GATE Solid State — Crystal Structures, Packing and Density
Solid state is one of the most calculable areas of the GATE Chemistry syllabus. Almost every question reduces to three relationships: how the atomic radius is tied to the edge length, how many formula units sit in one cell, and how those two give the density. If you can derive the packing fractions rather than recall them, you can rebuild the topic from scratch in the exam hall. This article does the derivations and then works the standard numericals in full. A separate Knowledge Base article covers the IIT-JAM treatment of unit-cell counting and routine density problems, so the two do not repeat each other.
The three cubic cells and their geometry
Count lattice points the standard way: a corner atom is shared by 8 cells, a face atom by 2, an edge atom by 4, and a body-centre atom belongs entirely to its own cell.
| Cell | Atoms per cell, Z | Touching direction | a in terms of r | Coordination no. |
|---|---|---|---|---|
| Simple cubic | 8 × ⅛ = 1 | Along the edge | a = 2r | 6 |
| Body-centred cubic | 8 × ⅛ + 1 = 2 | Body diagonal, √3·a = 4r | a = 4r/√3 | 8 |
| Face-centred cubic | 8 × ⅛ + 6 × ½ = 4 | Face diagonal, √2·a = 4r | a = 2√2·r | 12 |
Hexagonal close packing has the same coordination number and efficiency as fcc; only the stacking differs, ABAB… against ABCABC… For ideal hcp the axial ratio is fixed by geometry:
Deriving the packing fraction
Packing fraction = volume occupied by the spheres ÷ volume of the cell. Both parts must be in terms of the same variable, and the a–r relation is what allows that.
Simple cubic. Z = 1, a = 2r.
Occupied volume = 1 × (4/3)πr³. Cell volume = (2r)³ = 8r³.
PF = (4/3)πr³ / 8r³ = π/6 = 0.5236 → 52.36%, empty space 47.64%.
Body-centred cubic. Z = 2, a = 4r/√3.
Occupied = 2 × (4/3)πr³ = (8/3)πr³.
Cell volume = (4r/√3)³ = 64r³ / 3√3.
PF = (8π/3)r³ × (3√3 / 64r³) = 8π√3 / 64 = π√3 / 8
= 3.1416 × 1.7321 ÷ 8 = 5.4414 ÷ 8 = 0.6802 → 68.02%, empty space 31.98%.
Face-centred cubic (and hcp). Z = 4, a = 2√2·r.
Occupied = 4 × (4/3)πr³ = (16/3)πr³.
Cell volume = (2√2·r)³ = 8 × 2√2 · r³ = 16√2 · r³.
PF = (16π/3) / (16√2) = π / (3√2) = 3.1416 ÷ 4.2426
= 0.7405 → 74.05%, empty space 25.95%.
74.05% is the maximum possible for identical hard spheres — which is why so many metals adopt ccp or hcp.
Density from the unit cell
One cell contains Z formula units of molar mass M, so its mass is ZM/NA, and its volume for a cubic cell is a³:
Unit discipline decides this question: 1 pm = 10⁻¹⁰ cm, so 361.5 pm is 3.615 × 10⁻⁸ cm, and cubing turns a 10× slip into a 1000× error.
Worked example 1 — density of an fcc metal. Copper crystallises fcc with a = 361.5 pm; M = 63.546 g mol⁻¹, Z = 4.
a = 3.615 × 10⁻⁸ cm
a³ = (3.615)³ × 10⁻²⁴ = 47.24 × 10⁻²⁴ = 4.724 × 10⁻²³ cm³
a³·NA = 4.724 × 10⁻²³ × 6.022 × 10²³ = 28.45 cm³ mol⁻¹
Z·M = 4 × 63.546 = 254.18 g mol⁻¹
ρ = 254.18 / 28.45 = 8.93 g cm⁻³
Cross-check by the reverse route. Molar volume = M/ρ = 63.546 / 8.93 = 7.116 cm³ mol⁻¹. Volume per atom = 7.116 / (6.022 × 10²³) = 1.181 × 10⁻²³ cm³, and with 4 atoms per cell the cell volume must be 4 × 1.181 × 10⁻²³ = 4.725 × 10⁻²³ cm³ — the same a³ we started from, so the working is consistent. The radius follows as well: r = a/(2√2) = 361.5 / 2.828 = 127.8 pm.
Worked example 2 — going backwards to find Z. A metal of M = 51.996 g mol⁻¹ has a cubic cell with a = 288.4 pm and ρ = 7.20 g cm⁻³. Which cubic lattice is it?
Rearrange: Z = ρ·a³·NA / M
a³ = (2.884)³ × 10⁻²⁴ = 23.99 × 10⁻²⁴ = 2.399 × 10⁻²³ cm³
a³·NA = 2.399 × 10⁻²³ × 6.022 × 10²³ = 14.45 cm³ mol⁻¹
Z = 7.20 × 14.45 / 51.996 = 104.0 / 51.996 = 2.00 → Z = 2, so body-centred cubic
Then r = √3·a/4 = 1.7321 × 288.4 / 4 = 499.5 / 4 = 124.9 pm. Z must come out near a whole number; if it does not, the arithmetic or the unit conversion is wrong.
Worked example 3 — an ionic solid. Rock salt (NaCl) is fcc with 4 formula units per cell; a = 564.0 pm, M = 58.44 g mol⁻¹.
a³ = (5.640)³ × 10⁻²⁴ = 179.41 × 10⁻²⁴ = 1.7941 × 10⁻²² cm³
a³·NA = 1.7941 × 10⁻²² × 6.022 × 10²³ = 108.03 cm³ mol⁻¹
Z·M = 4 × 58.44 = 233.76
ρ = 233.76 / 108.03 = 2.16 g cm⁻³
Z = 4 here means four NaCl units — 4 Na⁺ and 4 Cl⁻ — so M must be the formula mass, not an atomic mass.
Holes, radius ratio and structure type
In a close-packed array of N anions there are exactly N octahedral holes and 2N tetrahedral holes. Which holes the cations occupy is set largely by the radius ratio r₊/r₋.
| r₊/r₋ | Hole occupied | Coordination | Example ratio |
|---|---|---|---|
| 0.155–0.225 | Trigonal planar | 3 | — |
| 0.225–0.414 | Tetrahedral | 4 | ZnS: 60/184 = 0.33 |
| 0.414–0.732 | Octahedral | 6 | NaCl: 102/181 = 0.56 |
| 0.732–1.000 | Cubic | 8 | CsCl: 174/181 = 0.96 |
Always take ionic radii from the data table supplied with the question, because a radius quoted for 6-coordination is not the 4-coordinate value. Treat the radius ratio as a guide, not a law: it fails where the bonding has significant covalent character.
One naming trap: caesium chloride is not body-centred cubic. The body-centre ion is a different species, so the lattice is primitive cubic with a two-ion basis. The rock salt, zinc blende, fluorite and antifluorite structure types are described in the companion IIT-JAM article.
Diffraction: Bragg's law and systematic absences
Worked example 4. A first-order reflection from the (111) planes of a cubic crystal appears at θ = 20.0° with λ = 154.0 pm. Find a.
sin 20.0° = 0.3420
d = λ / (2 sin θ) = 154.0 / (2 × 0.3420) = 154.0 / 0.6840 = 225.1 pm
√(1² + 1² + 1²) = √3 = 1.7321
a = d × √3 = 225.1 × 1.7321 = 389.9 pm
Reflection conditions are a standard one-mark item: for a primitive cubic lattice all hkl are allowed; for body-centred, only h + k + l even; for face-centred, only h, k, l all odd or all even. That is why the first observed lines differ between bcc and fcc powders, and how the lattice is assigned from the pattern.
- pm → cm conversion. Use 1 pm = 10⁻¹⁰ cm, then cube. This single step causes more wrong answers in this topic than any concept.
- Using an atomic mass where the formula mass is required in ionic density problems.
- Mixing up the touching direction: edge for sc, body diagonal for bcc, face diagonal for fcc. Everything else follows from this one choice.
- Forgetting that Z need not be an integer for a defective crystal — non-stoichiometric solids are exactly where fractional occupancy appears.
- Assuming the radius-ratio rule always predicts the structure. It is a guide; say so, and cite covalent character as the exception.
Question types to expect
| Style | What you must produce |
|---|---|
| Numerical (NAT) | ρ from a and Z, or a or Z from ρ; r from a; d from θ |
| Structure assignment | Lattice type from systematic absences or from Z |
| Counting | Formula from fractional site occupancies; number of holes filled |
| Conceptual | Schottky vs Frenkel defects and their effect on density |
| Derivation | Packing fraction or the a–r relation, shown step by step |
Take the syllabus and question pattern for your session from the current official notification.
The concepts are short; the arithmetic is where time goes. Cubing a number in 10⁻⁸ cm, multiplying by Avogadro's number and dividing by a molar mass is exactly the kind of exponent-heavy chain that goes wrong under pressure. Run each step through the scientific calculator and confirm the powers of ten before you commit to an answer.
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