Crystal Structures - Lecture 4 Chapter 10

Crystal Structures

Introduction to crystal structures
- Crystal structures are fundamental in general chemistry and materials science, defining how atoms, ions, or molecules are packed in a crystalline solid.
- Their arrangement dictates many macroscopic properties of materials, such as density, hardness, electrical conductivity, optical properties, and melting point.
- Understanding these structures is crucial for designing new materials with specific desired characteristics.

Types of Crystal Structures

Importance of unit cells in crystal structures.
- Definition of Unit Cell: The smallest repeating unit in a crystal structure, analogous to a single brick in a large wall. It represents the fundamental building block that, when repeated in three dimensions, generates the entire crystal lattice. Understanding the unit cell allows for the determination of macroscopic properties from microscopic arrangements.
Three main types of cubic unit cells:

Simple Cubic (SC)
- Visualization: A cube where identical atoms are located at each corner, touching along the edges.
- Atoms in unit cell: 1 atom
  - Explanation: Each corner atom is shared by 8 adjacent unit cells, so each contributes $\frac{1}{8}$ of an atom to the unit cell. With 8 corners:
    $\text{Total atoms} = 8 \times \frac{1}{8} = 1 \text{ atom}$
- Edge length (L) relation to atomic radius (r): $L = 2r$ (atoms touch along the cube edges).
- Coordination number: 6 (each atom is surrounded by 6 nearest neighbors).
- Packing efficiency: $\frac{\pi}{6} \approx 52.36\%$ (relatively low packing density).
- Examples: Polonium is the only element that exhibits a simple cubic structure under certain temperature and pressure conditions.
Body-Centered Cubic (BCC)
- Structure: Atoms at each corner plus an additional identical atom located exactly at the center of the cube.
- Atoms in unit cell: 2 atoms
  - Explanation: The 8 corner atoms contribute $\frac{1}{8}$ each, and the central atom is entirely within the unit cell (contributing 1 full atom):
    $\text{Total atoms} = (8 \times \frac{1}{8}) + 1 = 1 + 1 = 2 \text{ atoms}$
- Edge length (L) relation to atomic radius (r): $4r = L\sqrt{3} \implies L = \frac{4r}{\sqrt{3}}$ (atoms touch along the body diagonal).
- Coordination number: 8 (the central atom touches all 8 corner atoms).
- Packing efficiency: $\frac{\sqrt{3}\pi}{8} \approx 68\%$ (higher packing density than SC).
- Examples: Iron, sodium, potassium, chromium, tungsten, and molybdenum commonly crystallize in a BCC structure.
Face-Centered Cubic (FCC) - also known as Cubic Close-Packed (CCP)
- Structure: Atoms at each corner, with additional identical atoms centered on each of the 6 faces of the cube.
- Atoms in unit cell: 4 atoms
  - Explanation: The 8 corner atoms contribute $\frac{1}{8}$ each, and each of the 6 face-centered atoms is shared by 2 adjacent unit cells (contributing $\frac{1}{2}$ each):
    $\text{Total atoms} = (8 \times \frac{1}{8}) + (6 \times \frac{1}{2}) = 1 + 3 = 4 \text{ atoms}$
- Edge length (L) relation to atomic radius (r): $4r = L\sqrt{2} \implies L = \frac{4r}{\sqrt{2}} = 2\sqrt{2}r$ (atoms touch along the face diagonal).
- Coordination number: 12 (each atom is in contact with 12 nearest neighbors).
- Packing efficiency: $\frac{\pi}{3\sqrt{2}} \approx 74\%$ (the highest possible packing efficiency for identical spheres).
- Examples: Copper, silver, gold, aluminum, nickel, and platinum exhibit an FCC structure.

Edge Length of a Unit Cell

Importance of knowing the edge length for calculations.
The edge length (L) of a unit cell is critical for calculating various properties such as atomic radius, molar volume, and most importantly, the density of crystalline materials. The relationship between L and the atomic radius (r) varies for each type of unit cell (SC, BCC, FCC) as described above.

X-ray Diffraction

Definition and Purpose
- X-ray diffraction (XRD) is a powerful non-destructive technique used to analyze the atomic and molecular structure of crystalline materials. It works by shining X-rays onto a crystal, causing the X-rays to scatter off the electrons of the atoms within the crystal. This scattering produces a diffraction pattern that is unique to the crystal's atomic arrangement and interatomic distances.
- XRD helps to measure the interplanar spacing (d) between crystal layers, which allows for the accurate determination of the edge lengths of unit cells and ultimately leads to the derivation of Bragg's Law.
Bragg's Law (Equation): $n\lambda = 2d\sin\theta$
- Explanation of each term in the equation:
  - $n$ : The order of the reflected diffraction (an integer value: 1, 2, 3, …), representing the number of wavelengths in the path difference.
  - $\lambda$ : The wavelength of the incident X-ray beam.
  - $d$ : The distance between parallel planes of atoms (interplanar spacing) in the crystal lattice.
  - $\theta$ : The angle of incidence at which the X-ray beam strikes the crystal planes.

Derivation of Bragg's Law

Conceptual understanding:
- Imagine X-rays interacting with successive layers of atoms in a crystal, spaced apart by distance $d$ . When an X-ray beam strikes a crystal lattice, some rays penetrate and reflect from deeper layers.
- For constructive interference to occur (meaning the reflected waves are in phase and reinforce each other, producing a detectable signal), the difference in path length between waves reflected from adjacent parallel planes must be an integer multiple of the X-ray wavelength.
- By analyzing the geometry, if one ray reflects from the top plane and another from the second plane (distance $d$ below), the extra distance traveled by the second ray is $2d\sin\theta$ .
- To achieve constructive interference, this path length difference must equal a whole number of wavelengths: $2d\sin\theta = n\lambda$ .
Additional context and applications of X-ray diffraction:
- XRD is used extensively in material science, chemistry, and physics for phase identification (identifying unknown crystalline substances), determining crystal orientation, measuring lattice parameters, assessing crystallite size, and analyzing defects and strain in materials.

Sample Problem: Calculation of Iron Density in FCC Structure

Given:
- Atomic radius of iron (r): 124 pm = $1.24 \times 10^{-8} \text{ cm}$
- Crystal structure: Face-Centered Cubic (FCC)
- Atomic weight of Fe: $55.845 \text{ g/mol}$
- Avogadro's number: $6.022 \times 10^{23} \text{ atoms/mol}$
Determine edge length (L):
- For an FCC structure, atoms touch along the face diagonal, so the edge length formula is: $L = 2\sqrt{2}r$
- Edge length calculation: $L = 2\sqrt{2} \times 124 \text{ pm} = 350.72 \text{ pm}$
- Convert to cm: $L = 350.72 \times 10^{-10} \text{ cm} = 3.5072 \times 10^{-8} \text{ cm}$
Calculate volume of the unit cell:
$\text{Volume}_{\text{unit cell}} = L^3 = (3.5072 \times 10^{-8} \text{ cm})^3 = 4.316 \times 10^{-23} \text{ cm}^3$
Determine mass of the unit cell:
- An FCC unit cell contains 4 atoms. Therefore, the mass is:
  $\text{Mass}_{\text{unit cell}} = 4 \text{ atoms} \times \frac{55.845 \text{ g/mol}}{6.022 \times 10^{23} \text{ atoms/mol}} = 3.709 \times 10^{-22} \text{ g}$
Final Density Calculation:
- Density formula: $\text{Density} = \frac{\text{Mass}}{\text{Volume}}$
- Resulting density: $\text{Density} = \frac{3.709 \times 10^{-22} \text{ g}}{4.316 \times 10^{-23} \text{ cm}^3} = 8.60 \text{ g/cm}^3$

Conclusion

Recap of key points covered in the session, including the types of cubic unit cells (SC, BCC, FCC), their properties, and the importance of X-ray diffraction and Bragg's Law in characterizing crystal structures.
Encouragement for students to engage with additional problems related to crystal structures and calculations to solidify their understanding.
Mention of follow-up resources or examples planned for further exploration of content, such as hexagonal close-packed (HCP) structures or more complex unit cells.