Common Crystal Structures

Introduction to Metallic Crystal Structures

Crystal Structures describe the three-dimensional arrangement of atoms in solid materials. In metals, atoms are typically arranged in one of three common structures: Simple Cubic (SC), Body-Centered Cubic (BCC), or Face-Centered Cubic (FCC). Each structure has distinct properties including coordination numbers, packing efficiency, and mechanical characteristics.

🎬 Jmol Animation Focus Points

  • Unit Cell Rotation: 360° view to understand 3D arrangement
  • Atom Insertion: Show how atoms fit into lattice positions
  • Coordination Sphere: Highlight nearest neighbors
  • Packing Demonstration: Spheres touching along specific directions
  • Layer-by-Layer Building: Construct structures step by step

Cubic Crystal Structures

Simple Cubic (SC)

🧊 Like Ice Cubes Stacked

Lattice Points: 8 corners × 1/8 = 1 atom/unit cell

Coordination Number: 6

Packing Efficiency: 52.36%

Atoms Touch: Along cube edges

Relationship: a = 2r (where r = atomic radius)

Examples: Polonium (Po) - only known metal

SC Calculations

Volume of Unit Cell: V = a³

Volume of Atoms: 1 × (4/3)πr³

Packing Efficiency: (4/3)πr³/a³ = π/6 ≈ 0.524

Density: ρ = M/(N_A × a³)

Body-Centered Cubic (BCC)

🎯 Atom in Center + Corners

Lattice Points: 8 corners × 1/8 + 1 center = 2 atoms/unit cell

Coordination Number: 8

Packing Efficiency: 68.02%

Atoms Touch: Along body diagonal

Relationship: a = 4r/√3

Examples: α-Iron, Chromium, Tungsten

BCC Calculations

Body Diagonal: 4r = a√3

Volume of Unit Cell: V = a³

Volume of Atoms: 2 × (4/3)πr³

Packing Efficiency: √3π/8 ≈ 0.680

Density: ρ = 2M/(N_A × a³)

Face-Centered Cubic (FCC)

💎 Atoms on All Faces

Lattice Points: 8 corners × 1/8 + 6 faces × 1/2 = 4 atoms/unit cell

Coordination Number: 12

Packing Efficiency: 74.05%

Atoms Touch: Along face diagonal

Relationship: a = 2√2r

Examples: Copper, Gold, Aluminum, Silver

FCC Calculations

Face Diagonal: 4r = a√2

Volume of Unit Cell: V = a³

Volume of Atoms: 4 × (4/3)πr³

Packing Efficiency: π/(3√2) ≈ 0.740

Density: ρ = 4M/(N_A × a³)

Comparison of Cubic Structures

Structure Atoms/Unit Cell Coordination Number Packing Efficiency Relationship Common Metals
Simple Cubic 1 6 52.36% a = 2r Polonium
BCC 2 8 68.02% a = 4r/√3 Fe, Cr, W
FCC 4 12 74.05% a = 2√2r Cu, Au, Al

Close Packing Arrangements

Close Packing represents the most efficient ways to pack identical spheres in three dimensions, achieving maximum space utilization. There are two main types: Hexagonal Close Packing (HCP) and Cubic Close Packing (CCP), both with 74.05% packing efficiency.
Layer Arrangements in Close Packing

Layer A (First Layer)

🔴 Spheres in hexagonal array

Each sphere touches 6 others

Creates triangular gaps

Layer B (Second Layer)

🟡 Spheres sit in A-layer gaps

Two types of gaps remain

Choice determines structure

HCP: A-B-A-B...

🔵 Third layer above A positions

Hexagonal unit cell

ABAB... stacking

CCP: A-B-C-A-B-C...

🟢 Third layer in new C positions

Cubic unit cell (FCC)

ABCABC... stacking

Hexagonal Close Packing (HCP)

🏠 Like Hexagonal Pencils

Stacking Sequence: ABAB...

Unit Cell: Hexagonal

Lattice Parameters: a = b ≠ c, γ = 120°

c/a Ratio: 1.633 (ideal)

Atoms per Unit Cell: 6

Coordination Number: 12

Examples: Mg, Zn, Ti, Co

Cubic Close Packing (CCP/FCC)

💎 Same as FCC Structure

Stacking Sequence: ABCABC...

Unit Cell: Face-Centered Cubic

Lattice Parameters: a = b = c, all angles 90°

Relationship: a = 2√2r

Atoms per Unit Cell: 4

Coordination Number: 12

Examples: Cu, Au, Al, Ni

Both HCP and CCP have identical packing efficiency:
Packing Efficiency = π/(3√2) = 74.05%

Interstitial Sites

Interstitial Sites are empty spaces (holes) between atoms in crystal structures where smaller atoms can be accommodated. The size and number of these sites depend on the crystal structure and are crucial for understanding alloy formation and ionic compounds.

Types of Interstitial Sites

Tetrahedral Holes

Coordination: 4

Size Ratio: r_hole/r_sphere = 0.225

Number in FCC: 8 per unit cell

Number in HCP: 2 per atom

Location: Between 4 close-packed spheres

Octahedral Holes

Coordination: 6

Size Ratio: r_hole/r_sphere = 0.414

Number in FCC: 4 per unit cell

Number in HCP: 1 per atom

Location: Between 6 close-packed spheres

Cubic Holes

Coordination: 8

Size Ratio: r_hole/r_sphere = 0.732

Number in BCC: 1 per unit cell

Location: Center of cubic arrangement

Structure Tetrahedral Holes Octahedral Holes Cubic Holes Applications
Simple Cubic - - 1 CsCl structure
BCC 24 6 - Interstitial alloys
FCC 8 4 - NaCl, ZnS structures
HCP 12 6 - Wurtzite structure

Coordination Numbers and Geometry

Coordination Number Determination

Definition: Number of nearest neighbor atoms surrounding a central atom

Step-by-Step Method:

  1. Identify Central Atom: Choose reference atom
  2. Measure Distances: Calculate all interatomic distances
  3. Find Nearest Neighbors: Identify shortest equal distances
  4. Count Neighbors: Total number at nearest distance
  5. Verify Geometry: Check coordination polyhedron
Coordination Number Geometry Examples in Crystals Typical Structures
4 Tetrahedral ZnS, SiO₂ Covalent crystals
6 Octahedral NaCl, MgO Simple cubic, ionic
8 Cubic CsCl, BCC metals Body-centered
12 Cuboctahedral FCC, HCP metals Close-packed

Packing Efficiency Calculations

Packing Efficiency (Atomic Packing Factor) is the fraction of space occupied by atoms in a unit cell, calculated as the ratio of volume occupied by atoms to the total unit cell volume.
General Formula:
Packing Efficiency = (Number of atoms × Volume of one atom) / Volume of unit cell
APF = (n × 4πr³/3) / V_unit cell

🧮 Detailed Calculation Examples

1. Simple Cubic (SC)
  • Atoms touch along cube edge: a = 2r
  • Unit cell volume: V = a³ = (2r)³ = 8r³
  • Number of atoms: 1
  • Volume of atoms: 1 × (4πr³/3) = 4πr³/3
  • APF = (4πr³/3) / 8r³ = π/6 = 0.524 = 52.4%
2. Body-Centered Cubic (BCC)
  • Atoms touch along body diagonal: 4r = a√3
  • Therefore: a = 4r/√3
  • Unit cell volume: V = a³ = (4r/√3)³ = 64r³/(3√3)
  • Number of atoms: 2
  • Volume of atoms: 2 × (4πr³/3) = 8πr³/3
    • 2. Body-Centered Cubic (BCC)
    • Atoms touch along body diagonal: 4r = a√3
    • Therefore: a = 4r/√3
    • Unit cell volume: V = a³ = (4r/√3)³ = 64r³/(3√3)
    • Number of atoms: 2
    • Volume of atoms: 2 × (4πr³/3) = 8πr³/3
    • APF = (8πr³/3) / (64r³/(3√3)) = √3π/8 = 0.680 = 68.0%
    3. Face-Centered Cubic (FCC)
    • Atoms touch along face diagonal: 4r = a√2
    • Therefore: a = 4r/√2 = 2√2r
    • Unit cell volume: V = a³ = (2√2r)³ = 16√2r³
    • Number of atoms: 4
    • Volume of atoms: 4 × (4πr³/3) = 16πr³/3
    • APF = (16πr³/3) / (16√2r³) = π/(3√2) = 0.740 = 74.0%
    4. Hexagonal Close Packing (HCP)
    • Base area: A = (3√3/2)a² (hexagonal)
    • Height: c = (4/3)√(2/3)a for ideal packing
    • Unit cell volume: V = (3√3/2)a² × c
    • Number of atoms: 6
    • APF = π/(3√2) = 0.740 = 74.0% (same as FCC)

Density Calculations

Density Formula:
ρ = (n × M) / (N_A × V_unit cell)
where: n = atoms per unit cell, M = atomic mass, N_A = Avogadro's number

Density Calculation Steps:

  1. Identify Structure: Determine crystal structure type
  2. Count Atoms: Calculate effective atoms per unit cell
  3. Find Unit Cell Volume: Use lattice parameter relationships
  4. Apply Formula: Substitute values into density equation
  5. Check Units: Ensure consistent units (g/cm³)

Example: Density of Copper (FCC)

  • Given: Cu has FCC structure, a = 3.61 Å, M = 63.55 g/mol
  • Atoms per unit cell: n = 4 (FCC)
  • Unit cell volume: V = a³ = (3.61 × 10⁻⁸)³ cm³ = 4.70 × 10⁻²³ cm³
  • Density: ρ = (4 × 63.55) / (6.022 × 10²³ × 4.70 × 10⁻²³)
  • Result: ρ = 8.96 g/cm³ (experimental: 8.96 g/cm³) ✓

Mechanical Properties and Structure Relationship

Structure Slip Systems Ductility Strength Examples & Properties
FCC 12 High Moderate Cu, Al - very ductile, good conductors
BCC 48 Moderate High Fe, Cr - strong, less ductile
HCP 3-6 Low High Mg, Zn - brittle, anisotropic
Structure-Property Relationships:
• More slip systems → Higher ductility (FCC > BCC > HCP)
• Higher packing efficiency → Better malleability
• Close-packed structures → Metallic bonding optimization
• Coordination number affects mechanical response

Real-World Applications

FCC Metals (High Ductility):
Cu - Electrical wiring Al - Aircraft structures Au - Jewelry, electronics Ni - Alloys, catalysts
BCC Metals (High Strength):
Fe - Construction steel Cr - Stainless steel W - Light bulb filaments V - Steel alloys
HCP Metals (Specialized Uses):
Mg - Lightweight alloys Zn - Galvanizing Ti - Aerospace, medical Co - Magnetic alloys

Phase Transformations

Many metals undergo structural transformations with temperature or pressure changes, leading to different crystal structures and properties.
Common Phase Transformations:
Fe: BCC (α) ⇌ FCC (γ) at 912°C Ti: HCP (α) ⇌ BCC (β) at 882°C Co: HCP ⇌ FCC at 417°C Sn: Diamond ⇌ β-Sn at 13.2°C

Volume Changes During Transformation

Fe (BCC → FCC): Volume decreases by ~1%

Significance: Controls heat treatment of steel

Application: Quenching and tempering processes

Jmol Visualization Techniques

🎬 Advanced Animation Sequences

1. Structure Building Animation

  • Start with single atom at origin
  • Add nearest neighbors one by one
  • Show coordination sphere formation
  • Complete unit cell construction
  • Extend to show multiple unit cells

2. Packing Efficiency Demonstration

  • Show spheres as space-filling models
  • Gradually reduce sphere size to show voids
  • Color-code occupied vs empty space
  • Calculate and display efficiency percentage

3. Interstitial Site Visualization

  • Display host structure with reduced opacity
  • Highlight interstitial positions with different colors
  • Show size relationships with probe atoms
  • Animate insertion and removal of interstitial atoms

4. Coordination Environment

  • Select central atom and highlight
  • Draw lines to nearest neighbors
  • Show coordination polyhedron formation
  • Rotate to view from different angles

Problem-Solving Strategies

JEE Problem Approach:

  1. Identify Structure Type: SC, BCC, FCC, or HCP
  2. Determine Key Relationships: a vs r, coordination numbers
  3. Count Atoms Correctly: Use sharing factors for different positions
  4. Apply Appropriate Formulas: Density, packing efficiency, etc.
  5. Check Dimensional Analysis: Ensure units are consistent
  6. Verify Reasonableness: Compare with known values

Common JEE Problem Types

  • Density Calculations: Given lattice parameter, find density
  • Radius Calculations: From density and structure, find atomic radius
  • Unit Cell Determinations: From X-ray data, identify structure
  • Packing Efficiency: Calculate and compare different structures
  • Coordination Numbers: Determine from structure type
  • Interstitial Sites: Count and calculate occupancy

Quick Reference Formulas

Structure a vs r Atoms/Cell CN APF Density
SC a = 2r 1 6 π/6 M/(N_A·a³)
BCC a = 4r/√3 2 8 √3π/8 2M/(N_A·a³)
FCC a = 2√2r 4 12 π/(3√2) 4M/(N_A·a³)
HCP a = 2r 6 12 π/(3√2) 6M/(N_A·V_hex)
Key Points for Mastery:
• Understand 3D visualization: Use Jmol to develop spatial reasoning
• Master calculation methods: Packing efficiency and density formulas
• Know structure-property relationships: How crystal structure affects properties
• Practice interstitial site identification: Critical for understanding ionic compounds
• Connect to real applications: Why certain metals have specific structures
• Memorize key relationships: a vs r formulas for quick problem solving
• Understand close packing: Foundation for many advanced concepts
• Use systematic approach: Step-by-step problem-solving methodology