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Chapter 2 • Theory & Derivations

Chemical Crystallography I: Lattices, Unit Cells & Close-Packed Topologies

The space lattice and atomic basis, 7 crystal systems, 14 Bravais space lattices, geometry of close-packing (hcp vs ccp), tetrahedral and octahedral interstitial voids, Pauling radius ratio rules, atomic packing factors, and interstitial solid solutions.

§2.1 The Space Lattice, Atomic Basis & Primitive vs Non-Primitive Unit Cells

The fundamental mathematical description of crystalline matter rests upon the distinction between the abstract geometrical space lattice and the physical crystal structure.

The Mathematical Definition of a Crystal

A crystal structure is formed when an identical group of atoms, ions, or molecules—termed the basis or motif—is attached identically to every mathematical point of a three-dimensional space lattice:

\[\text{Crystal Structure} = \text{Space Lattice} + \text{Atomic Basis}\]

A mathematical space lattice is an infinite, periodic array of discrete points in space defined by translation vectors:

\[\mathbf{T} = u\mathbf{a} + v\mathbf{b} + w\mathbf{c}, \quad u, v, w \in \mathbb{Z}\]

where $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are non-coplanar primitive translation vectors. The atomic environment viewed from any lattice point $\mathbf{r}$ is identical to that viewed from $\mathbf{r}' = \mathbf{r} + \mathbf{T}$.

Primitive vs Non-Primitive (Conventional) Unit Cells

A unit cell is a parallelepiped defined by three basis vectors $\mathbf{a}, \mathbf{b}, \mathbf{c}$ with lengths $a, b, c$ and interaxial angles $\alpha = \angle(\mathbf{b}, \mathbf{c})$, $\beta = \angle(\mathbf{a}, \mathbf{c})$, and $\gamma = \angle(\mathbf{a}, \mathbf{b})$.

1. Primitive Cell ($P$): Contains exactly one lattice point per cell ($Z_{\text{lattice}} = 1$). A primitive cell represents the minimal volume unit that tiles all space under translational operations without gaps or overlaps.

2. Wigner-Seitz Primitive Cell: The unique, geometrically symmetric primitive cell constructed by drawing lines from a chosen lattice point to all adjacent lattice points and constructing normal bisecting planes. The region enclosed by these planes forms the Wigner-Seitz cell, which exhibits the full point group symmetry of the lattice.

3. Conventional (Non-Primitive) Unit Cells: Cells chosen with larger volumes ($Z_{\text{lattice}} > 1$) in order to display the full rotational and mirror symmetry of the crystal system:

  • Body-Centered ($I$, Innenzentriert): Additional lattice point at $(1/2, 1/2, 1/2) \implies Z_{\text{lattice}} = 8(1/8) + 1 = 2$.
  • Face-Centered ($F$, Flächenzentriert): Additional lattice points at $(1/2, 1/2, 0)$, $(1/2, 0, 1/2)$, $(0, 1/2, 1/2) \implies Z_{\text{lattice}} = 8(1/8) + 6(1/2) = 4$.
  • Base-Centered ($C, A, \text{ or } B$): Additional lattice points on one pair of opposite faces \implies Z_{\text{lattice}} = 8(1/8) + 2(1/2) = 2$.

§2.2 The 7 Crystal Systems & 14 Bravais Space Lattices in Three Dimensions

By combining spatial translational periodicity with rotational and inversion symmetry, Auguste Bravais demonstrated in 1848 that there exist precisely 14 distinct space lattices in three dimensions, grouped into 7 crystal systems.

The 7 Crystal Systems

The 7 crystal systems are classified by their essential characteristic symmetry elements, which impose geometric constraints upon unit cell edge lengths ($a, b, c$) and interaxial angles ($\alpha, \beta, \gamma$):

1. Cubic (Isometric): Essential symmetry: $4$ threefold rotation axes along body diagonals ($4C_3$).

  • Constraints: $a = b = c$, $\alpha = \beta = \gamma = 90^\circ$.
  • Bravais Lattices ($3$): Primitive ($P$), Body-Centered ($I$), Face-Centered ($F$).

2. Tetragonal: Essential symmetry: $1$ fourfold rotation or rotary-inversion axis ($1C_4$ or $1\bar{4}$).

  • Constraints: $a = b \neq c$, $\alpha = \beta = \gamma = 90^\circ$.
  • Bravais Lattices ($2$): Primitive ($P$), Body-Centered ($I$).

3. Orthorhombic: Essential symmetry: $3$ mutually perpendicular twofold axes or mirror planes ($3C_2$ or $3\sigma_v$).

  • Constraints: $a \neq b \neq c$, $\alpha = \beta = \gamma = 90^\circ$.
  • Bravais Lattices ($4$): Primitive ($P$), Body-Centered ($I$), Face-Centered ($F$), Base-Centered ($C$).

4. Hexagonal: Essential symmetry: $1$ sixfold rotation or rotary-inversion axis ($1C_6$ or $1\bar{6}$).

  • Constraints: $a = b \neq c$, $\alpha = \beta = 90^\circ, \gamma = 120^\circ$.
  • Bravais Lattice ($1$): Primitive ($P$).

5. Trigonal (Rhombohedral): Essential symmetry: $1$ threefold rotation axis ($1C_3$ or $1\bar{3}$).

  • Constraints: $a = b = c$, $\alpha = \beta = \gamma \neq 90^\circ < 120^\circ$ (in rhombohedral axes) or $a = b \neq c, \gamma = 120^\circ$ (in hexagonal axes).
  • Bravais Lattice ($1$): Rhombohedral ($R$).

6. Monoclinic: Essential symmetry: $1$ twofold rotation axis or mirror plane ($1C_2$ or $1m$).

  • Constraints: $a \neq b \neq c$, $\alpha = \gamma = 90^\circ, \beta \neq 90^\circ$ (unique $b$-axis setting).
  • Bravais Lattices ($2$): Primitive ($P$), Base-Centered ($C$).

7. Triclinic: Essential symmetry: Only identity ($1$) or inversion center ($\bar{1}$).

  • Constraints: $a \neq b \neq c$, $\alpha \neq \beta \neq \gamma \neq 90^\circ$.
  • Bravais Lattice ($1$): Primitive ($P$).

§2.3 Close-Packing of Hard Spheres: Hexagonal (hcp) vs Cubic (ccp) Topologies

Many elemental metals, noble gases, and ionic anion sublattices crystallize in dense structures derived from the closest packing of identical hard spheres in three dimensions.

Close-Packed Two-Dimensional Layers (Layer A)

In a single two-dimensional close-packed plane, each sphere is in contact with $6$ coplanar neighbors in a hexagonal arrangement. The surface area of the hexagonal unit rhombus containing $1$ sphere is:

\[A_{\text{rhombus}} = a^2 \sin(60^\circ) = (2R)^2 \frac{\sqrt{3}}{2} = 2\sqrt{3} R^2\]

The area occupied by the circular cross-section is $A_{\text{circle}} = \pi R^2$. The 2D packing fraction is:

\[\eta_{2\text{D}} = \frac{\pi R^2}{2\sqrt{3} R^2} = \frac{\pi}{2\sqrt{3}} \approx 0.9069 \quad (90.69\%)\]

Stacking in the Third Dimension: Voids B and C

Within a close-packed layer $A$, there exist two sets of triangular interstitial dips pointing in opposite directions:

  • Set $B$: Triangular depressions pointing "upward".
  • Set $C$: Triangular depressions pointing "downward".

A second close-packed layer can be placed in either set $B$ or set $C$. Choosing set $B$ establishes layer sequence $AB$. When adding the third layer, two distinct stacking choices emerge:

1. Hexagonal Close-Packing (hcp): Stacking sequence $ABABAB\dots$

  • Spheres of the third layer sit directly above the spheres of the first layer $A$.
  • Symmetry: Hexagonal system, space group $P6_3/mmc$.
  • Ideal Axial Ratio: For touching hard spheres:
\[\left(\frac{c}{a}\right)_{\text{ideal}} = \sqrt{\frac{8}{3}} = 2\sqrt{\frac{2}{3}} \approx 1.63299\]
  • Examples: $\text{Mg}, \text{Ti}, \text{Zr}, \text{Zn}, \text{Cd}, \text{Co}$.

2. Cubic Close-Packing (ccp / Face-Centered Cubic fcc): Stacking sequence $ABCABC\dots$

  • Spheres of the third layer occupy void set $C$, resting above neither $A$ nor $B$.
  • Symmetry: Cubic system, face-centered Bravais lattice ($Fm\bar{3}m$).
  • Close-packed $\{111\}$ planes intersect at $70.53^\circ$, giving four equivalent close-packed directions $\langle 110 \rangle$.
  • Examples: $\text{Cu}, \text{Ag}, \text{Au}, \text{Al}, \text{Ni}, \text{Pb}, \text{Pt}$.

Both hcp and ccp achieve the identical maximum atomic packing factor:

\[\text{APF} = \frac{\pi}{3\sqrt{2}} \approx 0.74048 \quad (74.05\%)\]

and each sphere has a coordination number of $\text{C.N.} = 12$ ($6$ coplanar, $3$ above, $3$ below).

§2.4 Geometry, Coordinates & Radii of Tetrahedral and Octahedral Voids

The interstitial cavities remaining between close-packed spheres host cations in ionic lattices (such as $\text{NaCl}, \text{ZnS}, \text{CaF}_2$) and interstitial atoms in alloys.

1. Tetrahedral Interstitial Voids ($T_d$)

A tetrahedral void is bounded by four mutually touching host spheres arranged at the vertices of a regular tetrahedron.

  • In the fcc unit cell (lattice parameter $a = 2\sqrt{2}R$), the cell can be subdivided into 8 smaller octants of edge length $a/2$.
  • The center of each octant is surrounded by 4 spheres (one corner atom and three adjacent face-centered atoms) forming a regular tetrahedron.
  • Fractional Coordinates: The 8 tetrahedral voids reside at:
\[\left( \frac{1}{4}, \frac{1}{4}, \frac{1}{4} \right), \left( \frac{3}{4}, \frac{1}{4}, \frac{1}{4} \right), \left( \frac{1}{4}, \frac{3}{4}, \frac{1}{4} \right), \left( \frac{1}{4}, \frac{1}{4}, \frac{3}{4} \right), \dots \text{ (8 sites per cell)}\]
  • Radius Ratio Derivation:

Distance from corner $(0,0,0)$ to octant center $(a/4, a/4, a/4)$ is:

\[d = \sqrt{\left(\frac{a}{4}\right)^2 + \left(\frac{a}{4}\right)^2 + \left(\frac{a}{4}\right)^2} = \frac{a\sqrt{3}}{4}\]

Substitute $a = 2\sqrt{2} R$:

\[R + r_{\text{tet}} = \frac{2\sqrt{2} R \sqrt{3}}{4} = \frac{\sqrt{6}}{2} R = \sqrt{\frac{3}{2}} R \implies \frac{r_{\text{tet}}}{R} = \sqrt{\frac{3}{2}} - 1 \approx 0.225\]

2. Octahedral Interstitial Voids ($O_h$)

An octahedral void is bounded by six host spheres arranged at the vertices of a regular octahedron.

  • In the fcc unit cell:

1. Body-Center Void: Located at $(1/2, 1/2, 1/2)$, coordinated by 6 face-centered spheres at distance $a/2$.

2. Edge-Center Voids: Located at the center of all 12 unit cell edges: $(1/2, 0, 0), (0, 1/2, 0), \dots$ Each edge-center is shared by 4 adjacent unit cells.

  • Total Count per Cell:
\[N_{\text{oct}} = 1\text{ (body)} + 12 \times \frac{1}{4}\text{ (edges)} = 1 + 3 = 4 \text{ voids per cell}\]
  • Radius Ratio Derivation:

Along any unit cell edge, two host spheres of radius $R$ at $(0,0,0)$ and $(1,0,0)$ sandwich an interstitial void of radius $r_{\text{oct}}$:

\[2R + 2r_{\text{oct}} = a = 2\sqrt{2} R \implies R + r_{\text{oct}} = \sqrt{2} R \implies \frac{r_{\text{oct}}}{R} = \sqrt{2} - 1 \approx 0.414\]

§2.5 Void Stoichiometry & Void-to-Atom Ratios in Close-Packed Lattices

A universal stoichiometric relationship governs interstitial sites in all close-packed lattices, regardless of whether the stacking is cubic ($ABC$) or hexagonal ($AB$).

The Universal Stoichiometric Rule

For any close-packed assembly containing $N$ host spheres:

\[\text{Number of Octahedral Voids} = N\]
\[\text{Number of Tetrahedral Voids} = 2N\]
\[\text{Total Interstitial Voids} = 3N\]

Rigorous Proof for the FCC / CCP Lattice

In a conventional face-centered cubic unit cell:

1. Host Atoms ($N$):

\[N = 8 \times \frac{1}{8} \text{ (corners)} + 6 \times \frac{1}{2} \text{ (faces)} = 1 + 3 = 4 \text{ atoms per cell}\]

2. Octahedral Voids:

  • $1$ at cell body-center $(1/2, 1/2, 1/2)$ $\implies 1 \times 1 = 1$.
  • $12$ at edge-centers $\implies 12 \times (1/4) = 3$.
  • Total $N_{\text{oct}} = 1 + 3 = 4 = N$.

3. Tetrahedral Voids:

  • $1$ inside each of the 8 constituent octants $\implies 8 \times 1 = 8 = 2N$.

Structural Archetypes Derived from Interstitial Filling

  • Rock Salt (NaCl): $100\%$ of octahedral voids occupied in ccp anion array ($6:6$ coordination).
  • Zinc Blende (ZnS): $50\%$ of tetrahedral voids occupied alternately in ccp anion array ($4:4$ coordination).
  • Fluorite (CaF$_2$): $100\%$ of tetrahedral voids occupied in ccp cation array ($8:4$ coordination).
  • Nickel Arsenide (NiAs): $100\%$ of octahedral voids occupied in hcp anion array ($6:6$ coordination).
  • Wurtzite (ZnS): $50\%$ of tetrahedral voids occupied alternately in hcp anion array ($4:4$ coordination).
  • Cadmium Iodide (CdI$_2$): $50\%$ of octahedral voids occupied in alternate layers of hcp iodide ($6:3$ coordination).

§2.6 Pauling's Radius Ratio Rules & Geometric Coordination Limits

In 1929, Linus Pauling formulated geometric principles governing ionic crystal coordination polyhedra. Pauling's First Rule states that a coordinated polyhedron of anions is formed about each cation, with the coordination number dictated by the radius ratio $\rho = r_+ / r_-$.

Derivation of Critical Radius Ratios

  • Linear Coordination (C.N. = 2):
\[\frac{r_+}{r_-} \ge 0\]
  • Trigonal Planar Coordination (C.N. = 3):
\[\frac{r_+ + r_-}{r_-} = \frac{2}{\sqrt{3}} \implies \frac{r_+}{r_-} = \frac{2}{\sqrt{3}} - 1 \approx 0.155\]
  • Tetrahedral Coordination (C.N. = 4):
\[\frac{r_+ + r_-}{r_-} = \sqrt{\frac{3}{2}} \implies \frac{r_+}{r_-} = \sqrt{\frac{3}{2}} - 1 \approx 0.225\]
  • Octahedral Coordination (C.N. = 6):
\[\frac{r_+ + r_-}{r_-} = \sqrt{2} \implies \frac{r_+}{r_-} = \sqrt{2} - 1 \approx 0.414\]
  • Cubic Coordination (C.N. = 8):
\[\frac{r_+ + r_-}{r_-} = \sqrt{3} \implies \frac{r_+}{r_-} = \sqrt{3} - 1 \approx 0.732\]
  • Cuboctahedral / Close-Packed (C.N. = 12):
\[\frac{r_+}{r_-} = 1.000\]

§2.7 Atomic Packing Fraction (APF) & Theoretical Mass Density

The efficiency with which atoms fill space in a crystal lattice dictates mechanical density, vacancy migration barriers, and compressibility.

Mathematical Definition of the Atomic Packing Fraction

The Atomic Packing Fraction (APF) is:

\[\text{APF} = \frac{N_{\text{atoms}} \times V_{\text{sphere}}}{V_{\text{cell}}} = \frac{N \left( \frac{4}{3}\pi R^3 \right)}{V_{\text{cell}}}\]
  • Simple Cubic (sc): $N = 1, a = 2R \implies \text{APF} = \frac{\pi}{6} \approx 52.36\%$.
  • Body-Centered Cubic (bcc): $N = 2, a\sqrt{3} = 4R \implies \text{APF} = \frac{\pi\sqrt{3}}{8} \approx 68.02\%$.
  • Face-Centered Cubic (fcc): $N = 4, a\sqrt{2} = 4R \implies \text{APF} = \frac{\pi}{3\sqrt{2}} \approx 74.05\%$.
  • Diamond Cubic: $N = 8, a\sqrt{3} = 8R \implies \text{APF} = \frac{\pi\sqrt{3}}{16} \approx 34.01\%$.

Theoretical Mass Density Derivation

\[\rho_{\text{theoretical}} = \frac{\text{Mass of Unit Cell}}{\text{Volume of Unit Cell}} = \frac{Z \times M}{N_A \times V_{\text{cell}}}\]

where $Z$ is the number of formula units per unit cell, $M$ is the molar mass ($\text{g/mol}$), $N_A = 6.02214 \times 10^{23}\text{ mol}^{-1}$, and $V_{\text{cell}}$ is in $\text{cm}^3$.

§2.8 Interstitial Alloys, Hydrides & The Hume-Rothery Rules

When host metal lattices incorporate solute elements, solid solutions form via interstitial or substitutional mechanisms.

Interstitial Solid Solutions & Hägg's Rule

Stable interstitial phases form when:

\[\frac{r_{\text{solute}}}{r_{\text{solvent}}} < 0.59\]
  • Carbon in Iron (Steel Metallurgy):
  • Austenite ($\gamma$-Fe, fcc): Large symmetric octahedral voids ($r_{\text{oct}} = 0.53\text{ \AA}$) allow up to $2.14\text{ wt}\%$ carbon solubility at $1147^\circ\text{C}$.
  • Ferrite ($\alpha$-Fe, bcc): Tetragonally distorted octahedral voids with a compressed axis opening of only $0.19\text{ \AA}$ severely restrict carbon solubility to $< 0.022\text{ wt}\%$.

Substitutional Solid Solutions & Hume-Rothery Rules

Extensive solid solubility requires:

1. Size Factor: Atomic radius mismatch $\le 15\%$.

2. Crystal Structure: Identical space lattice and crystal symmetry.

3. Electronegativity: Minimal electronegativity difference to avoid intermetallic precipitation.

4. Valence Electron Concentration ($e/a$): Characteristic ratios govern phase stability:

  • $\alpha$-phase (fcc solid solution): $e/a < 1.36$
  • $\beta$-phase (bcc): $e/a = 21/14 = 1.50$
  • $\gamma$-phase (complex cubic): $e/a = 21/13 \approx 1.62$
  • $\epsilon$-phase (hcp): $e/a = 21/12 = 1.75$
Medium Example 2.1: Exact Derivation of the Tetrahedral Interstitial Void Radius

Consider four identical hard spheres of radius $R$ packed in contact at the vertices of a regular tetrahedron.

  1. Inscribe the tetrahedron within a cube of edge length $L$. Relate $L$ to $R$.
  2. Derive the distance from the cube center to any of the four sphere centers.
  3. Prove that $r_{\text{tet}} / R = \sqrt{3/2} - 1 \approx 0.2247$.

Step 1: Inscription in a Cube

The tetrahedron vertices lie on alternating corners of a cube of edge $L$: $(0,0,0), (L,L,0), (L,0,L), (0,L,L)$. Tetrahedron edge length is the face diagonal: $d = L\sqrt{2} = 2R \implies L = R\sqrt{2}$.

Step 2: Distance from Center to Vertex

The centroid of the tetrahedron is at the body center $(L/2, L/2, L/2)$:

\[D = \frac{L\sqrt{3}}{2} = \frac{(R\sqrt{2})\sqrt{3}}{2} = R \sqrt{\frac{3}{2}}\]

Step 3: Radius Ratio

\[R + r_{\text{tet}} = R \sqrt{\frac{3}{2}} \implies \frac{r_{\text{tet}}}{R} = \sqrt{\frac{3}{2}} - 1 \approx 1.22474 - 1 = 0.2247 \approx 0.225\]
Easy Example 2.2: Exact Derivation of the Octahedral Interstitial Void Radius

Six identical hard spheres of radius $R$ surround an octahedral void.

  1. Relate the equatorial square edge $S$ to $R$.
  2. Using the square diagonal, derive the void radius $r_{\text{oct}}$.
  3. Prove that $r_{\text{oct}} / R = \sqrt{2} - 1 \approx 0.4142$.

Step 1: Equatorial Geometry

Edge length of square: $S = 2R$.

Step 2: Diagonal Geometry

Square diagonal: $D = S\sqrt{2} = 2\sqrt{2} R$. Along diagonal: $D = 2R + 2r_{\text{oct}} = 2(R + r_{\text{oct}})$.

Step 3: Radius Ratio

\[2(R + r_{\text{oct}}) = 2\sqrt{2} R \implies R + r_{\text{oct}} = R\sqrt{2} \implies \frac{r_{\text{oct}}}{R} = \sqrt{2} - 1 \approx 0.4142\]
Medium Example 2.3: Geometric Proof of Atomic Packing Factors for FCC and BCC Lattices

Prove analytically:

  1. $\text{APF}_{\text{fcc}} = \frac{\pi}{3\sqrt{2}} \approx 74.05\%$.
  2. $\text{APF}_{\text{bcc}} = \frac{\pi\sqrt{3}}{8} \approx 68.02\%$.
  3. Explain Kepler's conjecture on sphere packing.

Step 1: FCC Proof

\[a\sqrt{2} = 4R \implies a = 2\sqrt{2} R, \quad V_{\text{cell}} = 16\sqrt{2} R^3, \quad N = 4\]
\[\text{APF}_{\text{fcc}} = \frac{4 \times \frac{4}{3}\pi R^3}{16\sqrt{2} R^3} = \frac{\pi}{3\sqrt{2}} \approx 0.74048 \quad (74.05\%)\]

Step 2: BCC Proof

\[a\sqrt{3} = 4R \implies a = \frac{4R}{\sqrt{3}}, \quad V_{\text{cell}} = \frac{64 R^3}{3\sqrt{3}}, \quad N = 2\]
\[\text{APF}_{\text{bcc}} = \frac{2 \times \frac{4}{3}\pi R^3}{\frac{64 R^3}{3\sqrt{3}}} = \frac{\pi\sqrt{3}}{8} \approx 0.68017 \quad (68.02\%)\]

Step 3: Kepler Conjecture

Kepler conjectured in 1611 that no packing of identical spheres in 3D can exceed $\frac{\pi}{3\sqrt{2}} \approx 74.05\%$, proven rigorously by Thomas Hales in 1998.

Hard Example 2.4: Analytical Proof of the Ideal HCP Axial Ratio

In an ideal hcp lattice of touching spheres of radius $R$:

  1. Determine the interplanar height $h$ between layer $A$ and layer $B$.
  2. Prove that the ideal axial ratio is $c/a = \sqrt{8/3} \approx 1.633$.
  3. Interpret $c/a = 1.856$ in zinc.

Step 1: Interplanar Height

Distance from base vertex to triangle centroid: $r_{\text{centroid}} = a/\sqrt{3}$. Pythagorean theorem:

\[h^2 + \left(\frac{a}{\sqrt{3}}\right)^2 = a^2 \implies h^2 = \frac{2}{3}a^2 \implies h = a\sqrt{\frac{2}{3}}\]

Step 2: Total Cell Height $c$

\[c = 2h = 2a\sqrt{\frac{2}{3}} \implies \frac{c}{a} = 2\sqrt{\frac{2}{3}} = \sqrt{\frac{8}{3}} \approx 1.63299\]

Step 3: Interpretation of Zinc

Zinc's large $c/a = 1.856$ indicates strong in-plane covalent bonding and weak interlayer cohesion.

Easy Example 2.5: Theoretical Density Calculation for Metallic Copper

Copper (fcc, $M = 63.546\text{ g/mol}$, $R = 1.278\text{ \AA}$):

  1. Calculate the lattice parameter $a$.
  2. Calculate the theoretical density $\rho$.
  3. Compare with experimental density $8.96\text{ g/cm}^3$.

Step 1: Lattice Parameter

\[a = 2\sqrt{2} R = 2(1.41421)(1.278 \times 10^{-8}\text{ cm}) = 3.6148 \times 10^{-8}\text{ cm}\]

Step 2: Theoretical Density

\[V_{\text{cell}} = a^3 = 4.7235 \times 10^{-23}\text{ cm}^3\]
\[\rho = \frac{4 \times 63.546}{(6.02214 \times 10^{23})(4.7235 \times 10^{-23})} = 8.936\text{ g/cm}^3 \approx 8.94\text{ g/cm}^3\]

Step 3: Comparison

Matches experimental $8.96\text{ g/cm}^3$ within $0.25\%$.

Medium Example 2.6: Fractional Coordinates of Interstitial Voids in FCC

For an fcc unit cell of edge $a$:

  1. Tabulate the coordinates of all 4 octahedral voids.
  2. Tabulate the coordinates of all 8 tetrahedral voids.
  3. Calculate the distance between nearest-neighbor tetrahedral voids.

Step 1: Octahedral Voids

  • 1 body-center: $(1/2, 1/2, 1/2)$
  • 12 edge-centers: $(1/2, 0, 0), (0, 1/2, 0), \dots$ (weight $1/4$) $\implies$ Total 4 voids.

Step 2: Tetrahedral Voids

  • 8 octant centers: $(1/4, 1/4, 1/4), (3/4, 1/4, 1/4), \dots$ (weight 1) $\implies$ Total 8 voids.

Step 3: Separation

Distance between $(1/4, 1/4, 1/4)$ and $(3/4, 1/4, 1/4)$ is $\Delta x = a/2$.

Medium Example 2.7: Radius Ratio Threshold for Cubic Coordination

Derive the critical minimum radius ratio $r_+ / r_-$ for 8-fold cubic coordination.

Derivation

Anion contact along cube edge: $L = 2r_-$. Cation-anion contact along body diagonal: $2(r_+ + r_-) = L\sqrt{3} = 2r_-\sqrt{3}$.

\[r_+ + r_- = r_-\sqrt{3} \implies \frac{r_+}{r_-} = \sqrt{3} - 1 \approx 0.73205\]
Hard Example 2.8: Carbon Interstitial Site Distortion in Austenite vs Ferrite

Austenite (fcc Fe, $a = 3.589\text{ \AA}$) vs Ferrite (bcc Fe, $a = 2.866\text{ \AA}$):

  1. Calculate octahedral void radius in fcc iron.
  2. Calculate compressed opening of the octahedral void in bcc iron.
  3. Explain why carbon solubility in austenite is $100\times$ higher than in ferrite.

Step 1: FCC Void Radius

\[R_{\text{fcc}} = 1.269\text{ \AA} \implies r_{\text{oct, fcc}} = a/2 - R = 1.795 - 1.269 = 0.526\text{ \AA}\]

Step 2: BCC Compressed Void Radius

\[R_{\text{bcc}} = 1.241\text{ \AA} \implies r_{\text{oct, bcc}} = a/2 - R = 1.433 - 1.241 = 0.192\text{ \AA}\]

Step 3: Carbon Solubility Explanation

Carbon ($r = 0.77\text{ \AA}$) induces $300\%$ strain along the compressed axis in ferrite vs $46\%$ in austenite. The severe strain energy suppresses carbon solubility in ferrite to $< 0.02\%$, while austenite dissolves up to $2.14\%$. Trapping carbon by quenching produces tetragonally distorted martensite.

Easy Example 2.9: Hume-Rothery Electron Concentration in Brass

In the Cu-Zn system ($v_{\text{Cu}} = 1, v_{\text{Zn}} = 2$):

  1. Calculate atomic percentage of Zn at the $\alpha$-brass limit ($e/a = 1.36$).
  2. Verify $e/a$ for $\beta$-brass ($\text{CuZn}$) and $\gamma$-brass ($\text{Cu}_5\text{Zn}_8$).

Step 1: Alpha-Brass Limit

\[e/a = 1 + x = 1.36 \implies x = 0.36 \implies 36\text{ at}\%\text{ Zn}\]

Step 2: Beta and Gamma Brass

  • $\beta$-brass ($\text{CuZn}$): $e/a = (1+2)/2 = 1.50 = 21/14$.
  • $\gamma$-brass ($\text{Cu}_5\text{Zn}_8$): $e/a = (5(1) + 8(2))/13 = 21/13 \approx 1.62$.