JAM Solid State — Unit Cells and Density Calculations
Solid state is one of the few IIT-JAM topics where you can reach a certain answer in under two minutes, provided you are fluent with two things: counting how much of an atom belongs to a unit cell, and handling the picometre-to-centimetre conversion inside the density formula. This article drills exactly those. The geometric derivations of the packing fractions, the radius-ratio table and Bragg's law are in the companion GATE-level article linked at the end.
Counting atoms in a unit cell
An atom on a shared position belongs partly to the neighbouring cells, and the share is fixed by how many cells meet there.
| Position | Shared between | Contribution to one cell | How many such positions |
|---|---|---|---|
| Corner | 8 cells | 1/8 | 8 |
| Edge centre | 4 cells | 1/4 | 12 |
| Face centre | 2 cells | 1/2 | 6 |
| Body centre | 1 cell | 1 | 1 |
That gives the three cubic cells at once: simple cubic Z = 8 × 1/8 = 1; body-centred Z = 8 × 1/8 + 1 = 2; face-centred Z = 8 × 1/8 + 6 × 1/2 = 4. An end-centred cell (two opposite faces occupied) gives Z = 2.
Worked example 1 — formula from site occupancy. These are the fastest marks in the topic, and they are pure counting.
(a) A at all corners, B at all face centres.
A = 8 × 1/8 = 1; B = 6 × 1/2 = 3 → formula AB₃
(b) A at corners and at the body centre, B at all face centres.
A = 8 × 1/8 + 1 = 2; B = 6 × 1/2 = 3 → formula A₂B₃
(c) A at corners, B at the body centre, and the atoms at two opposite corners are
missing.
A = 6 × 1/8 = 0.75; B = 1 → ratio 0.75 : 1 = 3 : 4 → formula A₃B₄
(d) A at corners, B at edge centres, C at the body centre.
A = 8 × 1/8 = 1; B = 12 × 1/4 = 3; C = 1 → formula AB₃C
Write the counts as numbers first, then assemble the formula.
The density equation, and how to use it both ways
ρ in g cm⁻³ · Z = formula units per cell · M in g mol⁻¹ · a in cm · NA = 6.022 × 10²³ mol⁻¹
The conversion that decides the question: 1 pm = 10⁻¹⁰ cm, 1 Å = 10⁻⁸ cm. Because a is cubed, a factor-of-ten slip becomes a factor of a thousand — and a metal is never 0.008 or 8000 g cm⁻³, so use that absurdity as a free check.
Worked example 2 — density of a bcc metal. Sodium crystallises in a body-centred cubic lattice with a = 429.0 pm. M(Na) = 22.99 g mol⁻¹.
Z for bcc = 2
a = 429.0 pm = 4.290 × 10⁻⁸ cm
a³ = (4.290)³ × 10⁻²⁴ = 78.95 × 10⁻²⁴ = 7.895 × 10⁻²³ cm³
a³ × NA = 7.895 × 10⁻²³ × 6.022 × 10²³ = 47.55 cm³ mol⁻¹
Z × M = 2 × 22.99 = 45.98 g mol⁻¹
ρ = 45.98 / 47.55 = 0.967 g cm⁻³
Bonus part that is usually asked next: the metallic radius. In bcc the atoms touch
along the body diagonal, so √3·a = 4r:
r = √3 × 429.0 / 4 = 1.7321 × 429.0 / 4 = 743.1 / 4 = 185.8 pm
Worked example 3 — running the formula backwards for M. A metal crystallises fcc with a = 408.6 pm and density 10.5 g cm⁻³. Find its molar mass.
Rearranging: M = ρ·a³·NA / Z, with Z = 4 for fcc.
a = 4.086 × 10⁻⁸ cm
a³ = (4.086)³ × 10⁻²⁴ = 68.21 × 10⁻²⁴ = 6.821 × 10⁻²³ cm³
a³ × NA = 6.821 × 10⁻²³ × 6.022 × 10²³ = 41.08 cm³ mol⁻¹
ρ × 41.08 = 10.5 × 41.08 = 431.3 g mol⁻¹ (this is the mass of 4 moles of atoms)
M = 431.3 / 4 = 107.8 g mol⁻¹
That matches silver. The same rearrangement finds Z when M and ρ are given — and Z must land on a whole number, which is a built-in check on the arithmetic.
Packing efficiency and empty space
Quote these instantly; the derivations are in the GATE-level companion article.
| Arrangement | Z | a in terms of r | Coordination no. | Packing | Empty space |
|---|---|---|---|---|---|
| Simple cubic | 1 | a = 2r | 6 | 52.36% | 47.64% |
| Body-centred cubic | 2 | a = 4r/√3 | 8 | 68.02% | 31.98% |
| Face-centred cubic (ccp) | 4 | a = 2√2·r | 12 | 74.05% | 25.95% |
| Hexagonal close packed | 6 (hexagonal cell) | c/a = 1.633 | 12 | 74.05% | 25.95% |
ccp and hcp have identical efficiency and coordination; only the stacking differs, ABCABC… against ABAB…
Voids, and formulae built from them
An octahedral void is surrounded by 6 spheres and is the larger; a tetrahedral void is surrounded by 4 and is smaller. Most ionic-formula questions are this counting rule in disguise.
Worked example 4 — formula from void occupancy. Oxide ions form a ccp array. Cations A occupy one-eighth of the tetrahedral voids and cations B occupy one-half of the octahedral voids. Find the formula.
Take N = number of O²⁻ ions.
Tetrahedral voids = 2N; A fills 1/8 of them → A = 2N × 1/8 = N/4
Octahedral voids = N; B fills 1/2 of them → B = N/2
Ratio A : B : O = N/4 : N/2 : N. Multiply through by 4: A : B : O = 1 : 2 : 4
Formula: AB₂O₄
Check the charges are consistent: four O²⁻ carry −8, so A and B must supply +8 between them — for example A as +2 and the two B as +3 each. A formula that cannot be balanced for charge signals an arithmetic slip.
The standard structure types in words: rock salt — fcc anions, every octahedral void filled, 6:6, four formula units per cell; zinc blende — fcc anions, half the tetrahedral voids, 4:4; fluorite — fcc cations, all tetrahedral voids filled by anions, 8:4; antifluorite — the ions exchanged, as in Na₂O.
Defects, and their effect on density
- Schottky — a cation and an anion are both missing, so the crystal stays neutral but holds fewer ions in the same volume: the density decreases. Common when the ions are similar in size, as in NaCl and KCl.
- Frenkel — an ion (usually the smaller cation) moves from its lattice site into an interstitial site. Nothing leaves the crystal, so the density is unchanged. Common when the cation is much smaller, as in AgCl and ZnS.
- Metal excess with anion vacancies — an electron fills the vacancy to keep neutrality. That trapped electron is an F-centre and absorbs visible light, which is why non-stoichiometric alkali halides are coloured.
- Impurity defects — each Sr²⁺ replacing Na⁺ in NaCl creates one cation vacancy, because two Na⁺ must leave to balance the charge.
Both Schottky and Frenkel defects raise electrical conductivity, by giving ions a route to move.
- The pm → cm conversion. 1 pm = 10⁻¹⁰ cm, then cube. This is the single largest source of wrong answers in the topic.
- Using an atomic mass where a formula mass is needed. For NaCl, Z = 4 means four NaCl units, so M = 58.44, not 22.99.
- Taking the wrong touching direction: edge for simple cubic, body diagonal for bcc, face diagonal for fcc.
- Saying a Frenkel defect lowers the density. Nothing leaves the crystal, so mass and volume are both unchanged.
- Confusing tetrahedral and octahedral counts. Tetrahedral voids are twice as numerous and are the smaller ones.
Exam relevance
| Question style | Method |
|---|---|
| Numerical: density from a and Z | ρ = ZM/(a³NA), watch the units |
| Numerical: M, a or Z from ρ | The same equation rearranged; Z must be an integer |
| Radius from edge length | a = 2r, 4r/√3 or 2√2·r as appropriate |
| Formula from site or void occupancy | Fractional-share counting |
| Percentage empty space | Quote the packing fraction and subtract from 100 |
| Conceptual: defect type | Effect on density, neutrality and conductivity |
Check the current official notification for the syllabus and pattern for your session.
The method is short; the arithmetic is the risk. Cubing an edge length in 10⁻⁸ cm, multiplying by Avogadro's number and dividing by a molar mass is exactly where a correct method turns into a wrong answer. Do those steps in the scientific calculator and use the molar mass tool to get M right for ionic formulae before you start.
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