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Finding the Lattice Type from the
density
PROBLEM Al has density = 2.699 g/cm3 and Al
radius = 143 pm. Verify that Al is FCC.
SOLUTION
1. Calculate the unit cell volume
V = (d)3
Edge distance comes from face diagonal.
Diagonal distance(d) = r √ 8
Finding the Lattice Type
PROBLEM Al has density = 2.699 g/cm3 and Al radius
= 143 pm. Verify that Al is FCC.
d=r√8
d3 = volume
SOLUTION
V = (d)3 and face diagonal = r √ 8
Therefore, d = r √ 8 = 404 pm
In centimeters, edge = 4.04 x 10-8 cm
So, V of unit cell = d3= (4.04 x 10-8 cm)3
V = 6.62 x 10-23 cm3
Finding the Lattice Type
PROBLEM Al has density = 2.699 g/cm3 and Al radius
= 143 pm. Verify that Al is FCC.
SOLUTION
2. Use V and density to calculate mass of
unit cell from Density = Mass/ Volume
Mass = density • volume
= (6.62 x 10-23 cm3)(2.699 g/cm3)
= 1.79 x 10-22 g/unit cell
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11/1/2013
Finding the Lattice Type
PROBLEM Al has density = 2.699 g/cm3 and Al
radius = 143 pm. Verify that Al is FCC.
SOLUTION
3. Calculate number of Al per unit cell
from mass of unit cell.
Mass 1 Al atom =
26.98 g
1 mol
•
mol
6.022 x 10 23 atoms
1 atom = 4.480 x 10-23 g, so
1.79 x 10 -22 g
1 atom
•
= 3.99 Al atoms/unit cell
unit cell
4.480 x 10 -23 g
BCC = 2 atoms per unit cell
4r
d
d√2
An element has a body-centered cubic shape and has
a density of 3.47 g/cm3 and a radius of 200. pm. What
element is 3this?
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D = 3.47 g/cm
r = 200. pm = 2.00 x 10 cm
(4r)2 = 2d2 + d2
16r2 = 3d2
d=?(edge length)
4.62 x 10-8 cm = d
d3 = V = 9.86 x 10-23 cm3/cell
3.47 g
1
cm3
9.86 x 10-23 cm3 1 cell
1 cell
2 atoms
6.02 x 1023 atoms= 103 g/mol
1 mole
Rh = Rhodium
Finding the Density
• Iridium has a FCC structure with a unit cell edge (d)
= 383.3 pm. Calculate the density of Ir.
V = d3 = (383.3 pm)3 = 5.631 x 107 pm3
This is the volume of 4 atoms so one atom =
5.631 x 107 pm3/ 4 = 1.408x107 pm3 = 1.408x10-23cm3
D = m/v mass = 192.2 g /mol
D = 192.2 g 1 mol
1 atom =
23
1 mol 6.02 x 10 atoms 1.408x10-23cm3
D = 22.68 g/cm3
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11/1/2013
Metals and Their Properties
• Because metals float on a sea of their valence
electrons (see figure 10.18) when hammered on their
slide rather than fracture.
• This gives metals the properties of malleability and
ductility.
• Also, the highly mobile valence electrons explain
the high conductivity of both heat and electricity
found in metals.
• Bonding in metals is both strong and nondirectional.
Alloys
• Alloys – a substance that contains a mixture of
elements and has metallic properties.
• Substitutional Alloy – some of the host metal
atoms are replaced with different metals atoms
of similar size.
– Brass = 2/3 copper atoms and 1/3 zinc atoms
• Intersistial Alloy- small atoms are present in the
spaces between atoms (interstices).
– Steel = carbon occupy some of the holes in iron
• These cause changes in properties to the iron,
which is relatively soft – the directional
bonding caused by the carbon makes steel
stronger.
Simple Ionic Compounds
Lattices of many simple
ionic solids are built by
taking a SC or FCC lattice
of ions of one type and
placing ions of opposite
charge in the holes in the
lattice.
CsCl has a SC lattice of Cs+ ions
with Cl- in the center.
1 unit cell has 1 Cl- ion
plus
(8 corners)(1/8 Cs+ per corner)
= 1 net Cs+ ion.
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Simple Ionic Compounds
Salts with formula
MX can have
SC structure —
but not salts
with formula
MX2 or M2X
Two Views of CsCl Unit
Cell
Either arrangement leads to formula of 1 Cs+ and 1 Cl- per unit cell
NaCl Construction
FCC lattice of Cl- with
Na+ in holes
Na+ in
octahedral
holes
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11/1/2013
Octahedral Holes
in FCC Lattice
The Sodium Chloride
Lattice
Many common salts have FCC arrangements of
anions with cations in OCTAHEDRAL
HOLES — e.g., salts such as CA = NaCl
• FCC lattice of anions ---->
----> 4 A-/unit cell
• C+ in octahedral holes --->
---> 1 C+ at center
+ [12 edges • 1/4 C+ per edge]
= 4 C+ per unit cell
Comparing NaCl and CsCl
• Even though their formulas have one cation
and one anion, the lattices of CsCl and NaCl
are different.
• The different lattices arise from the fact that a
Cs+ ion is much larger than a Na+ ion.
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Common Ionic Solids
Titanium
dioxide, TiO2
There are 2 net
Ti4+ ions and 4
net O2- ions
per unit cell.
Common Ionic Solids
• Zinc sulfide, ZnS
• The S2- ions are in
TETRAHEDRAL
holes in the Zn2+
FCC lattice.
• This gives 4 net
Zn2+ ions and 4 net
S2- ions.
Common Ionic
Solids
• Fluorite or CaF2
• FCC lattice of Ca2+ ions
• This gives 4 net Ca2+
ions.
• F- ions in all 8
tetrahedral holes.
• This gives 8 net F- ions.
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Network Solids
Network solids are substances composed of
covalently bonded atoms.
Most network solids are very hard and rigid and are
characterized by high melting and boiling points.
Diamond
Graphite
Amorphous Solids
• Glass and plastics are an example of a
substance that has no definite crystal
shape or is amorphous.
• Amorphous solids occur when there is
no regular repeating crystalline shape,
unlike liquids, however, they have
strong forces of attraction that restrict
molecular movement.
• Therefore, they are considered solids.
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