Lectures 23–25 • Autumn 2026

Crystal Structures
From Bonds to Properties

Why do atoms arrange themselves the way they do? We start from the chemical bond, treat atoms as hard spheres, pack them as densely as possible, find the holes left between them, fill those holes with ions using Pauling’s rules, and finally ask what the resulting symmetry allows a crystal to do: conduct, deform, or turn a squeeze into a voltage.

Bonds
Packing
Voids
Ions
Properties
In the background: close-packed layers A, B and C sliding into an ABC stack, with small atoms hopping between the holes.
Where We Left Off

From Symmetry to Real Structures

So far we have described crystals abstractly: symmetry operations, lattices, space groups, Wyckoff positions, and how diffraction reads them. In Lectures 20–22 we saw how real crystals deviate from perfection. Now we ask the chemical question: which arrangement does a given set of atoms actually choose, and why?

The question

Copper is FCC, iron is BCC at room temperature, NaCl is rock salt and ZnS is zinc blende. A space group tells us the symmetry of each structure, but not why nature picked it.

The answer, in steps

The type of bond decides whether atoms can be treated as hard spheres. Sphere sizes, charges and stoichiometry then decide how they pack, which holes get filled and how many neighbours each atom has.

The payoff

Once the structure is known, its point group decides which properties are even possible: a centre of symmetry forbids piezoelectricity, pyroelectricity and ferroelectricity outright.

23–25
Lectures 23–25

From Bonds to Packed Spheres, Their Holes, and Properties

One chain of reasoning runs through all three lectures. Every link is explained below with something you can rotate, drag or play.

  • 1Bonding: types of bond, bond energy versus \(k_BT\), the interatomic potential
  • 2Packing: CN and packing fraction; SC, BCC, FCC and HCP; density
  • 3Voids: tetrahedral, octahedral and cubic holes, their number and size
  • 4Pauling’s rules for ionic crystals
  • 5Ionic structures: NaCl, CsCl, ZnS, wurtzite, spinel, perovskite
  • 6Structure → properties: Neumann’s principle, piezo-, pyro- and ferroelectrics, microstructure
1
Part 1 · Bonding

What Holds Atoms Together

The bond decides two things that matter for structure: how strong it is (compared with the thermal energy \(k_BT\)) and whether it has a direction. Non-directional bonds let us model atoms as hard spheres; directional bonds fix the bond angles instead.

Part 1 · Bonding

Types of Bonds

There are three strong (primary) bonds and a family of weak (secondary) ones. Each little animation shows the mechanism; the numbers are bond energies per atom, ion or molecule from Callister, Materials Science and Engineering, Table 2.3.

Ionic

  • An electron is transferred (Na → Cl), making Na+ and Cl−; the Coulomb attraction between the ions holds them.
  • Non-directional: the field of an ion is spherically symmetric, so the CN is set by the ion sizes (typically 4 to 8).
  • NaCl 3.3 eV, MgO 5.2 eV. The doubly charged ions of MgO bind much more strongly.
  • Hard, brittle, high melting, electrically insulating.

Covalent

  • Neighbours share electron pairs; the pair sits between the two nuclei.
  • Strongly directional: bond angles fix the CN. C and Si use four bonds at 109.5°, so CN = 4.
  • Si 4.7 eV, C (diamond) 7.4 eV
  • Very hard and high melting, yet the structures are open (low packing fraction).

Metallic

  • Positive ion cores sit in a “sea” of delocalised valence electrons that belong to the whole crystal.
  • Non-directional, so the cores pack densely: CN 8 or 12.
  • Hg 0.7, Al 3.4, Fe 4.2, W 8.8 eV: an enormous range.
  • Ductile (planes can slide without breaking specific bonds) and electrically conducting.

Secondary (vdW, H-bond)

  • Attraction between electric dipoles: fluctuating ones (van der Waals, even in Ar) or permanent ones (H-bonds in H2O, NH3).
  • Weak: Ar 0.08, Cl2 0.32 eV; H-bond: NH3 0.36, H2O 0.52 eV.
  • Hold molecules together in molecular solids and between polymer chains.
Watch the dipoles in the last panel: the δ+ end of one molecule and the δ− end of the next stay lined up as the charges fluctuate. That correlation is the whole origin of the van der Waals attraction.
Part 1 · Bonding

How Strong Are Bonds? Compare with \(k_BT\)

A bond energy only means something next to the energy that thermal motion supplies, \(k_BT\). At 300 K, \(k_BT = 8.617\times10^{-5}\ \text{eV/K} \times 300\ \text{K} = 0.026\ \text{eV}\). Drag the temperature and watch which substances would have boiled.

Bond energy (eV), log scale

Data: Callister, Table 2.3 (per atom, ion or molecule). Bars turn pale once \(15\,k_BT\) exceeds the bond energy, the rule of thumb for boiling explained in the next section.

T (K)300
kBT
0.026 eV
15 kBT (boiling cutoff)
0.39 eV
Strong bonds: several eV. Ionic and covalent bonds are comparably strong (3–7 eV, roughly 130–290 times \(k_BT\) at room temperature); metallic bonds range from 0.7 eV (Hg) to 8.8 eV (W).
Weak bonds: about 3–20 \(k_BT\). Ar (≈ 3 \(k_BT\)) boils at 87 K. Cl2, NH3 and H2O all sit at 12–20 \(k_BT\), yet only H2O is liquid at 300 K.
Phase is set by free energy. Boiling needs only \(k_BT \approx E_b/15\): the entropy of the gas does the rest. Melting needs even less, since the liquid keeps most of its bonds. Dissolving: solvent bonds plus entropy must win.
Part 1 · Bonding

Why Are Cl2 and NH3 Gases at Room Temperature?

Divide each bond energy by the thermal energy at the boiling point, \(k_BT_b\). Something remarkable happens: the ratio is about the same for every bond type, from argon to tungsten. Click any row to see it on the bar.

SubstanceEb (eV)Tb (K)kBTb (meV)Eb/kBTbAt 300 K

Bond energies: Callister, Table 2.3. Boiling points at 1 atm: CRC Handbook. \(k_B = 8.617\times10^{-5}\) eV/K.

Eb / kBTb for every substance
All eight lie between 11 and 18.
One rule for every bond type. From Ar (87 K) to W (5800 K), every substance boils when \(k_BT\) reaches about 1/11 to 1/18 of its bond energy.
Why not at Eb ≈ kBT?

Boiling is decided by the Gibbs free energy, \(G = H - TS\), not by energy alone. The vapour costs enthalpy (the bonds must be broken) but gains an enormous entropy, because each molecule can now be anywhere in a much larger volume. Trouton’s rule says this entropy gain is roughly the same for most liquids: \(\Delta S_\text{vap} \approx 10.5R\). So the vapour wins at \(T_b = \Delta H_\text{vap}/\Delta S_\text{vap}\), long before \(k_BT\) reaches \(E_b\).

\[ G_\text{gas} < G_\text{liquid} \iff T\,\Delta S_\text{vap} > \Delta H_\text{vap} \]
At 300 K the cutoff is about 0.4 eV

\(15\,k_BT \approx 15 \times 0.026 \approx 0.4\) eV. Cl2 (0.32 eV) and NH3 (0.36 eV) fall just below the cutoff and are gases. H2O (0.52 eV) is just above, so it stays liquid. The extra strength of the hydrogen bond is the only reason oceans exist.

Part 1 · Bonding · Animation

Energy vs Separation: The Interatomic Potential

Bring two atoms together. Far apart they attract; pushed too close, their electron clouds overlap and they repel violently. The sum of the two gives a potential well \(U(r)\), and everything about the bond can be read from its shape.

U(r) = attraction + repulsion
Depth E0
r0
Temperature
Bond energy
Stiffness U″(r0)
Mean spacing
1
Two competing forces. Attraction dominates at large \(r\); electron-cloud repulsion rises steeply at short \(r\).
2
Equilibrium spacing \(r_0\). The net force \(F = -dU/dr = 0\) at the minimum of \(U(r)\). The bond energy is the depth of the well.
3
Shape of the well. Depth → melting point; curvature → stiffness (elastic modulus); asymmetry → thermal expansion. Raise the temperature: the atom vibrates further, and because the outer wall is softer, its average position moves out.
Deep well

More energy is needed to pull the atoms apart, so the solid melts at a higher temperature. W (8.8 eV) melts at 3695 K; Hg (0.7 eV) is liquid at room temperature.

Sharp well

For small displacements \(U \approx U(r_0) + \tfrac12 U''(r_0)(r-r_0)^2\): a spring with constant \(U''(r_0)\). A narrow, sharply curved well means a stiff bond and a high elastic modulus.

Lopsided well

The repulsive wall is steeper than the attractive tail. A vibrating atom spends more of its time on the far side, so the mean spacing grows with temperature: thermal expansion. A perfectly parabolic well would show none.

2
Part 2 · Packing

Atoms as Hard Spheres

If the bond has no preferred direction, the only thing that matters is fitting as many neighbours as possible around each atom. Crystal structures then become a geometry problem: how do you pack spheres?

Part 2 · Packing

From Bonds to Hard Spheres

Look again at the potential: the repulsive wall is so steep that atoms almost never get closer than \(r_0\). That is the justification for modelling them as incompressible spheres of fixed radius.

Rotate me

Dense or Directional?

drag to rotate
Non-directional bonds: ionic, metallic, van der Waals

The steep repulsive wall in \(U(r)\) acts like a hard surface, so each atom or ion is modelled as an incompressible sphere of fixed radius. The structure is then the densest arrangement that the sphere sizes, charges and stoichiometry allow, giving a high CN (typically 6 to 12). The left view shows one atom with its 12 touching neighbours in FCC.

Directional bonds: covalent

Bond angles, not packing, decide the arrangement. In diamond each C bonds to exactly 4 neighbours at 109.5°. CN = 4 and the packing fraction is only \(\pi\sqrt3/16 = 0.34\): an open structure, despite very strong bonds.

Hard-sphere structures are decided by: ionic or atomic radii · charges on ions · stoichiometry · coordination number. Pauling’s rules (Part 4) turn these into predictions.
Part 2 · Packing · Animation

Coordination Number and Packing Fraction: Start in 2D

Two numbers describe any packing. Slide the rows of a square array sideways and let them settle into the hollows; watch both numbers change.

square array
Row shift
Coordination number
4
Packing fraction (area)
0.785
1 Coordination number (CN): the number of nearest neighbours touching an atom.
2 Packing fraction (PF): the volume (or, in 2D, area) of the atoms in a cell divided by the volume (area) of the cell.
3 Close-packed layer: CN 6, PF \(\pi/2\sqrt3 = 0.907\) in 2D. It has two kinds of triangular hollows, pointing up and down: b and c. They are the seeds of everything in 3D.
Part 2 · Metals · 3D

Simple Cubic and Body-Centred Cubic

Now stack in 3D. For each cell we need three things: the direction along which the spheres touch (this links \(a\) to \(r\)), the number of atoms per cell (corners count ⅛, body centre counts 1), and the resulting packing fraction.

drag to rotate
Part 2 · Close Packing · 3D

Stacking Close-Packed Layers: ABAB or ABCABC

Build your own stack. Start with layer A, then place the next layer into one set of hollows. The third layer is where the choice is made. A layer can never sit directly on an identical layer (no AA), because its spheres would land on top of each other instead of in the hollows.

emptydrag to rotate
Sequence (bottom → top)
Add layers to begin
1 Layers A and B. Layer B sits in one set of hollows (b) of layer A. Every close-packed structure starts like this; choosing the other set (c) is the same thing seen from a rotated viewpoint.
2 Third layer: two choices. Over the c-hollows of A gives a new position C: ABCABC… = cubic close packing (fcc). Directly over A gives ABAB… = hcp.
3 Both are close packed. CN = 12 and PF = 0.7405 either way, because each sphere touches 6 in its own layer, 3 above and 3 below. Mistakes in the sequence are stacking faults (Lecture 21).
Part 2 · Close Packing · 3D

FCC Is Cubic Close Packing: Finding the ABC Layers

The ABC layers are hidden inside the FCC cube, tilted. Every atom whose fractional coordinates add up to the same integer, \(x + y + z = n\), lies on the same (111) plane. Colour the atoms by plane and pull the planes apart.

FCC celldrag to rotate
Separate layers
1
Close-packed planes. In FCC the close-packed layers are the {111} planes. There are 4 such families: (111), (1̅11), (11̅1), (111̅).
2
Stacking direction. Along the body diagonal ⟨111⟩, the planes \(x+y+z = 0, 1, 2, 3\) (fractional coordinates) are A, B, C, A. Plane 1, for example, holds the corners (1,0,0), (0,1,0), (0,0,1) and the face centres (½,½,0), (½,0,½), (0,½,½). The corner atom at the origin is a one-atom A layer; the opposite corner closes the next A.
3
Why it matters. Four families of close-packed planes give many slip systems (12 of type {111}⟨110⟩), so FCC metals are ductile (Lecture 21). HCP has only one close-packed plane, the basal plane.
Part 2 · Metals

Packing Fraction of FCC (ccp)

In FCC the corner atoms do not touch each other; the face-centre atom pushes them apart. Contact runs along the face diagonal. Step through the derivation, then rotate the cell to count the 12 neighbours.

A cube face: atoms touch along the face diagonal
1
Contact along the face diagonal: \(\sqrt2\,a = 4r \;\Rightarrow\; a = 2\sqrt2\,r\)
2
Atoms per cell: \(n = 8\times\tfrac18\ (\text{corners}) + 6\times\tfrac12\ (\text{faces}) = 4\)
3
Packing fraction: \(\text{PF} = \dfrac{4\times\tfrac43\pi r^3}{(2\sqrt2\,r)^3} = \dfrac{\pi}{3\sqrt2} = 0.7405\)
4
Coordination number: 12 = 4 neighbours in each of the three {100} planes through an atom.
5
Examples: Cu, Al, Ni, Ag, Au, Pt, Pb, γ-Fe (912–1394 °C).
3D

Count the 12 Neighbours

drag to rotate

Pick a {100} plane through the face-centred atom: each one holds 4 of its neighbours.

Part 2 · Metals · 3D

The Hexagonal Close-Packed (HCP) Cell

The hexagonal prism is drawn because it shows the 6-fold symmetry, but it is three primitive cells. The true unit cell is the 120° rhombic prism, holding 2 atoms.

drag to rotate
1
Lattice + basis. A hexagonal P lattice with a 2-atom basis: one atom of layer A at (0,0,0) and one of layer B at (⅓,⅔,½). HCP is not a Bravais lattice, because the A and B atoms have differently oriented surroundings.
2
Counting atoms. Prism: 12 corners × ⅙ + 2 face centres × ½ + 3 inside = 6 atoms = 3 primitive cells × 2.
3
Ideal geometry. \(a = 2r\), \(c/a = \sqrt{8/3} = 1.633\), CN 12, PF 0.7405 (the same as FCC).
A corner of a hexagonal prism is shared by 6 prisms (3 around it in the plane × 2 above and below), hence ⅙. The top and bottom face centres are shared by 2.
Part 2 · Metals

HCP: Deriving c/a and the Packing Fraction

The B atom rests in the hollow of three touching A atoms. Find how high it sits and the whole cell follows.

Top view
Side view
1
Top view: the B atom (gold) sits over the centroid of three touching A atoms. Its horizontal distance from each is \(t = \tfrac23\times\text{median} = \tfrac23\cdot\tfrac{\sqrt3}{2}(2r) = \dfrac{2r}{\sqrt3}\).
2
Side view: B touches each A atom (distance \(2r\)), so its height is \(h = \sqrt{(2r)^2 - t^2} = 2\sqrt{2/3}\,r\), the height of the A–B tetrahedron.
3
The next A layer is another \(h\) up: \(c = 2h = 4\sqrt{2/3}\,r\) and \(a = 2r\) ⇒ \(c/a = \sqrt{8/3} = 1.633\).
4
Base hexagon: area \(= 6\times\tfrac{\sqrt3}{4}(2r)^2 = 6\sqrt3\,r^2\); volume \(V = 6\sqrt3\,r^2\times c = 24\sqrt2\,r^3\).
5
\(n = 6\) atoms per prism ⇒ \(\text{PF} = \dfrac{6\times\tfrac43\pi r^3}{24\sqrt2\,r^3} = \dfrac{\pi}{3\sqrt2} = 0.7405\).
Real HCP metals

How Ideal Are They?

Click a bar. Most HCP metals lie close to the ideal 1.633; Zn and Cd are stretched along c, so their atoms are not ideal hard spheres.

Part 2 · Metals

Four Metallic Structures at a Glance

Click a row to load the structure. The Wyckoff column is a reminder of the ITC route: put one atom on the listed site, apply the centring translations and site operations, and the whole cell follows.

StructureAtoms / cella in terms of rCNPFSpace group, WyckoffExamples
Simple cubic1a = 2r60.524Pm3̅m (221), 1aα-Po
BCC2a = 4r/√380.680Im3̅m (229), 2aα-Fe, Cr, Mo, W, Na, K
FCC (ccp)4a = 2√2 r120.740Fm3̅m (225), 4aCu, Al, Ni, Ag, Au, γ-Fe
HCP6 per prism (2 per primitive cell)a = 2r, c = 1.633a120.740P63/mmc (194), 2cMg, Ti, Zn, Co, Zr
Packing fraction
drag to rotate
Close packing is the densest. FCC and HCP reach the maximum packing fraction for equal spheres, \(\pi/(3\sqrt2) = 0.7405\) (Kepler’s conjecture, proved by Hales). Most metals adopt FCC, HCP or BCC (PF 0.68 to 0.74).
Part 2 · Worked Example

Does the Hard-Sphere Model Give the Right Density?

Copper is FCC with atomic radius \(r = 0.128\) nm and molar mass \(A = 63.55\) g mol−1. Predict its density and compare it with the measured 8.96 g cm−3. Then try other metals.

\[ \rho = \frac{n\,A}{V_c\,N_A} \]
1
\(n = 4\) atoms per cell (FCC)
2
\(a = 2\sqrt2\,r = 2\sqrt2\times0.128\ \text{nm} = 0.362\ \text{nm}\)
\(\Rightarrow\; V_c = a^3 = 4.74\times10^{-23}\ \text{cm}^3\)
3
\(\rho = \dfrac{4\times63.55\ \text{g mol}^{-1}}{4.74\times10^{-23}\ \text{cm}^3\times6.022\times10^{23}\ \text{mol}^{-1}}\)
4
\(\rho \approx 8.90\ \text{g cm}^{-3}\) (measured 8.96 g cm−3: within 1%)
Consistency check. The atomic radius (itself deduced from X-ray lattice parameters) and the FCC structure reproduce the measured density to within 1%.
Run it backwards. Measured density + structure → lattice parameter or atomic radius.
Ionic solids too. Same method: count formula units per cell (4 for NaCl) and use the formula mass.
Calculator

Density from Structure

n
a (nm)
Vc (cm³)
ρ (g cm−3)
–

For HCP the calculator assumes ideal \(c/a = 1.633\), so \(V = 24\sqrt2\,r^3\) per 6-atom prism.

3
Part 3 · Interstitial Voids

The Holes Between the Spheres

Even the densest packing leaves 26% of space empty. That space is not random: it comes as a small number of well-defined holes (interstitial voids) with fixed shapes, sizes and positions. Small atoms dissolve in them, and ionic crystals are built by filling them.

Part 3 · Interstitial Voids · 3D

Holes Between Close-Packed Layers

Put layer B on layer A and look between them. Only two kinds of hole appear, named after the polyhedron formed by the spheres around them.

drag to rotate
1
Tetrahedral void. Enclosed by 4 spheres: 3 in one layer + 1 in the next. Its centre lies a quarter of the way up the tetrahedron. A small hole.
2
Octahedral void. Enclosed by 6 spheres: 3 below + 3 above, the two triangles rotated by 60°. Its centre lies halfway between the layers. A larger hole.
3
The third layer over the octahedral voids occupies the new position C: ABC stacking, FCC.
4
The third layer over the tetrahedral voids sits directly above layer A: AB stacking, HCP. In both cases the voids are the same kinds; only their arrangement differs.
Every pair of close-packed layers has, per sphere in a layer, 1 octahedral void and 2 tetrahedral voids (one pointing up, one pointing down). That count holds for FCC and HCP alike.

Where exactly does layer B sit? Two fixed views

The same two layers seen from directly above and from 35° above the layer plane. Layer B is drawn see-through so the A atoms underneath stay visible. Step through to find each kind of void.

Part 3 · Interstitial Voids · 3D

Voids in the FCC Cell: Where Each One Is Formed

The same two holes, now located in the cubic cell. Click any void marker to see the polyhedron of atoms around it.

drag to rotate · click a void
Count per cell
Click a void.
1
Octahedral: 4 per cell. The body centre (enclosed by the 6 face-centre atoms) + 12 edge centres × ¼. Wyckoff 4b. 4 voids for 4 atoms: 1 per atom.
2
Tetrahedral: 8 per cell. One in each ⅛-cube, at (¼,¼,¼) and equivalents, enclosed by 1 corner + 3 face-centre atoms. Wyckoff 8c. 2 per atom.
3
Shared voids. An edge-centre octahedron is completed by atoms of neighbouring cells, so only ¼ of it belongs to this cell. Click an edge-centre void to see its octahedron sticking out of the cell.
Part 3 · Interstitial Voids · 3D

Voids in the HCP Cell: Where Each One Is Formed

In HCP the voids line up in columns along c. Toggle each kind and click a marker to see its polyhedron.

drag to rotate · click a void
Count per prism
Click a void.
1
Octahedral: 6 per prism. Over the C positions (neither A nor B) at \(z = c/4\) and \(3c/4\), between 3 A and 3 B atoms. They form columns of face-sharing octahedra along c. Wyckoff 2a.
2
Tetrahedral: 12 per prism. Face-sharing pairs along the A columns (at \(3c/8\), \(5c/8\)) and along the B columns (at \(c/8\), \(7c/8\)). Wyckoff 4f.
3
Per atom: 1 octahedral + 2 tetrahedral, with \(r/R = 0.414\) and \(0.225\): exactly the same as FCC, because both are built from the same pairs of close-packed layers.

Heights measured from the bottom A layer (A at 0 and c, B at c/2).

Part 3 · Interstitial Voids

How Big Are the Holes? The Largest Sphere That Fits

Grow a guest sphere in each hole and watch for the moment it touches its neighbours. Host atoms have radius \(R\); the guest has radius \(r\).

Octahedral void: along a cube edge
Guest r/R
Along the edge, the void touches both corner atoms: \(a = 2R + 2r\), and \(a = 2\sqrt2\,R\) (FCC contact) \(\Rightarrow\; r/R = \sqrt2 - 1 = 0.414\)
Tetrahedral void: corner → (¼,¼,¼) along ⟨111⟩
Guest r/R
Corner atom to void centre: \(R + r = \sqrt3\,a/4 = \sqrt3(2\sqrt2R)/4 = \sqrt{3/2}\,R \;\Rightarrow\; r/R = \sqrt{3/2} - 1 = 0.225\)
Per atom in FCC and HCP: 1 octahedral void (\(r = 0.414R\)) + 2 tetrahedral voids (\(r = 0.225R\)). Remember these two numbers: they reappear as Pauling’s critical radius ratios in Part 4.
Part 3 · Interstitial Voids · 3D

Voids in the BCC Cell: More of Them, but Smaller and Distorted

BCC is less densely packed than FCC, so you might expect bigger holes. The opposite is true: the empty space is split into many small, irregular holes. Click a void to measure its neighbours.

drag to rotate · click a void
Selected void
Click a void.
1
Octahedral: 6 per cell. Face centres (6 × ½) + edge centres (12 × ¼). The nearest atoms are 2 at \(a/2\) (along one axis) and 4 at \(a/\sqrt2\): a squashed octahedron. \(r = a/2 - R\) with \(a = 4R/\sqrt3\) ⇒ \(r/R = 2/\sqrt3 - 1 = 0.155\). Wyckoff 6b.
2
Tetrahedral: 12 per cell. 4 on each face × 6 faces × ½, e.g. at (½,¼,0). All 4 neighbours at \(\sqrt5\,a/4\), so \(R + r = \sqrt5\,a/4\) ⇒ \(r/R = \sqrt{5/3} - 1 = 0.291\). Wyckoff 12d.
3
Both distorted. The polyhedra are not regular (the tetrahedron has two edges of \(a\) and four of \(\sqrt3a/2\)). In BCC the tetrahedral void is the larger one.
Part 3 · Why It Matters

Why Void Size Matters: Carbon in Iron

Steel is iron with a little carbon in the voids. Carbon is too big for every hole in iron, so how much dissolves depends on which structure offers the least-bad hole.

Drawn to scale

Hole Sizes in Iron (R = 0.124 nm)

Solubility follows the largest hole. Up to 2.14 wt% C dissolves in γ-Fe (FCC), but only 0.022 wt% in α-Fe (BCC), even though BCC is the less densely packed structure.
Where C actually sits in α-Fe. In the octahedral sites, even though they are smaller: C only has to push apart the 2 nearest Fe atoms along one cube axis.
Animated

From BCC to Tetragonal Martensite

BCC iron, (010) projection
wt% C
c/a ≈ 1 + 0.045 × wt% C
1.000
Structure
BCC

All the C atoms sit in octahedral sites on edges parallel to one axis (vertical here). Each pushes its two Fe neighbours apart along that axis only. This one-axis strain is why martensite is tetragonal (body-centred tetragonal) rather than cubic.

Part 3 · Interstitial Voids · 3D

Voids in the Simple Cubic Cell: Only One Kind

Simple cubic is so open that its single hole is almost as large as an atom.

drag to rotate
1 Cubic void: 1 per cell. At (½,½,½), Wyckoff 1b, enclosed by all 8 corner atoms at \(\sqrt3a/2\). 1 per atom.
2 Size. \(\sqrt3a/2 = R + r\) with \(a = 2R\) ⇒ \(r/R = \sqrt3 - 1 = 0.732\): the largest hole of any of the four structures.
3 No tetrahedral or octahedral voids. The cubic hole is the only one. In CsCl, Cs+ fills it in a simple-cubic Cl− array (switch to the CsCl view).
Part 3 · Summary

Voids in SC, BCC, FCC and HCP at a Glance

StructureAtoms per cellOctahedral voids per cell (per atom)r/RWyckoffTetrahedral voids per cell (per atom)r/RWyckoff
SC1none; the only hole is the cubic void: 1 per cell (1 per atom), r/R = 0.732, Wyckoff 1bnone
BCC26 (3), distorted0.1556b12 (6), distorted0.29112d
FCC44 (1)0.4144b8 (2)0.2258c
HCP6 per prism6 (1)0.4142a12 (2)0.2254f
Largest guest that fits, r/R
Which voids do ions fill? NaCl: all octahedral voids of FCC Cl−. ZnS: half the tetrahedral voids of FCC S2−. CsCl: the cubic void of SC Cl−. (Part 5.)
And small atoms in metals. H, C, N and B dissolve interstitially in these voids (Lecture 20); how much dissolves depends on the hole size.
In-Class Exercise

Your Turn: Aluminium and Magnesium

Type your answers and press Check. Answers within about 2% are accepted. The worked solutions are below.

Question 1

Aluminium is FCC with \(a = 0.405\) nm (\(A = 26.98\) g mol−1). Find (a) the atomic radius \(R\), (b) the density, (c) the radius of the largest atom that fits in an octahedral void, and in a tetrahedral void.

0 / 4
?Show the worked answer

(a) Contact along the face diagonal: \(R = \sqrt2\,a/4 = 0.143\) nm.

(b) \(\rho = \dfrac{4\times26.98}{(0.405\times10^{-7}\ \text{cm})^3\times6.022\times10^{23}} = 2.70\) g cm−3 (measured 2.70).

(c) Octahedral: \(r = 0.414R = 0.059\) nm; tetrahedral: \(r = 0.225R = 0.032\) nm.

Question 2

Magnesium is HCP with \(c/a = 1.624\). Compared with ideal hard spheres (1.633), is the structure slightly stretched or slightly squashed along c?

Pick one
?Show the worked answer

\(c/a = 1.624 < 1.633\): slightly squashed along c.

The atoms are not perfect hard spheres; real bonding makes the in-plane and out-of-plane spacings differ a little. Zn and Cd (\(c/a \approx 1.86\)–\(1.89\)) are strongly stretched.

Lecture 23 · Summary

Take-Aways: Bonds, Packing and Voids

Bonds

Ionic and metallic bonding is non-directional, so atoms behave as hard spheres. Covalent bonds are directional and give open structures.

Packing

SC 0.52 · BCC 0.68 · FCC = HCP 0.74. FCC and HCP differ only in stacking: ABC vs AB.

Close-packed voids

FCC and HCP: per atom 1 octahedral (\(r = 0.414R\)) + 2 tetrahedral (\(r = 0.225R\)) voids.

BCC and SC voids

BCC: 3 octahedral (0.155R) + 6 tetrahedral (0.291R) per atom, both distorted. SC: 1 cubic void (0.732R).

Why it matters

Hole sizes control which atoms dissolve interstitially (C in γ-Fe vs α-Fe) and which ions fit where (next).

4
Part 4 · Lecture 24 · Pauling’s Rules

Ionic Crystals: Pauling’s Approach

Linus Pauling (1929) condensed the structures of ionic crystals into five rules. They are not laws, but they predict the coordination of most ionic solids from nothing more than ionic radii and charges.

Part 4 · Pauling’s Rules

Assumptions and the Five Rules

Assumptions
  • Ions are hard spheres with fixed ionic radii and formal charges (Na+, Cl−, Ti4+ …).
  • Anions are usually larger: they form the framework, and each cation sits in a coordination polyhedron of anions.
  • Electrical neutrality must hold locally, around every ion, not only for the crystal as a whole.

A cation (+) inside its polyhedron of anions (−)

Part 4 · Rule 1 · 3D Animation

Rule 1: The Radius Ratio Sets the Cation Coordination Number

A cation wants as many anion neighbours as possible (more attraction), but it must touch them. Shrink the cation and the anions close in until they touch each other; shrink it further and it rattles inside its cage, and the structure is unstable.

drag to rotate
r / R
1
Radius ratio. \(\rho = r_\text{cation}/R_\text{anion}\). Each CN is stable down to a critical \(\rho\).
2
At the critical ratio the anions touch the cation and each other. Octahedron: \(\sqrt2 - 1 = 0.414\).
3
Use it as a guide. Real CNs also depend on covalency and on the radius table used; check borderline cases.
Part 4 · Rule 1

Critical Radius Ratios: Where They Come From

At the critical ratio, the distance from the cation centre to an anion centre, \(R + r\), equals the circumradius of a polyhedron whose edge is \(2R\) (anions touching). Click a row to load it into the 3D model above at exactly the critical ratio.

Cation CNPolyhedronContact condition (anions just touching)Stable range of r/RExample
2Linear—< 0.155—
3Triangle\(R + r = 2R/\sqrt3\)0.155 – 0.225—
4Tetrahedron\(R + r = \sqrt{3/2}\,R\)0.225 – 0.414ZnS (zinc blende)
6Octahedron\(R + r = \sqrt2\,R\)0.414 – 0.732NaCl, MgO
8Cube\(R + r = \sqrt3\,R\)0.732 – 1.0CsCl, CaF2
12Cuboctahedron (close packing)\(R + r = 2R\)≥ 1.0Ba2+ in BaTiO3
Reading the table. If \(r/R\) falls in a range, that CN is the highest one whose polyhedron the cation can still fill. A larger CN would leave the cation rattling; a smaller one wastes possible attractive neighbours.
Same numbers as the voids. 0.414 and 0.225 are exactly the octahedral and tetrahedral hole sizes in close packing (Part 3), and 0.732 is the cubic hole of SC.
Part 4 · Rule 2

Rule 2: Electrostatic Valence Fixes the Anion CN

A cation of charge \(z_c\) with \(\text{CN}_c\) neighbours sends a fraction of its charge along each bond: the electrostatic bond strength. For local neutrality, the bonds arriving at every anion must exactly cancel its charge.

Drawn flat; in 3D the cations around the anion form a polyhedron (an octahedron for NaCl).

\[ s = \frac{z_\text{cation}}{\text{CN}_\text{cation}} \qquad \sum_\text{bonds to anion} s = |z_\text{anion}| \]
\[ \Rightarrow\; \text{CN}_\text{anion} = \frac{|z_\text{anion}|}{s} \quad\text{(one kind of cation)} \]
Calculator
Result
CrystalCation CN (rule 1)sAnion CN
NaCl61/61 ÷ 1/6 = 6
CsCl81/81 ÷ 1/8 = 8
ZnS42/4 = 1/22 ÷ 1/2 = 4
MgO62/6 = 1/32 ÷ 1/3 = 6
CaF282/8 = 1/41 ÷ 1/4 = 4
Part 4 · Rule 3 · 3D Animation

Rule 3: Corners Are Shared More Readily than Edges or Faces

Neighbouring polyhedra must share something (every anion belongs to several of them). Watch how the two cations at the centres approach each other as the polyhedra share a corner, then an edge, then a face.

drag to rotate
Cation–cation distance (d = cation–anion distance)
1 The rule. Shared edges, and especially shared faces, lower the stability of an ionic structure.
2 Why. Sharing pulls the two cations closer: 2.00d → 1.41d → 1.15d for octahedra (2.00d → 1.15d → 0.67d for tetrahedra), so the cation–cation repulsion rises.
3 Not absolute. The effect is weak for low-charge cations: the NaCl6 octahedra in rock salt share all 12 edges. It is strong for Si4+, Ti4+.
Part 4 · Lecture 25 · Rules 4 and 5

Rules 4 and 5

4

High-charge, low-CN cations avoid sharing

In a crystal with different cations, those with high valence and small CN tend not to share polyhedron elements with each other. Example: in silicates, SiO4 tetrahedra share only corners, never edges or faces.

corner-sharing SiO4 tetrahedra (a chain silicate)
5

Parsimony

The number of essentially different kinds of constituents (sites and polyhedra) in a crystal tends to be small. Example: perovskite BaTiO3 has just two cation environments, TiO6 octahedra and BaO12 cuboctahedra. Try the explorer below.

two environments in BaTiO3
How to use the five rules
1

Rule 1 → the cation polyhedron (CN of the cation)

2

Rule 2 → how many polyhedra meet at each anion (CN of the anion)

3

Rules 3–5 → how the polyhedra link: corners preferred, few kinds of sites

4

Then look up the space group: the ITC page gives every ion position from a few Wyckoff sites

Part 4 · Rule 5 · Interactive

Rule 5 in Practice: Count the Kinds of Site

Parsimony means economy: a crystal tends to use as few kinds of local environment as it can. What is counted is not the number of elements, but the number of essentially different coordination polyhedra each ion sits in. Pick a structure, predict, then reveal.

Predict

Distinct sites?crystallographic (Wyckoff) sites, all ions
Cation environments?kinds of polyhedron, counted per cation

Sites are not kinds

Kyanite has 16 distinct sites and sillimanite only 7, yet kyanite is the more parsimonious: all four of its Al sites are AlO6 octahedra. “Essentially different” means a different polyhedron, not a slightly distorted copy of the same one.

Why it usually holds

Rules 1–4 pick one best polyhedron for each ion and one best way of linking it. Repeating that best choice everywhere gives the lowest energy; every extra kind of site puts some ions in a worse one.

When it gives way

Something outside the hard-sphere picture tips the balance: crystal-field energy of d electrons in Fe3O4, or temperature and pressure among the Al2SiO5 polymorphs. Rule 5 is a tendency, the weakest of the five.

Use Rule 5 as a plausibility check, not a prediction: if a proposed structure puts the same ion into several different polyhedra, look for the reason (d-electron energy, pressure, temperature, covalency) before believing it.
5
Part 5 · Ionic Structures

Close-Packed Anions + Filled Holes

Most simple ionic crystals can be read as a close-packed array of the big ions with the small ions sitting in some fraction of the octahedral or tetrahedral holes. Pauling’s rules tell us which holes, and the space group gives every position.

Part 5 · Ionic Structures

Many Ionic Structures = Close-Packed Anions + Filled Holes

Click a row to build that structure in the viewer. Colours: anions are the large spheres; cations are the small ones.

StructureClose-packed arrayHoles filledCN (cation : anion)Examples
Rock salt (NaCl)Cl− fccall octahedral6 : 6NaCl, MgO, NiO, FeO
Zinc blende (ZnS)S2− fcc½ tetrahedral4 : 4ZnS, GaAs, CdTe, SiC (3C)
Wurtzite (ZnS)S2− hcp½ tetrahedral4 : 4ZnO, GaN, AlN
Spinel (MgAl2O4)O2− fcc⅛ tetrahedral (Mg) + ½ octahedral (Al)Mg 4, Al 6 : O 4MgAl2O4, Fe3O4, NiFe2O4
Perovskite (BaTiO3)Ba2+ + O2− ccp¼ octahedral (by Ti4+)Ti 6, Ba 12 : O 6BaTiO3, SrTiO3, PbTiO3
Fluorite (CaF2)Ca2+ fccall tetrahedral (by F−)8 : 4CaF2, ZrO2 (cubic), UO2
Corundum (Al2O3)O2− hcp⅔ octahedral6 : 4Al2O3, Cr2O3, Fe2O3
CsCl is the exception: Cl− is simple cubic (not close packed); Cs+ fills the cubic void (CN 8). In perovskite the large Ba2+ joins O2− in building the close-packed layers. In fluorite it is the cations that are close packed.
drag to rotate · click an ion
Selected ion
Click an ion to see its neighbours.
Part 5 · Rock Salt

Rock Salt (NaCl): Apply Rules 1 and 2

Before building anything, predict. The SVG is a (100) layer of NaCl drawn to scale; drag the ratio to see what changes when the cation is too small.

1
Rule 1. \(r(\text{Na}^+) = 1.02\) Å, \(r(\text{Cl}^-) = 1.81\) Å ⇒ \(r/R = 0.56\), between 0.414 and 0.732 ⇒ Na+ CN = 6 (octahedron).
2
Rule 2. \(s = 1/6\) per Na–Cl bond; Cl− needs \(\sum s = 1\) ⇒ Cl− CN = 6 (an octahedron too).
3
Check sizes. Ions touch along the cube edge: \(a = 2(1.02 + 1.81) = 5.66\) Å; measured 5.64 Å ✓
4
Anions apart. Cl–Cl distance \(= a/\sqrt2 = 3.99\) Å \(> 2\times1.81 = 3.62\) Å: the anions do not touch, as rule 1 requires for \(r/R > 0.414\) ✓
r / R

A (100) layer drawn to scale: ions touch along the rows, Cl− do not touch each other. 4 neighbours in the plane + 1 above + 1 below = 6.

Part 5 · Rock Salt · 3D

Rock Salt: Cl− FCC with Na+ in Every Octahedral Void

drag to rotate · click an ion
1
Positions. Cl− at 4a (0,0,0) forms the FCC array; Na+ at 4b (½,½,½) sits in its octahedral voids.
2
4 NaCl per cell. Two interpenetrating FCC arrays, shifted by (½,0,0). Either array alone is a perfect FCC lattice.
3
Coordination 6 : 6. NaCl6 octahedra share all their edges with neighbouring octahedra: allowed for a +1 cation (rule 3 is weak at low charge).
Selected ion
Click an ion.
Part 5 · Rock Salt · ITC

NaCl from Its ITC Page: Fm3̅m (No. 225)

The space group symbol already tells us most of the story. F: face-centred, so every position comes with the translations (0,0,0)+ (0,½,½)+ (½,0,½)+ (½,½,0)+. m3̅m: full cubic point symmetry (Oh). Apply the translations one at a time.

Fm3̅mOh5   m3̅m   Cubic   No. 225
Coordinates: (0,0,0)+  (0,½,½)+  (½,0,½)+  (½,½,0)+
IonWyckoffSite sym.RepresentativeAll 4 positions in the cell
Cl−4am3̅m0, 0, 0
Na+4bm3̅m½, ½, ½
Reading it. Cl− on 4a + the four F-translations → the Cl− FCC array. Na+ on 4b: (½,½,½) plus the same translations → the body centre and three edge centres (each edge centre is shared by 4 cells).
Same structure type. MgO, NiO, FeO, CaO, LiF, KCl, and carbides such as TiC.
Why 4a and 4b? Both are special positions of the highest site symmetry, as expected for ions sitting at the centres of perfect octahedra.
start: one Cl⁻ at 4adrag to rotate
Part 5 · Caesium Chloride · 3D

Caesium Chloride: CN 8 : 8, and Not BCC

drag to rotate · click an ion
1
Rule 1. \(r(\text{Cs}^+)/r(\text{Cl}^-) = 1.74/1.81 = 0.96 > 0.732\) ⇒ Cs+ CN 8 (cube).
2
Rule 2. \(s = 1/8\); Cl− needs \(\sum s = 1\) ⇒ Cl− CN 8.
3
ITC: Pm3̅m (221). Cl− 1a (0,0,0), Cs+ 1b (½,½,½). A simple cubic lattice with a 2-ion basis. Also CsBr, CsI, β-brass (CuZn).
Why not BCC? In BCC the body centre is a lattice translation: the atom there is identical to the corner atom. Here Cs+ ≠ Cl−, so (½,½,½) is not a translation of the crystal. Press “Make both ions identical” to see the BCC case.
Part 5 · Zinc Blende · 3D

Zinc Blende (ZnS): Zn2+ in Half the Tetrahedral Voids

drag to rotate · click an ion
1
Rule 1. \(r(\text{Zn}^{2+}) = 0.60\) Å (CN 4), \(r(\text{S}^{2-}) = 1.84\) Å ⇒ \(r/R = 0.33\) ⇒ CN 4 (tetrahedron).
2
Rule 2. \(s = 2/4 = \tfrac12\); S2− needs \(\sum s = 2\) ⇒ S2− CN 4.
3
Space group F4̅3m (216). S2− 4a (0,0,0), Zn2+ 4c (¼,¼,¼). The sites 4d (the other half of the tetrahedral voids) and 4b (the octahedral voids) stay empty.
Inversion test
Press “Try inversion”: every ion moves through the centre of the cell to the opposite side.
F4̅3mNo. 216
WyckoffSite sym.RepresentativeWhat sits there in ZnS
4a4̅3m0, 0, 0S2− (the fcc array)
4b4̅3m½, ½, ½empty (octahedral voids)
4c4̅3m¼, ¼, ¼Zn2+ (half of the tetrahedral voids)
4d4̅3m¾, ¾, ¾empty (the other half)
Either labelling works. Putting Zn at 4a and S at 4c gives the same structure, with the origin shifted by (¼,¼,¼).
Relation to diamond. Make both sites carbon: the structure gains extra symmetry and becomes diamond, Fd3̅m (No. 227), 8 atoms per cell.
No inversion centre. Filling only every other tetrahedral void removes the centre of symmetry: point group 4̅3m. So ZnS is non-centrosymmetric and piezoelectric, while NaCl (m3̅m) cannot be. Structure decides properties (Part 6).
Part 5 · Wurtzite · 3D

Wurtzite: the HCP Cousin of Zinc Blende

drag to rotate · click an ion
1
Recipe. S2− hcp; Zn2+ in the half of the tetrahedral voids that all point the same way. Space group P63mc (186): S and Zn both on 2b, Zn raised by \(u \approx 3/8\) of c.
2
Zinc blende vs wurtzite. The same ZnS4 tetrahedra and CN 4 : 4 (rules 1 and 2 as before); the only difference is ABC vs AB stacking of the S layers. SiC forms many such stacking variants (polytypes).
3
Polar c axis. Every tetrahedron points up: point group 6mm, piezoelectric and pyroelectric. AlN and ZnO films are used as piezoelectrics (e.g. RF filters in phones).
Part 5 · Spinel · 3D

Spinel AB2O4: Two Kinds of Holes Filled at Once

The spinel cell is big: 32 O2− + 8 Mg2+ + 16 Al3+ = 56 ions. Build it up one kind of ion at a time.

drag to rotate · click an ion
1
Recipe. O2− ccp (32 per cell). A in ⅛ of the tetrahedral voids (8a), B in ½ of the octahedral voids (16d). Fd3̅m (227): 8 AB2O4 = 56 ions per cell.
2
Rule 2 works. Each O2−: 1 Mg2+ (\(s = 2/4 = \tfrac12\)) + 3 Al3+ (\(s = 3/6 = \tfrac12\)) = 2 ✓. Rule 1 is unreliable here: \(r(\text{Al}^{3+})/r(\text{O}^{2-}) = 0.38\) suggests CN 4, yet Al3+ is octahedral.
3
Normal vs inverse. Normal: MgAl2O4, ZnFe2O4. Inverse, B3+[A2+B3+]O4: Fe3O4, NiFe2O4. Spinel ferrites are ferrimagnetic: inductor and transformer cores.
Selected ion
Click an O²⁻ ion to check rule 2.
Part 5 · Perovskite

Perovskite BaTiO3: Apply Rules 1 and 2

1
Rule 1 for Ti4+. \(r(\text{Ti}^{4+}) = 0.605\) Å, \(r(\text{O}^{2-}) = 1.40\) Å ⇒ \(r/R = 0.43\) ⇒ CN 6 (octahedron).
2
Rule 1 for Ba2+. \(r(\text{Ba}^{2+}, \text{CN }12) = 1.61\) Å ⇒ \(r/R = 1.15 \ge 1\) ⇒ CN 12. (With the smaller 1.36 Å radius quoted in some tables, \(r/R = 0.97\) would suggest 8: radii depend on CN.)
3
Rule 2 for O2−. \(s(\text{Ba}) = 2/12 = 1/6\); \(s(\text{Ti}) = 4/6 = 2/3\). Need \(\sum s = 2\): \(4\times\tfrac16 + 2\times\tfrac23 = 2\) ✓ ⇒ each O2− has 4 Ba2+ (a square) + 2 Ti4+ (linear).
4
Rules 3–5. TiO6 octahedra share only corners (the high-charge Ti4+ are kept apart); only two kinds of cation site.
Cell viewed along [100]

Ba2+ at the corners, O2− at the edge midpoints, Ti4+ at the centre (one more O2− directly above and below it).

Part 5 · Perovskite · 3D

Perovskite: a Network of Corner-Sharing TiO6 Octahedra

drag to rotate · click an ion
1
Pm3̅m (221), cubic. Ba2+ 1a (0,0,0); Ti4+ 1b (½,½,½); O2− 3c (0,½,½), (½,0,½), (½,½,0).
2
Coordination. Ti 6 (octahedron), Ba 12 (cuboctahedron), O 6 (4 Ba + 2 Ti).
3
A hint of what is coming. Tolerance factor \(t = \dfrac{r_\text{Ba} + r_\text{O}}{\sqrt2\,(r_\text{Ti} + r_\text{O})} \approx 1.06 > 1\): Ti4+ has room to shift off-centre (ferroelectricity, Part 6).
Tolerance factor calculator
Part 5 · Summary

Six Ionic Structures at a Glance

Click a row to load it. Keep an eye on the point-group column: it decides which properties a crystal can have, which is where we go next.

StructureCation CNAnion CNSpace groupWyckoff sitesPoint groupAlso adopted by
NaCl (rock salt)66Fm3̅m (225)Cl 4a, Na 4bm3̅mMgO, NiO, LiF, TiC
CsCl88Pm3̅m (221)Cl 1a, Cs 1bm3̅mCsBr, CsI, β-brass (CuZn)
ZnS (zinc blende)44F4̅3m (216)S 4a, Zn 4c4̅3mGaAs, InP, CdTe, 3C-SiC
ZnS (wurtzite)44P63mc (186)S 2b, Zn 2b (u ≈ 3/8)6mmZnO, GaN, AlN, CdS
MgAl2O4 (spinel)Mg 4, Al 64Fd3̅m (227)Mg 8a, Al 16d, O 32em3̅mFe3O4, NiFe2O4, ZnFe2O4
BaTiO3 (cubic perovskite)Ti 6, Ba 126Pm3̅m (221)Ba 1a, Ti 1b, O 3cm3̅mSrTiO3, PbTiO3, KNbO3
The recipe. Rule 1 → cation CN; rule 2 → anion CN; rules 3–5 → linking; space group + Wyckoff sites → every ion position.
drag to rotate
In-Class Exercise

Your Turn: Predict the Structure

Use the predictor to run rules 1 and 2 on any binary ionic compound, then check the two exercise answers.

Rules 1 + 2 predictor
r/R
cation CN
s
anion CN

KCl is a good borderline case: \(r/R = 1.38/1.81 = 0.76\) predicts CN 8, yet KCl adopts rock salt. Radius ratios are a guide, not a law.

1Magnesium oxide, MgO: \(r(\text{Mg}^{2+}) = 0.72\) Å, \(r(\text{O}^{2-}) = 1.40\) Å. Predict the CN of each ion and the structure type.

\(r/R = 0.72/1.40 = 0.51\) → between 0.414 and 0.732 → Mg2+ CN 6.

\(s = 2/6 = \tfrac13\); O2− needs 2 → O2− CN = 6.

6 : 6 with octahedra → rock salt structure, Fm3̅m (as observed; \(a = 4.21\) Å).

2Calcium fluoride, CaF2: \(r(\text{Ca}^{2+}, \text{CN 8}) = 1.12\) Å, \(r(\text{F}^-, \text{CN 4}) = 1.31\) Å. Predict the CN of Ca2+ and of F−.

\(r/R = 1.12/1.31 = 0.85\) → above 0.732 → Ca2+ CN 8 (a cube of F−).

\(s = 2/8 = \tfrac14\); F− needs 1 → F− CN = 4 (a tetrahedron of Ca2+).

Ca2+ fcc with F− in all the tetrahedral voids: the fluorite structure. Load “Fluorite” in the first viewer of this part to see it.

6
Part 6 · Structure → Properties

What the Symmetry Allows

Everything in this course so far pays off here. The point group of a crystal sets hard limits on its physical properties: how many numbers you need to describe its conductivity, and whether it can be piezoelectric, pyroelectric or ferroelectric at all.

Part 6 · 3D Animation

Neumann’s Principle: Properties Inherit the Crystal’s Symmetry

A second-rank property such as electrical conductivity links two vectors, \(\mathbf J = \sigma\,\mathbf E\), and can be drawn as an ellipsoid whose semi-axes are its principal values. Pick a crystal system: its symmetry decides how much freedom the ellipsoid has.

drag to rotate
Independent components of a 2nd-rank property
1
E direction
Angle between E and J
1
The principle. Every symmetry operation of the crystal’s point group must leave each of its physical properties unchanged.
2
2nd-rank properties (\(\sigma\), \(\kappa\), \(\varepsilon\), thermal expansion): 1 independent value (cubic), 2 (tetragonal, hexagonal, trigonal), 3 (orthorhombic), 4 (monoclinic), 6 (triclinic).
Careful with “cubic = isotropic”

A 4th-Rank Property: Young’s Modulus

Cubic crystals are isotropic for 2nd-rank properties only. Elasticity is 4th rank: cubic crystals need 3 constants (\(C_{11}, C_{12}, C_{44}\)) and are elastically anisotropic. The surface shows Young’s modulus in every direction, \(1/E(\mathbf n) = S_{11} - 2\left(S_{11} - S_{12} - \tfrac12 S_{44}\right)(n_1^2n_2^2 + n_2^2n_3^2 + n_3^2n_1^2)\).

E ⟨100⟩
E ⟨111⟩
Zener A

\(A = 2C_{44}/(C_{11}-C_{12})\); \(A = 1\) means isotropic. Elastic constants at room temperature from standard tables.

drag to rotate · red = stiff, blue = compliant
Part 6 · Structure → Properties

Why a Centre of Symmetry Switches Off Piezo- and Pyroelectricity

Inversion takes every vector \(\mathbf r\) to \(-\mathbf r\). Watch what it does to a polarization arrow, and to the coordination polyhedra of NaCl and ZnS.

P
1 Neumann + inversion. If the crystal has a centre of symmetry, every property must be unchanged by inversion: \(\mathbf P = -\mathbf P\), so \(\mathbf P = 0\).
2 Rank rule. Under inversion a polar property tensor of rank \(n\) is multiplied by \((-1)^n\). Odd rank (1: pyroelectric \(\mathbf P\); 3: piezoelectric \(d_{ijk}\)) must vanish; even rank (2: \(\sigma, \varepsilon, \kappa\); 4: elastic) survives.
3 Consequence. Of the 32 point groups, the 11 centrosymmetric ones can never be piezoelectric, pyroelectric or ferroelectric. NaCl, CsCl, spinel and cubic BaTiO3 (all m3̅m): none of these.
3D

Invert a Polyhedron

drag to rotate
r → λ r
Drag λ from +1 to −1
Rank rule: pick a property
Part 6 · Structure → Properties

The 32 Point Groups, Sorted by Property

Each split below removes the groups that cannot show the next property. Click a box in the tree to light up its point groups.

Click a box in the tree or any point group.
centrosymmetric (11)432: non-centrosymmetric but not piezoelectricpiezoelectric, not polar (10)polar = pyroelectric (10)
Part 6 · Structure → Properties

Which of Our Crystals Qualify?

CrystalPoint groupCentre of symmetry?PiezoelectricPyroelectric (polar)Ferroelectric
NaCl, CsCl, MgAl2O4m3̅myes———
ZnS (zinc blende), GaAs4̅3mno✓——
α-quartz (SiO2)32no✓——
ZnO, GaN, AlN (wurtzite)6mmno✓✓— (P cannot be switched)
BaTiO3 at room temperature4mmno✓✓✓
BaTiO3 T (°C)
Symmetry decides what is possible. Only point groups in the right class can show the property; how large it is depends on bonding and composition.
Polycrystals average out. Randomly oriented grains cancel piezoelectricity: a ferroelectric ceramic must be poled (domains aligned by a strong field). Try it in the microstructure section below.
Part 6 · 3D Animation

BaTiO3: a Ferroelectric You Can Switch

Cool the crystal through its Curie temperature and the Ti4+ ion settles off-centre, giving each cell a dipole. Then push it back and forth with an electric field and trace the hysteresis loop.

drag to rotate
free energy F(P)
P–E loop
T (°C)
Field E
Phase
Point group
1
Symmetry lowering. Pm3̅m (cubic, centrosymmetric) → P4mm (tetragonal, polar) on cooling through \(T_c \approx 120\) °C. Then orthorhombic (≈ 5 °C) and rhombohedral (≈ −90 °C).
2
The off-centre Ti4+. Tolerance factor \(t \approx 1.06 > 1\): the TiO6 cage is roomy, so Ti4+ settles off-centre. \(P_s \approx 26\ \mu\text{C cm}^{-2}\) at room temperature.
3
Uses. BaTiO3: the dielectric in multilayer ceramic capacitors (MLCCs). PZT, a related perovskite: sensors, actuators, ultrasound.

The well and the loop come from a simple Landau model, \(F = \alpha(T - T_c)P^2 + \beta P^4 - EP\), in arbitrary units. Real BaTiO3 has a first-order transition, but the picture is the same.

Part 6 · Microstructure → Properties

Same Crystal, Different Behaviour

The crystal structure sets what is possible; the microstructure (Lectures 20–22) tunes how much of it you get.

Feature (Lectures 20–22)How it changes propertiesExample
Grain sizeGrain boundaries block slip: \(\sigma_y = \sigma_0 + k_y d^{-1/2}\) (Hall–Petch). They also scatter phonons.Fine-grained steels; nanostructured thermoelectrics with low thermal conductivity
PrecipitatesDislocations must cut through them or loop around them: precipitation hardeningAge-hardened Al–Cu aircraft alloys
Solutes and point defectsSolid-solution strengthening; dopants and vacancies set ionic and electronic conductionY3+ in ZrO2 makes O2− vacancies: an oxygen-ion conductor
Texture (preferred orientation)Grains aligned → the polycrystal keeps the single crystal’s anisotropyGrain-oriented Fe–Si sheet for transformer cores
Pores and cracksLower strength, stiffness and thermal conductivitySintered ceramics; thermal-barrier coatings
Ferroelectric domainsPoling aligns domains; domain-wall motion adds to the piezoelectric responsePoled PZT ceramics
Animated

Poling a Ferroelectric Ceramic

as sintered: random
Poling field
net polarization0%

Each grain’s polarization can only point along its own allowed directions. A strong field switches each grain to the allowed direction closest to the field; most of that alignment remains when the field is removed.

Wrap-Up

The Course in One Picture

Structure (and how we characterise it) ↔ properties ↔ processing ↔ performance: the materials science tetrahedron from Lecture 0. Every block of this course fills in one part of it.

Summary

Take-Aways: Ionic Crystals and Properties

Pauling’s rules

Radius ratio → cation CN; bond strengths → anion CN; corners shared rather than edges or faces; few kinds of site.

Ionic structures

Close-packed anions + filled holes: NaCl (all oct.), zinc blende and wurtzite (½ tet., ABC vs AB), spinel (⅛ tet. + ½ oct.), perovskite.

Neumann’s principle

A property has at least the point-group symmetry: cubic crystals are isotropic for 2nd-rank properties.

Inversion

Centrosymmetric crystals cannot be piezo-, pyro- or ferroelectric. 20 piezoelectric classes, 10 polar, ferroelectrics among the polar.

Microstructure

Grain size, precipitates, solutes, texture, pores and domains tune properties on top of the crystal structure.

Practice · Systematic Absences

Revision Problem (XRD): Which Reflections Does Tungsten Show?

Tungsten is BCC, space group Im3̅m (No. 229), with W on Wyckoff 2a. Using the reflection conditions on its ITC page, list the reflections present in its powder pattern. Click each family to mark it present or absent, then check.

General condition (I-centring). hkl: \(h + k + l = 2n\). It applies to every Im3̅m crystal, whichever sites are occupied.
Special-position conditions. Extra conditions listed for a special site (e.g. 8c at (¼,¼,¼)) matter only if atoms sit there. W fills only 2a, so \(h + k + l = 2n\) is the only rule.
Note: 8c (¼,¼,¼) is not an interstitial void: it lies on the body diagonal where the corner and body-centre atoms touch. The BCC voids are 6b (octahedral) and 12d (tetrahedral).
In order of increasing \(h^2 + k^2 + l^2\) (cubic: hkl permutable)
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Peak positions for Cu Kα1 (λ = 1.5406 Å), a = 3.165 Å. Heights are not to scale.

Practice Assessment

Did It Sink In? - 23-25

60 questions on bonds and kBT, packing and density, interstitial voids, Pauling’s rules, the common ionic structures, and how symmetry decides piezo-, pyro- and ferroelectricity.

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