JEENEETClass 12

Crystal Structures & Unit Cells — Chemistry Mnemonic

Target: mnemonic for crystal structures unit cells chemistry

Why is this hard to memorize?

The solid state chapter tests your understanding of crystal structures — how atoms pack together and what properties this creates. The three cubic unit cells are: Simple Cubic (SC, 1 atom/cell, 52.4% packing), Body-Centered Cubic (BCC, 2 atoms/cell, 68%), and Face-Centered Cubic (FCC, 4 atoms/cell, 74%). FCC has the same packing efficiency as HCP. NEET and JEE test: atom counting in unit cells (corner=1/8, edge=1/4, face=1/2, body=1), coordination numbers, packing efficiency calculations, void types (tetrahedral and octahedral), and the relationship a = 2√2r for FCC.

Classic mnemonics you should know

The Atom Count
"Corner=1/8, Edge=1/4, Face=1/2, Body=1 — "CEFB = 8-4-2-1""

Each corner atom is shared by 8 unit cells (1/8). Edge by 4 cells (1/4). Face by 2 cells (1/2). Body center belongs fully to one cell (1). SC: 8×(1/8) = 1. BCC: 8×(1/8) + 1 = 2. FCC: 8×(1/8) + 6×(1/2) = 4.

The Packing Efficiency
"SC=52%, BCC=68%, FCC/HCP=74% — "52-68-74, that's the law""

Packing efficiency = volume of atoms / volume of unit cell × 100. SC is least efficient (big gaps). FCC/HCP are most efficient (closest packing). BCC is in between.

The Coordination Numbers
"SC=6, BCC=8, FCC=12 — "6-8-12 step up""

Coordination number = number of nearest neighbours. SC: each atom touches 6 others. BCC: 8 nearest neighbours. FCC: 12 nearest neighbours (the maximum for equal spheres).

The complete list

  1. Simple Cubic (SC): 1 atom, CN=6, 52.4%
  2. Body-Centered Cubic (BCC): 2 atoms, CN=8, 68%
  3. Face-Centered Cubic (FCC): 4 atoms, CN=12, 74%
  4. HCP: 6 atoms, CN=12, 74%
  5. Corner = 1/8, Face = 1/2, Body = 1
  6. Tetrahedral voids: 2n per FCC
  7. Octahedral voids: n per FCC

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Frequently asked questions

How do I count atoms in a unit cell?

Add up contributions: corner atoms × 1/8, edge atoms × 1/4, face atoms × 1/2, body atoms × 1. For BCC: 8 corners × 1/8 = 1 atom + 1 body center × 1 = 1 atom → total = 2. For FCC: 8 × 1/8 + 6 × 1/2 = 1 + 3 = 4 atoms.

What is the difference between tetrahedral and octahedral voids?

Tetrahedral void: formed when 3 atoms in one layer and 1 atom in adjacent layer form a tetrahedron. Octahedral void: formed when 3 atoms in one layer and 3 in the adjacent form an octahedron. In FCC: n atoms give 2n tetrahedral voids and n octahedral voids. Smaller atoms fill tetrahedral voids, larger atoms fill octahedral.

What is the relationship between edge length (a) and radius (r) in FCC?

In FCC, atoms touch along the face diagonal. Face diagonal = 4r = a√2, so a = 2√2r. For BCC: atoms touch along body diagonal: 4r = a√3, so a = 4r/√3. For SC: atoms touch along edge: 2r = a. These relationships are essential for density calculations.

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