In Cubic Close Packing (CCP), spheres are arranged in layers following an ABCABC… stacking sequence. Layer A is repeated after every two different intermediate layers B and C.
Each sphere is in contact with 12 nearest neighbours, giving a coordination number = 12. The packing efficiency is 74%. CCP is structurally identical to the FCC lattice — they are the same arrangement viewed differently.
In the FCC lattice, atoms are present at each of the 8 corners and at the centre of each of the 6 faces of the cube.
Atoms per unit cell:
Corner: 8 × 1/8 = 1 + Face: 6 × 1/2 = 3 = 4 atoms per unit cell
Relation: a = 2√2 · r, where a = edge length, r = atomic radius. Packing efficiency = 74%.
In Hexagonal Close Packing (HCP), spheres are stacked in an ABABAB… sequence — the third layer is directly above the first layer. Each unit cell consists of a hexagonal prism structure.
Atoms per unit cell:
Corner: 12 × 1/6 = 2 + Face: 2 × 1/2 = 1 + Body: 3 (fully inside) = 6 atoms per unit cell
Coordination number = 12. Packing efficiency = 74% (same as CCP/FCC). Ideal c/a ratio = 1.633.
🔺 Tetrahedral voids = 2n = 2 × 6 = 12 🔷 Octahedral voids = n = 6
| Property | CCP | FCC | HCP |
|---|---|---|---|
| Stacking Sequence | ABCABC… | ABCABC… | ABABAB… |
| Atoms per Unit Cell | 4 | 4 | 6 |
| Coordination Number | 12 | 12 | 12 |
| Packing Efficiency | 74% | 74% | 74% |
| Tetrahedral Voids | 8 (= 2n) | 8 (= 2n) | 12 (= 2n) |
| Octahedral Voids | 4 (= n) | 4 (= n) | 6 (= n) |
| Unit Cell Shape | Cubic | Cubic | Hexagonal Prism |
| Examples | Cu, Ag, Au | Cu, Ag, Au | Mg, Zn, Ti |
If number of atoms in close-packed structure = n
🔺 Tetrahedral Voids = 2n → CCP/FCC: 8 | HCP: 12
Each tetrahedral void is surrounded by 4 spheres at corners of a tetrahedron. In FCC, they lie along body diagonals at 1/4 distance from each corner.
🔷 Octahedral Voids = n → CCP/FCC: 4 | HCP: 6
Each octahedral void is surrounded by 6 spheres. In FCC: 1 at body centre + 12 edge centres × 1/4 = 3 → Total = 4.
The formula of an ionic compound is determined by identifying which fraction of voids are occupied by cations. Anions (X) form the close-packed framework; cations (M) sit in the voids.
Tetrahedral voids = 8 | Octahedral voids = 4
✦ All tetrahedral voids filled: M : X = 8 : 4 = 2 : 1 → M2X
✦ Half tetrahedral voids filled: M : X = 4 : 4 = 1 : 1 → MX ZnS (Zinc Blende)
✦ All octahedral voids filled: M : X = 4 : 4 = 1 : 1 → MX NaCl (Rock Salt)
✦ Half octahedral voids filled: M : X = 2 : 4 = 1 : 2 → MX2 CaF2 (Fluorite)
✦ 2/3 octahedral voids filled: M : X = 8/3 : 4 = 2 : 3 → M2X3 Al2O3
Tetrahedral voids = 12 | Octahedral voids = 6
✦ All octahedral voids filled: M : X = 6 : 6 = 1 : 1 → MX
✦ Half tetrahedral voids filled: M : X = 6 : 6 = 1 : 1 → MX
✦ 1/3 octahedral voids filled: M : X = 2 : 6 = 1 : 3 → MX3
📌 Key Rule: Tet. voids = 2n | Oct. voids = n | Fraction occupied by M × total voids = no. of M atoms per unit cell → Ratio M : X = empirical formula