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Chapter 3 • Theory & Derivations

Chemical Crystallography II: Macroscopic & Microscopic Symmetry

Crystallographic symmetry operations, proof of the crystallographic restriction theorem, 32 point groups, stereographic projections, translational symmetry (glide planes and screw axes), the 230 space groups, and solid-state polymorphic transformations.

§3.1 Macroscopic Symmetry Operations & The Crystallographic Restriction Theorem

The external morphological faces of well-formed macroscopic crystals exhibit rigorous symmetry described by crystallographic point groups.

The Crystallographic Restriction Theorem

Unlike free molecules in the gas phase, which can exhibit arbitrary rotational symmetry (such as $C_5$ in ferrocene or $C_7$ in tropylium), periodic crystals in two and three dimensions can possess only 1-, 2-, 3-, 4-, and 6-fold rotational symmetry. Fivefold ($C_5$) and eightfold or higher ($C_{\ge 7}$) rotation axes are strictly forbidden by translational periodicity.

Mathematical Proof:

Consider a 1D row of identical lattice points along a translation vector $\mathbf{a}$ of length $a$. Let two neighboring lattice points be $A$ and $B$, separated by distance $a$. Assume a counterclockwise rotation by angle $\alpha$ about an axis perpendicular to the plane passing through $A$ maps the lattice onto itself. Then point $B$ rotates to $B'$. Similarly, rotating clockwise by $\alpha$ about an axis passing through $B$ maps point $A$ to $A'$. Because the crystal lattice is translationally periodic, the vector connecting $A'$ and $B'$ must itself be a lattice translation vector parallel to $AB$, having length equal to an integer multiple $m$ of the primitive translation $a$:

\[\overline{A'B'} = m a, \quad m \in \mathbb{Z}\]

Projecting the segments onto the line $AB$:

\[\overline{A'B'} = a + 2 a \sin(\alpha - 90^\circ) = a - 2 a \cos\alpha = m a\]

Dividing throughout by $a$:

\[1 - 2\cos\alpha = m \implies \cos\alpha = \frac{1 - m}{2} = \frac{N}{2}, \quad N \in \mathbb{Z}\]

Because the cosine function is strictly bounded by $-1 \le \cos\alpha \le +1$:

\[-1 \le \frac{N}{2} \le +1 \implies N \in \{-2, -1, 0, +1, +2\}\]

Evaluating the permitted angles $\alpha$ and rotation orders $n = 360^\circ / \alpha$:

  1. $N = -2 \implies \cos\alpha = -1 \implies \alpha = 180^\circ \implies n = 2$ (twofold rotation).
  2. $N = -1 \implies \cos\alpha = -1/2 \implies \alpha = 120^\circ \implies n = 3$ (threefold rotation).
  3. $N = 0 \implies \cos\alpha = 0 \implies \alpha = 90^\circ \implies n = 4$ (fourfold rotation).
  4. $N = +1 \implies \cos\alpha = +1/2 \implies \alpha = 60^\circ \implies n = 6$ (sixfold rotation).
  5. $N = +2 \implies \cos\alpha = +1 \implies \alpha = 360^\circ \implies n = 1$ (identity rotation).

Fivefold rotation requires $\cos(72^\circ) = (\sqrt{5}-1)/4 \approx 0.3090$, which is not a half-integer ($N/2$). Hence, pentagonal tiles cannot tile Euclidean 2D space without gaps or overlaps.

§3.2 Inversion Centers, Mirror Planes & Rotary-Inversion Operations

Crystallographic point symmetry comprises proper rotations and improper (roto-inversion) operations.

Proper Rotations ($n$)

A proper rotation of order $n$ ($n \in \{1, 2, 3, 4, 6\}$) rotates an object by angle $\alpha = 2\pi/n$ about an axis, preserving enantiomeric handedness (congruent transformation).

Center of Inversion ($\bar{1}$ or $i$)

An inversion center maps point $(x, y, z)$ through the origin to $(-x, -y, -z)$:

\[\mathbf{r}' = -\mathbf{r}\]

Crystals possessing an inversion center are centrosymmetric and cannot exhibit polar physical phenomena such as piezoelectricity, pyroelectricity, or optical second-harmonic generation (SHG).

Mirror Planes ($m$ or $\sigma$)

A reflection plane maps coordinates across a plane. A mirror plane normal to the $z$-axis maps $(x, y, z) \to (x, y, -z)$. In Hermann-Mauguin notation, a mirror plane is denoted by $m$, equivalent to a twofold rotary-inversion axis:

\[m \equiv \bar{2}\]

Rotary-Inversion Axes ($\bar{n}$)

Combines a proper rotation by $2\pi/n$ with an immediate inversion through the origin:

  1. $\bar{1}$: Inversion center ($i$).
  2. $\bar{2}$: Equivalent to a mirror plane $m$ perpendicular to the axis.
  3. $\bar{3}$: A threefold rotation followed by inversion, equivalent to $3 + \bar{1}$ (a threefold axis with a center of inversion).
  4. $\bar{4}$: An essential fourfold rotary-inversion axis (characteristic of tetrahedral and tetragonal groups such as $\text{CuFeS}_2$ chalcopyrite). It consists of two operations: $\bar{4}^1$ and $\bar{4}^3$, containing an embedded twofold axis $\bar{4}^2 = 2$.
  5. $\bar{6}$: A sixfold rotation followed by inversion, equivalent to a threefold rotation with a perpendicular mirror plane ($3/m$).

§3.3 The 32 Crystallographic Point Groups & Stereographic Projections

Combining the 5 proper rotation axes ($1, 2, 3, 4, 6$) with inversion and reflection generates precisely 32 crystallographic point groups (crystal classes) distributed among the 7 crystal systems.

Stereographic Projection Principles

A stereographic projection represents 3D angular relationships of crystal faces and symmetry elements on a 2D plane:

  1. The crystal is placed at the center of a reference sphere.
  2. Normals to crystal faces (poles) radiate outward, intersecting the sphere surface.
  3. Points in the northern hemisphere ($z > 0$) are projected onto the equatorial plane by connecting them to the south pole; their intersections are marked with a solid dot ($\bullet$).
  4. Points in the southern hemisphere ($z < 0$) are projected to the north pole and marked with an open circle ($\circ$).
  5. Mirror planes perpendicular to the projection plane appear as straight diametral lines; mirror planes inclined to the plane appear as circular arcs.

Distribution of the 32 Point Groups Across Crystal Systems

  • Triclinic ($2$): $1$ (non-centrosymmetric), $\bar{1}$ (centrosymmetric).
  • Monoclinic ($3$): $2, m, 2/m$.
  • Orthorhombic ($3$): $222, mm2, mmm$.
  • Tetragonal ($7$): $4, \bar{4}, 4/m, 422, 4mm, \bar{4}2m, 4/mmm$.
  • Trigonal ($5$): $3, \bar{3}, 32, 3m, \bar{3}m$.
  • Hexagonal ($7$): $6, \bar{6}, 6/m, 622, 6mm, \bar{6}m2, 6/mmm$.
  • Cubic ($5$): $23, m\bar{3}, 432, \bar{4}3m, m\bar{3}m$.

Centrosymmetric vs Non-Centrosymmetric Groups

  • Centrosymmetric ($11$ groups): Contain $\bar{1}$. Include $\bar{1}, 2/m, mmm, 4/m, 4/mmm, \bar{3}, \bar{3}m, 6/m, 6/mmm, m\bar{3}, m\bar{3}m$.
  • Non-Centrosymmetric ($21$ groups): Lack inversion. Of these, 20 groups are piezoelectric (polar or chiral), and 10 groups are polar / pyroelectric ($1, 2, m, mm2, 4, 4mm, 3, 3m, 6, 6mm$), exhibiting spontaneous electric polarization along a unique polar axis.

§3.4 Hermann-Mauguin and Schoenflies Notations & Crystal Habits

Two notation systems are used internationally to denote crystallographic symmetry:

Hermann-Mauguin vs Schoenflies Notation

1. Schoenflies Notation: Widely used in molecular spectroscopy and quantum chemistry:

  • $C_n$: Cyclic group of order $n$.
  • $C_{nv}$: Cyclic group with vertical mirror planes.
  • $C_{nh}$: Cyclic group with horizontal mirror plane.
  • $D_n$: Dihedral group ($C_n$ plus $n$ perpendicular $C_2$ axes).
  • $D_{nh}, D_{nd}$: Dihedral groups with horizontal or diagonal mirror planes.
  • $T, T_h, T_d$: Tetrahedral cubic groups.
  • $O, O_h$: Octahedral cubic groups.

2. Hermann-Mauguin (International) Notation: Standard in solid-state chemistry and X-ray crystallography:

  • Direct axes are denoted by numbers: $1, 2, 3, 4, 6$.
  • Inversion axes are denoted with a bar: $\bar{1}, \bar{3}, \bar{4}, \bar{6}$.
  • Mirror planes are denoted by $m$.
  • A slash ($/$) indicates a mirror plane perpendicular to a rotation axis: $2/m, 4/m, 6/m$.
  • Successive symbols indicate symmetry along conventional crystallographic directions:
  • Orthorhombic ($mmm$): Mirror planes perpendicular to $\mathbf{a}, \mathbf{b}, \mathbf{c}$.
  • Tetragonal ($4/mmm$): $4/m$ along $c$; $m$ along $a, b$; $m$ along $[110]$.
  • Cubic ($m\bar{3}m$): $m$ along $\langle 100 \rangle$; $\bar{3}$ along body diagonals $\langle 111 \rangle$; $m$ along face diagonals $\langle 110 \rangle$.

Macroscopic Crystal Forms and Habits

A crystal form $\{hkl\}$ is a set of crystal faces related by the symmetry operations of the point group.

  • Open Forms: Faces do not enclose space (e.g., pedions [1 face], pinacoids [2 parallel faces], prisms, pyramids). Must be combined with other forms to create a closed crystal.
  • Closed Forms: Faces completely enclose space (e.g., octahedron $\{111\}$, cube $\{100\}$, rhombic dodecahedron $\{110\}$, tetrahedron).
  • Crystal Habit: The external morphological shape developed during growth (e.g., acicular [needle-like], platy [tabular], dendritic, prismatic), dictated by relative face growth velocities under differing supersaturation and impurity conditions.

§3.5 Translational Symmetry Elements: Axial, Diagonal & Diamond Glide Planes

When point symmetry operations are combined with fractional translations of the crystal space lattice, new microscopic symmetry elements arise that operate only at the atomic scale: glide planes and screw axes.

The Glide Plane Operation

A glide plane combines a mirror reflection across a plane with a simultaneous translation $\mathbf{t}$ parallel to that plane by a fraction of a unit cell vector:

\[\mathbf{r}' = \mathbf{R}_m \mathbf{r} + \mathbf{t}\]

Applying the operation twice corresponds to two reflections (identity) plus two translations $2\mathbf{t}$:

\[2\mathbf{t} = \mathbf{T}_{\text{lattice}} \implies \mathbf{t} = \frac{1}{2} \mathbf{T}_{\text{lattice}}\]

Types of Glide Planes in Hermann-Mauguin Notation

1. Axial Glide Planes ($a, b, c$):

Translation is exactly half of a primitive unit cell vector parallel to the glide plane:

  • $a$-glide: Reflection across plane, followed by translation $\mathbf{t} = \mathbf{a}/2$.
  • $b$-glide: Translation $\mathbf{t} = \mathbf{b}/2$.
  • $c$-glide: Translation $\mathbf{t} = \mathbf{c}/2$.

2. Diagonal Glide Plane ($n$):

Translation is half the face diagonal of the cell:

  • For an $n$-glide perpendicular to $c$: $\mathbf{t} = \frac{\mathbf{a} + \mathbf{b}}{2}$.

3. Diamond Glide Plane ($d$):

Translation is one-quarter of a face or body diagonal, occurring in centered lattices (such as the diamond cubic lattice $Fd\bar{3}m$):

  • $\mathbf{t} = \frac{\mathbf{a} \pm \mathbf{b}}{4}$ or $\frac{\mathbf{a} + \mathbf{b} + \mathbf{c}}{4}$.

Glide planes leave no invariant macroscopic face angles and are therefore invisible in morphological mineralogy. However, their translation components introduce destructive phase interference in diffracted X-ray beams, generating systematic absences in $(hkl)$ reflections that allow crystallographers to determine space groups unambiguously.

§3.6 Screw Axes ($n_m$) & Microscopic Helical Translations

A screw axis combines a proper rotation by angle $2\pi/n$ with a fractional translation parallel to the rotation axis.

Mathematical Formulation

A screw axis is denoted by the symbol:

\[n_m, \quad n \in \{2, 3, 4, 6\}, \quad m \in \{1, 2, \dots, n-1\}\]

The operation consists of:

  1. Rotation by $2\pi/n$ about the axis.
  2. Translation parallel to the axis by fractional vector:
\[\mathbf{t} = \frac{m}{n} \mathbf{T}\]

Applying the $n_m$ operation $n$ times yields an overall rotation of $n \times (2\pi/n) = 2\pi$ (identity) and a total translation of $n \times (m/n)\mathbf{T} = m\mathbf{T}$, which is an integer lattice translation.

The Eleven Screw Axes

  • Twofold:
  • $2_1$: Rotation $180^\circ$, translation $\mathbf{t} = \mathbf{c}/2$.
  • Threefold:
  • $3_1$: Rotation $120^\circ$, translation $\mathbf{t} = \mathbf{c}/3$ (right-handed helix).
  • $3_2$: Rotation $120^\circ$, translation $\mathbf{t} = 2\mathbf{c}/3$ (left-handed enantiomorph of $3_1$).
  • Fourfold:
  • $4_1$: Rotation $90^\circ$, translation $\mathbf{t} = \mathbf{c}/4$ (right-handed).
  • $4_2$: Rotation $90^\circ$, translation $\mathbf{t} = 2\mathbf{c}/4 = \mathbf{c}/2$ (neutral).
  • $4_3$: Rotation $90^\circ$, translation $\mathbf{t} = 3\mathbf{c}/4$ (left-handed enantiomorph of $4_1$).
  • Sixfold:
  • $6_1$ and $6_5$: Enantiomorphic pair with translations $\mathbf{c}/6$ and $5\mathbf{c}/6$.
  • $6_2$ and $6_4$: Enantiomorphic pair with translations $2\mathbf{c}/6$ and $4\mathbf{c}/6$.
  • $6_3$: Translation $3\mathbf{c}/6 = \mathbf{c}/2$.

Enantiomorphism and Chiral Crystals

Chiral molecules (such as L-amino acids and D-sugars) cannot crystallize in space groups possessing inversion centers, mirror planes, or glide planes. They crystallize exclusively in the 65 Sohncke space groups, which contain only proper rotations and screw axes ($n_m$). Quartz ($\alpha$-quartz) crystallizes as enantiomorphic pairs in space groups $P3_1 21$ (right-handed) and $P3_2 21$ (left-handed), exhibiting macroscopic optical rotation.

§3.7 The 230 Space Groups: Derivation, Symbols & Asymmetric Units

The complete symmetry of any crystalline solid is rigorously described by one of the 230 crystallographic space groups, derived independently in 1891 by Arthur Schoenflies and Evgraf Fedorov.

Construction of Space Groups

A space group $\mathcal{G}$ is the infinite discrete group of all symmetry operations that map a 3D periodic crystal onto itself. It is mathematically an extension of the translational group $\mathcal{T}$ (14 Bravais lattices) by a point group $\mathcal{P}$ (32 point groups):

  • Symmorphic Space Groups (73): Contain only point group operations and pure lattice translations (no glide planes or screw axes).
  • Non-Symmorphic Space Groups (157): Contain glide planes and/or screw axes with fractional translations.

Anatomy of Hermann-Mauguin Space Group Symbols

The standard full international symbol specifies:

1. First Character: Bravais lattice centering type:

  • $P$ (Primitive), $I$ (Body-centered), $F$ (Face-centered), $C, A, B$ (Base-centered), $R$ (Rhombohedral).

2. Subsequent Characters: Symmetry elements along principal crystallographic directions:

  • Monoclinic ($P2_1/c$):
  • $P$: Primitive lattice.
  • $2_1/c$: Along the unique $b$-axis, a $2_1$ screw axis is perpendicular to a $c$-glide plane.
  • Orthorhombic ($Pnma$):
  • $P$: Primitive lattice.
  • $n$: $n$-glide plane perpendicular to $a$.
  • $m$: Mirror plane perpendicular to $b$.
  • $a$: $a$-glide plane perpendicular to $c$.
  • Cubic ($Fm\bar{3}m$, Rock Salt):
  • $F$: Face-centered cubic lattice.
  • $m$: Mirror plane perpendicular to $\langle 100 \rangle$.
  • $\bar{3}$: Threefold rotary-inversion axis along body diagonals $\langle 111 \rangle$.
  • $m$: Mirror plane perpendicular to face diagonals $\langle 110 \rangle$.

The Asymmetric Unit and Wyckoff Positions

  • Asymmetric Unit: The minimal fraction of unit cell volume from which the entire crystal can be generated by applying all space group symmetry operations.
  • Wyckoff Positions: Sites within the unit cell classified by their site symmetry:
  • General Position: A point $(x, y, z)$ having no site symmetry ($1$). Its multiplicity equals the order of the space group.
  • Special Positions: Points lying on symmetry elements (such as mirror planes, rotation axes, or inversion centers). Their site symmetry is higher, and their multiplicity is a fraction of the general position.

§3.8 Solid-State Structural Relationships: Polymorphism, Polytypism & Isomorphism

The structural response of solid materials to temperature, pressure, and chemical substitution is categorized into four fundamental crystallographic relationships:

1. Polymorphism

The ability of a chemical compound of fixed stoichiometry to exist in two or more distinct crystal structures (e.g., $\text{CaCO}_3$ as calcite [trigonal] vs aragonite [orthorhombic]; $\text{TiO}_2$ as rutile, anatase, and brookite).

  • Enantiotropic Polymorphism: Phase transition is thermodynamically reversible at a specific transition temperature ($T_{\text{tr}}$) and pressure ($P_{\text{tr}}$) where $\Delta G = 0$.
  • Monotropic Polymorphism: One polymorph is thermodynamically stable under all conditions; other polymorphs are metastable and convert irreversibly to the stable phase.

2. Polytypism

A special one-dimensional sub-case of polymorphism in which distinct structures arise solely from different close-packed stacking sequences of identical two-dimensional modular layers:

  • Silicon Carbide (SiC): More than $250$ distinct polytypes exist:
  • $3C$: Cubic Zinc Blende stacking ($ABCABC\dots$, space group $F\bar{4}3m$).
  • $4H$: Hexagonal stacking ($ABCB\dots$, space group $P6_3 mc$).
  • $6H$: Hexagonal stacking ($ABCACB\dots$, space group $P6_3 mc$, repeat distance $c = 15.1\text{ \AA}$).
  • $15R$: Rhombohedral stacking with 15-layer repeat ($c = 37.8\text{ \AA}$).

3. Isomorphism

Distinct chemical compounds possessing identical crystal structures, space groups, and closely similar unit cell dimensions (discovered by Eilhard Mitscherlich in 1819).

  • Examples:
  • $\text{K}_2\text{SO}_4$, $\text{K}_2\text{SeO}_4$, and $\text{K}_2\text{CrO}_4$ (all orthorhombic $Pnma$).
  • Alums: $M^I M^{III}(\text{SO}_4)_2 \cdot 12\text{H}_2\text{O}$ (where $M^I = \text{K}^+, \text{NH}_4^+, \text{Rb}^+$ and $M^{III} = \text{Al}^{3+}, \text{Cr}^{3+}, \text{Fe}^{3+}$).

Isomorphous salts readily form continuous substitutional solid solutions and undergo epitaxy.

4. Allotropy

Polymorphism occurring in elemental substances:

  • Carbon: Diamond ($sp^3$ covalent network), graphite ($sp^2$ layered hexagonal sheets), fullerenes ($\text{C}_{60}$ molecular crystals), carbon nanotubes, and graphene.
  • Tin ($\text{Sn}$): Grey tin ($\alpha\text{-Sn}$, diamond cubic semiconductor) converts below $13.2^\circ\text{C}$ to white tin ($\beta\text{-Sn}$, body-centered tetragonal ductile metal), a volume expansion of $27\%$ responsible for "tin pest".
Medium Example 3.1: Mathematical Proof of the Crystallographic Restriction Theorem

Consider a 2D periodic lattice with lattice translation vector $\mathbf{a}$ of length $a$.

  1. Using the four-point collinear construction (points $A, B$ on the lattice line rotating by $\pm \alpha$ to $A', B'$), derive the equation:
\[\cos\alpha = \frac{1 - m}{2} = \frac{N}{2}, \quad N \in \mathbb{Z}\]
  1. Deduce all allowed rotational orders $n = 360^\circ / \alpha$.
  2. Prove that a 5-fold rotation axis cannot exist in a periodic 2D crystal lattice.

Step 1: Geometry of the Construction

Let $A$ and $B$ be two adjacent lattice points separated by vector $\mathbf{a}$ with $|\mathbf{a}| = a$. Rotate counterclockwise by angle $\alpha$ about $A$ to obtain lattice point $B'$. Rotate clockwise by angle $\alpha$ about $B$ to obtain lattice point $A'$. Because the transformed points are lattice points, the vector $A'B'$ must be parallel to $AB$ and equal to an integer multiple of $a$:

\[\overline{A'B'} = m a, \quad m \in \mathbb{Z}\]

Projecting onto the line $AB$:

\[\overline{A'B'} = a - 2a\cos\alpha\]

Equating:

\[a - 2a\cos\alpha = m a \implies 1 - 2\cos\alpha = m \implies \cos\alpha = \frac{1 - m}{2}\]

Setting $N = 1 - m \in \mathbb{Z}$:

\[\cos\alpha = \frac{N}{2}\]

Step 2: Permitted Solutions

Since $-1 \le \cos\alpha \le +1$:

\[-1 \le \frac{N}{2} \le +1 \implies N \in \{-2, -1, 0, +1, +2\}\]
  • $N = -2 \implies \cos\alpha = -1 \implies \alpha = 180^\circ \implies n = 2$.
  • $N = -1 \implies \cos\alpha = -1/2 \implies \alpha = 120^\circ \implies n = 3$.
  • $N = 0 \implies \cos\alpha = 0 \implies \alpha = 90^\circ \implies n = 4$.
  • $N = +1 \implies \cos\alpha = +1/2 \implies \alpha = 60^\circ \implies n = 6$.
  • $N = +2 \implies \cos\alpha = +1 \implies \alpha = 360^\circ \implies n = 1$.

Step 3: Fivefold Rotation Impossibility

For $n = 5$, $\alpha = 72^\circ$:

\[\cos(72^\circ) = \frac{\sqrt{5}-1}{4} \approx 0.309017\]

To exist in a lattice, $2\cos(72^\circ) = \frac{\sqrt{5}-1}{2} \approx 0.618034$ must be an integer. Since $\sqrt{5}$ is irrational, this is impossible.

Medium Example 3.2: Stereographic Projection of Point Group 4/mmm

Point group $4/mmm$ ($D_{4h}$) is the full holohedral symmetry of the tetragonal crystal system.

  1. List all symmetry operations belonging to point group $4/mmm$. What is the order of the group?
  2. Construct the stereographic projection of the general pole $(hkl)$ with $h > k > 0$ and $l > 0$.
  3. What is the multiplicity of this general form $\{hkl\}$?

Step 1: Symmetry Operations

The international symbol $4/mmm$ indicates:

  • A principal fourfold axis with a perpendicular mirror plane: $4/m \implies E, C_4^1, C_4^2 (= C_2), C_4^3, i, \sigma_h, S_4^1, S_4^3$ (8 operations).
  • Two mirror planes containing the $c$-axis parallel to $a$ and $b$: $2\sigma_v$ and two perpendicular twofold axes $2C_2'$ (4 operations).
  • Two mirror planes along diagonal directions $[110]$ and $[1\bar{1}0]$: $2\sigma_d$ and two diagonal twofold axes $2C_2''$ (4 operations).

Total operations: $16$. The order of the group is $h = 16$.

Step 2: Stereographic Projection Coordinates

Start with pole $(hkl)$ in the first octant:

  1. Fourfold rotation about $z$ replicates the pole 4 times in the upper hemisphere:

$(h, k, l), (-k, h, l), (-h, -k, l), (k, -h, l)$ $\implies 4$ solid dots ($\bullet$).

  1. Diagonal mirror planes reflect these 4 poles across the diagonals:

$(k, h, l), (-h, k, l), (-k, -h, l), (h, -k, l)$ $\implies 4$ additional solid dots (total 8 in northern hemisphere).

  1. Horizontal mirror plane $\sigma_h$ reflects all 8 poles to the southern hemisphere:

8 open circles ($\circ$) directly superimposed beneath the 8 solid dots.

Step 3: Multiplicity of Form

Multiplicity equals the total number of symmetry-equivalent faces:

\[\text{Multiplicity} = 8 \text{ (upper)} + 8 \text{ (lower)} = 16\]

The general form is a ditetragonal dipyramid $\{hkl\}$ with 16 faces.

Hard Example 3.3: Complete Space Group Deconstruction: P2_1/c

Space group $P2_1/c$ (No. 14, unique axis $b$) is the most common space group in organic and coordination chemistry, describing over $30\%$ of all solved molecular crystal structures.

  1. State the crystal system and Bravais lattice type.
  2. Identify the orientation and operation of the screw axis and glide plane.
  3. List the 4 symmetry-equivalent general positions $(x, y, z)$ generated in the unit cell.
  4. Prove that $P2_1/c$ is centrosymmetric and find the coordinates of the inversion centers.

Step 1: Crystal System and Bravais Lattice

  • Crystal System: Monoclinic ($a \neq b \neq c, \alpha = \gamma = 90^\circ, \beta \neq 90^\circ$).
  • Bravais Lattice: Primitive ($P$).

Step 2: Symmetry Elements

  • $2_1$ screw axis: Oriented along the unique $b$-axis $[010]$:

Rotation by $180^\circ$ about $y$ followed by translation $\mathbf{b}/2$.

  • $c$-glide plane: Perpendicular to the $b$-axis (in the $xz$-plane):

Reflection across $y = 0$ followed by translation $\mathbf{c}/2$.

Step 3: Derivation of the Four General Positions

Start with a general point $(x, y, z)$:

1. Identity ($1$):

\[(x, y, z)\]

2. Screw Axis ($2_1$ along $[010]$ passing through $(0, y, 1/4)$):

Inverts $x$ and $z$, adds $1/2$ translation along $y$:

\[\left(-x, y + \frac{1}{2}, -z + \frac{1}{2}\right)\]

3. Inversion ($i$ at origin $(0,0,0)$):

\[(-x, -y, -z)\]

4. $c$-Glide Plane (perpendicular to $b$):

Reflects $y$, adds $1/2$ translation along $z$:

\[\left(x, -y + \frac{1}{2}, z + \frac{1}{2}\right)\]

These 4 positions define the general position multiplicity $Z = 4$.

Step 4: Centrosymmetry Proof

Combining the $2_1$ screw axis $( -x, y+1/2, -z+1/2 )$ with the $c$-glide $( x, -y+1/2, z+1/2 )$: The composite operation transforms $(x, y, z) \to (-x, -y, -z)$, which is an inversion center! Inversion centers are located at the 8 standard centers of symmetry: $(0,0,0), (1/2, 0, 0), (0, 1/2, 0), (0, 0, 1/2), (1/2, 1/2, 0), (1/2, 0, 1/2), (0, 1/2, 1/2), (1/2, 1/2, 1/2)$.

Medium Example 3.4: Wyckoff Multiplicities and Site Invariants in Fm-3m

The space group of the Rock Salt ($\text{NaCl}$) crystal structure is $Fm\bar{3}m$ (No. 225).

  1. State the order of the general position in $Fm\bar{3}m$.
  2. Sodium ions occupy Wyckoff position $4a$: $(0, 0, 0)$. Determine their site symmetry.
  3. Chloride ions occupy Wyckoff position $4b$: $(1/2, 1/2, 1/2)$. Determine their site symmetry.
  4. Verify that the ratio of Wyckoff site multiplicities reproduces the $1:1$ stoichiometry of $\text{NaCl}$.

Step 1: General Position Multiplicity

In $Fm\bar{3}m$, the point group is $m\bar{3}m$ of order 48. The face-centered lattice $F$ adds 4 lattice points per cell:

\[\text{General Position Multiplicity} = 4 \times 48 = 192\]

Step 2: Site Symmetry of Sodium (Wyckoff 4a)

Sodium is at $(0, 0, 0)$. Lattice centering generates the 4 equivalent positions: $(0, 0, 0), (0, 1/2, 1/2), (1/2, 0, 1/2), (1/2, 1/2, 0) \implies \text{Multiplicity} = 4$. Site symmetry order:

\[\text{Site Symmetry Order} = \frac{192}{4} = 48\]

The site symmetry is $m\bar{3}m$ ($O_h$, full octahedral symmetry).

Step 3: Site Symmetry of Chloride (Wyckoff 4b)

Chloride is at $(1/2, 1/2, 1/2)$. Lattice centering generates 4 equivalent positions: $(1/2, 1/2, 1/2), (1/2, 0, 0), (0, 1/2, 0), (0, 0, 1/2) \implies \text{Multiplicity} = 4$. Site symmetry is also $m\bar{3}m$ ($O_h$).

Step 4: Stoichiometry

\[\frac{\text{Multiplicity}(4a)}{\text{Multiplicity}(4b)} = \frac{4}{4} = 1:1\]

Each unit cell contains 4 sodium ions and 4 chloride ions, giving $Z = 4$ formula units of $\text{NaCl}$.

Medium Example 3.5: Polytypic Stacking Sequences and Enthalpy in Silicon Carbide

Silicon carbide ($\text{SiC}$) exhibits polytypism with stacking periods ranging from 3 to hundreds of layers.

  1. Decode the Ramsdell notation for polytypes $3C, 4H, 6H, 15R$.
  2. Write the layer stacking sequences for $3C$ and $4H$ using $A, B, C$ notation.
  3. Calculate the hexagonality percentage ($h$-percentage) for $3C$ and $4H$ polytypes.
  4. Explain why the enthalpy difference between distinct SiC polytypes is less than $1\text{ kJ/mol}$.

Step 1: Ramsdell Notation

  • $3C$: $3$-layer repeat, Cubic lattice ($F\bar{4}3m$).
  • $4H$: $4$-layer repeat, Hexagonal lattice ($P6_3 mc$).
  • $6H$: $6$-layer repeat, Hexagonal lattice ($P6_3 mc$).
  • $15R$: $15$-layer repeat, Rhombohedral lattice ($R3m$).

Step 2: Stacking Sequences

  • $3C$: $ABCABC\dots$ (all cubic stacking).
  • $4H$: $ABCBABCB\dots$ (alternating cubic and hexagonal stacking).

Step 3: Hexagonality Percentage ($h$)

A layer is designated $h$ if its neighboring layers are identical (e.g., $ABA$), and $c$ if its neighbors are different (e.g., $ABC$):

  • In $3C$: Stacking is $c c c \implies h = 0\%$.
  • In $4H$: Stacking is $h c h c \implies 2$ hexagonal layers out of 4:
\[h\text{-percentage} = \frac{2}{4} \times 100\% = 50\%\]

Step 4: Physical Rationale

In all SiC polytypes, nearest-neighbor and next-nearest-neighbor coordination spheres are identical: every Si atom is tetrahedrally bonded to 4 carbon atoms at $1.89\text{ \AA}$, and every C is tetrahedrally bonded to 4 Si atoms. Structural differences arise only in third-nearest-neighbor arrangements ($> 4\text{ \AA}$ away). Because electrostatic and covalent energies are dominated by nearest neighbors, energy differences between polytypes are tiny ($< 0.5\text{ kJ/mol}$), allowing temperature and growth kinetics to stabilize hundreds of polytypes.

Easy Example 3.6: Mitscherlich's Law of Isomorphism & Solid Solution Formation

Eilhard Mitscherlich formulated the Law of Isomorphism in 1819.

  1. State Mitscherlich's original law and its modern crystallographic interpretation.
  2. Given that potassium dihydrogen phosphate ($\text{KH}_2\text{PO}_4$, KDP) and potassium dihydrogen arsenate ($\text{KH}_2\text{AsO}_4$, KDA) are isomorphous in tetragonal space group $I\bar{4}2d$:

Predict whether they will form solid solutions $\text{KH}_2(\text{P}_{1-x}\text{As}_x)\text{O}_4$.

  1. How did isomorphism historically assist J. J. Berzelius in establishing correct atomic weights?

Step 1: Definition of Isomorphism

  • Mitscherlich's Law: Substances of analogous chemical constitution crystallize in the same crystalline form with identical face angles.
  • Modern Interpretation: Isomorphous substances share the identical space group, identical Wyckoff site topology, and closely matched unit cell parameters ($\Delta a/a < 10\%$).

Step 2: KDP and KDA Solid Solutions

Both KDP and KDA crystallize in tetragonal space group $I\bar{4}2d$ with tetrahedral $\text{PO}_4^{3-}$ and $\text{AsO}_4^{3-}$ anions:

  • KDP: $a = 7.45\text{ \AA}, c = 6.97\text{ \AA}$
  • KDA: $a = 7.63\text{ \AA}, c = 7.16\text{ \AA}$

The lattice mismatch is $\Delta a/a = (7.63 - 7.45)/7.45 = 2.4\% \ll 15\%$. They form a complete, continuous substitutional solid solution $\text{KH}_2(\text{P}_{1-x}\text{As}_x)\text{O}_4$ across all compositions $0 \le x \le 1$.

Step 3: Historical Atomic Weight Determination

In the 1820s, chemical formulas were uncertain (e.g., water was thought to be $\text{HO}$). Berzelius observed that potassium sulfate and potassium selenate were isomorphous. Since sulfate was known to contain $\text{SO}_4$, selenate must contain $\text{SeO}_4$. By measuring the mass ratio of sulfur to selenium in isomorphous crystals, Berzelius deduced correct relative atomic weights.

Hard Example 3.7: Matrix Algebra Representation of Glide Plane Operations

In coordinate space, any crystallographic symmetry operation is represented by an affine transformation:

\[\mathbf{r}' = \mathbf{W} \mathbf{r} + \mathbf{w}\]

where $\mathbf{W}$ is a $3 \times 3$ rotation matrix and $\mathbf{w}$ is a $3 \times 1$ translation column vector.

  1. Write the matrix $\mathbf{W}$ and translation vector $\mathbf{w}$ for an $a$-glide plane perpendicular to the $c$-axis located at height $z = 0$.
  2. Apply the operation twice to prove that it produces an integer lattice translation along $\mathbf{a}$.
  3. What is the determinant $\det(\mathbf{W})$ for a glide plane?

Step 1: Matrix and Translation Vector

For an $a$-glide plane perpendicular to $c$ ($z = 0$):

  • Reflection across $z = 0$ maps $(x, y, z) \to (x, y, -z)$.
  • Translation along $x$ by $a/2$ adds $1/2$ to $x$.
\[\mathbf{W} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix}, \quad \mathbf{w} = \begin{pmatrix} 1/2 \\ 0 \\ 0 \end{pmatrix}\]

Operating on point $\mathbf{r} = (x, y, z)^T$:

\[\mathbf{r}' = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} + \begin{pmatrix} 1/2 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} x + 1/2 \\ y \\ -z \end{pmatrix}\]

Step 2: Applying Twice

Apply the affine transformation again to $\mathbf{r}'$:

\[\mathbf{r}'' = \mathbf{W} \mathbf{r}' + \mathbf{w} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix} \begin{pmatrix} x + 1/2 \\ y \\ -z \end{pmatrix} + \begin{pmatrix} 1/2 \\ 0 \\ 0 \end{pmatrix}\]
\[\mathbf{r}'' = \begin{pmatrix} x + 1/2 + 1/2 \\ y \\ -(-z) \end{pmatrix} = \begin{pmatrix} x + 1 \\ y \\ z \end{pmatrix} = \mathbf{r} + \mathbf{a}\]

Two consecutive operations yield identity reflection ($\mathbf{W}^2 = \mathbf{I}$) plus a full unit cell translation along $\mathbf{a}$.

Step 3: Determinant

\[\det(\mathbf{W}) = (1)(1)(-1) = -1\]

The determinant is $-1$, confirming that a glide plane is an improper, enantiomorph-inverting transformation.

Hard Example 3.8: Screw Axis Operator Algebra and Systematic Absence Genesis

Consider a $2_1$ screw axis along the $c$-axis $[001]$ passing through the origin.

  1. Write the transformation of coordinates $(x, y, z)$ under this $2_1$ operation.
  2. The crystallographic structure factor is:
\[F_{hkl} = \sum_j f_j \exp[2\pi i (h x_j + k y_j + l z_j)]\]

Prove that the presence of the $2_1$ screw axis forces the structure factor $F_{00l}$ to vanish whenever $l$ is odd ($l = 2n + 1$).

Step 1: Coordinate Transformation

A $2_1$ screw axis along $z$ rotates by $180^\circ$ about $z$ and translates by $c/2$:

\[(x, y, z) \longrightarrow \left(-x, -y, z + \frac{1}{2}\right)\]

Step 2: Systematic Absence Proof for $(00l)$ Reflections

For an arbitrary atom $j$ at $(x_j, y_j, z_j)$, the $2_1$ screw axis generates a symmetry-paired atom $j'$ at $(-x_j, -y_j, z_j + 1/2)$ with identical atomic scattering factor $f_j$. Evaluate the structure factor for $(00l)$ reflections ($h = 0, k = 0$):

\[F_{00l} = \sum_j f_j \left[ \exp(2\pi i l z_j) + \exp\left(2\pi i l (z_j + 1/2)\right) \right]\]

Factor out $\exp(2\pi i l z_j)$:

\[F_{00l} = \sum_j f_j \exp(2\pi i l z_j) \left[ 1 + \exp(\pi i l) \right]\]

Recall Euler's identity: $\exp(\pi i l) = (-1)^l$:

\[1 + \exp(\pi i l) = 1 + (-1)^l\]
  • If $l$ is even ($l = 2n$):
\[1 + (-1)^{2n} = 1 + 1 = 2 \implies F_{00l} = 2 \sum_j f_j \exp(2\pi i l z_j) \neq 0\]
  • If $l$ is odd ($l = 2n + 1$):
\[1 + (-1)^{2n+1} = 1 - 1 = 0 \implies F_{00l} = 0\]

Thus, all $(00l)$ reflections with $l = \text{odd}$ undergo destructive interference and are systematically absent!

Medium Example 3.9: Thermodynamics of Enantiotropic Phase Transitions: Grey vs White Tin

Tin undergoes an enantiotropic allotropic phase transition:

\[\alpha\text{-Sn (grey, diamond cubic)} \rightleftharpoons \beta\text{-Sn (white, bct metallic)}\]

Thermodynamic data at the equilibrium transition temperature $T_{\text{tr}} = 13.2\text{ }^\circ\text{C} = 286.35\text{ K}$:

  • Enthalpy of transition: $\Delta H_{\text{tr}} = +2.18\text{ kJ/mol}$
  • Densities: $\rho(\alpha\text{-Sn}) = 5.765\text{ g/cm}^3$, $\quad \rho(\beta\text{-Sn}) = 7.310\text{ g/cm}^3$
  • Molar mass of tin: $M = 118.71\text{ g/mol}$
  1. Calculate the entropy of transition $\Delta S_{\text{tr}}$ at $286.35\text{ K}$.
  2. Calculate the molar volume change $\Delta V_{\text{tr}} = V_{\text{m}}(\beta) - V_{\text{m}}(\alpha)$ in $\text{cm}^3/\text{mol}$ and $\text{m}^3/\text{mol}$.
  3. Using the Clapeyron equation $dP/dT = \Delta H_{\text{tr}} / (T \Delta V_{\text{tr}})$, calculate the pressure dependence of the transition temperature $dT/dP$ in $\text{K/bar}$. Does increasing pressure stabilize grey tin or white tin?

Step 1: Entropy of Transition

At thermodynamic phase equilibrium ($T = T_{\text{tr}}$), $\Delta G_{\text{tr}} = 0$:

\[\Delta S_{\text{tr}} = \frac{\Delta H_{\text{tr}}}{T_{\text{tr}}} = \frac{2180\text{ J/mol}}{286.35\text{ K}} = +7.613\text{ J/(mol}\cdot\text{K)}\]

Step 2: Molar Volume Change

Molar volumes:

\[V_{\text{m}}(\alpha) = \frac{M}{\rho(\alpha)} = \frac{118.71\text{ g/mol}}{5.765\text{ g/cm}^3} = 20.5915\text{ cm}^3/\text{mol}\]
\[V_{\text{m}}(\beta) = \frac{M}{\rho(\beta)} = \frac{118.71\text{ g/mol}}{7.310\text{ g/cm}^3} = 16.2394\text{ cm}^3/\text{mol}\]

Volume change:

\[\Delta V_{\text{tr}} = V_{\text{m}}(\beta) - V_{\text{m}}(\alpha) = 16.2394 - 20.5915 = -4.3521\text{ cm}^3/\text{mol} = -4.3521 \times 10^{-6}\text{ m}^3/\text{mol}\]

The metallic $\beta$-Sn phase is $21.1\%$ denser than the diamond-cubic $\alpha$-Sn phase.

Step 3: Clapeyron Slope and Pressure Effect

From the Clapeyron equation:

\[\frac{dP}{dT} = \frac{\Delta H_{\text{tr}}}{T \Delta V_{\text{tr}}} = \frac{2180\text{ J/mol}}{(286.35\text{ K})(-4.3521 \times 10^{-6}\text{ m}^3/\text{mol})} = \frac{2180}{-1.2462 \times 10^{-3}} = -1.7493 \times 10^6\text{ Pa/K}\]

Converting to $dT/dP$:

\[\frac{dT}{dP} = -\frac{1}{1.7493 \times 10^6}\text{ K/Pa} = -5.716 \times 10^{-7}\text{ K/Pa}\]

Since $1\text{ bar} = 10^5\text{ Pa}$:

\[\frac{dT}{dP} = -5.716 \times 10^{-7} \times 10^5 = -0.05716\text{ K/bar} = -57.2\text{ K/kbar}\]

Because $dT/dP < 0$, applying hydrostatic pressure lowers the transition temperature, stabilizing the denser metallic $\beta$-Sn phase down to lower temperatures (consistent with Le Chatelier's principle).