Solid State: Crystal Systems and Close Packed Structures (JEE Advanced)

“Classification of solids, crystalline state, seven crystal systems (cell parameters a, b, c, α, β, γ), close packed structure of solids (cubic and hexagonal), packing in fcc, bcc and hcp lattices; Nearest neighbours, ionic radii and radius ratio, point defects”

Source: JEE Advanced 2026 Information Bulletin, Annexure-I. Closest JEE Main topic: Elastic behaviour, stress-strain relationship.

JEE Main focuses on mechanical properties of solids (stress-strain), but JEE Advanced requires you to understand the atomic-level architecture of crystals, packing efficiency calculations, and radius ratio rules for ionic compounds. This is a scoring topic if you master the 3D visualization and numerical relationships.

• Unit cell is the smallest repeating unit. Seven crystal systems differ in edge lengths (a, b, c) and angles (α, β, γ). Cubic is simplest: a = b = c, α = β = γ = 90°.
• Close packing: atoms touch along specific directions. In fcc and hcp, packing efficiency is 74%. In bcc, it is 68%. Simple cubic is only 52%.
• Coordination number = number of nearest neighbours. For fcc and hcp it is 12, for bcc it is 8, for simple cubic it is 6.
• In fcc: atoms at corners (8 × 1/8 = 1) and face centers (6 × 1/2 = 3), total = 4 atoms per unit cell. In bcc: corners + 1 body center = 2 atoms.
• Radius ratio (r+/r−) determines which void cations occupy in ionic crystals. For octahedral voids: 0.414 to 0.732. For tetrahedral voids: 0.225 to 0.414.
• Point defects: Schottky (missing cation-anion pair, density decreases), Frenkel (ion moves to interstitial site, density unchanged). Know which defect occurs in which type of compound.

• Packing efficiency = (Volume of atoms in unit cell / Volume of unit cell) × 100
• For fcc: a = 2√2 × r, where r is atomic radius
• For bcc: a = (4r)/√3
• Density ρ = (Z × M) / (a³ × Nₐ), where Z = atoms per unit cell, M = molar mass, Nₐ = Avogadro number
• Number of octahedral voids = Z (number of atoms in unit cell)
• Number of tetrahedral voids = 2Z
• For fcc: Z = 4, coordination number = 12, packing efficiency = 74%
• For bcc: Z = 2, coordination number = 8, packing efficiency = 68%

• Confusing the relationship between edge length and atomic radius for different structures. In fcc atoms touch along face diagonal, in bcc they touch along body diagonal.
• Forgetting to account for atom contributions at corners (1/8), edges (1/4), faces (1/2), and body center (1) when counting atoms per unit cell.
• Mixing up Schottky and Frenkel defects. Schottky occurs in compounds with similar-sized ions (like NaCl), Frenkel in compounds with large size difference (like AgBr, ZnS).
• Using wrong void positions. Octahedral voids are at edge centers and body center in fcc. Tetrahedral voids are located along body diagonals at 1/4 distance from corners.

Question. Silver crystallizes in fcc lattice. The density of silver is 10.5 g/cm³ and its atomic mass is 108 g/mol. Calculate the edge length of the unit cell and the atomic radius of silver. (Nₐ = 6.022 × 10²³)

Solved twice independently; both solutions agree.