Metals-Review
• Properties of metals?
~lustrous surface (shiny) [so are silicon, iodine, fools gold (pyrite, FeS2): metalloid,
non-metal, inorganic compound]
~dense
[Li density 1/2 that of H2O; osmium - 40 x H2O]
~hard
[alkalis soft, Hg is a liquid!]
~malleable (can be flattened, deformed)
~ductile (can be pulled into wires)
~ high thermal conductivity
[but some of the TMs are quite brittle]
[and Hg is a liquid!]
[diamond (non-metal) has highest of any element]
~ high 3D electrical conductivity
[graphite (non-metal) high 2D version]
…even though two orders of magnitude difference between worst(plutonium) and
best(silver) metal conductivities under ambient conditions, plutonium conductivity is
105 higher than best conducting nonmetallic element
• Bonding in metals?
~ essentially covalent, but with the bonding (valence) electrons delocalized (free to
move) throughout the bulk metal structure. (e.g. electron-sea model)
~ in the solid state, the bulk structure consists of ordered arrays of atoms: these are
crystalline materials with lattice structures.
H&S, Fig. 2.10, p. 43
Radii
• van der Waals radius:
•
•
•
•
rvdw > rcov
distance = 2rvdw
• covalent radius:
internuclear separation (d)
•
1/2(d) = rcov
•
can only know if:
-element X has a stable form containing X-X single bond
-experiment is carried out to measure d, such as diffraction techniques;
e.g. X-ray, neutron, or electron diffraction
e.g. P4, S6 or S8, H2, F2, & can actually get alkali metals as M2 in the vapour phase,
but rcov not actually too meaningful for the alkali elements, which are metallic, and
their compounds, which are most often ionic and solid
• metallic radius:
What does a metallic solid look like?
1
Types of Solids
crystal
glass
atomic
arrangement
order
Regular
Random
Long Range
Short Range
name
Crystalline
Amorphous
concept
Ordered
Disordered
Which of the following images shows a substance that
would have long range order at the atomic level?
A
B
C
2
Crystals
Temperature dependence of metallic structure
“Napoleon’s retreat from Moscow”
by Adolph Northen (see Wikipedia entry
on 1812 French invasion of Russia.
3
Early Crystallography
7 crystal systems
4
3 ways to fill a cubic crystal
5
Metallic Crystal Structures of the Elements
6
Lattice Structures of Metallic Solids
• Despite our ability to describe the covalent bonding in metals using molecular orbital
theory (more later), it is easiest to describe the structures by considering the atoms
as hard spheres
~we can place them in ordered arrays just like so many gumballs
• Many different ordered arrangements are possible: we use packing models to
describe these crystal lattices
• Simplest possible arrangement: simple cubic packing
~ 1st layer: atoms are placed side by side
~ 2nd layer: atoms are placed directly on
top of those in the previous layer
Bottom layer
• Next simplest crystal lattice arises from body-centred cubic
~ take two layers of four atoms from the
simple cubic structure, insert one more
atom into the interstitial hole in the
middle.
next layer
H&S, Fig. 6.6, p. 151
Lattice Structures of Metallic Solids (cont’d)
• Unit cell: smallest repeating unit of a lattice structure
~can construct an infinite lattice using all the information contained in a unit cell
Bottom layer
1
4
2
3
• Coordination number: number of other atoms “touching” each atom in a lattice
~ each atom is surrounded by four other
~ each atom is touched by four other
atoms in the plane, one above, and one
atoms above, and four below. Coord # =
below. Coord # = 6
8
• Total number of spheres in each unit cell?
• Relative packing efficiencies (or % occupancy)
7
What fraction of the sphere is in the unit cell?
Lattice Structures of Metallic Solids (cont’d)
• Total number of spheres in each unit cell?
simple cubic
~ the corners of the unit cell cube are at
the nucleus of each atom
~ 8 corners
~1/8 of each sphere is inside the unit cell
8 x (1/8) = 1 sphere (atom) per unit cell
body-centred cubic
~ have again cut a cube from the centre
of 8 atoms
~ 8 corners plus one whole sphere inside
the unit cell
8 x (1/8) + 1 = 2 spheres per unit cell
8
What is the volume of a cube with side length = 2r?
length=width=height
What is the volume of a sphere with radius = r
9
Lattice Structures of Metallic Solids (cont’d)
• Total number of spheres in each unit cell?
simple cubic
body-centred cubic
• Relative packing efficiencies (% occupancy) of the different lattices:
~ radius of each sphere = r
~ edge length of unit cell = 2r
~volume of cell = 8r3
~volume of sphere = 4/3πr3 = ~4.19r3
% occupancy = 1 sphere/unit cell
= (~4.19r3/8r3) x 100%
= ~52%
calculations later
= ~68%
H&S, Ch6; RC&O, Ch4
rFe
rFe
abcc
afcc
Assume the radius of Fe is the same, regardless of crystal structure. Therefore
rFe=rFe . However abcc ≠ afcc.
10
A review of triangles
1
a2 +b2 = c2
if a and b are equal
a2 +a2 = c2 and
therefore c=√2 a
c
a
b
3
2
a
d
a
a
c
c
a
a
a2 +a2 = c2 and
therefore c=√2 a
a2 +c2 = d2
and a2 +a2 = c2
therefore a2 +a2 + a2 = d2
or d=√3 a
FCC/CCP
BCC
d
a
a
c
a
a
c
d=√3 a
c=√2 a
d
c
4 atoms/cell
c=4rFe =√2aFCC
2 atoms/cell
d=4rFe =√3aBCC
11
The calculations (BCC)
ρ= 7.86 g/cm 3 (BCC) aw=55.847 g/mole
ρ=
(grams per cell)
=
(atoms/cell)(atomic weight)
NA
(volume of cell)
a3
(2 atoms/cell)(55.847 grams/mol)
6.022*1023 atoms/mol
a3=g/ρ =
7.86
=2.36*10-23 cm 3
grams/cm3
aBCC = 2.86* 10-8 cm
4rFe =√3aBCC
rFe =√3aBCC
4
rFe =1.24*10-8 cm
The calculations (FCC)
ρ= ? g/cm 2 (FCC) aw=55.847 g/mole
rFe =1.24*10-8 cm
ρ=(grams per cell)/a3 (volume of cell)
4rFe =√2aFCC
aFCC= 4rFe
=
4*(1.24*10-8 cm)
√2
aFCC
=3.51*10-8
√2
cm
(4 atoms/cell)(55.847 grams/mol)
ρ=
6.022*1023 atoms/mol
(3.51*10-8 cm)3
= 8.57 grams/cm3
12
What is the atomic radius of Barium?
ρ= 7.86 g/cm 2 (BCC) aw=55.847 g/mole
ρ=g(grams per cell)/a3 (volume of cell)
BCC
d
d
a
c
d=√3 a
2 atoms/cell
d=4rFe =√3aBCC
The calculations (BCC)
ρ= 3.50 g/cm 2 (BCC) aw=137.33 g/mole
ρ=(grams per cell)/a3 (volume of cell)
(2 atoms/cell)(137.33 grams/mol)
a3=g/ρ =
6.022*1023 atoms/mol
=2.36*10-23 cm 3
3.50grams/cm3
aBCC = 5.07* 10-8 cm
4rBa =√3aBCC
rBa =√3aBCC
4
rBa=2.19*10-8 cm or 2.19*10-10 m or 219 pm or 2.19 A
13
What is the packing efficiency of a BCC?
BCC
d
d
a
c
d=√3 a
2 atoms/cell
d=4rFe =√3aBCC
d
2 atoms/cell
d=4rFe =√3aBCC
~ radius of each sphere = r
~ diagonal length of unit cell = r+2r+r = 4r
edge length x (3)1/2 = 4r
~ edge length = 4r/~1.73 = ~2.31r
~ volume of cell = ~12.33r3
% occupancy = 2 sphere/unit cell
= 2(~4.19r3/12.33r3) x 100%
= ~68%
14
EXTRA: Close Packed Lattice Structures
• here are ccp/fcc unit cell diagrams from our four different sources:
• important to remember the layer diagram describes mechanism for packing whole
spheres, while the unit cell’s function is only to provide the minimum positional
information necessary to reproduce the bulk lattice.
Lattice Structures of Metallic Solids
• cubic close-packed (face-centred cubic)
~ the corners of the unit cell cube are at
the nuclei of 8 atoms
~ the faces of the unit cell cube cut four
atoms in half
~ thus inside the unit cell are
8 x (1/8) + 6 x (1/2) atoms
= 1 + 3 atoms
= 4 atoms per unit cell
• hexagonal close-packed
“ccp” or “fcc”
~ the hexagonal corners of the unit cell cube are at the
nuclei of 12 atoms
~ the two hexagonal faces of the unit cell cube cut two
atoms in half
~ three atoms are completely within the unit cell
~ thus inside the unit cell are
12 x (1/6) + 2 x (1/2) + 3 atoms
= 2+ 1 + 3 atoms
= 6 atoms per unit cell
“hcp”
15
The calculations (FCC)
ρ= ? g/cm 3 (FCC) aw=55.847 g/mole
rFe =1.24*10-8 cm
ρ=(grams per cell)/a3 (volume of cell)
4rFe =√2aFCC
aFCC= 4rFe
=
4*(1.24*10-8 cm)
√2
√2
aFCC =3.51*10-8 cm
(4 atoms/cell)(55.847 grams/mol)
6.022*1023 atoms/mol
ρ=
(3.51*10-8 cm)3
= 8.57 grams/cm3
The calculations (BCC)
ρ= 3.50 g/cm 2 (BCC) aw=137.33 g/mole
ρ=(grams per cell)/a3 (volume of cell)
(2 atoms/cell)(137.33 grams/mol)
a3=g/ρ =
6.022*1023 atoms/mol
=2.36*10-23 cm 3
3.50grams/cm3
aBCC = 5.07* 10-8 cm
4rBa =√3aBCC
rBa =√3aBCC
4
rBa=2.19*10-8 cm or 2.19*10-10 m or 219 pm or 2.19 A
16
7 crystal systems
17
Metallic Crystal Structures of the Elements
Temperature dependence of metallic structure
“Napoleon’s retreat from Moscow”
by Adolph Northen (see Wikipedia entry
on 1812 French invasion of Russia.
18
To know
•
•
•
•
•
•
Structure: FCC, BCC, SC (be able to draw)
Spheres per unit cell: FCC, BCC, SC
Atomic radius change: hex, FCC, BCC, SC
Number of nearest neighbours: hex, FCC, BCC, SC.
Packing Efficiency: hex, FCC, BCC, SC.
Should know that there are 7 crystal systems and 14
Bravais lattices, but only need to reproduce the cubic
ones.
• Density  atomic radius etc.. calculations
• Note that FCC (face centered cubic) and CCP (cubic
close packing) are the same.
Crystal
structure
packing
efficiency
reproduce
structure
SC
Spheres per Number of
unit cell
nearest
neighbours
1
6
52%
yes
BCC
2
8
68%
yes
FCC
4
12
74%
yes
hexagonal
n/a
12
74%
no
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Metallic radii
• to calculate the metallic radius of an element, need to know the lattice type and the
density of the metal.
• metallic radius, rmetal: half of the distance between the nearest neighbour atoms in a
solid state metallic lattice
• to compare within a periodic sequence of elements, need rmetal for a consistent
number of nearest neighbours.
e.g. atoms in bcc structures are eight-coordinate, each at distance x (=2rmetallic),
although there are six more neighbours at distances of 1.15x. This gives packing
efficiency of 68%, lower than close packing efficiencies of 74% (12 coordinate).
~ ratio of interatomic distances for bcc vs cp forms of the same metal is 0.97:1.00
~corresponds to a change of coordn # from 8 to 12.
• rmetal varies with coordination number:
coordination number
12
relative radius
1.00
8
0.97
6
0.96
4
0.88
~ some values listed for rmetal with coordination number of 12 are estimated, since not
all metals actually adopt close-packed structures (See table 6.2)
• for the alkali metals:
~all adopt body-centred cubic lattices
~rmetal in Table 6.2 reported for 12-coordinate atom, in Table 11.1 for 8coordinate (see K: rmetal12 = 235 pm, rmetal8 = 227)
~regardless, trends are as we would predict based on Zeff arguments.
Ionic radii relative to metallic (or covalent) radii (in Å)
Brown, LeMay, Bursten & Murphy “Chemistry The Central Science” 11th Ed., Pearson 2009, Fig. 7.8, p. 263
20
Lattice structures of ionic solids
• Things to keep in mind when thinking about packing arrangements for ionic solids:
(1) Ions are assumed to be hard, charged spheres: incompressible, nonpolarizable
~ in fact there is usually at least some degree of covalency in all ionic compounds,
but a hard sphere model works well for most “classic” ionic compounds
(2) Ions will surround themselves with the maximum number of ions of opposite
charge: this is especially important for cations
~ primary cause of variable ionic packing arrangements is the relative size of cation
vs anion: how close is radius ratio r+/r- to 1?
~ i.e. how many anions can surround a cation, given the relative sizes and the
unwillingness of anions to be squished together?
RC&O, Table 5.5, p. 83
(3) The cation to anion ratio throughout the lattice must reflect the actual chemical
composition of the salt.
e.g. in CaCl2 the lattice will include an array of anions interspersed with an array of
half the number of cations.
Salt
Salt Satyagraha,1930
21
Sodium chloride
Brown, LeMay, Bursten & Murphy “Chemistry The Central Science” 11th Ed., Pearson 2009, Fig. 11.35, p. 461
Cesium chloride
rion
(Cs+) =
170 pm
r+/r– = 0.939
rion(Cl–) = 181 pm
22
Solid-state structures CaF2
Solid-state structures CaF2
23
Fluorite and antifluorite structures
MX2
Red = cation, Green = anion
M2X
Green = cation, Red = anion
Zinc Blende/Diamond
H&S, Fig 6.19, p. 169
24
Gauging the strength of ionic bonds
H&S Fig. 12.5, p316
Strength of Ionic Bonds
Brown, LeMay, Bursten & Murphy “Chemistry The Central Science” 11th Ed., Pearson 2009, Fig. 11.35, p. 461
25
Calculating lattice energies
• Lattice energies can be calculated for ionic solids, and these values can be useful,
especially for understanding the role of lattice structure in hypothetical compounds.
• Obviously, the calculation takes into account the coulombic (charge) interactions of
the ions in the lattice: not just the charges of the ions’ nearest neighbours, but also the
influence of charges further away in the lattice.
• Coulombic contributions to lattice energy can be described using convergent
mathematical series, in which the Madelung constant (A) varies depending on the
lattice type.
Measuring lattice energies?
• Very difficult to measure lattice energy directly
H&S Fig. 6.24, p175
26
Measuring lattice energies?
• Pretty hard to measure a lattice energy directly. (See definition again!)
• Just as we discussed for ionization energies and electron affinities, the internal
energy change at 0K can be approximated by the enthalpy change (at room
temperature) for the analogous process.
• Lattice enthalpy values can be measured indirectly using thermochemical cycles:
Born-Haber cycles.
Hess’s Law: for a
For n >1 must sum IEs
reaction carried
out in a series of
steps, H for the
overall reaction will
equal the sum of
enthalpy changes
for the individual
steps
H&S Fig. 6.24, p175
Trends in the periodic table
http://www.meta-synthesis.com/webbook/35_pt/pt2.html
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