Crystals, Lattices & Unit Cells

The dimensional characteristics of a unit cell are its three edge lengths (a, b, c) and the three interfacial angles (α, β, γ). For the triclinic system — the least symmetric of the seven crystal systems — these are:

• Edge lengths: a ≠ b ≠ c (all three unequal)
• Interfacial angles: α ≠ β ≠ γ ≠ 90° (all three unequal, and none is a right angle)

Because there is no symmetry restriction at all, only one Bravais lattice is possible in this system — the primitive (P) lattice. Any attempt to centre a triclinic cell (body, face or base) simply produces an equivalent, smaller primitive cell, so centred varieties are not counted separately.

Examples: copper sulphate pentahydrate (CuSO₄·5H₂O), potassium dichromate (K₂Cr₂O₇).

Tetragonal class: a = b ≠ c and α = β = γ = 90° — two edges equal, the third different, and all three angles right angles. (It is like a cube stretched or squeezed along one axis.)

Bravais lattices possible: two —

1. Primitive (simple) tetragonal, P — lattice points only at the 8 corners.
2. Body-centred tetragonal, I — corners + one point at the body centre.

A face-centred or base-centred tetragonal cell can always be redrawn as a smaller primitive or body-centred cell, so they are not counted as separate Bravais lattices.

Atoms per unit cell (Z) for body-centred tetragonal:

This is the crystallographic restriction theorem: only 1-, 2-, 3-, 4- and 6-fold rotation axes can exist in a crystal.

Proof in simple words: a crystal must fill space by repeating its unit cell with no gaps. Take a row of lattice points with spacing a. Rotating the lattice by an angle θ about a lattice point must carry lattice points onto lattice points. The new translation generated is 2a·cosθ, and for the lattice to map onto itself this must be an integral multiple of a:

Checking each possible rotation:

• n = 1 (θ = 360°): 2cosθ = 2 ✔
• n = 2 (θ = 180°): 2cosθ = −2 ✔
• n = 3 (θ = 120°): 2cosθ = −1 ✔
• n = 4 (θ = 90°): 2cosθ = 0 ✔
• n = 5 (θ = 72°): 2cosθ = 0.618 ✘ — not an integer
• n = 6 (θ = 60°): 2cosθ = 1 ✔

So a 5-fold rotation cannot map a lattice onto itself — trying to tile space with regular pentagons always leaves gaps. (The same test also rules out 7-fold, 8-fold and higher axes.)

Orthorhombic system: a ≠ b ≠ c, α = β = γ = 90°. All edges unequal, all angles right angles. It has the maximum number of Bravais lattices — four:

1. Primitive (P): lattice points only at the 8 corners → Z = 8 × ⅛ = 1
2. Base-centred (C): corners + 2 face centres (top and bottom faces) → Z = 1 + 2 × ½ = 2
3. Body-centred (I): corners + 1 body centre → Z = 1 + 1 = 2
4. Face-centred (F): corners + all 6 face centres → Z = 1 + 6 × ½ = 4

Lattice diagram (what to draw in the exam): four small boxes side by side — an empty box (P), a box with dots on the top and bottom faces (C), a box with one dot in the centre (I), and a box with dots on all six faces (F). Write “a ≠ b ≠ c, all angles 90°” beside each.

Examples: rhombic sulphur, KNO₃, MgSO₄·7H₂O.

The statement is correct — justify it.

Miller indices (hkl) do not describe one single face; they describe a family of parallel, equally spaced planes. Any plane parallel to a given (hkl) plane — at any distance from the origin — has the same indices, because the indices come from the ratios of the reciprocal intercepts, and parallel planes cut the axes in the same ratio.

Mathematically, the whole family is hx/a + ky/b + lz/c = m, where m = 0, ±1, ±2, … — every integer m is another plane of the same family, spaced d_hkl apart.

Example: in a cubic crystal all six cube faces belong to the family {100} — (100), (010), (001) and their negatives (1̄00), (01̄0), (001̄). One symbol, a whole class of equivalent faces. (Round brackets (hkl) = one plane; curly brackets {hkl} = the full family.)

Why it matters: in X-ray diffraction, every plane of the family reflects together — the Bragg reflection labelled (200) really comes from the whole {200} family of planes.

Step 1 — intercepts from Miller indices. Intercepts are proportional to the reciprocals of the indices:

So the planes cut the x-axis at a/2, the y-axis at −b, and are parallel to the z-axis.

Step 2 — the first two planes nearest the origin. A “class” (family) of planes (hkl) is the whole set of parallel, equally spaced lattice planes hx/a + ky/b + lz/c = m, where m = 0, ±1, ±2, … Each integer m is one plane of the family, and its perpendicular distance from the origin is |m|·d_hkl. The two planes most near to the origin are therefore m = +1 and m = −1 (both at distance d_2̄10 from the origin, on opposite sides of it).

• Plane 1 (m = +1): 2x/a − y/b = 1 → intercepts (a/2, −b, ∞)
• Plane 2 (m = −1): 2x/a − y/b = −1 → intercepts (−a/2, +b, ∞)

Note: halving the intercepts to (a/4, −b/2, ∞) would give 2x/a − y/b = 4 — that is the fourth plane, not the second. And (a/4, −b/2, ∞) as “half intercepts” would mean m = 1/2, which is not a lattice plane at all (it passes through no lattice points).

Step 3 — depiction. Draw the crystal axes. For plane 1, mark a/2 on the +x axis and b on the −y axis, join the marks with a straight line and extend it parallel to the z-axis. For plane 2, mark a/2 on the −x axis and b on the +y axis and draw the parallel plane on the opposite side of the origin. Both planes run parallel to the z-axis and sit equidistant from the origin — this picture shows what “a class of planes” really means.

Method: intercepts on (x, y, z) → take reciprocals → clear fractions. A bar over an index means the intercept is on the negative axis. The six faces form the family {100}:

• Front face — intercepts (a, ∞, ∞) → reciprocals (1, 0, 0) → (100); back face → (1̄00)
• Right face — intercepts (∞, b, ∞) → (0, 1, 0) → (010); left face → (01̄0)
• Top face — intercepts (∞, ∞, c) → (0, 0, 1) → (001); bottom face → (001̄)

Opposite faces differ only by the bar — (100) and (1̄00) are parallel faces of the same family. Note that for a simple cubic cell the (200) planes are a different family (half the spacing), but the six faces of the cube itself are all {100} type.

Diagram: draw a cube with the x, y, z axes coming out of one corner, and write each index on its face — (100) on front, (1̄00) on back, (010) on right, (01̄0) on left, (001) on top, (001̄) on bottom. This picture shows the “class of faces” idea clearly: one family {100}, six members.

Intercepts are the cell edges divided by the corresponding Miller index (intercept = edge ÷ index):

• x-intercept = a/h = 6.8/2 = 3.4 nm
• y-intercept = b/k = 8.6/0 = ∞ — the index 0 means the plane never cuts the y-axis, i.e. it runs parallel to the y-axis
• z-intercept = c/l = 4.6/1 = 4.6 nm

Answer: the (201) plane cuts the x-axis at 3.4 nm and the z-axis at 4.6 nm, and is parallel to the y-axis.

Part 1 — void space in fcc.

An fcc unit cell contains Z = 4 atoms. The atoms touch along the face diagonal, so 4r = √2·a, i.e. r = a/(2√2). Volume occupied by the 4 spheres:

Packing efficiency = V_atoms/a³ = π/(3√2) ≈ 3.1416/4.2426 ≈ 0.74 (74%).

(This empty space is shared between 8 tetrahedral and 4 octahedral voids per cell — the basis of ionic structures like NaCl and spinels.)

Part 2 — Miller indices from intercepts. Intercepts: ½, 2, ³⁄₂. Take reciprocals:

Multiply by 6 (the LCM of the denominators) to clear fractions:

Part 1 — interplanar distance (orthorhombic).

An (hkl) plane makes intercepts a/h, b/k, c/l on the three (mutually perpendicular) axes, so its equation is

The perpendicular distance of a plane px + qy + rz = 1 from the origin is 1/√(p²+q²+r²). Here p = h/a, q = k/b, r = l/c, giving

For the cubic case (a = b = c) this reduces to the familiar d = a/√(h²+k²+l²).

Part 2 — coordination number in 3-D close packing.

In both hcp and ccp every sphere sits in a close-packed layer and touches:

• 6 neighbours in its own layer (hexagonal arrangement),
• 3 spheres in the layer directly below (sitting in the hollows),
• 3 spheres in the layer directly above.

12 is the highest coordination number possible for equal spheres — this is why hcp/ccp are called closest packing.

A cube belongs to point group O_h with 48 symmetry operations in total:

Proper rotations (24):

• E — identity (1)
• 3 C₄ axes through opposite face centres → C₄, C₄², C₄³ each = 9 operations
• 4 C₃ axes through opposite body diagonals → 8 operations
• 6 C₂ axes through mid-points of opposite edges → 6 operations

Total: 1 + 9 + 8 + 6 = 24 proper rotations.

Improper operations (24): centre of inversion i (1); 9 mirror planes — 3 σ_h (parallel to faces) + 6 σ_d (diagonal); 3 S₄ axes (6 operations) and 4 S₆ axes (8 operations). Total: 1 + 9 + 6 + 8 = 24.

24 + 24 = 48 operations — the highest symmetry of any crystal shape, which is why cubic crystals (NaCl, diamond) show such regular habits.

Octahedron comment (2024 Q1(d)): the regular octahedron is the dual of the cube — joining the six face-centres of a cube gives a perfect octahedron. Faces and vertices swap roles, but every symmetry operation survives, so the octahedron has exactly the same elements and the same O_h group (the C₄ axes now pass through opposite vertices, and the C₃ axes through opposite face centres).

Derivation: consider a parallel beam of X-rays of wavelength λ striking a set of parallel lattice planes spaced d apart, at glancing angle θ (angle with the plane, not the normal).

• The ray reflected from the second plane travels further than the ray reflected from the first plane.
• From the geometry, the extra path on the way in is d·sinθ and the extra path on the way out is another d·sinθ.
• So the total path difference = 2d·sinθ.

For constructive interference (a bright diffraction spot), this path difference must be a whole number of wavelengths:

This is Bragg's equation. (Draw two horizontal planes, the incident and reflected rays, and mark the two d·sinθ segments — the diagram earns marks.)

Conditions for validity:

1. λ must be comparable to the interplanar spacing (λ ≈ d, i.e. the X-ray region, ~0.1–2 Å). If λ ≫ d, sinθ would exceed 1 and no diffraction occurs.
2. The crystal must be reasonably perfect — planes regularly spaced over many unit cells — so that the reflected waves stay in phase.
3. Since sinθ ≤ 1, we need nλ ≤ 2d: only orders satisfying this can be observed for a given crystal and wavelength.

KCl truly has the rock-salt structure — two interpenetrating fcc sub-lattices (K⁺ and Cl⁻), exactly like NaCl. But X-rays are scattered by electrons, and the scattering power of an ion (its scattering factor) depends on its electron count:

• K⁺ has 18 electrons; Cl⁻ also has 18 electrons — they are isoelectronic.
• So every lattice point scatters X-rays identically, and the K⁺ and Cl⁻ sub-lattices become indistinguishable to the X-ray beam.
• The reflections that would distinguish the two sub-lattices (e.g. (111)) vanish, and the diffraction pattern looks like that of a simple cubic lattice of half the cell edge.

In NaCl this does not happen, because Na⁺ (10 e⁻) and Cl⁻ (18 e⁻) scatter very differently, so the true fcc pattern is seen.

An ordinary (optical) diffraction grating works because it has regularly spaced scattering lines with spacing comparable to the wavelength of light. A crystal satisfies the same two requirements, in three dimensions:

• Regular spacing: atoms/ions sit on a periodic 3-D lattice, giving sets of parallel planes with interplanar spacings of the order of 1 Å.
• Matching wavelength: X-rays have wavelengths of the same order (~0.1–2 Å), so the condition λ ≈ d is satisfied.
• Each lattice plane reflects a tiny fraction of the beam; the reflected waves interfere constructively only at the Bragg angles (nλ = 2d·sinθ), giving sharp spots instead of a diffuse scatter.

Because the periodicity extends along all three axes, the crystal diffracts in three dimensions — unlike a ruled grating, which is one-dimensional. This is the basis of X-ray crystallography (von Laue, 1912; Bragg father and son, 1913), the technique that revealed the structure of NaCl, diamond, DNA and proteins.

In Bragg's equation nλ = 2d·sinθ, the integer n = 1, 2, 3, … is the order of reflection.

Meaning: it counts how many whole wavelengths fit into the path difference (2d·sinθ) between waves reflected from successive lattice planes. The n = 1 reflection is the first order (usually the strongest); n = 2 is the second order, and so on.

Significance:

• Higher orders appear at larger glancing angles (sinθ = nλ/2d) and are progressively weaker, so usually only the first few orders are observed.
• Since sinθ ≤ 1, only orders with n ≤ 2d/λ can occur — this sets a hard limit on which reflections are observable for a given crystal and wavelength (e.g. with λ = 1.54 Å and d = 2.8 Å, only n = 1, 2, 3 are possible).
• The set of observed orders is a fingerprint used to identify the crystal structure.

From Bragg's law nλ = 2d·sinθ, the glancing angle can never exceed 90°, so sinθ ≤ 1. Rearranging,

For the first order (n = 1), the minimum interplanar spacing that can diffract radiation of wavelength λ is therefore

Interpretation: lattice planes closer together than half the wavelength cannot produce any diffraction pattern with that radiation — the required sinθ would exceed 1, which is impossible. To resolve finer spacings you must use shorter-wavelength (harder) X-rays.

Example: with Cu-Kα radiation (λ = 1.54 Å), no plane with d < 0.77 Å can diffract.

For first-order reflection (n = 1), Bragg's law gives the smallest spacing that can diffract as d_min = λ/2 (because sinθ ≤ 1). Substituting the given wavelength:

So plane spacings smaller than ≈ 0.84 Å cannot be measured with 1.67 Å X-rays — the reflection would need sinθ > 1, which is impossible. This is a direct numerical application of the d_min = λ/2 result (asked as theory in 2024 Q3(a)(iii)).

Diffraction can only occur when the wavelength is comparable to the spacing being probed (λ ≈ d). The interplanar spacings in crystals are of the order of 1 Å (10⁻¹⁰ m).

• Radio waves have wavelengths from millimetres to kilometres — roughly a million to ten billion times larger than atomic spacings.
• From Bragg's law, sinθ = nλ/2d: with λ ≫ d, the required sinθ would be far greater than 1, which is impossible — so no diffraction pattern can form.
• The radio wave simply “does not see” the individual atoms; the crystal looks like a continuous medium to it.

This is exactly why von Laue (1912) chose X-rays (λ ~ 0.1–2 Å, matching atomic spacings) to prove that crystals diffract — and why visible light (λ ~ 5000 Å) is equally useless for crystal-structure determination.

Step 1 — relate the lattice parameters. Molar volume V_m ∝ a³ (one formula unit per rock-salt cell edge), so

Step 2 — apply Bragg's law. For the (200) plane, d = a/2. With the same wavelength and order (n = 1), nλ = 2d·sinθ gives sinθ ∝ 1/d ∝ 1/a. Hence

Check: the KCl cell is larger, so its (200) planes are further apart (d larger) — Bragg's law then demands a smaller glancing angle. 5.4° < 5.9° ✔ — the answer is physically sensible.

Derivation: start from Bragg's equation

The glancing angle θ can at most be 90°, so the maximum value of sinθ is 1. The largest integer n that can satisfy the equation is therefore

(strictly, the largest integer ≤ 2d/λ).

Physical meaning: the path difference between successive planes can never exceed twice the interplanar spacing (2d) — the waves would have to travel “more than there is”. So only a finite number of orders exists for any crystal–wavelength combination.

Worked illustration: with Cu-Kα X-rays (λ = 1.54 Å) and the NaCl (200) planes (d ≈ 2.82 Å), n_max = 2 × 2.82/1.54 ≈ 3.66, so only the 1st, 2nd and 3rd orders can be observed.

Practical note: intensity falls off rapidly with n (atomic scattering factor decreases and thermal vibration smears higher orders), so in real spectra usually only the first two or three orders have measurable intensity — the higher ones are “allowed but invisible”.

For a cubic crystal the spacing of the (hkl) family is

For d = a/√7 we would need integers h, k, l with

But 7 cannot be written as the sum of three integer squares. The squares available below 7 are only 0, 1 and 4 (the next, 9, already exceeds 7):

• 4 + 1 + 1 = 6 (too small)
• 4 + 4 + 0 = 8 (too big)
• 1 + 1 + 1 = 3; 4 + 1 + 0 = 5; 4 + 0 + 0 = 4 …

No combination gives 7. Hence no (hkl) plane has spacing a/√7 — proved.

Step 1 — equation of the (hk) line. By definition of Miller indices, the (hk) line makes intercepts a/h on the X-axis and a/k on the Y-axis. In intercept form:

Step 2 — distance from the origin. The perpendicular distance of the line px + qy = c from the origin is |c|/√(p²+q²). Here p = h, q = k, c = a, so

Step 3 — successive planes. The whole family is hx + ky = m·a (m = 0, ±1, ±2, …); each member is parallel to the first and the perpendicular separation between neighbours is exactly the distance found above — a/√(h²+k²). Proved.

Note: this is the 2-D version of the cubic formula d = a/√(h²+k²+l²) — setting l = 0 (no third dimension) gives the same result.

Step 1 — find Z (atoms per unit cell) from the density.

Convert: a = 6.18 Å = 6.18×10⁻⁸ cm, so a³ = (6.18)³×10⁻²⁴ = 236.0×10⁻²⁴ = 2.36×10⁻²² cm³. M = 132.9 g mol⁻¹.

Z = 2 → the lattice is body-centred cubic (bcc).

Step 2 — radius from the bcc geometry. In bcc the atoms touch along the body diagonal: 4r = √3·a.

Step 1 — find the cube edge a. For fcc, Z = 4:

Step 2 — closest approach. In fcc the nearest neighbours touch along the face diagonal: nearest-neighbour distance = a/√2 = 2r.

So two neighbouring Al atoms are 2.86 Å apart (centre to centre) — this is also twice the metallic radius of Al (r ≈ 1.43 Å).

Step 1 — find the cube edge a. In fcc the nearest neighbours touch along the face diagonal, so nearest-neighbour distance = a/√2:

Step 2 — density. For fcc, Z = 4:

The experimental density of silver is 10.5 g cm⁻³ — the small difference comes from rounding; the method (nearest-neighbour → a → density) is what earns full marks. (Repeated in 2019 as a 4-mark question.)

Step 1 — find the cube edge a. For bcc, Z = 2:

Step 2 — atomic radius. In bcc the atoms touch along the body diagonal: 4r = √3·a.

So the europium atom has a metallic radius of about 1.98 Å — a large value, as expected for a lanthanide (Eu is famously the largest of the lanthanides in the metallic state because it is divalent).

Step 1 — volume of one unit cell. Orthorhombic: V = a·b·c.

Convert: 1 nm = 10⁻⁹ m, so 1 nm³ = 10⁻²⁷ m³:

Step 2 — mass of one unit cell (mass = density × volume):

Step 3 — molar mass. The cell holds 66 insulin molecules, so one molecule weighs m_cell/66; multiply by Avogadro's number:

Molar mass of insulin ≈ 3.6×10³ g mol⁻¹ (≈ 3600 g mol⁻¹).

Why this question matters: this is exactly how X-ray crystallography was historically used to determine molecular weights of proteins — Dorothy Hodgkin's insulin work is the classic example. The method is the same density formula, just with molecules instead of atoms.