

Fe0.85-0.95O

Fe0.85O

2x + 3(0.85-x) = 2

Fe2+x ; Fe3+

0.85-x

2x + 2.55 –3x = 2

-x = -0.55

Fe0.85-0.95O

Fe0.85O

2x + 3(0.85-x) = 2

Fe2+x ; Fe3+

0.85-x

2x + 2.55 –3x = 2

x = 0.55

Fe0.85-0.95O

Fe0.85O

2x + 3(0.85-x) = 2

Fe2+x ; Fe3+

0.85-x

2x + 2.55 –3x = 2

x = 0.55

(Fe2+0.55, Fe3+

0.30 )O

(Fe2+0.55, Fe3+

0.30 )O

Fe0.85O

(Fe2+0.85, Fe3+

0.10 )O

Fe0.95O

Thermodynamics of Crystals

Na+ Cl- ionic bond

Na+

Account for ionic attractions

and repulsions based on the distance

of the ions and their charges.

- +r

- +r

The energy of this pair depends

on coulombic attraction and

repulsion.

Ep = - z1z2e2 b+r rn

- +r

The energy of this pair depends

on coulombic attraction and

repulsion.

Ep = - z1z2e2 b+r rn

attraction term(decreases energy)

- +r

The energy of this pair depends

on coulombic attraction and

repulsion.

Ep = - z1z2e2 b+r rn

attraction term(decreases energy)

Repulsive term(increases energy)

- +r

Ep = - z1z2e2 b+r rn

z = charge number

- +r

Ep = - z1z2e2 b+r rn

z = charge numbere = electron charge

- +r

Ep = - z1z2e2 b+r rn

z = charge numbere = electron charger = internuclear separation

- +r

Ep = - z1z2e2 b+r rn

z = charge numbere = electron charger = internuclear separationb, n are repulsion constants

- +r

Ep = - z1z2e2 b+r rn

z = charge numbere = electron charger = internuclear separationb, n are repulsion* constants

* repulsion due to physical contact, notcoulombic repulsion

Ep = - z1z2e2 b+r rn

The lattice energy for a mole of

NaCl can be evaluated by multiplying

the energy by No and including a

factor that accounts for all ion-ion

interactions.

Ep = -

U = NoAz1z2e2 B+

r rn

z1z2e2 b+r rn

-

Ep = -

U = NoAz1z2e2 B+

r rn

z1z2e2 b+r rn

Lattice energy

-

Ep = -

U = NoAz1z2e2 B+

r rn

z1z2e2 b+r rn

Lattice energyAvagadro’s number

-

Ep = -

U = NoAz1z2e2 B+

r rn

z1z2e2 b+r rn

Lattice energyAvagadro’s number

Madelung constant

-

Repeat S&P pg 80

When an individual ion is considered

in a cubic lattice, there is a group of

oppositely charged ions at a given

distance followed by a group of like

charged ions at a longer distance.

If r = a in NaCl then there are 6 Cl- at

distance a from Na+.

If r = a in NaCl then there are 6 Cl- at

distance a from Na+.

There are 12 Na+ at a distance of 2 a

from the initial Na+.

a

a

2 a

Madelung constant for NaCl

Potential energy for nearest neighbors = -6e2

a

Potential energy for next-nearest = 12e2

2 a

Madelung constant for NaCl

Potential energy for nearest neighbors = -6e2

a

Potential energy for next-nearest = 12e2

2 a

e2

a- 6 12 8+ - + .........

1 32

Madelung constant for NaCl

Potential energy for nearest neighbors = -6e2

a

Potential energy for next-nearest = 12e2

2 a

e2

a- 6 12 8+ - + .........

1 32 1.75

U = NoAz1z2e2 B+

a an

B = Az1z2e2

nan-1

-

B =

U = NoAz1z2e2 +

a an-

U = NoAz1z2e2 B+

a an-

Az1z2e2

nan-1

Az1z2e2

nan-1

B =

U = NoAz1z2e2 B+

a an-

Az1z2e2

nan-1

U = -NoAz1z2e2

a

U = NoAz1z2e2 +

a an-

Az1z2e2

nan-1

1- 1n

U = -NoAz1z2e2

a1- 1

n

n varies from 9 to 12; it is determined

from the compressibility of the material

U = -NoAz1z2e2

a1- 1

n

Ucalc Uexp kJ/mol

NaCl 770 770

KF 808 803

NaH 845 812

Where do experimental values for U

come from?

E

E

K(g) K+(g) + e- 419 kJ/mol

K(g) K+(g) + e- 419 kJ/mol

First ionizationenergy

K(g) K+(g) + e- 419 kJ/mol

First ionizationenergy

Cl(g) + e- Cl-(g) 349 kJ/mol

K(g) K+(g) + e- 419 kJ/mol

First ionizationenergy

Cl(g) + e- Cl-(g) 349 kJ/mol

Electron affinity

K(s) K(g) Hsublimation

K(s) K(g) Hsublimation

½ Cl2(g) Cl(g) Hdissociation

K(s) K(g) Hsublimation

½ Cl2(g) Cl(g) Hdissociation

K(g) K+(g) + e- 419 kJ/mol

Cl(g) + e- Cl-(g) 349 kJ/mol

K(s) K(g) Hsublimation

½ Cl2(g) Cl(g) Hdissociation

K(g) K+(g) + e- 419 kJ/mol

Cl(g) + e- Cl-(g) 349 kJ/mol

K+(g) + Cl-

(g) KCl(s) U

K+(g) + Cl-

(g) KCl(s) -U

K(g) + Cl(g) K(s) + ½ Cl2(g)

Hsub+ ½ D

I -A-e +e - Hf

K+(g) + Cl-

(g) KCl(s) -U

K(g) + Cl(g) K(s) + ½ Cl2(g)

Hsub+ ½ D

I -A-e +e - Hf

Born-Haber Cycle

K+(g) + Cl-

(g) KCl(s) -U

K(g) + Cl(g) K(s) + ½ Cl2(g)

Hsub+ ½ D

I -A-e +e - Hf

Born-Haber Cycle

Only term not from experiment

K+(g) + Cl-

(g) KCl(s) -U

K(g) + Cl(g) K(s) + ½ Cl2(g)

Hsub+ ½ D

I -A-e +e - Hf

Born-Haber Cycle

Only term not from experiment

U = - Hf + Hsub+ ½ D + I - A

Born-Haber Cycle

U = - Hf + Hsub+ ½ D + I - A

NaCl -414 109 113 490 347

kJ/mol

Born-Haber Cycle

U = - Hf + Hsub+ ½ D + I - A

NaCl 779 -414 109 113 490 347

kJ/mol

Born-Haber Cycle

U = - Hf + Hsub+ ½ D + I - A

NaCl 779 -414 109 113 490 347NaBr -377 109 96 490 318

kJ/mol

Born-Haber Cycle

U = - Hf + Hsub+ ½ D + I - A

NaCl 779 -414 109 113 490 347NaBr 754 -377 109 96 490 318NaI -322 109 71 490 297

kJ/mol

Born-Haber Cycle

U = - Hf + Hsub+ ½ D + I - A

NaCl 779 -414 109 113 490 347NaBr 754 -377 109 96 490 318NaI 695 -322 109 71 490 297

kJ/mol

Born-Haber Cycle

U = - Hf + Hsub+ ½ D + I - A

U = -NoAz1z2e2

a1- 1

n

Born-Haber Cycle

U = - Hf + Hsub+ ½ D + I - A

U = -NoAz1z2e2

a1- 1

n

Thermo(B-H) Theory

NaCl 779 795NaBr 754 757NaI 695 715

1. Construct a diagram for the Born-Haber

cycle for the various thermodynamic

properties associated with the formation

of magnesium chloride.

Homework problems for 10/3

continued

The important values are:

Hsub Mg 147.7 kJ/molIE1 Mg 737.7 kJ/molIE2 Mg 1450.7 kJ/molD Cl2 243 kJ/molA Cl 348.6 kJ/molHf MgCl2 -642 kJ/mol

continued

CsCl2. In the CsCl structure,

how many ions would be

included in the first

attractive term for the

Madelung constant.

continued

CsCl2. In the CsCl structure,

how many ions would be

included in the first

repulsive term for the

Madelung constant.