disorder in crystals
DESCRIPTION
Disorder in crystals. Disorder in crystals. All lattice points are not always the same. Apatite Ca 3 (PO 4 ) 2. Ca 2+. Apatite Ca 3 (PO 4 ) 2. Ca 2+ 0.98 Å. Apatite Ca 3 (PO 4 ) 2. Ca 2+ 0.98 Å. Sr 2+ 1.12 Å. Apatite Ca 3 (PO 4 ) 2. Ca 2+ 0.98 Å. - PowerPoint PPT PresentationTRANSCRIPT
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.