🔬 Solid State Chemistry II

Comprehensive Study of Cubic Solids, Crystal Packing, and Structural Analysis

📐 Types of Cubic Crystal Systems

🧊 Simple Cubic (SC)

Structure: Atoms are located only at the corners of the cube.

Coordination Number: 6

Atoms per Unit Cell: 1 (8 corners × 1/8 = 1)

Examples: α-Po (Polonium)

🎯 Body-Centered Cubic (BCC)

Structure: Atoms at corners + one atom at body center

Coordination Number: 8

Atoms per Unit Cell: 2 (8 corners × 1/8 + 1 center = 2)

Examples: α-Fe, Cr, W, Mo

💎 Face-Centered Cubic (FCC)

Structure: Atoms at corners + atoms at face centers

Coordination Number: 12

Atoms per Unit Cell: 4 (8 corners × 1/8 + 6 faces × 1/2 = 4)

Examples: Cu, Au, Ag, Al, Ni

📊 Packing Efficiency Comparison

Crystal System	Packing Efficiency (%)	Coordination Number	Atoms per Unit Cell	Common Examples
Simple Cubic	52.4%	6	1	α-Po
Body-Centered Cubic	68.0%	8	2	α-Fe, Cr, W
Face-Centered Cubic	74.0%	12	4	Cu, Au, Ag
Hexagonal Close Packing	74.0%	12	6	Zn, Mg, Co

🌟 Crystal Packing Types

🔄 ABAB Type Packing (Hexagonal Close Packing - HCP)

Structure: Two-layer repetition pattern

Sequence: A-B-A-B-A-B...

Coordination Number: 12

Packing Efficiency: 74.05%

Examples: Zn, Mg, Co, Cd, Be

Layer A

Layer B

🔺 ABCABC Type Packing (Face-Centered Cubic - FCC)

Structure: Three-layer repetition pattern

Sequence: A-B-C-A-B-C-A-B-C...

Coordination Number: 12

Packing Efficiency: 74.05%

Examples: Cu, Au, Ag, Al, Ni, Pt

Layer A

Layer B

Layer C

🕳️ Types of Voids in Crystal Structures

🔺 Tetrahedral Voids

Shape: Triangular pyramid

Coordination Number: 4

Size: Smaller voids

Number per sphere: 2

Radius ratio: 0.225 - 0.414

Examples: ZnS (zinc blende), CuCl

🔶 Octahedral Voids

Shape: Square bipyramid

Coordination Number: 6

Size: Larger voids

Number per sphere: 1

Radius ratio: 0.414 - 0.732

Examples: NaCl, MgO, FeO

🧮 Numerical Problems and Solutions

Problem 1: Density Calculation for FCC Structure

Given: Copper crystallizes in FCC structure with edge length a = 3.61 Å

Find: Density of copper (Atomic mass of Cu = 63.5 g/mol)

Step 1: Calculate volume of unit cell

V = a³ = (3.61 × 10⁻⁸)³ = 4.70 × 10⁻²³ cm³

Step 2: Number of atoms in FCC = 4

Z = 4 atoms per unit cell

Step 3: Apply density formula

ρ = (Z × M) / (N_A × V)

ρ = (4 × 63.5) / (6.022 × 10²³ × 4.70 × 10⁻²³)

Step 4: Calculate final density

ρ = 254 / 28.3 = 8.98 g/cm³

Problem 2: Packing Efficiency of BCC Structure

Given: Iron crystallizes in BCC structure

Find: Packing efficiency

Step 1: In BCC, atoms touch along body diagonal

Body diagonal = 4r = a√3

Therefore: r = a√3/4

Step 2: Volume of atoms in unit cell

V_atoms = 2 × (4/3)πr³ = 2 × (4/3)π × (a√3/4)³

V_atoms = (π√3/8) × a³

Step 3: Volume of unit cell

V_{unit cell} = a³

Step 4: Packing efficiency

η = (V_atoms / V_{unit cell}) × 100%

η = (π√3/8) × 100% = 68.0%

Problem 3: Void Calculations in FCC

Given: FCC structure with N atoms

Find: Number of tetrahedral and octahedral voids

Step 1: Tetrahedral voids

Number of tetrahedral voids = 2N

For N = 4 atoms: Tetrahedral voids = 8

Step 2: Octahedral voids

Number of octahedral voids = N

For N = 4 atoms: Octahedral voids = 4

Step 3: Total voids

Total voids = 8 + 4 = 12

📏 Packing Efficiency Calculations

🔢 Formula for Packing Efficiency

Packing Efficiency (η) = (Volume of atoms in unit cell / Volume of unit cell) × 100%

η = (Z × (4/3)πr³ / a³) × 100%

Where:

Z = Number of atoms per unit cell
r = Atomic radius
a = Edge length of unit cell

Simple Cubic

a = 2r

Z = 1

52.4%

Body-Centered Cubic

a = 4r/√3

Z = 2

68.0%

Face-Centered Cubic

a = 2√2r

Z = 4

74.0%

🔬 Advanced Concepts

🌡️ Thermal Effects on Crystal Structure

Thermal Expansion: Linear expansion coefficient α

ΔL = L₀ × α × ΔT

Debye Temperature: θ_D = (h/k) × (6π²N/V)^(1/3) × v_s

Where: v_s = average sound velocity

⚡ Electrical Properties

Conductivity: σ = n × e × μ

Where:

n = charge carrier density
e = elementary charge
μ = mobility

Band Gap Energy: E_g = hν = hc/λ

🎯 Real-World Applications

Semiconductor Industry: Silicon (Diamond cubic structure)

Metallurgy: Steel (BCC α-Fe ⟷ FCC γ-Fe transformation)

Catalysis: Platinum (FCC structure) for automotive catalysts

Superconductors: High-T_c cuprates with perovskite-related structures

📋 Summary Table: Key Relationships

Property	Simple Cubic	Body-Centered Cubic	Face-Centered Cubic
Lattice Parameter	a = 2r	a = 4r/√3	a = 2√2r
Nearest Neighbors	6	8	12
Second Nearest	12	6	6
Atomic Radius	r = a/2	r = a√3/4	r = a/(2√2)
Density Formula	ρ = M/(N_A × a³)	ρ = 2M/(N_A × a³)	ρ = 4M/(N_A × a³)

🔍 Miller Indices and Crystal Planes

📐 Miller Indices Notation

Definition: Miller indices (hkl) describe the orientation of crystal planes

Procedure:

Find intercepts on x, y, z axes
Take reciprocals
Clear fractions to get smallest integers
Enclose in parentheses (hkl)

d_hkl = a/√(h² + k² + l²) (for cubic systems)

Problem 4: Miller Indices Calculation

Given: A plane intercepts at (2a, 3a, 6a)

Find: Miller indices

Step 1: Intercepts = (2, 3, 6)

Step 2: Take reciprocals = (1/2, 1/3, 1/6)

Step 3: Clear fractions (multiply by 6)

Miller indices = (3 2 1)

🎯 X-Ray Diffraction and Bragg's Law

📡 Bragg's Law

nλ = 2d sin θ

Where:

n = order of diffraction (integer)
λ = wavelength of X-ray
d = interplanar spacing
θ = angle of incidence

Problem 5: X-Ray Diffraction

Given: Cu Kα radiation (λ = 1.54 Å), first-order diffraction at θ = 26.0°

Find: Interplanar spacing d_hkl

Step 1: Apply Bragg's law

nλ = 2d sin θ

Step 2: Substitute values (n = 1)

1 × 1.54 = 2d × sin(26.0°)

Step 3: Calculate

d = 1.54 / (2 × 0.4384) = 1.76 Å

⚛️ Defects in Crystal Structures

🔸 Point Defects

Vacancy: Missing atom from lattice site

Interstitial: Extra atom in interstitial site

Substitutional: Foreign atom replacing host atom

Frenkel Defect: Atom displaced to interstitial

Schottky Defect: Paired vacancy (cation + anion)

📏 Line Defects

Edge Dislocation: Extra half-plane of atoms

Screw Dislocation: Helical arrangement

Burgers Vector: Magnitude and direction of lattice distortion

Dislocation Density: ρ = L/V (length per unit volume)

🌡️ Thermodynamics of Crystal Formation

⚡ Lattice Energy

U = -NAMz⁺z⁻e²/4πε₀r₀ × (1 - 1/n)

Born-Landé Equation components:

M = Madelung constant
z⁺, z⁻ = charges on cation and anion
r₀ = nearest neighbor distance
n = Born exponent

Problem 6: Lattice Energy of NaCl

Given: r₀ = 2.82 Å, M = 1.748, n = 7.5

Find: Lattice energy

Step 1: Apply Born-Landé equation

U = -NAMz⁺z⁻e²/4πε₀r₀ × (1 - 1/n)

Step 2: Substitute values

z⁺ = +1, z⁻ = -1, e² = 2.31 × 10⁻¹⁹ J·m

Step 3: Calculate

U = -786 kJ/mol

🔬 Advanced Characterization Techniques

🔍 Powder X-Ray Diffraction

Debye-Scherrer Method

Crystallite size: D = Kλ/(β cos θ)

K = 0.9 (Scherrer constant)

β = peak broadening (radians)

🌊 Neutron Diffraction

Advantages:

Light element detection
Magnetic structure
Isotope distinction

λ = h/(m·v) (de Broglie)

⚡ Electron Diffraction

Selected Area Diffraction

λ = h/√(2m₀eV)

Very short wavelength

Surface sensitivity

🧪 Ionic Crystals and Radius Ratios

Structure Type	Radius Ratio (r⁺/r⁻)	Coordination Number	Examples
Linear	< 0.155	2	Rare
Triangular	0.155 - 0.225	3	B₂O₃
Tetrahedral	0.225 - 0.414	4	ZnS, SiO₂
Octahedral	0.414 - 0.732	6	NaCl, MgO
Cubic	0.732 - 1.000	8	CsCl, CaF₂

Problem 7: Radius Ratio Calculation

Given: LiF with r_Li⁺ = 0.76 Å, r_F⁻ = 1.33 Å

Find: Coordination number and structure type

Step 1: Calculate radius ratio

r⁺/r⁻ = 0.76/1.33 = 0.571

Step 2: Compare with table

0.414 < 0.571 < 0.732

Step 3: Determine structure

Structure: Octahedral (NaCl type)

Coordination Number: 6

🎨 Electronic Band Structure

⚡ Band Theory

Valence Band: Highest occupied energy band

Conduction Band: Lowest unoccupied energy band

Band Gap: Energy difference between VB and CB

E_g = E_CB - E_VB

Classification:

Metals: E_g = 0 eV (overlapping bands)
Semiconductors: E_g = 0.1 - 3.0 eV
Insulators: E_g > 3.0 eV

🔋 Common Band Gaps

Diamond (C): 5.5 eV (insulator)
Silicon (Si): 1.1 eV (semiconductor)
Germanium (Ge): 0.7 eV (semiconductor)
Gallium Arsenide (GaAs): 1.4 eV (semiconductor)
Zinc Oxide (ZnO): 3.2 eV (wide band gap)

🌟 Phase Transitions and Polymorphism

🔄 Types of Phase Transitions

Martensitic: Displacive, diffusionless transformation

Reconstructive: Bond breaking and reformation

Order-Disorder: Change in atomic arrangement

ΔG = ΔH - TΔS (Gibbs free energy)

At equilibrium: ΔG = 0, so T_transition = ΔH/ΔS

Problem 8: Phase Transition Temperature

Given: α-Fe → γ-Fe transition, ΔH = 837 J/mol, ΔS = 7.6 J/(mol·K)

Find: Transition temperature

Step 1: At equilibrium ΔG = 0

ΔH = TΔS

Step 2: Solve for T

T = ΔH/ΔS = 837/7.6

Step 3: Calculate

T = 110 K (-163°C)

🧮 Advanced Numerical Problems

Problem 9: Vacancy Concentration

Given: Copper at 1000°C, formation energy E_v = 0.9 eV

Find: Vacancy concentration

Step 1: Apply Boltzmann distribution

n_v/N = exp(-E_v/kT)

Step 2: Convert temperature

T = 1000 + 273 = 1273 K

Step 3: Calculate

kT = 8.617 × 10⁻⁵ × 1273 = 0.110 eV

n_v/N = exp(-0.9/0.110) = 1.2 × 10⁻⁴

Problem 10: Diffusion Coefficient

Given: Carbon in iron, D₀ = 2.0 × 10⁻⁶ m²/s, Q = 80 kJ/mol

Find: Diffusion coefficient at 900°C

Step 1: Apply Arrhenius equation

D = D₀ exp(-Q/RT)

Step 2: Convert values

T = 900 + 273 = 1173 K

RT = 8.314 × 1173 = 9.75 kJ/mol

Step 3: Calculate

D = 2.0 × 10⁻⁶ × exp(-80/9.75)

D = 6.1 × 10⁻¹⁰ m²/s

📊 Comprehensive Reference Tables

Crystal System	Lattice Parameters	Angles	Examples
Cubic	a = b = c	α = β = γ = 90°	NaCl, Diamond
Tetragonal	a = b ≠ c	α = β = γ = 90°	TiO₂, SnO₂
Orthorhombic	a ≠ b ≠ c	α = β = γ = 90°	BaSO₄, KNO₃
Hexagonal	a = b ≠ c	α = β = 90°, γ = 120°	Graphite, ZnO
Trigonal	a = b = c	α = β = γ ≠ 90°	Calcite, Quartz
Monoclinic	a ≠ b ≠ c	α = γ = 90° ≠ β	Gypsum, β-S
Triclinic	a ≠ b ≠ c	α ≠ β ≠ γ ≠ 90°	CuSO₄·5H₂O

Complete reference for cubic crystal structures, defects, phase transitions, and advanced characterization

Includes Miller indices, X-ray diffraction, electronic properties, and comprehensive numerical solutions