Packing Efficiency in Simple Cubic, BCC and FCC
Solid-state chemistry is an essential scoring topic in CBSE Class 12, NEET, JEE Main, and CUET examinations. Questions testing unit cell dimensions, atomic radii, and packing efficiency appear regularly in board exams and entrance tests. Mastering how particles fill cubic lattices enables students to solve volume, density, and void fraction problems accurately.
Exam details
- Source: practice problem written for this article (not a past-paper question)
- Relevant exams: CBSE Class 12, JEE Main, NEET, CUET
- Topic: Solid State Unit Cell Geometry
- Difficulty: Intermediate
What Is Packing Efficiency?
In crystalline solids, constituent atoms are modeled as hard spheres of radius r. When spheres pack into a crystal lattice, empty spaces called interstitial voids remain between them. Packing efficiency represents the percentage of total unit cell volume occupied by these constituent spheres.
Packing Efficiency Formula:
Packing Efficiency (η) = [ (Z × (4/3)πr³) / a³ ] × 100
Where:
Z = effective number of atoms in the unit cell (dimensionless rank)
r = radius of constituent spherical atom (pm or cm)
a = edge length of the cubic unit cell (pm or cm)
Void fraction (%) = 100% - Packing Efficiency (%)
Packing Efficiency in Simple Cubic Lattice (SC)
In a simple cubic cell, particles occupy only the eight corners. Each corner particle is shared equally among eight neighbouring unit cells, giving Z = 8 × (1/8) = 1 atom per cell. Spheres touch along the cube edge, so a = 2r.
Unit cell volume is V_cell = a³ = (2r)³ = 8r³. The single sphere occupies volume (4/3)πr³. Packing efficiency is η = [ ((4/3)πr³) / (8r³) ] × 100 = (π / 6) × 100 ≈ 52.36%, rounded to 52.4%. Unoccupied void space equals 100% - 52.4% = 47.6%.
Packing Efficiency in Body-Centered Cubic Lattice (BCC)
A body-centered cubic unit cell contains atoms at eight corners plus one atom at the cube centre. Corner atoms contribute 8 × (1/8) = 1 atom and the centre atom contributes 1 atom, giving Z = 2. Corner atoms touch the central atom along the cube body diagonal.
The body diagonal length is √3 × a = 4r, giving a = 4r / √3. Unit cell volume is a³ = 64r³ / (3√3). Volume of 2 spheres is 2 × (4/3)πr³ = (8/3)πr³. Evaluating efficiency gives η = [ ((8/3)πr³) / (64r³ / (3√3)) ] × 100 = (√3π / 8) × 100 ≈ 68.02%, rounded to 68.0%. Unoccupied void space is 32.0%.
Packing Efficiency in Face-Centered Cubic Lattice (FCC)
In face-centered cubic packing (FCC or CCP), atoms occupy eight corners and six face centres. Corners contribute 8 × (1/8) = 1 atom and face centres contribute 6 × (1/2) = 3 atoms, giving Z = 4. Spheres touch along the face diagonal.
The face diagonal length is √2 × a = 4r, giving a = 4r / √2 = 2√2 × r. Unit cell volume is a³ = (2√2 × r)³ = 16√2 × r³. Volume of 4 spheres is (16/3)πr³. Packing efficiency is η = [ ((16/3)πr³) / (16√2 × r³) ] × 100 = (π√2 / 6) × 100 ≈ 74.05%, rounded to 74.0%. Unoccupied void space is 26.0%.
Worked Step-by-Step Examples
Question: An element crystallises in a body-centered cubic (BCC) lattice with an edge length of 288 pm. Find the atomic radius of the element, the volume occupied by atoms in one unit cell, the total unit cell volume, and the percentage void space.
Step-by-step solution:
Step 1: In a BCC crystal, atoms touch along the body diagonal: √3 × a = 4r.
Radius: r = (√3 × a) / 4 = (1.73205 × 288 pm) / 4 = 124.71 pm.
Step 2: A BCC cell has Z = 2. Volume of 2 spheres:
V_atoms = 2 × (4/3) × π × r³ = (8/3) × 3.14159 × (124.71 pm)³ = 16,249,150 pm³.
Step 3: Total unit cell volume:
V_cell = a³ = (288 pm)³ = 23,887,872 pm³ = 2.389 × 10−23 cm³.
Step 4: Compute packing efficiency and void space:
η = (16,249,150 pm³ / 23,887,872 pm³) × 100 = 68.02% ≈ 68.0%.
Void space = 100% - 68.02% = 31.98% ≈ 32.0%.
Verification by second route: Using theoretical formula η = (√3π / 8) × 100 = (1.73205 × 3.14159 / 8) × 100 = 68.02%. Both routes confirm 68.0% packing.
Answer: Radius = 124.71 pm; unit cell volume = 2.389 × 10−23 cm³; void space = 32.0%.
Question: Copper crystallises in a face-centered cubic (FCC) lattice with an atomic radius of 127.8 pm. Determine the unit cell edge length in picometres, the unit cell volume in cm³, and verify that packing efficiency is 74.0%.
Step-by-step solution:
Step 1: In an FCC lattice, atoms touch along the face diagonal: √2 × a = 4r.
Edge length: a = 2√2 × r = 2 × 1.41421 × 127.8 pm = 361.47 pm.
Step 2: Convert edge length to centimetres: a = 361.47 × 10−10 cm = 3.6147 × 10−8 cm.
Step 3: Calculate unit cell volume:
V_cell = a³ = (3.6147 × 10−8 cm)³ = 4.723 × 10−23 cm³.
Step 4: An FCC cell has Z = 4 atoms. Volume of 4 atoms:
V_atoms = 4 × (4/3) × π × r³ = (16/3) × 3.14159 × (127.8 pm)³ = 34,976,500 pm³.
Step 5: Compute packing efficiency:
η = (34,976,500 pm³ / 47,229,000 pm³) × 100 = 74.06% ≈ 74.0%.
Verification by second route: Using η = (π√2 / 6) × 100 = (3.14159 × 1.41421 / 6) × 100 = 74.05% ≈ 74.0%. Calculated values agree with theory.
Answer: Edge length = 361.47 pm; unit cell volume = 4.723 × 10−23 cm³; packing efficiency = 74.0%.
Common Mistakes Students Make
- Swapping diagonal formulas: Using √2 × a = 4r for BCC or √3 × a = 4r for FCC is a common error. Remember BCC uses the body diagonal and FCC uses the face diagonal.
- Incorrect rank Z: Forgetting corner sharing (1/8) or face sharing (1/2) leads to incorrect Z values in volume calculations.
- Nearest neighbor distance vs radius: Nearest neighbor distance equals 2r. Forgetting to divide by 2 causes an erroneous radius.
- Unit conversion errors: Mixing picometres and centimetres without converting 1 pm = 10−10 cm leads to large power-of-ten mistakes.
- Confusing packing fraction with void space: Writing 74% when an exam asks for void space instead of 26% causes loss of marks.
Exam Relevance and Preparation Strategy
Understanding cubic crystal systems is essential for competitive examinations. The table below outlines how this topic is assessed.
| Examination | Question Focus | Key Skills Required |
|---|---|---|
| CBSE Class 12 Boards | Derivations of packing efficiency | Writing step-by-step proofs and substituting rank Z correctly. |
| JEE Main | Numerical problems linking edge length, radius, and density | Applying diagonal contact equations and unit conversions. |
| NEET | Direct formula recall and void percentages | Identifying coordination numbers and void values rapidly. |
| CUET | Comparative lattice properties and matching questions | Ranking packing efficiency (FCC > BCC > SC) accurately. |
Explore real crystal structures in the Crystal Lab, then test yourself on previous-year questions.
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