Packing Efficiency of SC, BCC and FCC Unit Cells (Class 12)
Class 12 solid state asks a simple question: when atoms are packed as hard spheres, how much of the space do they really fill? The answer is called packing efficiency, and it depends only on the arrangement, not on the metal. Learning three numbers (52.4%, 68.0% and 74.0%) and the reason behind each one is enough to score full marks on this part of the chapter.
Definition and formula
= (Z × (4/3)πr3) ÷ a3 × 100
- Z = number of atoms that belong to one unit cell (corner atom = 1/8, face atom = 1/2, body-centre atom = 1).
- r = radius of the atom, treated as a hard sphere.
- a = edge length of the cubic unit cell.
- Void fraction = 100 − packing efficiency (the empty space).
The only extra step is to link r with a. That link comes from asking along which line do the spheres touch each other?
The three cubic arrangements
| Lattice | Z | Spheres touch along | Relation of r and a | Coordination number |
|---|---|---|---|---|
| Simple cubic (sc) | 1 | Edge | a = 2r | 6 |
| Body-centred cubic (bcc) | 2 | Body diagonal (length √3 a) | √3 a = 4r | 8 |
| Face-centred cubic (fcc) | 4 | Face diagonal (length √2 a) | √2 a = 4r | 12 |
Hexagonal close packing (hcp) and cubic close packing (ccp, the same as fcc) both have a coordination number of 12 and the same efficiency.
Deriving each efficiency
Simple cubic. Put a = 2r, so r = a/2 and Z = 1.
Volume of atom = (4/3)π(a/2)3 = πa3/6.
Efficiency = (πa3/6) ÷ a3 = π/6 = 0.5236 = 52.4%. Void = 47.6%.
Body-centred cubic. The body diagonal passes through three spheres, so √3 a = 4r, giving r = √3 a/4. Z = 2.
Volume of atoms = 2 × (4/3)π(√3 a/4)3 = π√3 a3/8.
Efficiency = π√3/8 = 0.6802 = 68.0%. Void = 32.0%.
Face-centred cubic. The face diagonal passes through three spheres, so √2 a = 4r, giving r = a/(2√2) = 0.35355 a. Z = 4.
Volume of one atom = (4/3)π(0.35355)3 a3 = 0.18512 a3.
Four atoms = 4 × 0.18512 = 0.7405 a3.
Efficiency = π/(3√2) = 74.05%. Void = 25.95%.
Summary table
| Arrangement | Packing efficiency | Empty space |
|---|---|---|
| Simple cubic | 52.4% | 47.6% |
| Body-centred cubic | 68.0% | 32.0% |
| fcc / ccp / hcp | 74.05% | 25.95% |
Many books round these to 52.4%, 68% and 74%. Both forms are accepted in board answers.
Numerical examples
Example 1. Iron crystallises in a bcc lattice with edge length a = 287 pm. Find the atomic radius.
r = √3 a/4 = 1.7321 × 287 ÷ 4 = 497.1 ÷ 4 = 124.3 pm.
Check: 4r = 497.1 pm, and √3 a = 1.7321 × 287 = 497.1 pm. The two agree.
Example 2. Aluminium has an fcc lattice with a = 405 pm. Find its atomic radius.
r = a/(2√2) = 405 ÷ 2.8284 = 143.2 pm.
Check: face diagonal = √2 × 405 = 572.8 pm, and 4r = 4 × 143.2 = 572.8 pm.
Example 3. A metal has atomic radius 128 pm. What would the edge length be if it packed as sc, bcc or fcc?
sc: a = 2r = 2 × 128 = 256 pm
bcc: a = 4r/√3 = 512 ÷ 1.7321 = 295.6 pm
fcc: a = 2√2 r = 2.8284 × 128 = 362.0 pm
The edge length is smallest for sc, but that does not mean sc is the best packing. Compare the cell volume per atom (a3 ÷ Z): sc 2563 ÷ 1 = 16.8 million pm3; bcc 295.63 ÷ 2 = 12.9 million pm3; fcc 362.03 ÷ 4 = 11.9 million pm3. The volume used per atom falls from sc to fcc, which is exactly what higher packing efficiency means.
Common mistakes
- Using the wrong touching line: in fcc the spheres touch along the face diagonal, not the body diagonal. In bcc it is the opposite.
- Writing √2 a = 2r for fcc. Along the face diagonal there is one corner atom radius, a full face-centre atom (2r) and another corner atom radius, so the length is 4r.
- Taking Z = 8 for corners instead of 1 (8 × 1/8).
- Forgetting to cube r when finding the volume of the sphere.
- Reporting the void percentage when the question asks for efficiency, or the other way round.
Where it appears
| Topic in Class 12 | Use of packing efficiency |
|---|---|
| Close-packed structures | Compare hcp, ccp and the less efficient cubic lattices |
| Calculation of density | Needs Z, a and molar mass; the Z values here are reused |
| Voids | Tetrahedral and octahedral voids exist in close packing |
For the exact marks distribution and the current syllabus, check the latest official board notification.
The ratios involve √2 and √3 and π, so use the scientific calculator and keep at least four decimal places until the last step.
Open the Chemistry Calculator Suite →Explore real crystal structures in the Crystal Lab, then test yourself on previous-year questions.
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