Which statement about a lattice point in a crystal is correct?

A mathematical point with surroundings identical to every other lattice point; its motif can be one atom or a group of atoms

It is the centre of an atom, so in every crystal the number of lattice points per cell equals the number of atoms per cell

It is a position occupied by exactly one atom in every crystal structure, with different atoms sitting on different lattice points

A mathematical point with surroundings identical to every other lattice point; its motif can be one atom or a group of atoms

It is the point where two or more atoms touch, so a lattice point exists only where hard spheres are in contact with each other

Why can a periodic crystal not possess a five-fold rotation axis?

A 72° rotation cannot map a translation-periodic lattice onto itself: only angles with 1 − 2cos α an integer are allowed, i.e. 60°, 90°, 120°, 180°, 360°

Five identical atoms cannot be placed at equal distances around a central point in any dimension, so no five-fold arrangement can exist anywhere in matter, whether in a crystal, a molecule or a glass

A five-fold axis would force the unit cell to contain an odd number of lattice points, whereas every conventional lattice cell is built so that it contains an even number of them

A five-fold axis would require the unit cell to contain exactly five atoms, and no crystal structure in nature can have a unit cell holding that many atoms in total

According to the symmetry-based definition of the seven crystal systems, which of the following statements are correct?

A crystal is cubic only if it has four 3-fold rotation axes along the body diagonals of its cell

A hexagonal crystal is characterised by a single 6-fold rotation axis

A crystal is cubic only if it has four 3-fold rotation axes along the body diagonals of its cell

A crystal whose unit cell has a = b = c and α = β = γ = 90° must be cubic, whatever its actual symmetry

A hexagonal crystal is characterised by a single 6-fold rotation axis

A monoclinic crystal has three mutually perpendicular 2-fold axes

Which statement correctly distinguishes hexagonal close packing from cubic close packing of equal spheres?

Both have coordination number 12 and packing fraction 0.74; they differ only in the stacking of the close-packed layers, ABAB… in HCP and ABCABC… in CCP

Hexagonal packing has coordination number 12 whereas cubic packing has coordination number 8, because only the cubic stacking puts layers directly above one another

Cubic packing reaches a packing fraction of 0.74 whereas hexagonal packing reaches only about 0.68, since the hexagonal arrangement is a slightly distorted one

Both have coordination number 12 and packing fraction 0.74; they differ only in the stacking of the close-packed layers, ABAB… in HCP and ABCABC… in CCP

Hexagonal packing is ABCABC… stacking of its basal planes and cubic packing is ABAB… stacking of its {111} planes, but the two are otherwise identical

In an FCC crystal of hard spheres of radius R, the numbers per atom and radii of the voids are

One octahedral void of radius 0.414R and two tetrahedral voids of radius 0.225R

Two octahedral voids of radius 0.414R and one tetrahedral void of radius 0.225R

One octahedral void of radius 0.225R and two tetrahedral voids of radius 0.414R

Three octahedral voids of radius 0.155R and six tetrahedral voids of radius 0.291R

FCC iron (austenite) dissolves far more interstitial carbon than BCC iron (ferrite), although BCC has more voids per atom. Which explanation is correct?

BCC has more voids per atom but all are smaller (largest, tetrahedral, 0.291R) than the FCC octahedral void (0.414R), so carbon strains BCC far more

BCC iron has no voids at all between its atoms, because its coordination number is only 8 and its spheres are not close-packed, so no solute can enter

Carbon is negatively charged and can occupy only tetrahedral voids, which exist in FCC iron but are completely absent from every BCC structure

BCC iron has fewer voids per atom than FCC iron, so the number of available interstitial sites per atom is what limits the carbon that can dissolve

BCC has more voids per atom but all are smaller (largest, tetrahedral, 0.291R) than the FCC octahedral void (0.414R), so carbon strains BCC far more

A cutting-tool material consists of hard tungsten carbide particles embedded in a ductile cobalt matrix. It is best classified as

A composite, because two distinct phases are combined to give a property mix that neither provides alone

A ceramic, because tungsten carbide is a carbide, carbides are ceramic compounds, and the hard carbide particles dominate the cutting action

A composite, because two distinct phases are combined to give a property mix that neither provides alone

A polymer, because it is formed into shape by pressing and heating powders into a bonded mass, as thermoplastics are processed

A single-phase metallic solid solution in which tungsten, carbon and cobalt atoms all share one crystal lattice, with no separate particles

Which of the following statements about the ordered structures of carbon are correct?

Diamond can be described as an FCC lattice with a motif of two carbon atoms

Graphite layers are held together by weak van der Waals forces, which lets the layers slide over one another

In diamond every carbon atom has three nearest neighbours arranged in a plane

Diamond can be described as an FCC lattice with a motif of two carbon atoms

Every single-walled carbon nanotube is metallic, whatever its chiral indices (n, m)

Graphite layers are held together by weak van der Waals forces, which lets the layers slide over one another

Classification of Materials, Crystallography, Close-Packed Structures and Carbon

This chapter belongs to the Engineering Sciences (XE) paper. It answers the first four headings of Section XE2.1 of XE2 Materials Science, one of the nine optional sections, any two of which a candidate chooses in the exam. The syllabus reads: “Classification of Materials: Metals, ceramics, polymers and composites. Fundamentals of crystallography: Definitions of crystal, lattice and motif: Crystal=Lattice+Motif … Fourteen Bravais lattices … Miller and Miller-Bravais indices of crystallographic planes and directions. Close-packed crystal structures of elements: CCP, HCP and BCC structures; Stacking sequence of planes; Tetrahedral and octahedral voids. Crystalline and other ordered structures of carbon: Diamond and Graphite in terms of lattice and motif; Graphene and fullerene; Carbon nanotubes.” The chapter teaches how bonding sorts materials into four classes, how a crystal is a lattice plus a motif, the symmetry that gives 7 crystal systems and 14 Bravais lattices, the index notation for planes and directions, the packing of spheres in FCC, HCP and BCC, the voids between them, and the structures built from carbon. The numericals are planar density, packing factor, density from the lattice parameter, interplanar spacing, void radii, atom counts in fullerenes and nanotube diameter. Compounds, polymers and defects follow in the next two chapters.

1. Classifying materials: metals, ceramics, polymers and composites

Materials are grouped by the bonding between atoms, because bonding fixes how atoms arrange themselves and therefore how the material behaves. Metals have metallic bonding: positive ion cores in a sea of shared, delocalised electrons. The bond is non-directional, so atoms pack densely into close-packed or nearly close-packed crystals; the mobile electrons give high electrical and thermal conductivity and lustre, and planes can slide, which gives ductility. Ceramics are compounds of metals with non-metals (oxides, nitrides, carbides) or covalent solids such as diamond; ionic and covalent bonds are strong and directional, so ceramics are hard, stiff, brittle, high-melting and usually electrical insulators, and they may be crystalline or glassy. Polymers are long chain molecules, mostly carbon with hydrogen and other atoms, joined by strong covalent bonds along the chain but only weak secondary bonds between chains; hence low density, low stiffness, low softening temperature and easy shaping. Composites combine two or more distinct phases, a matrix and a reinforcement, to give a property that neither offers alone; examples are glass-fibre reinforced polymer, cemented carbide (WC particles in cobalt) and concrete.

The four classes at a glance (typical behaviour, not universal rules)
Class	Bonding	Usual structure	Typical behaviour
Metals	Metallic	Crystalline, close-packed (FCC, HCP, BCC)	Conducting, ductile, strong
Ceramics	Ionic and covalent	Crystalline or glassy	Hard, brittle, refractory, insulating
Polymers	Covalent chains, weak secondary links	Amorphous or semi-crystalline chains	Light, compliant, low softening temperature
Composites	Interfaces between unlike phases	Matrix plus reinforcement	Properties tailored by fraction and geometry

⚠️ “Typical” is not “always”

Exam statements are safe only when scoped. Indium tin oxide is a ceramic that conducts, doped polyacetylene is a polymer that conducts, and diamond, a covalent ceramic, is an excellent conductor of heat. Classify by bonding and structure first, and read a property as a tendency.

2. Crystal = lattice + motif, and the symmetry that translation allows

A crystal is a solid whose atoms repeat in space with long-range order. The repetition is captured by two separate ideas. The lattice is an infinite array of points in which every point has identical surroundings. The motif (basis) is the atom or group of atoms attached in the same way to every lattice point. Crystal = lattice + motif. Every lattice point is reached from the origin by a translation R = u a + v b + w c, with integers u, v, w and lattice vectors a, b, c.

A lattice point is not an atom. It is a mathematical point; the motif puts atoms on it, and one lattice point may carry one atom, several atoms or an offset group. Copper is the FCC lattice with a one-atom motif. Diamond is the FCC lattice with a two-atom motif at (0, 0, 0) and (¼, ¼, ¼). Magnesium is the simple hexagonal lattice with a two-atom motif at (0, 0, 0) and (⅓, ⅔, ½). Rock salt is the FCC lattice with Na⁺ at (0, 0, 0) and Cl⁻ at (½, 0, 0); CsCl is the simple cubic lattice with Cs⁺ and Cl⁻ at (0, 0, 0) and (½, ½, ½).

⚠️ Two classic slips

CsCl is not BCC: the corner ion and the centre ion are different, so they are not equivalent lattice points and the lattice is simple cubic with a two-ion motif. HCP is not a Bravais lattice: the two atoms of its motif sit in different environments, so no pure translation maps one onto the other.

A symmetry operation brings a structure into coincidence with itself: translation, rotation (n-fold means 360°/n), reflection in a mirror plane, inversion through a point, and their combinations. Translation is required of every lattice, and it limits the rotations. Take two neighbouring lattice points a apart and rotate each about the other, in opposite senses, by α. The two new points lie on a row parallel to the first with separation a|1 − 2cos α|, which must be a whole multiple of a. So 1 − 2cos α = m (an integer), giving cos α = 1, ½, 0, −½, −1 and α = 360°, 60°, 90°, 120°, 180°. Only 1-, 2-, 3-, 4- and 6-fold axes can occur. Five-fold axes are forbidden in a periodic crystal; the five-fold diffraction symmetry of quasicrystals goes with the absence of translational periodicity.

3. Seven crystal systems, unit cells and the 14 Bravais lattices

A crystal system is defined by its minimum symmetry; the cell shape then follows. A crystal whose cell happens to have a = b = c and 90° angles but which lacks the four 3-fold axes is not cubic.

Crystal systems and Bravais lattices (P primitive, I body-centred, F face-centred, C base-centred, R rhombohedral)
System	Minimum symmetry	Cell constraints	Lattices
Cubic	Four 3-fold axes along body diagonals	a = b = c; 90°, 90°, 90°	P, I, F (3)
Tetragonal	One 4-fold axis	a = b ≠ c; all 90°	P, I (2)
Orthorhombic	Three perpendicular 2-fold axes	a ≠ b ≠ c; all 90°	P, C, I, F (4)
Hexagonal	One 6-fold axis	a = b ≠ c; α = β = 90°, γ = 120°	P (1)
Trigonal (rhombohedral)	One 3-fold axis	a = b = c; α = β = γ ≠ 90°	R (1)
Monoclinic	One 2-fold axis (or one mirror plane)	a ≠ b ≠ c; α = γ = 90° ≠ β	P, C (2)
Triclinic	None (identity or inversion only)	a ≠ b ≠ c; α ≠ β ≠ γ	P (1)

A primitive cell has one lattice point: lattice points only at the corners, each shared by eight cells, so 8 × ⅛ = 1. A non-primitive (conventional) cell has more: body-centring adds one point (I, 2 per cell), face-centring adds six face points each shared by two cells (F, 4 per cell), base-centring adds two shared points (C, 2 per cell). Non-primitive cells are used because they display the full symmetry. The primitive cell of FCC is a rhombohedron of edge a/√2, interior angle 60° and volume a³/4; that of BCC has volume a³/2. Seven systems times four centrings would give 28 lattices, but many are duplicates or lose the defining symmetry: base-centred cubic has lost its 3-fold axes and is simply a simple tetragonal lattice, and face-centred tetragonal is the body-centred tetragonal lattice drawn on a larger cell. That leaves 14.

4. Miller and Miller-Bravais indices, and interplanar spacing

Plane (hkl): take the intercepts on a, b, c in units of the cell edges; take reciprocals; clear fractions to the smallest integers; write in round brackets. A plane parallel to an axis has intercept ∞, so index 0. A negative index (written with a bar over it in print, here a minus sign) means a negative intercept. {hkl} is the family of symmetry-equivalent planes.
Direction [uvw]: components of the vector (head minus tail) in units of a, b, c, reduced to the smallest integers, in square brackets; ⟨uvw⟩ is the family. In cubic crystals only, [hkl] is perpendicular to (hkl). A direction [uvw] lies in the plane (hkl) when hu + kv + lw = 0.
Spacing of the (hkl) planes in a cubic crystal: d = a/√(h² + k² + l²). A plane (hkl) cuts the axes at a/h, b/k, c/l.

Worked example. A plane meets the a axis at 2a, the b axis at 3b and is parallel to c. Intercepts 2, 3, ∞; reciprocals ½, ⅓, 0; multiply by 6: (3 2 0). For a = 0.400 nm, d = 0.400/√(9 + 4) = 0.400/3.606 = 0.111 nm. The same plane written with the intercepts as indices, (2 3 0), is the commonest error.

Hexagonal crystals use four axes: a₁, a₂, a₃ at 120° in the basal plane, and c. Planes are (hkil) with i = −(h + k); directions are [uvtw] with t = −(u + v). The redundant index makes equivalent planes look equivalent: the six prism faces are (1 0 −1 0), (0 1 −1 0), (−1 1 0 0) and their negatives, all of the family {1 0 −1 0}. A three-index direction [UVW] converts as u = (2U − V)/3, v = (2V − U)/3, t = −(u + v), w = W. So the a₁ direction [1 0 0] becomes [2 −1 −1 0] after clearing fractions, and the basal plane is (0 0 0 1).

⚠️ Spacing is not the same as atomic layer spacing

In FCC and BCC the layers of atoms parallel to a cube face are a/2 apart, which is d₂₀₀, not d₁₀₀ = a. The formula gives the spacing of the family you name; it does not say which of those planes are fully populated by atoms. Also remember that (hkl) is a plane, [hkl] a direction, and that perpendicularity of the two holds only in cubic crystals.

5. Close-packed structures: CCP (FCC), HCP and BCC

Treat atoms as hard spheres of radius R touching along the closest-packed direction. In a close-packed layer each sphere touches six others (2D packing fraction π/(2√3) = 0.907). A second layer B sits in half of the triangular hollows of layer A. The third layer has two choices: directly over the first, A, giving ABAB… = HCP; or over the hollows still uncovered, C, giving ABCABC… = CCP, which is the FCC lattice viewed along its [111] body diagonal. Close-packed planes are (0001) in HCP and {111} in FCC; close-packed directions are ⟨1 1 −2 0⟩ in HCP and ⟨110⟩ in FCC. Both have coordination number 12 and packing fraction 0.7405. The ideal HCP axial ratio is c/a = √(8/3) = 1.633; real HCP metals deviate from it.

Hard-sphere cubic and related structures (n = atoms per conventional cell)
Structure	CN	n	Cell edge in terms of R	APF
Simple cubic	6	1	a = 2R	π/6 = 0.524
BCC	8	2	a = 4R/√3 (⟨111⟩ touch)	π√3/8 = 0.680
FCC	12	4	a = 2√2 R (⟨110⟩ touch)	π/(3√2) = 0.740
HCP (hexagonal cell)	12	6	a = 2R, c = 1.633a	0.740
Diamond cubic	4	8	a = 8R/√3	π√3/16 = 0.340

Atomic packing factor APF = n (4πR³/3)/a³. For FCC: n = 4, a = 2√2 R, so APF = (16πR³/3)/(16√2 R³) = π/(3√2) = 0.7405. For BCC: n = 2, a = 4R/√3, APF = (8πR³/3)/(64R³/(3√3)) = π√3/8 = 0.680. BCC is not close-packed: it has 8 nearest neighbours plus 6 next-nearest only 15% farther, and its densest plane is {110}. Density ρ = nA/(V_c N_A), with A in g/mol, V_c the cell volume and N_A = 6.022 × 10²³ mol⁻¹. Iron (BCC, a = 0.2866 nm, A = 55.85 g/mol): ρ = 2 × 55.85/[6.022 × 10²³ × (2.866 × 10⁻⁸ cm)³] = 7.88 g/cm³.

Planar density is atoms (counted by the fraction lying inside the area) per unit area. FCC (111): the triangle of side a√2 has area (√3/2)a² and holds 3 × ⅙ + 3 × ½ = 2 atoms, so PD = 4/(√3 a²) = 2.31/a². BCC (110): the rectangle a × a√2 holds 4 × ¼ + 1 = 2 atoms, so PD = √2/a² = 1.41/a². Linear density along FCC ⟨110⟩ and BCC ⟨111⟩ is 1/(2R), the closest packing.

🎯 Why FCC iron is denser than BCC iron

For hard spheres of the same radius, the volume per atom is (2√2 R)³/4 = 5.657R³ in FCC and (4R/√3)³/2 = 6.158R³ in BCC, so BCC → FCC shrinks the volume by 8.1%. In real iron the change is much smaller, because the effective atomic radius is not the same in the two structures, which is why the equal-radius model is only an estimate.

6. Tetrahedral and octahedral voids

The gaps between touching spheres are voids. A tetrahedral void is surrounded by four atoms (CN 4); an octahedral void by six (CN 6). The void radius r is the largest sphere that fits while touching all its neighbours of radius R. In close-packed structures there is 1 octahedral and 2 tetrahedral void per atom, whether the stacking is FCC or HCP.

Voids in FCC/HCP and BCC
Structure	Void	Per atom	Per cubic cell	r/R
FCC (and HCP)	Tetrahedral	2	8	0.225
FCC (and HCP)	Octahedral	1	4	0.414
BCC	Tetrahedral	6	12	0.291
BCC	Octahedral (flattened)	3	6	0.155

Derivations. FCC octahedral voids sit at the cell centre and the 12 edge midpoints: 1 + 12 × ¼ = 4. Along a cell edge the sequence is atom, void, atom, so 2R + 2r = a = 2√2 R, giving r = (√2 − 1)R = 0.414R. FCC tetrahedral voids sit at (¼, ¼, ¼)-type points, 8 per cell; the distance from the corner atom is √3a/4 = R + r, giving r = (√6/2 − 1)R = 0.225R. In BCC, the octahedral voids lie at face centres and edge midpoints (3 + 3 = 6 per cell); the two nearest atoms (the corner atoms for an edge-midpoint void, two body-centre atoms for a face-centre void) are a/2 from the void, so R + r = a/2 with R = √3a/4, giving r = 0.155R. The tetrahedral voids at (½, ¼, 0)-type points (24 positions each shared by two cells, 12 per cell) are √5a/4 from the nearest atoms, giving r = 0.291R.

🎯 Why the voids matter

Small solute atoms such as carbon sit in voids. FCC iron (austenite) dissolves up to about 2 mass % carbon in its 0.414R octahedral voids; BCC iron (ferrite) dissolves only about 0.02 mass % at most because its voids, though more numerous, are all smaller (largest 0.291R). The same void radii, written as minimum radius ratios of 0.225, 0.414 and 0.732 for CN 4, 6 and 8, reappear as Pauling’s radius-ratio rule for ionic crystals in the next chapter.

7. Carbon: diamond, graphite, graphene, fullerene and nanotubes

Ordered forms of carbon
Form	Bonding	Lattice and motif	Key numbers
Diamond	sp³, four neighbours	FCC lattice; motif C at (0, 0, 0) and (¼, ¼, ¼); 8 atoms per cell	a = 0.3567 nm; C–C = √3a/4 = 0.154 nm; APF 0.34
Graphite	sp²: three σ bonds plus delocalised π electrons	Hexagonal lattice, 4-atom motif; honeycomb layers stacked ABAB	In-plane C–C 0.142 nm; layer spacing 0.335 nm (c ≈ 0.67 nm); van der Waals between layers
Graphene	sp²	One layer: 2D hexagonal lattice, 2-atom honeycomb motif	Lattice constant 0.246 nm
Fullerene C₆₀	sp²-like, curved	Closed molecular cage, a truncated icosahedron: 12 pentagons, 20 hexagons	60 atoms; diameter about 0.7 nm
Carbon nanotube	sp²	Graphene sheet rolled up; chirality (n, m)	d = a√(n² + nm + m²)/π, a = 0.246 nm

Properties follow from structure. Diamond has a rigid three-dimensional covalent network, so it is among the hardest known materials, an electrical insulator and, unusually for an insulator, an excellent thermal conductor. Graphite is strongly bonded within layers and weakly between them: it is anisotropic, conducts well along the layers through the π electrons, and shears easily across them, which is why it works as a solid lubricant. Both are carbon; they differ in lattice and motif and so in every property.

Fullerene cages. Each atom has three bonds, so E = 3V/2. With p pentagons and h hexagons, V = (5p + 6h)/3 and F = p + h. Euler’s relation V − E + F = 2 then gives p/6 = 2, so p = 12 for every closed cage, while h is free: C₆₀ has h = 20 (V = (60 + 120)/3 = 60) and C₇₀ has h = 25 (V = (60 + 150)/3 = 70). Nanotubes are specified by the chiral vector C = n a₁ + m a₂ of the graphene lattice: (n, n) armchair, (n, 0) zigzag, others chiral. The diameter is d = a√(n² + nm + m²)/π; for (10, 10), d = 0.246 × √300/π = 1.36 nm. Ignoring curvature effects, a tube is metallic when n − m is a multiple of 3 and semiconducting otherwise.

Key takeaways

Materials are classified by bonding: metallic (metals), ionic and covalent (ceramics), covalent chains with weak links between them (polymers), and multi-phase combinations (composites). Properties are tendencies, not laws.
Crystal = lattice + motif; a lattice point is a mathematical point, not an atom. Translation symmetry allows only 1-, 2-, 3-, 4- and 6-fold axes, and the minimum symmetry defines the 7 systems, which carry 14 Bravais lattices (3, 2, 4, 1, 1, 2, 1).
Planes (hkl): reciprocal intercepts cleared to integers; directions [uvw]: components; in cubic d = a/√(h² + k² + l²) and [hkl] ⟂ (hkl). Hexagonal: i = −(h + k) and t = −(u + v).
FCC (ABCABC) and HCP (ABAB) both have CN 12 and APF 0.74; BCC has CN 8 and APF 0.68; diamond cubic has CN 4 and APF 0.34. Density is nA/(V_c N_A).
Close-packed structures have 1 octahedral (0.414R) and 2 tetrahedral (0.225R) voids per atom; BCC has 3 octahedral (0.155R) and 6 tetrahedral (0.291R), all smaller.
Diamond is an FCC lattice with a two-atom motif; graphite is a hexagonal lattice with a four-atom motif; graphene is one honeycomb layer. Every closed fullerene has 12 pentagons, and nanotube diameter is a√(n² + nm + m²)/π.

Practice questions (18)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

Which statement about a lattice point in a crystal is correct?
1. It is the centre of an atom, so in every crystal the number of lattice points per cell equals the number of atoms per cell
2. It is a position occupied by exactly one atom in every crystal structure, with different atoms sitting on different lattice points
3. A mathematical point with surroundings identical to every other lattice point; its motif can be one atom or a group of atoms
4. It is the point where two or more atoms touch, so a lattice point exists only where hard spheres are in contact with each other
Show answer
Answer: C — A mathematical point with surroundings identical to every other lattice point; its motif can be one atom or a group of atoms
A lattice is a set of points with identical surroundings; the atoms come from the motif. Diamond has 8 atoms per cubic cell but its FCC lattice has only 4 lattice points per cell, which shows that atoms and lattice points need not be equal in number. The idea that a lattice point is an atom centre is the tempting error.
Why can a periodic crystal not possess a five-fold rotation axis?
1. A 72° rotation cannot map a translation-periodic lattice onto itself: only angles with 1 − 2cos α an integer are allowed, i.e. 60°, 90°, 120°, 180°, 360°
2. Five identical atoms cannot be placed at equal distances around a central point in any dimension, so no five-fold arrangement can exist anywhere in matter, whether in a crystal, a molecule or a glass
3. A five-fold axis would force the unit cell to contain an odd number of lattice points, whereas every conventional lattice cell is built so that it contains an even number of them
4. A five-fold axis would require the unit cell to contain exactly five atoms, and no crystal structure in nature can have a unit cell holding that many atoms in total
Show answer
Answer: A — A 72° rotation cannot map a translation-periodic lattice onto itself: only angles with 1 − 2cos α an integer are allowed, i.e. 60°, 90°, 120°, 180°, 360°
Translation symmetry fixes the allowed rotations through 1 − 2cos α = integer, giving only 1-, 2-, 3-, 4- and 6-fold axes. The distractors invent geometric or counting obstacles that do not exist: five atoms can sit around a point in a ring (as in a pentagon of a fullerene), and lattice cells can contain odd numbers of points, for example 1 in a primitive cell.
Which one of the following is NOT among the 14 Bravais lattices?
1. Face-centred orthorhombic
2. Base-centred cubic
3. Body-centred tetragonal
4. Base-centred monoclinic
Show answer
Answer: B — Base-centred cubic
Centring a pair of faces of a cubic cell destroys the four 3-fold axes, so the lattice is no longer cubic; it is just a simple tetragonal lattice (edge a/√2 in the plane, c = a). The other three are genuine: P, C, I and F orthorhombic all exist, body-centred tetragonal exists, and base-centred monoclinic exists.
According to the symmetry-based definition of the seven crystal systems, which of the following statements are correct?
1. A crystal is cubic only if it has four 3-fold rotation axes along the body diagonals of its cell
2. A crystal whose unit cell has a = b = c and α = β = γ = 90° must be cubic, whatever its actual symmetry
3. A hexagonal crystal is characterised by a single 6-fold rotation axis
4. A monoclinic crystal has three mutually perpendicular 2-fold axes
Show answer
Answer: A — A crystal is cubic only if it has four 3-fold rotation axes along the body diagonals of its cell; C — A hexagonal crystal is characterised by a single 6-fold rotation axis
Systems are defined by minimum symmetry: four 3-fold axes for cubic and one 6-fold axis for hexagonal. A metrically cubic cell does not make a crystal cubic if the symmetry is lower (it would then be, for example, tetragonal or orthorhombic with accidental equal edges). Three perpendicular 2-fold axes define orthorhombic; monoclinic needs only one 2-fold axis.
A plane in a cubic crystal cuts the x axis at 2a and the y axis at 3a and is parallel to the z axis. Its Miller indices are
1. (2 3 0)
2. (3 2 ∞)
3. (6 4 0)
4. (3 2 0)
Show answer
Answer: D — (3 2 0)
Intercepts 2, 3, ∞ give reciprocals ½, ⅓, 0; multiplying by 6 clears the fractions to (3 2 0). Writing the intercepts themselves, (2 3 0), is the commonest slip; (6 4 0) is not reduced to the smallest integers and would describe a different plane with half the spacing; ∞ is never written in an index because a parallel plane has index 0.
In the four-index Miller-Bravais notation for a hexagonal crystal, the a₁ axis direction, written [1 0 0] in three indices, becomes
1. [1 0 −1 0]
2. [2 −1 −1 0]
3. [1 1 −2 0]
4. [1 0 0 0]
Show answer
Answer: B — [2 −1 −1 0]
Use u = (2U − V)/3, v = (2V − U)/3, t = −(u + v), w = W with U, V, W = 1, 0, 0: u = ⅔, v = −⅓, t = −⅓, w = 0, which becomes [2 −1 −1 0] after multiplying by 3. [1 1 −2 0] is the a₁ + a₂ direction ([1 1 0] in three indices); [1 0 −1 0] points 30° away from a₁ (it is [2 1 0] in three indices); and [1 0 0 0] violates t = −(u + v).
Which statement correctly distinguishes hexagonal close packing from cubic close packing of equal spheres?
1. Hexagonal packing has coordination number 12 whereas cubic packing has coordination number 8, because only the cubic stacking puts layers directly above one another
2. Cubic packing reaches a packing fraction of 0.74 whereas hexagonal packing reaches only about 0.68, since the hexagonal arrangement is a slightly distorted one
3. Both have coordination number 12 and packing fraction 0.74; they differ only in the stacking of the close-packed layers, ABAB… in HCP and ABCABC… in CCP
4. Hexagonal packing is ABCABC… stacking of its basal planes and cubic packing is ABAB… stacking of its {111} planes, but the two are otherwise identical
Show answer
Answer: C — Both have coordination number 12 and packing fraction 0.74; they differ only in the stacking of the close-packed layers, ABAB… in HCP and ABCABC… in CCP
The two structures use the same close-packed layers; after layers A and B the third layer repeats A (HCP, ABAB…) or takes the new position C (CCP, ABCABC…). Local packing around an atom is the same in terms of the number of neighbours (12) and the packing fraction π/(3√2) = 0.7405 is the same. The statement that swaps the ABAB and ABCABC labels between the two structures is wrong.
In an FCC crystal of hard spheres of radius R, the numbers per atom and radii of the voids are
1. One octahedral void of radius 0.414R and two tetrahedral voids of radius 0.225R
2. Two octahedral voids of radius 0.414R and one tetrahedral void of radius 0.225R
3. One octahedral void of radius 0.225R and two tetrahedral voids of radius 0.414R
4. Three octahedral voids of radius 0.155R and six tetrahedral voids of radius 0.291R
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Answer: A — One octahedral void of radius 0.414R and two tetrahedral voids of radius 0.225R
The FCC cell has 4 atoms, 4 octahedral voids (1 + 12 × ¼) and 8 tetrahedral voids, so per atom there is 1 octahedral and 2 tetrahedral. The octahedral void is the larger, r = (√2 − 1)R = 0.414R; the tetrahedral is r = (√6/2 − 1)R = 0.225R. The 3 and 6 per atom with radii 0.155R and 0.291R are the BCC values.
FCC iron (austenite) dissolves far more interstitial carbon than BCC iron (ferrite), although BCC has more voids per atom. Which explanation is correct?
1. BCC iron has no voids at all between its atoms, because its coordination number is only 8 and its spheres are not close-packed, so no solute can enter
2. Carbon is negatively charged and can occupy only tetrahedral voids, which exist in FCC iron but are completely absent from every BCC structure
3. BCC iron has fewer voids per atom than FCC iron, so the number of available interstitial sites per atom is what limits the carbon that can dissolve
4. BCC has more voids per atom but all are smaller (largest, tetrahedral, 0.291R) than the FCC octahedral void (0.414R), so carbon strains BCC far more
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Answer: D — BCC has more voids per atom but all are smaller (largest, tetrahedral, 0.291R) than the FCC octahedral void (0.414R), so carbon strains BCC far more
BCC has 3 octahedral and 6 tetrahedral voids per atom, against 1 and 2 in FCC, so site count is not the limit. What matters is size: the largest BCC void is 0.291R, while the FCC octahedral void is 0.414R, so the carbon atom is accommodated with much less strain in FCC. The claim that BCC has no voids or fewer voids is false, and carbon sits in voids as a neutral solute atom.
For a cubic lattice of parameter a, with the spacing of the (hkl) planes given by d = a/√(h² + k² + l²), the ratio d₁₁₁/d₂₀₀ is
1. 0.866
2. 1.414
3. 1.155
4. 0.577
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Answer: C — 1.155
d₁₁₁ = a/√3 and d₂₀₀ = a/√4 = a/2, so d₁₁₁/d₂₀₀ = 2/√3 = 1.155. The value 0.866 is the inverse ratio, 1.414 comes from taking √2 for the sum of squares, and 0.577 is 1/√3, which is d₁₁₁/a on its own.
A cutting-tool material consists of hard tungsten carbide particles embedded in a ductile cobalt matrix. It is best classified as
1. A ceramic, because tungsten carbide is a carbide, carbides are ceramic compounds, and the hard carbide particles dominate the cutting action
2. A composite, because two distinct phases are combined to give a property mix that neither provides alone
3. A polymer, because it is formed into shape by pressing and heating powders into a bonded mass, as thermoplastics are processed
4. A single-phase metallic solid solution in which tungsten, carbon and cobalt atoms all share one crystal lattice, with no separate particles
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Answer: B — A composite, because two distinct phases are combined to give a property mix that neither provides alone
The ceramic carbide particles provide hardness and the metal matrix provides toughness, and the two remain separate phases, which is the definition of a composite. Calling it a ceramic ignores the cobalt; it is not a polymer, and it is not a single-phase solution because the carbide and the cobalt are distinct phases.
Which of the following statements about the ordered structures of carbon are correct?
1. In diamond every carbon atom has three nearest neighbours arranged in a plane
2. Diamond can be described as an FCC lattice with a motif of two carbon atoms
3. Every single-walled carbon nanotube is metallic, whatever its chiral indices (n, m)
4. Graphite layers are held together by weak van der Waals forces, which lets the layers slide over one another
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Answer: B — Diamond can be described as an FCC lattice with a motif of two carbon atoms; D — Graphite layers are held together by weak van der Waals forces, which lets the layers slide over one another
Diamond is sp³ with four tetrahedral neighbours and is the FCC lattice with two atoms at (0, 0, 0) and (¼, ¼, ¼); graphite’s weak interlayer van der Waals bonding gives easy shear. A nanotube is metallic only when n − m is a multiple of 3 (ignoring curvature effects), so semiconducting tubes also exist. Three neighbours in a plane describe graphite or graphene, not diamond.
In a cubic crystal, a direction [uvw] lies in the plane (hkl) when hu + kv + lw = 0. Which of the following directions lie in the (111) plane?
1. [1 1 0]
2. [1 −1 0]
3. [1 1 −2]
4. [1 0 0]
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Answer: B — [1 −1 0]; C — [1 1 −2]
For (111) the condition is u + v + w = 0. [1 −1 0] gives 1 − 1 + 0 = 0 and [1 1 −2] gives 1 + 1 − 2 = 0, so both lie in the plane. [1 1 0] gives 2 and [1 0 0] gives 1, so neither lies in it.
An FCC metal has lattice parameter a = 0.360 nm. Find the planar density of its (111) plane, counting each atom by the fraction of it lying inside the plane’s repeating area, in atoms per nm². Give the answer to two decimal places.
Numerical answer — type the value.
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Answer: 17.82
The (111) plane is a triangle of side a√2 = 0.5091 nm with area (√3/4)(a√2)² = (√3/2)a² = 0.11224 nm². It holds 3 × ⅙ + 3 × ½ = 2 atoms. PD = 2/0.11224 = 17.82 atoms/nm², which equals 4/(√3 a²) = 4/(1.7321 × 0.1296) = 17.82.
Silicon has the diamond cubic structure with lattice parameter a = 0.5431 nm. Take the atomic mass A = 28.09 g/mol and Avogadro’s number N_A = 6.022 × 10²³ mol⁻¹. Calculate its density in g/cm³, to two decimal places.
Numerical answer — type the value.
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Answer: 2.33
The diamond cubic cell holds 8 atoms (FCC lattice with a two-atom motif). V_c = (0.5431 × 10⁻⁷ cm)³ = 1.6019 × 10⁻²² cm³. ρ = nA/(V_c N_A) = 8 × 28.09/(1.6019 × 10⁻²² × 6.022 × 10²³) = 224.72/96.47 = 2.33 g/cm³. Using 4 atoms (treating it as plain FCC) would give half this value.
Iron changes from BCC to FCC on heating. Treat the atoms as hard spheres of the same radius R in both structures (BCC: a = 4R/√3, 2 atoms per cell; FCC: a = 2√2 R, 4 atoms per cell). By what percentage does the volume per atom decrease on going from BCC to FCC? Give the answer to one decimal place.
Numerical answer — type the value.
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Answer: 8.1
Volume per atom in BCC = (4R/√3)³/2 = 12.317R³/2 = 6.158R³. In FCC = (2√2 R)³/4 = 22.627R³/4 = 5.657R³. Decrease = (6.158 − 5.657)/6.158 = 0.0814, that is 8.1%. Real iron shrinks far less because the effective atomic radius changes with structure, so this is a hard-sphere estimate.
A closed fullerene cage has every carbon atom bonded to exactly three others, and its faces are 12 pentagons and 25 hexagons. How many carbon atoms does the cage contain?
Numerical answer — type the value.
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Answer: 70
Each atom is shared by three faces, so V = (5 × 12 + 6 × 25)/3 = (60 + 150)/3 = 70. Check with Euler’s relation: E = 3V/2 = 105 and F = 37, so V − E + F = 70 − 105 + 37 = 2. This is C₇₀.
A single-walled carbon nanotube has chiral indices (n, m) = (12, 6). Take the graphene lattice constant a = 0.246 nm and π = 3.1416, and use d = a√(n² + nm + m²)/π. Find its diameter in nm, to two decimal places.
Numerical answer — type the value.
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Answer: 1.24
n² + nm + m² = 144 + 72 + 36 = 252, and √252 = 15.875. d = 0.246 × 15.875/3.1416 = 3.905/3.1416 = 1.24 nm. Using n² + m² = 180 (leaving out the cross term) is the usual slip and would give 1.05 nm.