MTE 271‐002
Review Session
• This study guide has been prepared to help you in studying for the first quiz
you in studying for the first quiz.
• If you understand these concepts, you have a chance to do well.
Primary Atomic Bonds
• Covalent: – Shared outer shell electrons.
• Metallic: M t lli
– Shared sea of electrons. Non‐directional.
• Ionic: – Donation of valence electron (e‐) to balance charge.
• van der Waals / secondary: – Attraction between (+) and (‐) charged regions.
Attraction between (+) and ( ) charged regions
Know the expected types of bonding for typical materials
for typical materials
(Here are some examples)
Metals:
• Nickel ‐ metallic
• Iron ‐ metallic
Polar Molecule
Polar Molecule
• Water (H2O) ‐ secondary, van der Waals
Polymers
• Polypropylene ‐
P l
l
covalent along chain, secondary between
l
l
h i
d b
• Polyethylene ‐ covalent along chain, secondary between
Semiconductors
• Diamond ‐ covalent
Table 2.5 on Page 51 of your • Silicon ‐ covalent
text has a good summary of Ionic Crystals
Ionic Crystals
bonding for different • CsCl ‐ ionic
material types.
COORDINATION # AND IONIC RADII
Symmetry is determined by the r
/ ranion ((aka, r/R) ratio
y
y
y
/ )
cation /
Issue: How many anions can you arrange around a cation?
rcation
ranion
Coord #
< .155
2
ZnS
(zincblende)
Adapted from Fig. 12.4,
Callister 6e.
6e
.155-.225
3
.225-.414
.225
.414
4
NaCl
(sodium
chloride)
Adapted from Fig. 12.2,
.414-.732
6
.732-1.0
8
Adapted from Table
12.2, Callister 6e.
Callister 6e.
CsCl
(cesium
chloride)
Adapted from Fig. 12.3,
Callister 6e.
Take home message: For ionic bonding crystals, coordination (size of ions) impacts the stacking of the ions together to form the crystal structure (# of bonds formed)
Know how to calculate coordination number
b
• EX:
EX: what is the coordination number for CsCl
what is the coordination number for CsCl
if rCs+=0.165 nm and rCl‐ = 0.181 nm ?
• ANS:
rCs  0.165 nm ; RCl   0.181 nm
rCs
RCl 
00.165
165 nm

 0.91
0.181 nm
From the table on the previous page, CN = 8
Know how to calculate IPF
• EX: Using the radius information provided in the prior example problem, calculate the IPF for CsCl.
• ANS: IPF 
  # ions / u.c.  volume / atom 
i
i
IPF 
 volume of
i
u.c.
4
4
1 Cs ion     rCs3   1 Cl ion     rCl3 
3

 a3 
3
 ; a  2rCs  2rCl
3
4
4
1   (0.165 nm)3  1   (0.181 nm)3
2  0.165 nm   2  0.181 nm 
3

 0.399 nm
;
IPF  3
a
3
 a3 
substituting in calculated values
IFP 
0.019  0.025
 0.68
0 68
0.064
• Take a look at the next two slides for visualization purposes.
Watch Out! Different Ions. Must Modify APF/IPF Equations
3a
Rblue
a
a
2a
Close‐packed directions:
Adapted from Fig. 3.2(a), Callister 7e.
g
( ),
a
length = 2Rblue
bl + 2Rredd = 3 a
Must put in correct value for atoms atoms
unit cell
(SHOWN ON NEXT PAGE)
(SHOWN ON NEXT PAGE)
2
IPF = 4
3
 ( 3a/4)
a3
3
volume
atom
volume
unit cell
Watch Out! Different Ions. Must Modify APF/IPF Equations
3a
Rblue
a
a
2a
Close‐packed directions:
Adapted from Fig. 3.2(a), Callister 7e.
g
( ),
a
Atoms touch along the body
diagonal in this structure.
structure Thus
Thus,
diagonal  3a  2 Rblue  2 Rred
IPF =
IPF = 4 3
4 3
1  Rblue  1  Rred
3
3
a3
2 Rblue  2 Rred
a
3
Understand how to do this type of problem:
Calculate the number of Mg2+ vacancies produced by the solubility of 1 mol of
Al2O3 in 99 mol of MgO (see Figure 4.6,
4 6 page 105).
105) Why do these vacancies
form?
99 mol of anion (O) sites from MgO
+3 mol of anion (O) sites from Al2O3
102 mol of anion (O) sites in solid solution
99 mol of cation sites from MgO
+2 mol of cation sites from Al2O3
101 mol of cation sites in solid solution
(anions) – (cations) = (vacancies)
102 mol – 101 mol = 1 mol Mg2+ vacancies = 6.022  1024 vacancies
Vacancies form to balance charge.
Hume‐Rothery Rules
Hume‐
for Forming a Solid Solution
for Forming a Solid Solution
1. <  15% difference in atomic radii
2. The same crystal structure
3. Similar electronegativity (i.e., ability of an atom to attract an electron)
4. Same valence
If one or more of these rules are violated, only partial solubility is possible. Types of Imperfections:
There is no such thing as a perfect crystal
• Vacancy atoms
• Interstitial atoms
• Substitutional atoms
Point defects
[0‐D]
• Dislocations
Line defects
[1‐D]
• Grain Boundaries
Planar defects
[2‐D]
• Cracks, voids
Volume defects
[3‐D]
I would know the categories, different types, and how to describe them!
Zero – Dimensional Defects
Zero Dimensional Defects
(descriptions)
• A
A vacancy:
vacancy:
a vacant lattice site
• An interstitial defect:
the placement of an p
atom in a interstitial (in between regular lattice sites).
)
Substitutional Defect
Substitutional Defect
(descriptions)
• A
A substitutional
substitutional defect:
another type of atom ‘substitutes’ on a particular lattice site.
Defects in Ionic Crystals
(types and descriptions)
• In ionic crystals (such as NaCl), single vacancies can not occur. Why?
Because off the
B
h constraints
i off charge
h
neutrality,
li either
ih a
Frenkel defect OR a Schottky defect forms.
F k l defect:
Frenkel
d f
vacancy-interstitial
i
i i l combination.
bi i
pp
y charged
g ion vacancies.
Schottkyy defect: ppair of oppositely
Shottky Defect:
Frenkel Defect
Adapted from Fig.12.21, Callister & Adapted
from Fig.12.21, Callister &
Rethwisch 8e. (Fig. 12.21 is from W.G. Moffatt, G.W. Pearsall, and J. Wulff, The Structure and Properties of Materials, Vol. 1, Structure, John Wiley and Sons, Inc., p. 78.)
Know your Miller Indices and how to Know your Miller Indices and how to d
draw planes draw planes and directions
l
and directions
dd
• Don
Don’tt worry about hexagonal crystals.
worry about hexagonal crystals
• Just be familiar with cubic crystals.
b f ili
i h bi
l
• Don’t worry about x‐ray diffraction. We’ll save that for the next exam.