Crystal Systems — Unit Cell Mnemonic
Target: mnemonic for crystal systems unit cell BCC FCC chemistry
Why is this hard to memorize?
Crystalline solids have atoms arranged in a regular 3D pattern described by unit cells. The three cubic unit cells are: Simple Cubic (SC: 1 atom/cell, 52% packing), Body-Centred Cubic (BCC: 2 atoms/cell, 68% packing), and Face-Centred Cubic (FCC/CCP: 4 atoms/cell, 74% packing). The seven crystal systems range from cubic (most symmetric) to triclinic (least symmetric). JEE tests unit cell calculations (density, edge length, number of atoms), packing efficiency, coordination numbers, and void types (tetrahedral and octahedral).
Classic mnemonics you should know
"SC=1, BCC=2, FCC=4. "1-2-4" — count: corner(1/8×8=1), body(1), face(1/2×6=3)"
Simple cubic: 8 corners × 1/8 = 1 atom. BCC: 8 corners × 1/8 + 1 body centre = 2 atoms. FCC: 8 corners × 1/8 + 6 faces × 1/2 = 4 atoms. These counts are fundamental to all unit cell calculations.
"SC=52%, BCC=68%, FCC/HCP=74%. "FCC is the most efficient cubic packing""
Packing efficiency = volume of atoms / volume of unit cell × 100%. FCC and HCP both achieve 74% (closest packing). BCC is 68%. SC is only 52% (lots of empty space). Real metals use FCC (Cu, Al, Au) or BCC (Fe, W, Cr) or HCP (Zn, Mg, Ti).
"SC: CN=6. BCC: CN=8. FCC: CN=12. "6-8-12" increasing with packing"
Coordination number = number of nearest neighbours touching each atom. SC: 4 in same layer + 1 above + 1 below = 6. BCC: 8 atoms at body diagonals. FCC: 12 (the close-packed arrangement). Higher CN = more efficient packing.
The complete list
- SC: 1 atom, CN=6, 52% packing
- BCC: 2 atoms, CN=8, 68% packing
- FCC: 4 atoms, CN=12, 74% packing
- HCP: also 74% packing (same as FCC)
- Density: ρ = nM/(a³Nₐ)
- Tetrahedral void: 2 per atom in FCC
- Octahedral void: 1 per atom in FCC
- 7 crystal systems, 14 Bravais lattices
Frequently asked questions
How do I calculate the density of a crystal from its unit cell?
Formula: ρ = nM/(a³Nₐ). Where n = atoms per unit cell (1/2/4), M = molar mass (g/mol), a = edge length (cm), Nₐ = 6.022×10²³. For FCC copper: n=4, M=63.5, a=3.61×10⁻⁸ cm → ρ = 4×63.5/((3.61×10⁻⁸)³ × 6.022×10²³) = 8.92 g/cm³. Always use consistent units (usually CGS).
What is the relationship between edge length and atomic radius in each unit cell?
SC: atoms touch along edge → a = 2r. BCC: atoms touch along body diagonal → 4r = a√3 → r = a√3/4. FCC: atoms touch along face diagonal → 4r = a√2 → r = a√2/4. These relationships connect atomic radius to edge length and are essential for calculations. "Edge = 2r for SC, Face diagonal = 4r for FCC, Body diagonal = 4r for BCC."
What are tetrahedral and octahedral voids?
In close-packed structures (FCC/HCP), spaces between atoms form voids: Tetrahedral void: surrounded by 4 atoms (2 from each layer), shape of tetrahedron. 2 per atom → 8 per FCC unit cell. Octahedral void: surrounded by 6 atoms (3 from each layer), shape of octahedron. 1 per atom → 4 per FCC unit cell. Ionic compounds fill these voids: small cation in void, large anions form the lattice.
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