---
title: "Crystallography and X-ray Diffraction"
book: "University Chemistry — Year 3"
subject: chemistry
language: en
chapter: 9
exercises: 12
source: https://one-course.com/books/chemistry/4/en/chapter/9-crystallography-and-x-ray-diffraction
license: CC-BY-NC-SA-4.0
credit: "One Chemistry Book, One Course (one-course.com)"
---

# Chapter 9 — Crystallography and X-ray Diffraction

In 1912 a beam of X-rays sent through a crystal of copper sulfate left on a photographic plate a pattern of sharp spots: crystals diffract X-rays as a grating diffracts light, because the distances between their atoms are comparable to the X-ray wavelength. A year later William Lawrence Bragg, then a student, read from such patterns the structure of rock salt — and showed that the solid contains no $\ce{NaCl}$ molecules at all, only a lattice in which each ion has six neighbours of the other kind. Every crystal structure quoted in the Year 1 volume and in this book was found in this way. This chapter gives the geometry of lattices and of their symmetry, the condition for diffraction, the intensities that reveal the contents of the cell, and the powder and single-crystal methods used in every laboratory.

**You already know.**

The Year 1 volume described crystals by a lattice of nodes, a motif and a unit cell with its lattice parameters; counted the multiplicity of a cell; built the face-centred and body-centred cubic and the hexagonal close-packed structures and the rock-salt, caesium chloride, zinc-blende and fluorite types; and computed a density from a cell. [Chapter 4](https://one-course.com/books/chemistry/4/en/chapter/4-symmetry-and-point-groups#ch-b3-point-groups) defined [symmetry operations](https://one-course.com/books/chemistry/4/en/chapter/4-symmetry-and-point-groups#def-b3-point-groups-operation) and [point groups](https://one-course.com/books/chemistry/4/en/chapter/4-symmetry-and-point-groups#def-b3-point-groups-point-group).

![A benchtop powder diffractometer: the X-ray tube and the detector turn on a circle around the flat sample.](https://one-course.com/images/onecourse/chapters/chemistry-4/b3-x-ray-diffraction/img-0e6c78b5a644.jpg)

*A benchtop powder diffractometer: the X-ray tube and the detector turn on a circle around the flat sample.*

## 9.1 Crystal systems and Bravais lattices

A lattice can only have [symmetry operations](https://one-course.com/books/chemistry/4/en/chapter/4-symmetry-and-point-groups#def-b3-point-groups-operation) that map the lattice onto itself. That restricts them severely.

**Theorem 9.1 (The crystallographic restriction).**

The only rotation axes compatible with a lattice are of order 1, 2, 3, 4 and 6.

**Proof.** In a basis of lattice vectors, a symmetry rotation maps lattice vectors onto lattice vectors, so its matrix has integer entries and an integer trace. The trace does not depend on the basis; in an orthonormal basis with $z$ along the axis it is $1 + 2\cos\theta$. Hence $2\cos\theta$ is an integer between $-2$ and 2: $\cos\theta \in \{-1, -\frac12, 0, \frac12, 1\}$, $\theta = 180^\circ, 120^\circ, 90^\circ,
60^\circ, 0^\circ$: orders 2, 3, 4, 6, 1. A fivefold axis is impossible. ∎

**Definition 9.2 (Crystal system, Bravais lattice, lattice centring).**

Lattices are grouped by their symmetry into seven *crystal systems* (triclinic, monoclinic, orthorhombic, tetragonal, trigonal, hexagonal, cubic). A cell may hold lattice nodes only at its corners (primitive, P) or also at its centre (body-centred, I), at the centres of all faces (face-centred, F) or of one pair of faces (base-centred, C), or along a body diagonal (rhombohedral, R): this is its *lattice centring*. The 14 distinct combinations of a system and a centring are the *Bravais lattices*.

| system | cell constraints | centrings |
| --- | --- | --- |
| triclinic | none | P |
| monoclinic | $\alpha = \gamma = 90^\circ$ | P, C |
| orthorhombic | $\alpha = \beta = \gamma = 90^\circ$ | P, C, I, F |
| tetragonal | $a = b$, all angles $90^\circ$ | P, I |
| trigonal | $a = b = c$, $\alpha = \beta = \gamma \ne 90^\circ$ (rhombohedral axes) | R |
| hexagonal | $a = b$, $\alpha = \beta = 90^\circ$, $\gamma = 120^\circ$ | P |
| cubic | $a = b = c$, all angles $90^\circ$ | P, I, F |

![The three cubic Bravais lattices: primitive (nodes at the corners), body-centred (plus the centre, red) and face-centred (plus the six face centres, red). Their cells hold 1, 2 and 4 nodes.](https://one-course.com/images/onecourse/chapters/chemistry-4/b3-x-ray-diffraction/fig-7bb8553210e6.svg)

*The three cubic [Bravais lattices](#def-b3-x-ray-diffraction-crystal-system): primitive (nodes at the corners), body-centred (plus the centre, red) and face-centred (plus the six face centres, red). Their cells hold 1, 2 and 4 nodes.*

## 9.2 Lattice planes and the reciprocal lattice

**Definition 9.3 (Lattice plane, Miller indices, interplanar spacing).**

A *lattice plane* is a plane through at least three non-aligned nodes; it belongs to a family of parallel, equally spaced planes that contain every node. If the plane of the family nearest the origin cuts the axes at $a/h$, $b/k$, $c/l$, the integers $(hkl)$, without common factor, are its *Miller indices* (an index 0 for a plane parallel to an axis). The distance between adjacent planes is the *interplanar spacing* $d_{hkl}$.

**Proposition 9.4 (Spacings in orthogonal cells).**

For an orthorhombic cell, $\dfrac{1}{d_{hkl}^2} = \dfrac{h^2}{a^2} + \dfrac{k^2}{b^2} +
\dfrac{l^2}{c^2}$; for a cubic cell, $d_{hkl} = a/\sqrt{h^2 + k^2 + l^2}$.

**Proof.** The plane $hx/a + ky/b + lz/c = 1$ has normal vector $\mathbf n = (h/a, k/b, l/c)$ and lies at distance $1/|\mathbf n|$ from the origin; the parallel plane through the origin belongs to the family, so $d = 1/|\mathbf n|$. ∎

**Definition 9.5 (Reciprocal lattice).**

For a lattice of basis vectors $\mathbf a$, $\mathbf b$, $\mathbf c$ and cell volume $V$, the *reciprocal lattice* has basis vectors $\mathbf a^* = (\mathbf b\times\mathbf c)/V$, $\mathbf b^* = (\mathbf c\times\mathbf a)/V$, $\mathbf c^* = (\mathbf a\times\mathbf b)/V$, so that $\mathbf a\cdot\mathbf a^* = 1$ and $\mathbf a\cdot\mathbf b^* = 0$. Its node $\mathbf G_{hkl} = h\mathbf a^* + k\mathbf b^* + l\mathbf c^*$ is perpendicular to the planes $(hkl)$, and $|\mathbf G_{hkl}| = 1/d_{hkl}$.

## 9.3 Diffraction

**Theorem 9.6 (Bragg’s law).**

X-rays of wavelength $\lambda$ are reflected by a family of planes of spacing $d$ only at the glancing angles $\theta$ satisfying

$$
2d\sin\theta = n\lambda, \qquad n = 1, 2, \dots,
$$

*Bragg’s law*. The order $n$ is absorbed by writing the reflection $(nh\,nk\,nl)$ with spacing $d/n$.

**Proof.** Rays scattered by two adjacent planes, at equal angles of incidence and reflection $\theta$, differ in path by $2d\sin\theta$ (the two segments on either side of the normal through the lower plane). They reinforce when this difference is a whole number of wavelengths; with millions of planes, any other angle gives rays that cancel in pairs. ∎

![Bragg’s construction. Rays reflected by adjacent planes travel farther by the two red segments, d each; the waves reinforce when 2d is a whole number of wavelengths.](https://one-course.com/images/onecourse/chapters/chemistry-4/b3-x-ray-diffraction/fig-808391076054.svg)

*Bragg’s construction. Rays reflected by adjacent planes travel farther by the two red segments, $d\sin\theta$ each; the waves reinforce when $2d\sin\theta$ is a whole number of wavelengths.*

**Proposition 9.7 (Laue condition).**

With incident and scattered wave vectors $\mathbf k$ and $\mathbf k'$ of length $1/\lambda$, diffraction occurs when $\mathbf k' - \mathbf k = \mathbf G_{hkl}$, a vector of the [reciprocal lattice](#def-b3-x-ray-diffraction-reciprocal); this is equivalent to [Bragg’s law](#thm-b3-x-ray-diffraction-bragg).

**Proof.** Waves scattered by two nodes separated by a lattice vector $\mathbf R$ differ in phase by $2\pi(\mathbf k' - \mathbf k)\cdot\mathbf R$; they all reinforce if this is a multiple of $2\pi$ for every $\mathbf R$, i.e. if $\mathbf k' - \mathbf k$ is in the [reciprocal lattice](#def-b3-x-ray-diffraction-reciprocal) (by the definition $\mathbf a\cdot\mathbf a^* = 1$, $\mathbf a\cdot\mathbf b^* = 0$…). For elastic scattering, $|\mathbf k| = |\mathbf k'|$ and $|\mathbf k' - \mathbf k| = 2\sin\theta/\lambda$, where $2\theta$ is the angle between them; with $|\mathbf G| = 1/d$ this is $2d\sin\theta = \lambda$, and $\mathbf G$ is normal to the reflecting planes. ∎

**Definition 9.8 (Ewald sphere).**

The *Ewald sphere* has radius $1/\lambda$ and passes through the origin of the [reciprocal lattice](#def-b3-x-ray-diffraction-reciprocal), its centre at $-\mathbf k$ from the origin; a reflection occurs whenever a reciprocal-lattice node lies on it.

![The Ewald construction in two dimensions (the circle passes through the origin O of the reciprocal lattice). A node G lying on the circle gives a diffracted beam k' = k + G. Here the circle, of radius chosen for the drawing, passes through the node (-1, 2).](https://one-course.com/images/onecourse/chapters/chemistry-4/b3-x-ray-diffraction/fig-04a84c4f7243.svg)

*The Ewald construction in two dimensions (the circle passes through the origin $O$ of the [reciprocal lattice](#def-b3-x-ray-diffraction-reciprocal)). A node $\mathbf G$ lying on the circle gives a diffracted beam $\mathbf k' = \mathbf k + \mathbf G$. Here the circle, of radius chosen for the drawing, passes through the node $(-1, 2)$.*

## 9.4 Intensities and symmetry

[Bragg’s law](#thm-b3-x-ray-diffraction-bragg) gives the directions of the diffracted beams, which depend only on the lattice; their intensities depend on what is in the cell.

**Definition 9.9 (Atomic scattering factor, structure factor).**

The *atomic scattering factor* $f_j$ of an atom is the amplitude it scatters, in units of that of one electron; it equals its number of electrons in the forward direction and falls off at higher angles. The *structure factor* of reflection $hkl$ is

$$
F_{hkl} = \sum_jf_j\,\eu^{2\pi\iu(hx_j + ky_j + lz_j)},
$$

summed over the atoms of the cell at fractional coordinates $(x_j, y_j, z_j)$.

**Proposition 9.10 (Intensity).**

The intensity of a reflection is proportional to $|F_{hkl}|^2$.

**Partial proof.** Atom $j$ sits at $\mathbf r_j = x_j\mathbf a + y_j\mathbf b + z_j\mathbf c$; for a reflection, $(\mathbf k' - \mathbf k)\cdot\mathbf r_j = hx_j + ky_j + lz_j$, so its wave has the phase $2\pi(hx_j + ky_j + lz_j)$ relative to an atom at the origin, and amplitude $f_j$: the cell scatters the sum $F_{hkl}$, and every cell the same. An intensity is the square modulus of an amplitude. That multiple scattering can be neglected (the kinematic approximation) is admitted. ∎

**Definition 9.11 (Systematic absence).**

A *systematic absence* is a reflection whose [structure factor](#def-b3-x-ray-diffraction-structure-factor) vanishes for every crystal of a given [lattice centring](#def-b3-x-ray-diffraction-crystal-system) or symmetry, whatever its atoms.

**Proposition 9.12 (Absences of centred lattices).**

In a body-centred lattice, reflections with $h + k + l$ odd are absent; in a face-centred lattice, reflections with $h$, $k$, $l$ of mixed parity are absent.

**Proof.** I centring: every atom at $(x,y,z)$ has a copy at $(x + \frac12, y + \frac12, z + \frac12)$, whose phase factor differs by $\eu^{\iu\pi(h+k+l)} = (-1)^{h+k+l}$: $F = (1 +
(-1)^{h+k+l})F'$. F centring: copies at three face centres give the factor $1 +
(-1)^{h+k} + (-1)^{h+l} + (-1)^{k+l}$, equal to 4 if $h$, $k$, $l$ are all even or all odd and to 0 otherwise. ∎

**Example 9.13 (Rock salt and sylvite).**

In the rock-salt structure, cations at $(0,0,0)$ and anions at $(\frac12,0,0)$, each with the F translations: $F_{hkl} = 4[f_+ + f_-(-1)^{h+k+l}]$ for unmixed indices. For $\ce{NaCl}$, $f(\ce{Na+}) \approx 10$ and $f(\ce{Cl-}) \approx 18$: the all-odd reflections (111), (311) are weak but present. For $\ce{KCl}$, $\ce{K+}$ and $\ce{Cl-}$ both have 18 electrons: the all-odd reflections almost vanish, and the pattern looks like that of a primitive cubic lattice of half the cell edge (figure below).

![Computed powder patterns (Cu K_1; scattering factors taken equal to the numbers of electrons, so that only the absences and the broad trends of intensity are meaningful). Top: NaCl and KCl; the all-odd lines of KCl vanish. Bottom left: copper (F) shows (111), (200), (220), (311)…; tungsten (I) shows (110), (200), (211)…. Bottom right: the (200) line of NaCl for crystallites of 200 (black), 50 (blue) and 10 nm (red): smaller crystals give broader lines.](https://one-course.com/images/onecourse/chapters/chemistry-4/b3-x-ray-diffraction/fig-fe95d8f0ac88.svg)

![Computed powder patterns (Cu K_1; scattering factors taken equal to the numbers of electrons, so that only the absences and the broad trends of intensity are meaningful). Top: NaCl and KCl; the all-odd lines of KCl vanish. Bottom left: copper (F) shows (111), (200), (220), (311)…; tungsten (I) shows (110), (200), (211)…. Bottom right: the (200) line of NaCl for crystallites of 200 (black), 50 (blue) and 10 nm (red): smaller crystals give broader lines.](https://one-course.com/images/onecourse/chapters/chemistry-4/b3-x-ray-diffraction/fig-1e7ee1753bdc.svg)

*Computed powder patterns (Cu K$\alpha_1$; scattering factors taken equal to the numbers of electrons, so that only the absences and the broad trends of intensity are meaningful). Top: $\ce{NaCl}$ and $\ce{KCl}$; the all-odd lines of $\ce{KCl}$ vanish. Bottom left: copper (F) shows (111), (200), (220), (311)…; tungsten (I) shows (110), (200), (211)…. Bottom right: the (200) line of $\ce{NaCl}$ for crystallites of 200 (black), 50 (blue) and 10 nm (red): smaller crystals give broader lines.*

**Proposition 9.14 (Friedel’s law).**

In the absence of anomalous scattering, $|F_{hkl}| = |F_{\bar h\bar k\bar l}|$: a diffraction pattern looks centrosymmetric even when the crystal is not.

**Proof.** With real $f_j$, $F_{\bar h\bar k\bar l} = \sum f_j\eu^{-2\pi\iu(\dots)} = F_{hkl}^*$, of the same modulus. ∎

[Symmetry elements](https://one-course.com/books/chemistry/4/en/chapter/4-symmetry-and-point-groups#def-b3-point-groups-operation) that combine a rotation or a reflection with a fraction of a lattice translation exist only in crystals.

**Definition 9.15 (Screw axis, glide plane, space group).**

A *screw axis* $n_p$ combines a rotation by $2\pi/n$ with a translation of $p/n$ of the lattice repeat along the axis ($2_1$: half a turn and half a cell). A *glide plane* combines a reflection with a translation of half a lattice vector parallel to the plane ($a$, $b$, $c$ glides). The group of all the [symmetry operations](https://one-course.com/books/chemistry/4/en/chapter/4-symmetry-and-point-groups#def-b3-point-groups-operation) of a crystal, translations included, is its *space group*: there are 230. Its smallest part from which the whole crystal is generated by the operations is the *asymmetric unit*. A set of points left on themselves by a [subgroup](https://one-course.com/books/chemistry/4/en/chapter/4-symmetry-and-point-groups#def-b3-point-groups-class) of the operations is a *Wyckoff position*: general positions have no symmetry, special positions lie on [symmetry elements](https://one-course.com/books/chemistry/4/en/chapter/4-symmetry-and-point-groups#def-b3-point-groups-operation).

**Proposition 9.16 (Absences from a screw axis).**

A $2_1$ axis along $b$ makes the reflections $0k0$ with $k$ odd absent.

**Proof.** The axis maps $(x, y, z)$ to $(-x, y + \frac12, -z)$. For $h = l = 0$ both atoms have phase factors $\eu^{2\pi\iu ky}$ and $\eu^{2\pi\iu k(y + 1/2)} = (-1)^k\eu^{2\pi\iu ky}$: their sum vanishes for odd $k$. ∎

**Method 9.17 (Reading a space-group symbol).**

1. The first letter is the [lattice centring](#def-b3-x-ray-diffraction-crystal-system) (P, I, F, C, R).
2. The next symbols give the symmetry along the main directions; for the monoclinic system, the single direction is $b$ . A fraction bar means “with a plane perpendicular to this axis”.
3. *P* $2_1/c$ , the commonest [space group](#def-b3-x-ray-diffraction-space-group) of organic molecules: primitive, a $2_1$ axis along $b$ , and a $c$ -glide plane perpendicular to it; these generate centres of inversion. Absences: $0k0$ with $k$ odd, $h0l$ with $l$ odd.

![The space group P2_1/c seen along c (projection on the ab plane). Dotted lines: c-glide planes perpendicular to b; half-arrows: 2_1 screw axes along b (at height z = 1/4); small circles: centres of inversion. A general position at height +z generates three others; a comma marks a mirror-image copy (produced by the glide or the inversion); 1/2- means height 1/2 - z.](https://one-course.com/images/onecourse/chapters/chemistry-4/b3-x-ray-diffraction/fig-59a3d93a0298.svg)

*The [space group](#def-b3-x-ray-diffraction-space-group) *P*$2_1/c$ seen along $c$ (projection on the $ab$ plane). Dotted lines: $c$-glide planes perpendicular to $b$; half-arrows: $2_1$ screw axes along $b$ (at height $z = \frac14$); small circles: centres of inversion. A general position at height $+z$ generates three others; a comma marks a mirror-image copy (produced by the glide or the inversion); $\frac12{-}$ means height $\frac12 - z$.*

## 9.5 Powder and single-crystal methods

**Definition 9.18 (Powder diffraction pattern).**

A finely ground sample contains crystallites in all orientations: every family of planes finds some in reflecting position, and the diffracted beams form cones. The intensity recorded against $2\theta$ is its *powder diffraction pattern*, a fingerprint of the crystalline phase.

**Method 9.19 (Indexing a cubic powder pattern).**

1. Compute $\sin^2\theta$ for each line; for a cubic crystal $\sin^2\theta =  (\lambda^2/4a^2)N$ with $N = h^2 + k^2 + l^2$ .
2. Divide by the smallest value and look for the integer series: P gives $N = 1, 2, 3, 4, 5, 6, 8, \dots$ (never 7); I gives $2, 4, 6, 8, 10, \dots$ ; F gives $3, 4, 8, 11, 12, 16, \dots$
3. Choose the series that fits all lines; then $a = \lambda\sqrt N/(2\sin\theta)$ , best from high-angle lines, where an error in $\theta$ matters least.

**Method 9.20 (The number of formula units in the cell).**

With the molar mass $M$ of a formula unit and the measured density $\rho$, $Z =
\rho N_Aa^3/M$ (for a cubic cell) must be a whole number; it tests a proposed structure.

**Proposition 9.21 (Scherrer equation).**

Crystallites of mean size $L$ broaden a line by $\beta \approx K\lambda/(L\cos\theta)$ (full width at half maximum in radians of $2\theta$, $K \approx 0.9$).

**Proof.** *Admitted at this level.* ∎

Below about $100\,\mathrm{nm}$ the broadening becomes measurable; the Scherrer equation, derived in more advanced courses from the Fourier transform of a finite stack of planes, then estimates the size of nanocrystals ([Chapter 23](https://one-course.com/books/chemistry/4/en/chapter/23-inorganic-materials-and-nanomaterials#ch-b3-inorganic-materials)).

**Definition 9.22 (Phase problem, R factor).**

The intensities give $|F_{hkl}|$ but not the phases of the $F_{hkl}$; without them the [electron density](https://one-course.com/books/chemistry/4/en/chapter/3-computational-chemistry-hartreefock-and-dft#def-b3-computational-chemistry-dft) cannot be computed by inverting the Fourier series. This is the *phase problem*. A trial structure is refined by least squares until computed and observed moduli agree; the residual $R = \sum\big||F_o| - |F_c|\big|/\sum|F_o|$ is its *R factor* — a few per cent for a good structure.

For a single crystal, a diffractometer measures thousands of reflections; direct methods (statistical relations between the phases of strong reflections) give a first electron-density map for most small molecules, and refinement places every atom to a few thousandths of a nanometre. The results are deposited as CIF files in open databases, from which the lattice parameters quoted in this series are taken.

**In the lab — A powder measurement.**

The powder is ground in an agate mortar, pressed flat into a holder, and scanned from 5 to $90{}^{\circ}$ in $2\theta$ with Cu K$\alpha$ radiation; a nickel filter removes most of the K$\beta$ line. The pattern is compared with reference patterns from a database, and lattice parameters are refined with an internal standard such as silicon powder. X-ray instruments are enclosed and interlocked: the shutter cannot open when the door is.

**History — The Braggs, 1913.**

![](https://one-course.com/images/onecourse/chapters/chemistry-4/b3-x-ray-diffraction/img-05f0ef0b16e3.jpg)

*W. L. Bragg.*

Max von Laue proposed in 1912 that crystals diffract X-rays, and Walter Friedrich and Paul Knipping recorded the first pattern. William Lawrence Bragg, twenty-two years old, explained the spots as reflections from [lattice planes](#def-b3-x-ray-diffraction-miller) and, with his father William Henry Bragg, who built an X-ray spectrometer, solved the structures of $\ce{NaCl}$, $\ce{KCl}$ and diamond in 1913. Father and son shared the 1915 Nobel Prize in Physics.

## 9.6 Exercises

**Exercise 9.1 ★.**

For $\ce{NaCl}$ ($a = 564.06\,\mathrm{pm}$), compute $d_{111}$, $d_{200}$ and $d_{220}$.

**Solution of Exercise 9.1.**

$d = a/\sqrt N$: $d_{111} = 325.66\,\mathrm{pm}$, $d_{200} = 282.03\,\mathrm{pm}$, $d_{220} =
199.42\,\mathrm{pm}$.

**Exercise 9.2 ★.**

Copper is FCC with $a = 361.50\,\mathrm{pm}$. With Cu K$\alpha_1$ ($\lambda = 154.06\,\mathrm{pm}$), compute $2\theta$ of its first two reflections, after naming them.

**Solution of Exercise 9.2.**

FCC: the first reflections are (111) and (200). $\sin\theta = \lambda\sqrt N/2a$: $2\theta =
43.32{}^{\circ}$ and $50.45{}^{\circ}$.

**Exercise 9.3 ★.**

Which of the reflections 100, 110, 111, 200, 210, 211, 220 are present for a cubic P, I and F lattice?

**Solution of Exercise 9.3.**

P: all seven. I ($h + k + l$ even): 110, 200, 211, 220. F (unmixed): 111, 200, 220.

**Exercise 9.4 ★.**

Give the reciprocal basis of a two-dimensional rectangular lattice of sides $a$ and $b$, and draw the first reciprocal nodes.

**Solution of Exercise 9.4.**

$\mathbf a^* = \mathbf e_x/a$, $\mathbf b^* = \mathbf e_y/b$: a rectangular lattice of sides $1/a$ and $1/b$, the long side of the direct lattice becoming the short one.

**Exercise 9.5 ★★.**

A metal gives lines at $2\theta = 40.27$, 58.26, 73.20, 87.01 and $100.66{}^{\circ}$ with Cu K$\alpha_1$. Index the pattern, find the lattice type and $a$.

**Solution of Exercise 9.5.**

$\sin^2\theta$ ratios $1 : 2 : 3 : 4 : 5$, i.e. $N = 2, 4, 6, 8, 10$: body-centred (110, 200, 211, 220, 310). From the first line, $d_{110} = \lambda/2\sin(20.135{}^{\circ}) =
223.77\,\mathrm{pm}$ and $a = d\sqrt2 = 316.5\,\mathrm{pm}$: tungsten.

**Exercise 9.6 ★★.**

A crystal of $\ce{KCl}$ ($a = 629.29\,\mathrm{pm}$) has a measured density of $1.99\,\mathrm{g}/\mathrm{cm}^{3}$ (data of the exercise). Find $Z$.

**Solution of Exercise 9.6.**

$Z = \rho N_Aa^3/M = 1.99 \times 6.022\times10^{23} \times (6.2929\times10^{-8})^3/74.6 = 4.0$.

**Exercise 9.7 ★★.**

The (200) line of a $\ce{NaCl}$ nanopowder ($2\theta = 31.70{}^{\circ}$) is $0.40{}^{\circ}$ wide, the instrument contributing nothing. Estimate the crystallite size.

**Solution of Exercise 9.7.**

$\beta = 0.40{}^{\circ} = 6.98 \times 10^{-3}\,\mathrm{rad}$, $\cos\theta = \cos15.85{}^{\circ} = 0.962$: $L = 0.9 \times 0.15406/(6.98\times10^{-3} \times 0.962) = 21\,\mathrm{nm}$.

**Exercise 9.8 ★★.**

$\ce{CsCl}$ has $\ce{Cs+}$ at $(0,0,0)$ and $\ce{Cl-}$ at $(\frac12,\frac12,\frac12)$ in a P cell. With $f
\approx$ the number of electrons, compute $F_{100}$ and $F_{110}$ and the ratio of their intensities ([structure factor](#def-b3-x-ray-diffraction-structure-factor) only). Why is $\ce{CsCl}$ not body-centred?

**Solution of Exercise 9.8.**

$F_{hkl} = f_{\ce{Cs+}} + f_{\ce{Cl-}}(-1)^{h+k+l}$: $F_{100} = 54 - 18 = 36$, $F_{110} = 72$: intensities in the ratio 1 : 4. The centre is occupied by a different ion from the corners: the lattice is primitive (a body-centred lattice needs identical contents at both points), and 100 is present.

**Exercise 9.9 ★★.**

Show that a C-centred lattice (extra node at $(\frac12,\frac12,0)$) makes the reflections with $h + k$ odd absent.

**Solution of Exercise 9.9.**

Each atom has a copy shifted by $(\frac12,\frac12,0)$, multiplying $F$ by $1 +
\eu^{\iu\pi(h+k)} = 1 + (-1)^{h+k}$, zero for $h + k$ odd.

**Exercise 9.10 ★★★.**

Explain why the 1980s discovery of alloys whose diffraction patterns show sharp spots with tenfold symmetry did not contradict the crystallographic restriction, and what it changed in the definition of a crystal.

**Solution of Exercise 9.10.**

Quasicrystals have long-range order without periodicity: they have no lattice, so the restriction, which concerns lattices, does not apply to them. Their sharp diffraction spots led to defining a crystal as any solid with an essentially discrete diffraction pattern, periodic or not.

**Exercise 9.11 ★★★.**

Show that in *P*$2_1/c$ the [glide plane](#def-b3-x-ray-diffraction-space-group) makes the reflections $h0l$ with $l$ odd absent.

**Solution of Exercise 9.11.**

The $c$ glide maps $(x, y, z)$ to $(x, \frac12 - y, \frac12 + z)$. For $k = 0$ the two phase factors are $\eu^{2\pi\iu(hx + lz)}$ and $\eu^{2\pi\iu(hx + lz + l/2)} = (-1)^l
\eu^{2\pi\iu(hx+lz)}$: their sum vanishes for odd $l$.

**Exercise 9.12 ★★★.**

A pure enantiomer can never crystallise in a [space group](#def-b3-x-ray-diffraction-space-group) containing an inversion centre, a mirror or a [glide plane](#def-b3-x-ray-diffraction-space-group). Why? To which kind of [space groups](#def-b3-x-ray-diffraction-space-group) is it restricted, and what does Friedel’s law imply for determining its absolute configuration?

**Solution of Exercise 9.12.**

An inversion, a mirror or a glide turns a molecule into its mirror image, so a crystal containing one would contain both enantiomers. A pure enantiomer crystallises only in the 65 [space groups](#def-b3-x-ray-diffraction-space-group) built from rotations, screw axes and translations (Sohncke groups), such as *P*$2_12_12_1$. By Friedel’s law the pattern of one enantiomer equals that of the other; the absolute configuration is obtained from the small deviations caused by anomalous scattering of heavier atoms.

## 9.7 Problem: Which White Powder?

**Problem 9.1.**

Weekend problem — a powder pattern of an unknown cubic salt: spacings, indexing, the lattice parameter and the identity of the salt, and the size of its crystallites

An unknown white crystalline powder, known to be $\ce{NaCl}$, $\ce{KCl}$ or $\ce{KBr}$, gives with Cu K$\alpha_1$ radiation ($\lambda = 154.059\,\mathrm{pm}$) the lines (data of the problem, computed for this exercise):

| $2\theta$ / degree | 28.342 | 40.512 | 50.178 | 58.632 | 66.380 | 73.692 |
| --- | --- | --- | --- | --- | --- | --- |
| relative intensity | 100 | 92 | 38 | 20 | 61 | 49 |

No other line appears between 20 and $75{}^{\circ}$. Reference lattice parameters: $\ce{NaCl}$ $564.06\,\mathrm{pm}$, $\ce{KCl}$ $629.29\,\mathrm{pm}$, $\ce{KBr}$ $658.47\,\mathrm{pm}$. Molar masses: K 39.1, Na 23.0, Cl 35.5, Br $79.9\,\mathrm{g}/\mathrm{mol}$.

**Part I — Spacings.**

1. Compute $\theta$ and $d$ for the first line.
2. Compute $d$ for all lines.
3. Compute the ratios $\sin^2\theta/\sin^2\theta_1$ .
4. What series of integers do you find?
5. Could the pattern be that of a primitive cubic lattice? With which $a$ ?
6. For a primitive cubic lattice, which integer $N$ can never occur? Does the pattern tell the two interpretations apart so far?

**Part II — The real lattice.**

7. All three candidates are F lattices. Index the lines with the F series.
8. Deduce $a$ from the first line.
9. Which candidate fits?
10. For that salt, why are the all-odd reflections (111), (311) absent?
11. Would $\ce{NaCl}$ show (111)? $\ce{KBr}$ ?
12. Explain why the primitive interpretation of Part I is a trap.

**Part III — Density and contents.**

13. How many formula units does the cell contain?
14. Compute the density from the cell.
15. Give the coordination number of each ion.
16. Compute the shortest K–Cl distance.
17. What reflection would appear first above $75{}^{\circ}$ ?
18. Why are intensities less useful than positions for identifying a phase?

**Part IV — Refinement and crystallite size.**

19. Why is the lattice parameter best computed from a high-angle line?
20. Compute $a$ from the (420) line.
21. The (420) line is $0.25{}^{\circ}$ wide in $2\theta$ (instrumental width negligible). Estimate the crystallite size.
22. Would a crystallite size of $1\,\text{µ}\mathrm{m}$ broaden the line measurably?
23. State the result: the lattice parameter refined from the (420) line, and the identity of the powder.

**Solution of Problem 9.1.**

**1.** $\theta = 14.171{}^{\circ}$, $d = \lambda/2\sin\theta = 314.64\,\mathrm{pm}$. **2.** 314.64, 222.49, 181.66, 157.32, 140.71, $128.45\,\mathrm{pm}$. **3.** 1.000, 2.000, 3.000, 4.000, 5.000, 6.000. **4.** 1, 2, 3, 4, 5, 6. **5.** Yes, as (100), (110), (111), (200), (210), (211) of a primitive cubic cell with $a' = 314.6\,\mathrm{pm}$. **6.** $N = 7$ (and 15, 23…), which is not a sum of three squares; the first difference would come at the eighth line. Six lines cannot tell a P cell of edge $a'$ from an F cell of edge $2a'$. **7.** $N = 4, 8, 12, 16, 20, 24$: (200), (220), (222), (400), (420), (422). **8.** $a = 2d_{200} = 629.3\,\mathrm{pm}$. **9.** $\ce{KCl}$ ($629.29\,\mathrm{pm}$). **10.** $F = 4[f(\ce{K+}) - f(\ce{Cl-})]$ for all-odd indices, and the two ions have 18 electrons each: the amplitudes cancel. **11.** Yes for both: $\ce{Na+}$ (10) and $\ce{Cl-}$ (18), $\ce{K+}$ (18) and $\ce{Br-}$ (36) scatter differently; $\ce{NaCl}$’s (111) would be at $27.36{}^{\circ}$, $\ce{KBr}$’s at $23.38{}^{\circ}$, both inside the recorded range. **12.** X-rays see $\ce{K+}$ and $\ce{Cl-}$ as nearly identical; identical ions on a rock-salt arrangement form a simple cubic array of edge $a/2$, an apparent cell. **13.** Four $\ce{KCl}$. **14.** $\rho = 4 \times 74.6/(6.022\times10^{23} \times (6.293\times10^{-8})^3) =
1.99\,\mathrm{g}/\mathrm{cm}^{3}$. **15.** Six and six (octahedra of the other ion). **16.** $a/2 = 314.6\,\mathrm{pm}$. **17.** (440), $N = 32$, at $87.65{}^{\circ}$ (the all-odd (333)/(511) being absent). **18.** Intensities change with preferred orientation of the crystallites, absorption, the fall-off of scattering factors and thermal motion; positions depend only on the lattice. **19.** From $d = \lambda/2\sin\theta$, $\Delta d/d = -\cot\theta\,\Delta\theta$: the same angular error gives a smaller relative error at large $\theta$. **20.** $d_{420} = \lambda/2\sin33.190{}^{\circ} = 140.71\,\mathrm{pm}$, $a = d\sqrt{20} =
629.3\,\mathrm{pm}$. **21.** $L = 0.9 \times 0.15406/(4.36\times10^{-3} \times \cos33.19{}^{\circ}) = 38\,\mathrm{nm}$. **22.** No: $\beta = 0.0095{}^{\circ}$, far below a typical instrumental width. **23.** **$a = 629.3\,\mathrm{pm}$: the powder is potassium chloride**, its odd reflections silenced by two ions with the same number of electrons.