Solid State

Solid state links crystal geometry to macroscopic density. Advanced tests unit-cell packing, coordination and void occupancy, density calculations, and point defects with their consequences. The packing-fraction and Z-per-cell facts, plus the density formula, are directly examinable.

Key results

• Unit cells: simple cubic (Z=1), body-centred (Z=2), face-centred (Z=4). Atoms per cell counted by sharing (corner 1/8, face 1/2, body 1).
• Packing efficiency: sc 52.4%, bcc 68%, fcc/hcp 74% (closest packing).
• Relations: sc a = 2r; bcc √3 a = 4r; fcc √2 a = 4r.
• Density: ρ = Z·M/(a³·N_A).
• Voids: fcc has octahedral voids = N (radius 0.414r) and tetrahedral = 2N (radius 0.225r). Ionic radius ratio decides coordination.
• Defects: Schottky (missing pairs, lowers density), Frenkel (ion displaced, density unchanged); doping gives F-centres/semiconduction.

Cell	Z	Packing	a–r relation
Simple cubic	1	52.4%	a = 2r
BCC	2	68%	√3 a = 4r
FCC	4	74%	√2 a = 4r

Exam Tricks & Tips

• 🎯 Use ρ = ZM/(a³N_A) to link density, edge length and molar mass — the single most-used solid-state numerical; solve for whichever is unknown.
• 🎯 FCC and HCP both pack at 74% (closest packing); bcc 68%, simple cubic only 52.4% — memorise the order.
• 🎯 Schottky defects lower density (ions missing); Frenkel defects keep density constant (an ion just moves to a void) — a favourite discriminator.
• 🎯 Octahedral voids = number of anions; tetrahedral voids = twice that — occupancy fixes the formula of an ionic solid.
• 🎯 Radius ratio predicts coordination: 0.225–0.414 tetrahedral(4), 0.414–0.732 octahedral(6), 0.732–1 cubic(8).
• ❌ Common mistake: miscounting shared atoms — corner atoms contribute 1/8, face atoms 1/2, edge 1/4, body 1; always weight before summing Z.

Expected exam pattern

1 question, typically a density/edge-length numerical or a void-occupancy/defect conceptual (multiple-correct). Packing-efficiency comparisons appear as quick checks.

Quick recap

Z: sc 1, bcc 2, fcc 4. Packing: fcc/hcp 74%, bcc 68%, sc 52.4%. ρ = ZM/(a³N_A). FCC voids: octahedral N, tetrahedral 2N. Schottky lowers density, Frenkel does not. Radius ratio sets coordination.

Cover the answer, recall, then check. 11 cards on the solid state for JEE Advanced.

Q1. Atoms per unit cell for simple cubic, bcc, and fcc.
A1. 1, 2, and 4 respectively.

Q2. Packing efficiencies of sc, bcc, and fcc/hcp.
A2. 52.4%, 68%, and 74%.

Q3. Density formula for a crystal.
A3. ρ = Z·M/(a³·N_A).

Q4. Edge–radius relation for a bcc lattice.
A4. √3 a = 4r.

Q5. Edge–radius relation for an fcc lattice.
A5. √2 a = 4r.

Q6. Number of octahedral and tetrahedral voids in an fcc lattice of N atoms.
A6. Octahedral = N; tetrahedral = 2N.

Q7. Effect of a Schottky defect on density.
A7. Density decreases (equal numbers of cations and anions are missing).

Q8. Effect of a Frenkel defect on density.
A8. No change — an ion merely moves to an interstitial void.

Q9. Contribution of a corner atom and a face atom to the unit cell.
A9. Corner 1/8, face 1/2.

Q10. Radius-ratio range for octahedral coordination.
A10. 0.414 to 0.732 (coordination number 6).

Q11. What are F-centres?
A11. Anion vacancies occupied by trapped electrons, giving the crystal colour.

Solid state is a high-scoring, geometry-heavy topic: unit cells, packing efficiency, and density calculations reward spatial reasoning. JEE Advanced tests the cubic lattices quantitatively, void occupancy, and how defects tune real materials' properties.

Core concept: crystalline solids repeat a unit cell in three dimensions; counting atoms per cell and relating edge length to atomic radius gives density, packing efficiency and coordination number.

Core ideas

Beginner — unit cells and atom counting

Contribution rules: a corner atom counts 1/8, an edge atom 1/4, a face atom 1/2, a body-centre atom 1. So:

• Simple cubic (SCC): 8×1/8 = 1 atom, coordination number 6, packing 52%.
• Body-centred cubic (BCC): 8×1/8 + 1 = 2 atoms, CN 8, packing 68%.
• Face-centred cubic (FCC/CCP): 8×1/8 + 6×1/2 = 4 atoms, CN 12, packing 74% (the densest).
  Hexagonal close packing (HCP) also achieves 74%.

Intermediate — radius–edge relations and density

Relating atomic radius r to edge a: SCC a = 2r; BCC √3 a = 4r; FCC √2 a = 4r. Density ρ = (Z·M)/(a³·N_A), where Z = atoms per cell. This one formula, run forward or backward, answers most numericals — find density, or find M or a from density.

Advanced — voids, ionic solids and defects

Close packing leaves tetrahedral voids (2 per atom, radius ratio 0.225–0.414) and octahedral voids (1 per atom, ratio 0.414–0.732). Ionic structures place cations in voids sized by the radius ratio rule: NaCl (rock salt, FCC Cl⁻ with Na⁺ in all octahedral voids, 6:6), ZnS (zinc blende, cations in half the tetrahedral voids, 4:4), CaF₂ (fluorite, 8:4), CsCl (body-centred, 8:8). Defects: Schottky (missing ion pairs, lowers density, in high-CN ionics like NaCl) and Frenkel (ion displaced to an interstitial, density unchanged, in low-CN ionics like AgCl). Non-stoichiometric defects and doping create colour centres (F-centres) and semiconducting/magnetic behaviour.

Worked example

An element crystallises in FCC with edge length 400 pm and density 6.25 g/cm³. Find its molar mass (N_A = 6.022×10²³). For FCC, Z = 4. ρ = ZM/(a³N_A), so M = ρa³N_A/Z. a = 400 pm = 4×10⁻⁸ cm, a³ = 6.4×10⁻²³ cm³. M = (6.25 × 6.4×10⁻²³ × 6.022×10²³)/4 = (6.25 × 6.4 × 6.022×10⁰)/4 ≈ (240.8)/4 ≈ 60.2 g/mol. Getting Z right for the lattice type is the make-or-break step.

How JEE Advanced tests this

Density ↔ molar mass ↔ edge-length calculations for SCC/BCC/FCC; packing-efficiency and void-count reasoning; radius-ratio prediction of ionic structure and coordination number; formula determination from occupied voids (e.g. "cations fill 1/3 of octahedral voids → formula"); and Schottky vs Frenkel defect effects on density.

Exam tricks & shortcuts

• Atoms/cell: SCC 1, BCC 2, FCC 4. Density ρ = ZM/(a³N_A).
• Radius–edge: SCC 2r, BCC √3a=4r, FCC √2a=4r.
• Tetrahedral voids = 2×atoms, octahedral = 1×atoms.
• Schottky lowers density (NaCl); Frenkel keeps it (AgCl).

Key formulas

• Density of unit cell: d = ZM/(a³ N_A) (Z = atoms per cell).
• Z: simple cubic 1, BCC 2, FCC 4.
• Packing efficiency: SC 52.4%, BCC 68%, FCC/HCP 74%.
• Relations: BCC √3 a = 4r; FCC √2 a = 4r; SC a = 2r.
• Radius ratio determines coordination number; defects: Schottky (density ↓), Frenkel.

Unit-cell geometry connects edge length, atomic radius and density; packing efficiency ranks the lattice types.

Worked Example

Problem: An element crystallises in a face-centred cubic (FCC) lattice with edge length a = 400 pm and atomic mass M = 60 g/mol. Calculate its density. (N_A = 6.022 × 10²³ mol⁻¹.)

Solution:
For an FCC unit cell, the number of atoms per cell is Z = 4.
Convert the edge to cm: a = 400 pm = 4.00 × 10⁻⁸ cm, so a³ = (4.00 × 10⁻⁸)³ = 6.4 × 10⁻²³ cm³.

Density formula:
ρ = (Z M) / (a³ N_A)
= (4 × 60) / (6.4 × 10⁻²³ × 6.022 × 10²³)
= 240 / (38.54)
≈ 6.23 g/cm³.

Answer: The density of the element is ≈ 6.23 g/cm³.