4.3 Solid-State Chemistry: Crystal Systems, Unit Cells, and Band Theory

4.3 Solid-State Chemistry: Crystal Systems, Unit Cells, and Band Theory

Quick Answer: Solid-state chemistry investigates the three-dimensional crystalline arrangements of atoms, ions, and molecules and their resulting electronic band structures. Key board exam competencies include classifying crystals among the 7 crystal systems and 14 Bravais lattices; deriving unit cell relationships for simple cubic (sc), body-centered cubic (bcc), and face-centered cubic (fcc) lattices; calculating theoretical crystallographic densities ($\rho = ZM / N_A V_c$); determining hole distributions ($N$ octahedral holes and $2N$ tetrahedral holes per close-packed array of $N$ spheres); applying radius ratio rules ($r_+/r_-$) to predict structural prototypes ($NaCl, CsCl, ZnS, CaF_2$); solving Bragg's Law ($n\lambda = 2d\sin\theta$); and distinguishing conductors, insulators, and $n$-type vs. $p$-type semiconductors using electronic band theory.

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The 7 Crystal Systems and 14 Bravais Lattices

A crystal lattice is an infinite periodic array of points in space generated by repeating a fundamental parallelepiped termed the unit cell. The geometry of a unit cell is defined by six lattice parameters: three edge lengths ($a, b, c$) and three interaxial angles ($\alpha$ between $b$ and $c$, $\beta$ between $a$ and $c$, and $\gamma$ between $a$ and $b$).

All crystalline solids belong to one of 7 crystal systems, which in turn give rise to exactly 14 Bravais lattices when combined with the four centering operations (Primitive $P$, Body-centered $I$, Face-centered $F$, and Base-centered $C$):

Exam Note: The orthorhombic system is uniquely versatile because it is the only crystal system that accommodates all four Bravais centering types ($P, I, F, C$). Conversely, the triclinic system has the lowest symmetry ($C_1$ or $C_i$).

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Cubic Unit Cell Geometries, Packing Factors, and Density Calculations

In cubic crystals, atoms are modeled as hard, touching spheres of radius $r$. The number of net equivalent atoms per unit cell ($Z$) is calculated by accounting for fractional site sharing:

• Corner atom: shared by 8 adjacent unit cells $\rightarrow 1/8$ atom per corner.
• Edge-center atom: shared by 4 unit cells $\rightarrow 1/4$ atom per edge.
• Face-center atom: shared by 2 unit cells $\rightarrow 1/2$ atom per face.
• Body-center atom: unshared, residing entirely within the cell $\rightarrow 1$ atom.

The Atomic Packing Factor (APF) Derivation for FCC

$\text{APF} = \frac{V_\text{atoms}}{V_\text{cell}} = \frac{Z \times \frac{4}{3}\pi r^3}{a^3} = \frac{4 \times \frac{4}{3}\pi r^3}{(2\sqrt{2}r)^3} = \frac{\frac{16}{3}\pi r^3}{16\sqrt{2}r^3} = \frac{\pi}{3\sqrt{2}} \approx 0.7405\, (74.05\%)$

Worked Crystallographic Density Calculation

Problem: Metallic copper crystallizes in an fcc lattice with an atomic radius $r = 1.278\text{ \AA} = 1.278 \times 10^{-8}\text{ cm}$. The molar mass of copper is $M = 63.546\text{ g/mol}$. Calculate the theoretical crystallographic density of copper in $\text{g/cm}^3$.

Solution:

1. For an fcc unit cell, $Z = 4$.
2. Compute the unit cell edge length $a$: $a = 2\sqrt{2}r = 2 \times 1.4142 \times (1.278 \times 10^{-8}\text{ cm}) = 3.615 \times 10^{-8}\text{ cm}$
3. Calculate the unit cell volume $V_c$: $V_c = a^3 = (3.615 \times 10^{-8}\text{ cm})^3 = 4.724 \times 10^{-23}\text{ cm}^3$
4. Apply the crystallographic density formula: $\rho = \frac{Z \cdot M}{N_A \cdot V_c} = \frac{4 \times 63.546\text{ g/mol}}{(6.022 \times 10^{23}\text{ mol}^{-1}) \times (4.724 \times 10^{-23}\text{ cm}^3)} = \frac{254.184}{28.448} = 8.935\text{ g/cm}^3$ This value perfectly matches the experimentally measured density of copper ($8.94\text{ g/cm}^3$).

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Close Packing Systems and Interstitial Sites

Close packing of equal-sized spheres in two dimensions produces a hexagonal close-packed layer where each sphere touches 6 neighbors. Stacking these layers produces two distinct close-packed crystal systems with an identical, maximum theoretical packing efficiency of $74.05\%$ and a coordination number of 12:

1. Hexagonal Close-Packed (hcp): Stacking sequence follows an alternating $ABABAB\dots$ pattern. The third layer spheres lie directly above the first layer spheres. Space group $P6_3/mmc$; unit cell contains 2 atoms (primitive) or 6 atoms (hexagonal prism). Typical metals: $Mg, Zn, Ti, Co$.
2. Cubic Close-Packed (ccp / fcc): Stacking sequence follows a three-layer $ABCABC\dots$ pattern. The third layer occupies the alternate set of triangular interstitial depressions, and spheres repeat every fourth layer. Space group $Fm\bar{3}m$. Typical metals: $Cu, Ag, Au, Al$.

Interstitial Holes in Close-Packed Lattices

In any close-packed lattice containing $N$ spheres, there are exactly:

• $N$ Octahedral Holes: Formed by six spheres arranged at the vertices of a regular octahedron. In an fcc cell ($N = 4$), octahedral holes are located at the center of the unit cell (1 hole) and at the centers of the 12 edges ($12 \times 1/4 = 3$ holes), yielding a total of $1 + 3 = 4$ octahedral holes. Critical radius ratio: $\frac{r_h}{r_s} = \sqrt{2} - 1 \approx 0.414$.
• $2N$ Tetrahedral Holes: Formed by four spheres in a tetrahedral cluster. In an fcc cell ($N = 4$), there are $2 \times 4 = 8$ tetrahedral holes, located entirely inside the unit cell, one within each of the 8 smaller octants along the body diagonals. Critical radius ratio: $\frac{r_h}{r_s} = \sqrt{\frac{3}{2}} - 1 \approx 0.225$.

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Ionic Crystal Prototypes and Radius Ratio Rules

In binary ionic crystals ($M_x X_y$), structures are idealized as a close-packed or primitive lattice of larger anions ($X^{n-}$) with smaller cations ($M^{m+}$) occupying interstitial holes. The coordination number of the cation is governed by the radius ratio ($r_+ / r_-$):

Crucial Exam Trap: $CsCl$ is not body-centered cubic (bcc)! A true bcc lattice requires identical atoms at both the corner and body-center positions. Because $Cs^+$ and $Cl^-$ are distinct chemical species with different scattering factors, $CsCl$ consists of two interpenetrating primitive cubic lattices, possessing simple cubic Bravais symmetry.

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X-Ray Diffraction and Bragg's Law

X-ray diffraction (XRD) probes long-range periodic order in crystals because X-ray wavelengths ($\lambda \sim 0.5\text{--}2.5\text{ \AA}$) match interplanar lattice spacings ($d_{hkl}$).

Bragg's Law Formulation

$n\lambda = 2d_{hkl}\sin\theta$

Where:

• $n$ is the diffraction order (an integer, typically 1).
• $\lambda$ is the X-ray wavelength (e.g., $Cu\text{ } K_\alpha = 1.5418\text{ \AA}$).
• $d_{hkl}$ is the perpendicular distance between adjacent lattice planes characterized by Miller indices $(hkl)$.
• $\theta$ is the glancing Bragg angle between the incident beam and the crystal plane.

For a cubic crystal with edge length $a$, the interplanar spacing is:

$d_{hkl} = \frac{a}{\sqrt{h^2 + k^2 + l^2}}$

Systematic Absences (Selection Rules)

Destructive interference among centered lattice planes systematically eliminates certain reflections, allowing experimental determination of the Bravais lattice type from diffraction patterns:

• Primitive Cubic ($P$): All reflections allowed (all combinations of $h, k, l$ observed: $100, 110, 111, 200, 210, 211, \dots$).
• Body-Centered Cubic ($I$): Reflections observed only when the sum of indices is even: $h + k + l = 2n$ (e.g., $110, 200, 211, 220, \dots$). Reflections like $100, 111$ are absent.
• Face-Centered Cubic ($F$): Reflections observed only when $h, k, l$ are unmixed (either all odd or all even): $111, 200, 220, 311, 222, \dots$. Reflections like $110, 210$ are systematically absent.

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Band Theory of Solids: Conductors, Semiconductors, and Insulators

When $N$ atomic orbitals overlap across a crystal lattice, they split into $N$ discrete molecular orbitals. Because $N \sim 10^{23}$, the energy separation between adjacent levels becomes infinitesimally small ($\sim 10^{-22}\text{ eV}$), coalescing into continuous energy bands separated by forbidden band gaps ($E_g$):

• Valence Band (VB): The highest energy occupied band of electrons at absolute zero ($0\text{ K}$).
• Conduction Band (CB): The lowest unoccupied or partially filled energy band through which delocalized electrons travel.
• Fermi Level ($E_F$): The chemical potential of electrons; the highest occupied energy level at $0\text{ K}$, where the Fermi-Dirac occupation probability $f(E) = 0.5$.

Extrinsic Semiconductors: $n$-Type vs. $p$-Type Doping

Pure silicon ($Si$, Group 14) is an intrinsic semiconductor with modest room-temperature conductivity. Introducing trace impurities (parts-per-million) through doping dramatically alters its carrier concentration:

1. $n$-Type Semiconductors (Negative Carriers):
   • Doped with Group 15 pentavalent donor atoms ($P, As, Sb$).
   • Four of the five valence electrons form covalent bonds with the host silicon lattice, leaving one extra unbonded electron.
   • This extra electron occupies a discrete donor energy level ($E_d$) located just below the conduction band edge ($\Delta E_d \approx 0.045\text{ eV}$ for $P$ in $Si$).
   • At room temperature, thermal energy fully ionizes donor impurities, promoting electrons into the conduction band.
   • Electrons are the majority charge carriers; positive holes in the valence band are minority carriers. The Fermi level shifts upward toward the conduction band.
2. $p$-Type Semiconductors (Positive Carriers):
   • Doped with Group 13 trivalent acceptor atoms ($B, Al, Ga, In$).
   • The dopant has only three valence electrons, creating an electron deficiency (a "hole") in the covalent silicon network.
   • This creates a discrete, unoccupied acceptor energy level ($E_a$) located just above the valence band edge ($\Delta E_a \approx 0.045\text{ eV}$ for $B$ in $Si$).
   • Valence electrons are easily excited thermally into these acceptor levels, generating mobile positive holes in the valence band.
   • Holes are the majority charge carriers; electrons are minority carriers. The Fermi level shifts downward toward the valence band.