Crystals, Lattices & Unit Cells

The triclinic crystal system is the least symmetric of all seven systems. Its unit cell is described by:

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

There is only one Bravais lattice in this system — the primitive (P) lattice, because centring a triclinic cell always produces an equivalent smaller primitive cell. Example: CuSO₄·5H₂O.

Tetragonal class: a = b ≠ c and α = β = γ = 90°. Two edges equal, the third different, all angles right angles.

Bravais lattices possible: two —

1. Primitive (simple) tetragonal, P
2. Body-centred tetragonal, I

(Face-centred and base-centred tetragonal cells reduce to one of these two, so they are not counted separately.)

Atoms per unit cell (Z) for body-centred tetragonal: 8 corners × ⅛ + 1 body centre × 1 = 2 atoms.

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 angle θ about a lattice point must carry lattice points onto lattice points, so the new translation produced is 2a·cosθ, which must be an integral multiple of a:

For a 5-fold axis, θ = 72°, and 2cos72° = 0.618 — not an integer. So a 5-fold rotation cannot map a lattice onto itself. Trying to tile space with pentagons always leaves gaps.

Allowed values give θ = 360°, 180°, 120°, 90°, 60° — i.e. 1-, 2-, 3-, 4- and 6-fold axes only.

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

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

In the exam, draw four small boxes: an empty box (P), a box with dots on top/bottom faces (C), a box with a centre dot (I), and a box with dots on all faces (F). Label a ≠ b ≠ c beside each.

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 reciprocal intercepts, and parallel planes cut the axes in the same ratio.

Example: in a cubic crystal all six cube faces belong to the family {100} — (100), (010), (001) and their negatives. One symbol, a whole class of equivalent faces.

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.

The six faces of a cube form the family {100}. Their Miller indices are:

• Front face (cuts x at a): (100); back face: (1̄00)
• Right face (cuts y at b): (010); left face: (01̄0)
• Top face (cuts z at c): (001); bottom face: (001̄)

How to get them: take intercepts (e.g. a, ∞, ∞), take reciprocals (1/a, 0, 0), clear fractions → (100). A bar over the index means the intercept is on the negative axis.

Diagram: draw a cube, label the three axes x, y, z from one corner, and write each index on its face. Opposite faces differ only by the bar — this shows the “class of faces” idea clearly.

Intercepts = cell edge ÷ corresponding index:

• x-intercept = a/h = 6.8/2 = 3.4 nm
• y-intercept = b/k = 8.6/0 = ∞ (plane 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 runs parallel to the y-axis.

Part 1 — void space in fcc. Packing efficiency of fcc (ccp) = 74%. Therefore

(Proof sketch: 4 atoms per cell, 4r = √2·a; occupied volume 4 × ⁴⁄₃πr³ divided by a³ = π/(3√2) ≈ 0.74.)

Part 2 — Miller indices. Intercepts: ½, 2, ³⁄₂. Take reciprocals: 2, ½, ⅔. Multiply by 6 to clear fractions: (12 3 4).

Part 1 — interplanar distance (orthorhombic). For axes at right angles with edges a, b, c, the perpendicular distance of the (hkl) plane from the origin is:

Derivation: the plane x/(a/h) + y/(b/k) + z/(c/l) = 1 has normal direction (h/a², k/b², l/c²); the perpendicular distance from the origin gives the result above. For cubic (a = b = c) it reduces to d = a/√(h²+k²+l²).

Part 2 — coordination number in 3-D close packing. In both hcp and ccp, every sphere touches 6 neighbours in its own layer, 3 in the layer below and 3 in the layer above:

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

• Proper rotations (24): E (1); 3C₄ axes through opposite face centres → 9 operations; 4C₃ axes through body diagonals → 8; 6C₂ axes through opposite edge mid-points → 6. Total 1 + 9 + 8 + 6 = 24.
• Improper (24): inversion centre i (1); 3σ_h + 6σ_d mirror planes (9); 3S₄ (6) and 4S₆ (8) improper axes.

Octahedron comment (2024): 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, but every symmetry operation survives, so the octahedron has exactly the same elements and the same O_h group (C₄ axes now pass through opposite vertices, C₃ through opposite face centres).

Derivation: consider X-rays of wavelength λ striking parallel lattice planes spaced d apart at glancing angle θ. The ray reflected from the second plane travels an extra distance 2·d·sinθ compared with the ray from the first plane. For constructive interference (bright spot), this path difference must be a whole number of wavelengths:

Conditions for validity:

1. λ must be comparable to the interplanar spacing (λ ≈ d, i.e. X-ray region, ~1 Å).
2. The crystal must be reasonably perfect — planes regularly spaced over many unit cells.
3. Since sinθ ≤ 1, we need nλ ≤ 2d — only orders with λ ≤ 2d/n are observable.

KCl truly has the rock-salt (fcc) structure like NaCl. But X-rays are scattered by electrons, and the scattering power of an ion 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 two interpenetrating fcc sub-lattices become indistinguishable.

The diffraction pattern therefore looks like that of a simple cubic lattice (with a smaller effective cell). In NaCl this does not happen because Na⁺ (10 e⁻) and Cl⁻ (18 e⁻) scatter differently.

A diffraction grating needs regularly spaced scattering centres with spacing comparable to the wavelength of the radiation:

• In a crystal, atoms/ions sit on a periodic 3-D lattice with interplanar spacings of the order of 1 Å.
• X-rays have wavelengths of the same order (~0.1–2 Å).
• Each lattice plane reflects a tiny fraction of the beam; the reflected waves interfere constructively only at the Bragg angles.

Because the periodicity extends in all three directions, the crystal works as a three-dimensional grating — this is the basis of X-ray crystallography.

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 between successive planes. n = 1 is the first-order (strongest) reflection.
• Significance: higher orders appear at larger glancing angles and are progressively weaker; since sinθ ≤ 1, only orders with n ≤ 2d/λ can occur — this limits which reflections are observable for a given crystal and wavelength.

From Bragg's law nλ = 2d·sinθ, with sinθ ≤ 1:

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

Planes closer than λ/2 cannot produce any diffraction pattern with that radiation.

Using d_min = λ/2 for first-order reflection:

So plane spacings smaller than ≈ 0.84 Å cannot be measured with 1.67 Å X-rays.

Diffraction needs λ ≈ d (interplanar spacing ~1 Å). Radio waves have wavelengths from millimetres to kilometres — millions of times larger than atomic spacings.

From Bragg's law, sinθ = nλ/2d would far exceed 1, so no diffraction is possible. Only X-rays (λ ~ 1 Å) match the atomic scale — that is why von Laue's experiment used X-rays.

Step 1 — relate lattice parameters. Molar volume ∝ a³, so

Step 2 — Bragg's law for the same order and wavelength. For the (200) plane, d = a/2, so sinθ ∝ 1/a:

The larger KCl cell gives a smaller glancing angle, as expected.

From nλ = 2d·sinθ, since the maximum value of sinθ is 1:

So the highest observable order is the largest integer less than or equal to 2d/λ. Physically: the path difference can never exceed twice the interplanar spacing, so only finitely many orders exist — and in practice intensity falls off rapidly, so usually only the first few orders are seen.

For a cubic crystal,

For d = a/√7 we would need h² + k² + l² = 7 with integers h, k, l. But 7 cannot be written as the sum of three integer squares:

• Squares available: 0, 1, 4 (next, 9 > 7)
• 4 + 1 + 1 = 6; 4 + 4 = 8 > 7; 1 + 1 + 1 = 3 … no combination gives 7.

Hence no (hkl) plane has spacing a/√7. (Same logic rules out a/√15, a/√23, etc. — a favourite examiner trick.)

The (hk) line (plane in 2-D) makes intercepts a/h on X and a/k on Y. Its equation is

The perpendicular distance of this line from the origin is |−a|/√(h²+k²) = a/√(h²+k²). Successive parallel lines of the family are equally spaced by exactly this amount. This is the 2-D version of the cubic formula d = a/√(h²+k²+l²).

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

a = 6.18 Å = 6.18×10⁻⁸ cm; a³ = 2.36×10⁻²² cm³; M = 132.9 g mol⁻¹.

Z = 2 means a bcc lattice.

Step 2 — radius. In bcc, 4r = √3·a:

Step 1 — edge length. For fcc, Z = 4:

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

Step 1 — edge length. Nearest-neighbour distance in fcc = a/√2:

Step 2 — density.

(The experimental value is 10.5 g cm⁻³ — the method is what earns marks.)

Step 1 — edge length (bcc → Z = 2):

Step 2 — radius (bcc: 4r = √3·a):

Step 1 — volume of the unit cell:

Step 2 — mass of one unit cell:

Step 3 — molar mass (66 molecules per cell):

So the molar mass of insulin ≈ 3600 g mol⁻¹. (X-ray crystallography was historically used exactly this way to find molecular weights of proteins.)