Solid-State Chemistry: Lattices, Miller Planes, Bragg’s Law, Packing, Ionic Structures, Defects, Bands and Zeolites
1. Crystal systems, Bravais lattices, Miller indices and Bragg’s law
A crystal is a lattice of identical points with a basis of atoms attached to each. The unit cell is defined by a, b, c and α, β, γ, which fall into seven crystal systems — cubic, tetragonal, orthorhombic, hexagonal, trigonal (rhombohedral), monoclinic and triclinic — and, allowing body-, face- and base-centring, fourteen Bravais lattices. Cubic has three: primitive (P), body-centred (I) and face-centred (F).
Miller indices (hkl) label a family of planes: take the intercepts on the a, b, c axes in units of the cell edges, take reciprocals and clear fractions. A plane cutting at 2a, 3b and parallel to c has reciprocals ½, ⅓, 0 → (320). For a cubic lattice the spacing is d_hkl = a/√(h² + k² + l²). X-rays of wavelength λ reflect constructively from planes of spacing d when Bragg’s law nλ = 2d sin θ holds. Not every (hkl) appears: centring causes systematic absences. Primitive cubic shows all reflections (100, 110, 111, 200, …); body-centred only h + k + l even (110, 200, 211, 220, …); face-centred only h, k, l all odd or all even (111, 200, 220, 311, …). The ratios of sin²θ therefore identify the lattice type.
2. Crystal packing, voids, radius ratios and density
| Cell | Atoms per cell Z | Coordination number | Contact along | Packing fraction |
|---|---|---|---|---|
| simple cubic | 1 | 6 | edge: a = 2r | 0.524 |
| body-centred cubic | 2 | 8 | body diagonal: √3a = 4r | 0.680 |
| face-centred cubic (ccp) | 4 | 12 | face diagonal: √2a = 4r | 0.740 |
| hexagonal close-packed | 6 (hexagonal cell) | 12 | ABAB layers | 0.740 |
Cubic close packing (ABCABC) and hexagonal close packing (ABAB) both fill 74% of space, and each contains, per N spheres, N octahedral holes and 2N tetrahedral holes. The radius ratio r₊/r₋ predicts which hole a cation prefers in an anion array: 0.155–0.225 triangular (CN 3), 0.225–0.414 tetrahedral (CN 4), 0.414–0.732 octahedral (CN 6), above 0.732 cubic (CN 8). It is a rough guide — polarisation and covalency override it — but it explains why NaCl (0.56) is six-coordinate and CsCl (0.93) eight-coordinate. The density of a crystal follows from its cell: ρ = ZM/(N_A a³), and conversely a measured density fixes Z and so the cell type.
3. Ionic structures: AX, AX₂, ABX₃ and spinels
| Type | Description | Coordination | Z (formula units per cell) |
|---|---|---|---|
| rock salt NaCl (AX) | ccp anions, cations in all octahedral holes | 6:6 | 4 |
| caesium chloride CsCl (AX) | primitive cubic anions, cation at the body centre | 8:8 | 1 |
| zinc blende ZnS (AX) | ccp S, Zn in half the tetrahedral holes | 4:4 | 4 |
| wurtzite ZnS (AX) | hcp S, Zn in half the tetrahedral holes | 4:4 | 2 (hexagonal cell) |
| fluorite CaF₂ (AX₂) | ccp Ca²⁺, F⁻ in all tetrahedral holes | 8:4 | 4 |
| antifluorite Na₂O (A₂X) | ccp O²⁻, Na⁺ in all tetrahedral holes | 4:8 | 4 |
| rutile TiO₂ (AX₂) | distorted hcp O, Ti in half the octahedral holes | 6:3 | 2 |
| perovskite CaTiO₃ (ABX₃) | Ti at the cube centre in an O₆ octahedron, Ca at the corners, O at face centres | Ti 6, Ca 12 | 1 |
Spinels AB₂O₄ have a ccp oxide array with one-eighth of the tetrahedral and one-half of the octahedral holes filled. In a normal spinel (MgAl₂O₄) A²⁺ is tetrahedral and B³⁺ octahedral, written A[B₂]O₄; in an inverse spinel B[AB]O₄ half the B³⁺ ions take the tetrahedral sites and A²⁺ goes octahedral. Crystal-field theory predicts which, through the octahedral site preference energy (CFSE in Oh minus CFSE in Td): Fe₃O₄ is inverse, Fe³⁺[Fe²⁺Fe³⁺]O₄, because high-spin Fe³⁺ (d⁵) has no preference and Fe²⁺ (d⁶) gains by going octahedral; NiFe₂O₄ is inverse because Ni²⁺ (d⁸) has a large octahedral preference; Mn₃O₄ and Co₃O₄ are normal because Mn³⁺ (d⁴) and low-spin Co³⁺ (d⁶) strongly prefer octahedral sites.
4. Crystal defects
Above 0 K every crystal contains point defects, because a few vacancies raise the entropy more than they cost in enthalpy. A Schottky defect is a pair of cation and anion vacancies (keeping charge balance); it lowers the density and is typical of ionic solids with similar ion sizes and high coordination — NaCl, KCl, CsCl. A Frenkel defect moves an ion, usually the smaller cation, into an interstitial site, leaving a vacancy; the density is unchanged, and it is typical where the cation is small and polarising and the structure open — AgCl, AgBr (which also show Schottky defects), and ZnS. Non-stoichiometric defects change the composition: wüstite is really Fe₁₋ₓO, with some Fe³⁺ balancing the cation vacancies (metal deficiency); heating NaCl in sodium vapour gives Na₁₊ₓCl, whose extra electrons sit in anion vacancies as F-centres (colour centres) and make the crystal yellow. Deliberate aliovalent doping (Ca²⁺ in NaCl, P in Si) creates vacancies or carriers and controls ionic and electronic conductivity. Line defects (dislocations) and planar defects (grain boundaries) govern mechanical strength.
5. Band theory: metals, insulators and semiconductors
Combining the atomic orbitals of N atoms gives N molecular orbitals so closely spaced that they form continuous bands. In a metal the highest occupied band is only partly filled (sodium’s 3s band is half full) or two bands overlap (magnesium’s 3s and 3p), so electrons move into empty levels with infinitesimal energy; conductivity falls as temperature rises because lattice vibrations scatter the electrons. In an insulator a filled valence band is separated from an empty conduction band by a large band gap (diamond about 5.5 eV). In a semiconductor the gap is small (Si 1.1 eV, Ge 0.67 eV, GaAs 1.4 eV), so some electrons are thermally promoted and intrinsic conductivity rises steeply with temperature, as exp(−E_g/2kT). The absorption edge lies at λ = hc/E_g, about 1130 nm for silicon.
Extrinsic semiconductors are doped. A group-15 donor (P, As) in silicon supplies an extra electron in a level just below the conduction band — n-type, electrons as majority carriers; a group-13 acceptor (B, Ga) creates a level just above the valence band that captures electrons and leaves mobile holes — p-type. Non-stoichiometric oxides behave the same way: metal-excess ZnO₁₋ₓ is n-type and metal-deficient Ni₁₋ₓO (with Ni³⁺ acting as holes) is p-type.
6. Zeolites and their applications
Zeolites are crystalline aluminosilicates, Mₓ/ₙ[(AlO₂)ₓ(SiO₂)ᵧ]·zH₂O, whose frameworks of corner-sharing SiO₄ and AlO₄ tetrahedra enclose regular cages and channels of molecular size (about 0.3–1 nm). Each Al carries a negative charge balanced by an exchangeable cation Mⁿ⁺ in the pores, and Löwenstein’s rule forbids Al–O–Al links, so Si/Al ≥ 1. Structures are built from secondary units such as the sodalite (β) cage: zeolite A joins sodalite cages through four-rings, faujasite (X, Y) through six-rings into larger supercages, and ZSM-5 has ten-ring channels.
- Ion exchange — the pore cations exchange with those in solution: Na-zeolites soften water by taking up Ca²⁺ and Mg²⁺, and were used in detergents in place of phosphates.
- Molecular sieves — pore windows admit molecules below a cut-off size: 3A (K⁺ form) dries solvents by admitting only water, 4A (Na⁺) and 5A (Ca²⁺) separate linear from branched alkanes.
- Catalysis — the H⁺ forms are strong solid Brønsted acids; zeolite Y cracks heavy petroleum fractions, and shape-selective H-ZSM-5 converts methanol to gasoline-range hydrocarbons and favours p-xylene, whose slim shape alone diffuses out of the channels quickly.
Key takeaways
- Seven crystal systems, fourteen Bravais lattices; d_hkl = a/√(h² + k² + l²) for cubic; nλ = 2d sin θ; bcc reflections need h + k + l even, fcc all odd or all even.
- Z = 1, 2, 4 for sc, bcc, fcc; packing 52%, 68%, 74%; ρ = ZM/(N_A a³); N octahedral and 2N tetrahedral holes per N spheres; radius-ratio limits 0.225, 0.414, 0.732.
- NaCl 6:6, CsCl 8:8, ZnS 4:4, CaF₂ 8:4, TiO₂ 6:3, perovskite Ti 6/Ca 12; Fe₃O₄ and NiFe₂O₄ are inverse spinels, Mn₃O₄ and Co₃O₄ normal, by octahedral site preference.
- Schottky defects lower density (NaCl), Frenkel defects do not (AgCl); F-centres colour alkali halides; Fe₁₋ₓO is metal-deficient.
- Metals conduct worse when hot, semiconductors better; n-type from donors, p-type from acceptors; zeolites exchange ions, sieve molecules and catalyse shape-selectively.
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.
A first-order Bragg reflection of Cu Kα X-rays (λ = 154.06 pm) is observed at θ = 20.0°. What is the interplanar spacing, in pm, to the nearest whole number?
Numerical answer — type the value.
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Answer: 225
nλ = 2d sin θ with n = 1: d = 154.06/(2 × sin 20.0°) = 154.06/(2 × 0.3420) = 225.2 pm, i.e. 225. Using 2θ = 20° as θ gives about 444 pm; diffractometers report 2θ, so the angle must be halved first.What is the spacing of the (111) planes in a cubic crystal with a = 400 pm, in pm, to one decimal place?
Numerical answer — type the value.
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Answer: 230.9
d_hkl = a/√(h² + k² + l²) = 400/√3 = 230.9 pm. Dividing by 3 instead of √3 gives 133.3 pm; for (200) the spacing would be 200 pm.A plane intersects the crystal axes at 2a, 3b and is parallel to c. What are its Miller indices?
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Answer: A — (320)
Intercepts 2, 3, ∞; reciprocals ½, ⅓, 0; multiplying by 6 clears fractions to (320). Writing the intercepts themselves gives (23∞), and inverting the order of the reciprocals gives (230).The first (lowest-angle) reflection in the powder pattern of a body-centred cubic metal is
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Answer: A — (110)
For an I lattice only h + k + l even survives, so 100 and 111 are absent and the lowest allowed is 110, followed by 200, 211 and 220. A face-centred lattice would start with 111 and a primitive one with 100.Copper is face-centred cubic with a = 361.5 pm and molar mass 63.55 g/mol. What is its density, in g/cm³, to two decimal places? (N_A = 6.022 × 10²³ mol⁻¹)
Numerical answer — type the value.
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Answer: 8.94
ρ = ZM/(N_A a³) with Z = 4 and a = 3.615 × 10⁻⁸ cm: a³ = 4.724 × 10⁻²³ cm³, so ρ = 4 × 63.55/(6.022 × 10²³ × 4.724 × 10⁻²³) = 254.2/28.45 = 8.94 g/cm³ (measured 8.96). Taking Z = 2 halves it to 4.47.Sodium chloride has the rock-salt structure with a = 564 pm. What is its density, in g/cm³, to two decimal places? (M = 58.44 g/mol, N_A = 6.022 × 10²³ mol⁻¹)
Numerical answer — type the value.
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Answer: 2.16
Rock salt has 4 NaCl formula units per cell. a³ = (5.64 × 10⁻⁸ cm)³ = 1.794 × 10⁻²² cm³, so ρ = 4 × 58.44/(6.022 × 10²³ × 1.794 × 10⁻²²) = 233.8/108.0 = 2.16 g/cm³. Counting 8 ions as 8 formula units doubles the answer.Iron has a cubic cell with a = 286.6 pm and a density of 7.87 g/cm³ (M = 55.85 g/mol). How many atoms are there per unit cell? (N_A = 6.022 × 10²³ mol⁻¹)
Numerical answer — type the value.
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Answer: 2
Z = ρN_A a³/M = 7.87 × 6.022 × 10²³ × (2.866 × 10⁻⁸)³/55.85 = 7.87 × 6.022 × 10²³ × 2.354 × 10⁻²³/55.85 = 1.998 ≈ 2, so α-iron is body-centred cubic. Forgetting to cube the edge in cm gives nonsense.In a face-centred cubic metal with a = 361.5 pm, what is the atomic radius, in pm, to one decimal place?
Numerical answer — type the value.
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Answer: 127.8
In fcc the atoms touch along the face diagonal: √2a = 4r, so r = √2 × 361.5/4 = 127.8 pm. Using the body diagonal (√3a = 4r, the bcc condition) gives 156.5 pm, and the edge (a = 2r) 180.8 pm.In a cubic close-packed array of N spheres, how many tetrahedral holes are there per sphere?
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Answer: 2
An fcc cell has 4 spheres, 8 tetrahedral holes (one inside each small cube) and 4 octahedral holes (body centre plus 12 edge centres shared by 4): 2 tetrahedral and 1 octahedral per sphere. That is why fluorite fills all tetrahedral holes with twice as many F⁻ as Ca²⁺.What fraction of space is occupied in a body-centred cubic structure of identical spheres? Give the answer to two decimal places.
Numerical answer — type the value.
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Answer: 0.68
Z = 2 and √3a = 4r, so the fraction is 2 × (4/3)πr³/a³ = π√3/8 = 0.680, i.e. 0.68. Close packing gives 0.74 and simple cubic 0.52.How many formula units of CaF₂ are there in one unit cell of the fluorite structure?
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Answer: 4
Ca²⁺ occupies the fcc positions, 8 × ⅛ + 6 × ½ = 4, and the 8 F⁻ ions sit wholly inside in the tetrahedral holes, giving Ca₄F₈ = 4 CaF₂. Counting ions (12) instead of formula units is the usual slip.Magnetite, Fe₃O₄, is an inverse spinel. How many Fe³⁺ ions per formula unit occupy tetrahedral sites?
Numerical answer — type the value.
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Answer: 1
Inverse spinel B[AB]O₄ = Fe³⁺[Fe²⁺Fe³⁺]O₄: one Fe³⁺ tetrahedral, and Fe²⁺ plus the other Fe³⁺ octahedral. High-spin Fe³⁺ (d⁵) has no site preference, while Fe²⁺ (d⁶) gains CFSE by going octahedral. A normal spinel would put Fe²⁺ in the tetrahedral site and both Fe³⁺ octahedral, giving 0.In the perovskite structure CaTiO₃, what are the coordination numbers of Ti⁴⁺ and Ca²⁺ by oxide?
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Answer: A — Ti 6, Ca 12
With Ti at the cube centre and O at the six face centres, Ti is octahedral; each corner Ca is surrounded by the 12 face-centre oxygens of the eight cells that share it (12 edge-midpoint positions in the alternative setting). The large A-site cation is what needs 12-coordination.Using the radius ratio r(Na⁺)/r(Cl⁻) = 102/181, what coordination is predicted for Na⁺ in NaCl?
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Answer: A — Octahedral (6)
102/181 = 0.56 lies between 0.414 and 0.732, the octahedral range, matching the observed 6:6 rock-salt structure. Cubic eight-coordination needs a ratio above 0.732, as in CsCl (about 0.93).Which statements about crystal defects are correct?
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Answer: A — Schottky defects lower the density of the crystal; B — Frenkel defects are common in AgCl and AgBr; C — F-centres are electrons trapped in anion vacancies
Missing ion pairs lower the mass per volume; the small, polarisable Ag⁺ moves easily into interstitial sites; electrons in anion vacancies absorb visible light and colour the crystal. A Frenkel defect only relocates an ion, so both composition and density are unchanged.A semiconductor has a band gap of 2.48 eV. What is the wavelength of its absorption edge, in nm? (hc = 1240 eV nm)
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Answer: 500
λ = hc/E_g = 1240/2.48 = 500 nm: light of shorter wavelength is absorbed, longer is transmitted. Silicon (1.1 eV) has its edge near 1130 nm, in the infrared, which is why it looks opaque and grey.How does the electrical conductivity of an intrinsic semiconductor change as the temperature rises?
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Answer: A — It increases steeply, because more electrons are promoted across the gap
The number of carriers rises as exp(−E_g/2kT), which outweighs the extra lattice scattering. In a metal the carrier number is fixed and scattering alone acts, so conductivity falls with temperature — the practical test that tells the two apart.Which statements about doped semiconductors and zeolites are correct?
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Answer: A — Phosphorus-doped silicon is n-type; B — Boron-doped silicon has holes as majority carriers; C — Löwenstein’s rule forbids Al–O–Al linkages in zeolites
P supplies donor electrons (n-type); B creates acceptor levels and holes (p-type); zeolites never place two AlO₄ tetrahedra on one oxygen. 3A has windows of about 0.3 nm that admit water and exclude almost everything larger — the reverse of the last statement, which is why it dries solvents.