THE SOLID STATE (Ch 11.4)
(see also table
11.4, p. 475)
Many different ways to classify:
e.g. metals and non-metals. Non metallic solids usually
fall into three categories according to their bonding:
Al
CaCO3
C
examples
metals
Cu, Li, Al,
Os, Hg, U
ionic solids
NaCl,
BaCl2
network
solids
graphite
diamond,
SiO 2, BN
I2, ice,
PCl5, S8 ,
glucose
molecular
solids
S
Bonding and
structure
Metallic lattice positive metal
ions held
together by
delocalised
electrons
Ionic lattice
Covalent lattice
Covalent
molecules held
in lattice by
London forces,
hydrogen
bonding etc..
Conductivity
good
m.p./
b.p.
high
Very low
(except in
solution)
Very low
High
Very low
low
Very
high
strength
Malleable,
ductile
Hard,
rigid,
brittle
Very hard,
rigid,
brittle
rigid,
brittle
can also classify according to macro structure:
- e.g.
crystalline
metals, salts
amorphous
e.g. glass, wax,
polymers e.g.
poly(ethylene)
resins e.g. amber
In crystalline solids atoms or molecule are
arranged regularly giving solids with flat faces,
straight edges
Gemstones are hard, weather-resistant crystals
bonding in crystals can be:
ionic (e.g. KCl),
covalent (diamond)
hydrogen bonding (ice, glucose)
van der Waals’ forces only (iodine, solid Ar)
1
Crystal structure in metals (p.466)
-easiest to describe – all particles are single atoms of the same size
-atoms usually adopt a ‘close-packed’ arrangement
-close-packed structure: one in which atoms occupy the smallest total
volume leaving the least empty space between atoms
-two possibilities: hexagonal close packing and cubic close packing
-imagine we build up layers of atoms, step-by-step:
-the second layer (B) is added
so the atoms lie in the dimples
(holes) of the first (A)
the third layer (C) of
atoms can be overlayed
in two ways:
the atoms are
superimposed over
the atoms of the
bottom layer
(ABABAB pattern
of layers)
the atoms are
superimposed over
the dimples (holes)
of the bottom layer
(ABCABC pattern
of layers)
2
creates a hexagonal
close packed structure:
the atoms are
superimposed over
the atoms of the
bottom layer
(ABABAB pattern
of layers)
e.g. Mg, Zn
creates a cubic close
packed structure
the atoms are
superimposed over
the dimples (holes)
of the bottom layer
(ABCABC pattern
of layers)
e.g. Al, Cu, Ag, Au
Unit Cell – The smallest unit of a lattice that when repeated
gives the large scale structure of the crystal
-atoms at faces or corners of unit cell are shared with
neighbouring unit cells
in cubic close packed
structure atoms of four
layers (ABCA) contribute
to each unit cell:
ccp sometimes
known as facecentered cubic
(fcc), as there is
an atom at the
centre of each
face of the unit
cell
3
Coordination number – the number of neighbours surrounding each atom
In both close packed structures each atom touches - there are:
3 from the layer above
3 from the layer below
6 in the same layer.
Co-ordination number = 12
hcp
co-ordination
number = 12
ccp
co-ordination
number = 12
Q: How many atoms are in a fcc unit
cell of copper?
A: important to remember- all atoms in
fcc are shared with neighbouring unit
cell.
8x
1
8
=1
Face centre atoms: 6 x
1
2
=3
Corner atoms:
in fcc there are 4
atoms per unit cell
4
Q: The atomic radius of fcc Cu is 0.128 nm.
What’s the density of the metal in g/cm3?
A: density is an intensive property - same
value for unit cell as bulk Cu metal
b
using: ρ = m/v (density = mass/volume)
r
first we need to calculate the mass and
volume of the unit cell….
Mass of unit cell = 4 x 63.54 x 1.6605 x 10-24 = 4.220 x 10-22 g/cell
atoms per
unit cell
Volume = a3
Atomic wt.
Cu (Da)
1 a.m.u.
(g)
so we need to find a
calculate a from Pythagoras theorum:
a2 + a2 = b2
b = 4r (atoms touch each other along face
diagonal)
b
2a2 = (4r)2 = 16r2
r
a2 = 8r2
a = (8r2)½ sub r = 0.128 x 10-9 m
a = 3.62 x 10-10 m (0.36 nm)
V = a3 = 4.745 x 10-29 m3
⇒ V = 4.745 x 10-23 cm3
ρ = m/v = 4.220 x 10-22 g /4.744 x 10-23 cm3
= 8.889
= 8.9 g/cm3
pra
Ex 1 ctise
11.4 1.3, Q
s
0&
11.4
2
5
Q: Calculate the atomic radius of Ag, given
that the metal crystallises in a fcc
arrangement with a density of 10.5 g/cm3?
b
A: density is intensive property - same
value for unit cell as bulk Cu metal
using: ρ = m/v (density = mass/volume)
r
first we need to calculate the mass and
volume of the unit cell….
Mass of unit cell = 4 x 107.87 x 1.6605 x 10-24 = 7.164 x 10-22 g/cell
atoms per
unit cell
Volume = a3
Atomic wt.
Cu (Da)
1 a.m.u.
(g)
We need to find radius r
from Pythagoras theorum:
a2 = 8r2
⇒ r = (a2/8)½
b
mass (m) and density (ρ) of unit cell known
⇒ can find volume and a of unit cell
r
ρ = m/v
v = m/ρ = 7.164 x 10-22 g /(10.5 g/cm3)
v = 6.823 x 10-23 cm3
a = V-1/3 m3
a = (6.823 x 10-23) -1/3 = 4. 0862 x 10-8 cm
⇒ r = (a2/8)½
⇒ r = 1.445 x 10-8 cm
(0.144 nm)
6
Q: Crystalline silicon has a cubic structure with a unit cell edge
length of 543 pm. Given that the density of the solid is 2.33 g/cm3,
calculate the number of Si atoms per unit cell.
A: 8 Si atoms per unit cell
try
at hom
e
+
Non-close packed structures
-some metals are lower in energy if they leave more than minimum amount of
space between atoms
e.g. in Na,
K, Fe and U
e.g. body-centered cubic arrangement is not close-packed.
-Atoms touch each other along the body diagonal of the
cubic unit cell.
Q: How many atoms are there per unit cell in bcc?
A: 2 (1 at cell centre, 8 x 1 at corners)
8
Q: What is the co-ordination number in bcc?
A: 8
7
Q: Potassium (atomic radius 220 pm) crystallises in a body-centred
cubic (bcc) arrangement. Calculate:
a) The mass of the unit cell (Da)
b) the length of the unit cell, a
A: 2 atoms per unit cell
Mass = 2 x 39.10 Da = 78.2 Da
A: atoms touch each other along
body diagonal (corner to opposite
corner) of unit cell…...
from Pythagoras theorum:
a2 + a2 + a2 = c2
c = 4r (atoms touch each other)
3a2 = (4r)2 = 16r2
a = (16r2/3)½ sub r = 0.22 x 10-9 m
a = 508 pm
Q: Calculate the atomic radius of vanadium, given that the metal
crystallises in a bcc arrangement with a density
of 5.96 g/cm3?
try
at hom
e
A: a = 305 pm so therefore atomic radius
(i.e. c/4) = 132 pm
+
8
Metallic Bonding (p. 475)
- close packing of metals has important
consequences:
-atoms’ outer (valence) electrons are
delocalised
(can move freely throughout lattice)
-metal structure is array of positive ions
immersed in a sea of electrons
Metallic bonding:
delocalised electrons
-delocalised electrons:
-bind atoms strongly - strong high melting points (e.g. Fe 1530 ºC, W 3500 ºC)
-allow conduction of heat and electricity throughout solid (Ag is best conductor)
Metallic Bonding (p. 475)
-bonding between atoms is not directional
so metals are:
i) malleable - can be beaten into
thin sheets without fracturing
ii) ductile - can be drawn into
fine wires without fracturing
-close-packing of atoms (cf. stacked fruit)
explains metals’ high density:
as high as:
8 gcm-3 for Cu
22 gcm-3 for Os, Pt & Ir
9
Crystal structures of Ionic solids (p.472)
-ionic solids adopt similar crystals structures to metals
-BUT: cations and anions have different charges and different sizes
-Crystal structure maximises attraction between opposite-charge ions and
minimises repulsions
-larger ion (usually anion) forms close-packed arrangement and cations
fit into holes in lattice:
tetrahedral holes
octahedral holes
octahedral holes – 4 per unit cell
smaller than octahedral holes – 8 per unit cell
-which holes are occupied by the metal depends on the relative sizes of
the ions e.g. in fcc sphalerite (ZnS)
-Zn2+ (0.07 nm) is much smaller than sulphide (0.18 nm)
⇒ tetrahedral sites occupied
S2- ion
-four S2- ions per unit cell ⇒ only half the tetrahedral holes need be
occupied
- each anion surrounded by 4 cations
each cation surrounded by 4 anions
⇒ there are 4 formula units per unit cell
⇒ co-ordination is 4,4
10
Cl- ion
NaCl is also ccp. Here cations only
slightly smaller than anions so fit
into octahedral holes
- each anion surrounded by 6
cations and each cation surrounded
by 6 anions ⇒ co-ordination is 6,6
Unit cell of NaCl:
4 cations
4 anions
i.e. there are 4 formula units per unit cell
Unusual salt crystal types
….if cation is bigger than anion
e.g. CsCl
one formula unit (one anion and one cation) per unit cell . Co-ordination
is 8,8
11
Q:
A: 1 x Ba
1 x Ti
3xO
⇒ BaTiO3
Q: How many formula units are present in this unit cell of the
mineral rutile? How many Daltons does a unit cell weigh? What is
the salt’s formula?
(lies at the face of the cell)
A: 2 x Ti
4xO
⇒ two formula units of
TiO2 present
(unit cell mass = 159.8
Da);
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Network Solids (Covalent Crystals, p.474)
-bonding is covalent and in an infinite lattice – ‘one giant molecule’
-e.g. diamond, quartz, ruby, sapphire – brittle but very hard
diamond – a three-dimensional lattice of tetrahedral (sp3-hybridised)
carbon atoms
- lattice very rigid: the hardest substance known – used in drill-bits and
(powdered) as an abrasive
- good thermal conductor (rigidity transfers atomic vibrations)
but electrical insulator (no delocalised electrons)
graphite - a two-dimensional array of trigonal (sp2-hybridised)
carbon atoms –
- hexagonal rings - strong bonding within the layers but weak
bonding between them
graphite is soft (used in pencils and as lubricant).
Non-bonded electrons are delocalised throughout plane – graphite
conducts electricity
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