The Solid State — Crystal Systems, Close Packing, Radius Ratio and Point Defects

A crystal is a pattern repeated in three dimensions, and almost every question about one reduces to geometry you can draw: how many atoms a unit cell really owns, how close two of them can come, what fraction of space they fill, and which holes a smaller ion fits into. The chapter builds those four in order — unit cells, close packing and its voids, packing efficiency and density, then ionic radii and the radius ratio — and ends with the imperfections, because real crystals are never perfect and the type of defect decides whether the density changes. Every formula here comes from one picture of spheres touching along a line, so learn the line, not the formula.

Crystal Systems and Unit Cells

Crystalline solids have long-range order and sharp melting points; amorphous solids such as glass have only short-range order and soften over a range. A crystal's unit cell is described by three edge lengths a, b, c and three angles α, β, γ, and those six parameters sort every crystal into one of seven crystal systems. Placing lattice points only at corners gives a primitive cell; adding them at the body centre, the face centres or one pair of end faces gives the centred cells, fourteen Bravais lattices in all.

The seven crystal systems
System	Edges	Angles
Cubic	a = b = c	α = β = γ = 90°
Tetragonal	a = b ≠ c	α = β = γ = 90°
Orthorhombic	a ≠ b ≠ c	α = β = γ = 90°
Rhombohedral	a = b = c	α = β = γ ≠ 90°
Hexagonal	a = b ≠ c	α = β = 90°, γ = 120°
Monoclinic	a ≠ b ≠ c	α = γ = 90°, β ≠ 90°
Triclinic	a ≠ b ≠ c	α ≠ β ≠ γ ≠ 90°

How many atoms does a cubic cell own? A corner atom is shared by 8 cells, a face atom by 2, an edge atom by 4, and a body-centre atom by none. So the simple cubic cell owns 8 × 1/8 = 1, body-centred cubic (bcc) owns 1 + 1 = 2, and face-centred cubic (fcc) owns 1 + 6 × 1/2 = 4. This number, Z, is what enters the density formula.

Close Packing, Voids and Packing Efficiency

Equal spheres pack most tightly in layers where each sphere touches six in its own layer. The second layer sits in half the hollows of the first. The third can sit directly over the first (ABAB…, hexagonal close packing, hcp) or over the hollows the second did not use (ABCABC…, cubic close packing, ccp, which is the fcc lattice). Both give each sphere 12 nearest neighbours and fill 74% of space. In any close packing of N spheres there are N octahedral voids (surrounded by six spheres) and 2N tetrahedral voids (surrounded by four).

Where the spheres touch, and what follows
Cell	Spheres touch along	Relation	Z	Coordination number	Packing efficiency
Simple cubic	An edge	a = 2r	1	6	52.4%
Body-centred cubic	The body diagonal	√3 a = 4r	2	8	68%
Face-centred cubic (ccp)	A face diagonal	√2 a = 4r	4	12	74%

🎯 Worked reasoning: density from the unit cell

Density is the mass the cell owns over its volume: ρ = Z M/(N_A a³). For an fcc metal of molar mass 60 g mol⁻¹ and edge 400 pm: a = 4 × 10⁻⁸ cm, a³ = 6.4 × 10⁻²³ cm³, and ρ = 4 × 60/(6.022 × 10²³ × 6.4 × 10⁻²³) ≈ 6.23 g cm⁻³. Convert the edge to centimetres before cubing — leaving it in picometres is the most common way to get an absurd answer.

Ionic Radii and the Radius Ratio

In an ionic crystal the larger ion (usually the anion) forms the packing and the smaller one sits in its voids. A cation is stable in a void only if it is at least big enough to touch all the anions around it; if it rattles, the structure switches to a void with fewer neighbours. That geometry gives limiting values of the radius ratio r₊/r₋: below 0.225 a cation takes a linear or trigonal site; 0.225–0.414 tetrahedral (coordination 4); 0.414–0.732 octahedral (6); 0.732–1 cubic (8). Rock salt (6 : 6), CsCl (8 : 8), zinc blende (4 : 4) and fluorite (8 : 4) are the type structures to know.

🧠 Reading a formula off the voids

If anions form a ccp array (N of them) and cations fill a fraction f of the tetrahedral voids, there are 2fN cations per N anions. Two-thirds of the tetrahedral voids gives (4/3)N cations — a ratio of 4 : 3, so the formula is M₄X₃. One-third gives (2/3)N, a ratio of 2 : 3, so M₂X₃. Always count the 2N before applying the fraction.

Point Defects

Point defects are irregularities at single lattice sites. In ionic solids the two stoichiometric defects keep the formula unchanged. A Schottky defect is a pair of missing ions — one cation and one anion leave the crystal — so mass falls at the same volume and the density decreases; it is favoured when the two ions are of similar size (NaCl, KCl, CsCl). A Frenkel defect is an ion, usually the small cation, that leaves its site for an interstitial position; nothing leaves the crystal, so the density is unchanged; it is favoured when the ions differ greatly in size (ZnS, AgCl). AgBr shows both.

Non-stoichiometric defects change the formula slightly. In the metal-excess case, heating NaCl in sodium vapour leaves anion vacancies occupied by electrons — F-centres — which absorb visible light and colour the crystal yellow. In the metal-deficiency case, as in FeO, some Fe²⁺ become Fe³⁺ and cation sites stay empty to balance charge. Impurity defects arise when an ion of different charge is introduced: each Sr²⁺ that replaces Na⁺ in NaCl brings one cation vacancy with it to keep the crystal neutral.

⚠️ Schottky or Frenkel — ask whether anything left

Only a defect that removes matter from the crystal can lower its density. Schottky removes ion pairs; Frenkel only relocates an ion inside the crystal. If a question asks which defect lowers density, Frenkel is the distractor every time.

Key takeaways

Seven crystal systems from six cell parameters; a cubic cell owns 1 (simple), 2 (bcc) or 4 (fcc) atoms.
Spheres touch along the edge (a = 2r), the body diagonal (√3a = 4r) or the face diagonal (√2a = 4r); packing efficiencies are 52.4%, 68% and 74%.
N close-packed spheres create N octahedral and 2N tetrahedral voids; hcp and ccp both give 12 nearest neighbours.
Density ρ = ZM/(N_A a³), with a in centimetres for g cm⁻³.
Radius-ratio limits 0.225, 0.414 and 0.732 separate coordination numbers 4, 6 and 8.
Schottky defects lower density; Frenkel defects do not; F-centres colour metal-excess crystals; aliovalent impurities create vacancies.

Practice questions (8)

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.

The number of atoms effectively owned by one body-centred cubic unit cell is:
1. 1
2. 9
3. 4
4. 2
Show answer
Answer: D — 2
Eight corner atoms are each shared by eight cells, contributing 8 × 1/8 = 1, and the body-centre atom belongs wholly to the cell: 1 + 1 = 2. Nine counts every atom drawn in the picture as if none were shared, and 4 is the fcc count.
In an ionic solid the cation-to-anion radius ratio is 0.52. The cation most probably occupies a void of coordination number:
1. 6 (octahedral)
2. 4 (tetrahedral)
3. 8 (cubic)
4. 3 (trigonal)
Show answer
Answer: A — 6 (octahedral)
0.52 lies between 0.414 and 0.732, the range in which a cation just touches six anions around an octahedral hole, so the coordination number is 6, as in NaCl. Tetrahedral sites need 0.225–0.414 and cubic sites above 0.732; picking 8 assumes a bigger ratio means more neighbours without checking the band.
In a face-centred cubic lattice of edge length a, the distance between nearest-neighbour atoms is:
1. a
2. √3 a/2
3. a/√2
4. a/2
Show answer
Answer: C — a/√2
In fcc the atoms touch along a face diagonal of length √2 a, which holds a corner atom, the face-centre atom and another corner: the nearest-neighbour distance is half of it, √2 a/2 = a/√2 (equal to 2r). √3 a/2 is the bcc value, from the body diagonal, and a is the simple-cubic value.
In a cubic close-packed arrangement of N identical spheres, the number of tetrahedral voids is:
1. N/2
2. N
3. 4N
4. 2N
Show answer
Answer: D — 2N
Every close packing has N octahedral and 2N tetrahedral voids. In an fcc cell (Z = 4) the eight tetrahedral voids sit one inside each small cube of the cell, eight per four atoms, which is 2N. Answering N confuses them with the octahedral voids.
Anions X form a cubic close-packed array, and cations M occupy one-third of the tetrahedral voids. The formula of the compound is:
1. MX₃
2. M₂X₃
3. M₃X₂
4. MX
Show answer
Answer: B — M₂X₃
N anions give 2N tetrahedral voids; one-third of them is 2N/3 cations. The ratio M : X = 2/3 : 1 = 2 : 3, so the formula is M₂X₃. MX₃ applies the fraction to N instead of 2N, the usual slip, and M₃X₂ inverts the ratio.
Which of these ionic solids shows both Schottky and Frenkel defects?
1. AgBr
2. NaCl
3. ZnS
4. CsCl
Show answer
Answer: A — AgBr
Silver bromide is the standard case of both: Ag⁺ is small and mobile enough to move into interstitial sites (Frenkel), and ion-pair vacancies also occur (Schottky). NaCl and CsCl, with similar-sized ions, show mainly Schottky defects; ZnS, with a small cation in a large anion lattice, shows mainly Frenkel.
Sodium chloride is doped with 2 × 10⁻³ mol% of SrCl₂, each Sr²⁺ replacing an Na⁺. The number of cation vacancies created per mole of NaCl is about:
1. 2.4 × 10¹⁹
2. 1.2 × 10²¹
3. 1.2 × 10¹⁹
4. 6.0 × 10¹⁸
Show answer
Answer: C — 1.2 × 10¹⁹
One Sr²⁺ in place of one Na⁺ adds an extra positive charge, so one further Na⁺ site must stay empty: one vacancy per Sr²⁺. 2 × 10⁻³ mol% means 2 × 10⁻⁵ mol of SrCl₂ per mole of NaCl, giving 2 × 10⁻⁵ × 6.022 × 10²³ ≈ 1.2 × 10¹⁹ vacancies. Forgetting to divide the percentage by 100 gives 1.2 × 10²¹, and counting two vacancies per Sr²⁺ gives 2.4 × 10¹⁹.
A metal crystallises in a body-centred cubic lattice with edge length 300 pm. Its atomic radius is about:
1. 150 pm
2. 106 pm
3. 130 pm
4. 260 pm
Show answer
Answer: C — 130 pm
In bcc the atoms touch along the body diagonal: √3 a = 4r, so r = √3 × 300/4 ≈ 129.9 ≈ 130 pm. 150 pm uses the simple-cubic relation a = 2r, 106 pm the fcc relation √2 a = 4r, and 260 pm is the nearest-neighbour distance 2r rather than the radius.