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WAChemistrySyllabus dot point

How does x-ray crystallography reveal the three-dimensional arrangement of atoms in a solid?

Describe how x-ray crystallography uses the diffraction of x-rays by a crystal to determine three-dimensional structure

A focused answer to the WACE Year 12 Chemistry dot point on x-ray crystallography, how the diffraction of x-rays by the regular array of atoms in a crystal is used to determine bond lengths, bond angles and three-dimensional structure, with a worked example and common exam mistakes.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

X-ray crystallography is the technique that reveals the actual three-dimensional arrangement of atoms in a crystalline solid. Unlike mass spectrometry, infrared and NMR, which give pieces of evidence about molar mass, functional groups and connectivity, x-ray crystallography can deliver the complete structure including bond lengths and angles.

Why x-rays and crystals

X-rays are used because their wavelength (around $10^{-10}$ m) is comparable to the spacing between atoms in a crystal. When a wave meets a regularly repeating array with a similar spacing, it diffracts. The ordered, repeating arrangement of atoms in a crystal lattice acts as a three-dimensional diffraction grating for the x-rays.

How the structure is found

The x-rays scattered by the atoms interfere with one another, reinforcing in some directions and cancelling in others, producing a pattern of spots whose angles and intensities depend on the spacing and arrangement of the atoms. By measuring this pattern from many orientations and applying mathematical analysis, the positions of the atoms (and therefore bond lengths and bond angles) can be reconstructed.

Strengths and limitations

The great strength is that the technique gives the full three-dimensional structure, including precise bond lengths and angles, which the spectroscopic methods cannot. Its main limitation is that the substance must form a suitable single crystal; many compounds (and most liquids) cannot be crystallised easily, so the method is not always applicable. Growing a good crystal can be the hardest step.

Bragg's idea: why a pattern of spots forms

The diffraction pattern is not random. X-rays reflect off successive parallel planes of atoms in the crystal, and the reflected waves reinforce (constructive interference) only when the extra distance travelled by waves from deeper planes is a whole number of wavelengths. This is Bragg's condition, $n\lambda = 2d\sin\theta$ , where $d$ is the spacing between planes, $\theta$ the angle of the beam to the planes, $\lambda$ the x-ray wavelength and $n$ a whole number. The condition is met only at particular angles, which is why the pattern appears as sharp, discrete spots rather than a smear. By measuring the angles $\theta$ at which spots appear, the plane spacings $d$ can be calculated, and from many such measurements in different orientations the full atomic arrangement is reconstructed. At WACE level you are not expected to apply Bragg's equation in detail, but understanding that constructive interference at specific angles encodes the atomic spacing explains why the technique works and why a regular crystal is essential.

Why this matters

X-ray crystallography has determined the structures of countless molecules, from simple salts to DNA and proteins, and it underpins drug design and materials science. In the context of Unit 4 it completes the toolkit of instrumental analysis: it is the technique to reach for when a complete three-dimensional structure of a crystalline solid is needed.

Exam-style practice questions

Practice questions written in the style of SCSA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WACE 20226 marks(a) Explain why x-rays, rather than visible light, are used to probe the structure of a crystal. (b) State two pieces of structural information that x-ray crystallography provides that infrared and NMR spectroscopy do not. (c) An unknown is supplied as a clear liquid. Explain why x-ray crystallography cannot be used directly, and suggest what could be done.

Show worked answer →

A 6 mark question rewards the wavelength reasoning, the unique information, and the liquid limitation.

(a) Diffraction occurs strongly only when the wavelength of the radiation is comparable to the spacing between the diffracting objects. The spacing between atoms in a crystal is about $10^{-10}\ \text{m}$ , which matches the wavelength of x-rays but is far smaller than the wavelength of visible light (about $5 \times 10^{-7}\ \text{m}$ ), so only x-rays are diffracted by the atomic array.

(b) X-ray crystallography gives precise bond lengths and bond angles, and the full three-dimensional arrangement (geometry) of the atoms. Infrared identifies functional groups and NMR shows the number of atomic environments and connectivity, but neither gives exact bond lengths or the 3D shape.

(c) A liquid has no long-range ordered array of atoms, so it produces no sharp diffraction pattern and cannot be analysed directly. The substance would first need to be crystallised (cooled or evaporated to a well-ordered solid); if it cannot be crystallised, x-ray crystallography is not suitable and other techniques must be used.

Markers reward the wavelength-matching argument, bond lengths/angles and 3D geometry as the unique data, and the need for a crystalline (not liquid) sample.

WACE 20205 marksAn organic compound gives the following data: a mass spectrum molecular ion at

m/z = 46

, an infrared spectrum with a broad band near

3300\ \text{cm}^{-1}

and no strong band near

1700\ \text{cm}^{-1}

, and a single-crystal x-ray structure showing a C-O bond length of

0.143\ \text{nm}

. Explain how each technique contributes to identifying the compound as ethanol.

Show worked answer →

A 5 mark answer needs the contribution of each of the three techniques.

The mass spectrum molecular ion at $m/z = 46$ gives a molar mass of $46\ \text{g mol}^{-1}$ , consistent with $\text{C}_2\text{H}_6\text{O}$ (ethanol).

The infrared spectrum shows a broad band near $3300\ \text{cm}^{-1}$ , indicating an O-H group (an alcohol), and the absence of a strong band near $1700\ \text{cm}^{-1}$ rules out a carbonyl (so it is not an aldehyde, ketone or acid), pointing to ethanol rather than an isomer such as ethanal.

The x-ray structure gives a definite C-O single-bond length of $0.143\ \text{nm}$ (typical of a single C-O bond, not the shorter C=O of about $0.122\ \text{nm}$ ), confirming the alcohol contains a C-O single bond and fixing the three-dimensional geometry.

Together the molar mass, the O-H with no carbonyl, and the C-O single-bond length identify the compound as ethanol, $\text{CH}_3\text{CH}_2\text{OH}$ .

Markers reward the molar mass from MS, the O-H present and carbonyl absent from IR, and the C-O single-bond length from x-ray, combining to ethanol.