# Coordinated crystals

Explore the lattice and vector structure of this crystal.

## Problem

A certain crystal, $X$, is formed from two types of atom, $A$ and $B$. The atoms of $A$ are found at all the points, and only the points, with coordinates $(l, m, n)$ for any whole numbers $l, m, n$; the atoms of $B$ are found at all the points, and only the points, with coordinates $(l+0.5, m+0.5, n+0.5)$.

Think about the geometry of this crystal. Can you visualise its structure? Can you devise a clear pictorial representation? How simply can you describe its structure in words?

How close are the various $A$ and $B$ atoms to each other? What bond angles are formed?

What crystal structure does this represent?

Can you represent any other crystal structures in a similar way?

NOTES AND BACKGROUND

Developing an understanding of the symmetry properties of crystals leads to insights into many of the chemical and physical properties of chemicals. Due to the mathematical constraints of three-dimensional geometry, there are a limited number of possibilities for the symmetry structures. You can read more about this topic at http://en.wikipedia.org/wiki/Unit_cell#Unit_cell

## Getting Started

Focus on the atom at the origin and its nearest neighbours to begin
with.

Use the scalar product to work out the angles once you have visualised the structure.

Use the scalar product to work out the angles once you have visualised the structure.

## Student Solutions

Alex from Stoke on Trent Sixth Form College sent in this solution, using the natural language of vectors, translations and tesselations to engage with this problem and give a very clear visualisation -- Well done!

The translations need to map the atom $A$ at the origin to the atoms $A$ closest to it in the positive direction are $(1,0,0) (0,1,0) (0,0,1) (1,1,0) (0,1,1) (1,0,1)$ and $(1,1,1)$.

Together with the origin, these can be considered as the vertices of a cube.

This cube structure tessellates across the space such that each atom $A$ is a shared vertex of 8 cubes.

Applying the translation $(0.5, 0.5, 0.5)$ to each atom $A$ gives an atom $B$ and the set of $B$ atoms can be thought of as the set of points in the centre of each of the $A$ cubes. Similarly, each atom $A$ could be considered at the centre of a cube of atoms $B$. Thus the same structure is generated, because $B$ is also the shared vertices of tessellating cubes.

Each $A$ atom is the same distance from each of its surrounding $B$ atoms, as is $B$ from its $A$ atoms. From Pythagoras' theorem, this distance is $\frac{\sqrt{3}}{2}$.

This crystal structure represents caesium chloride.

Steve writes

The $A$ atoms form a cubic lattice with atoms at whole numbers. Shifting this lattice by half a unit in each of the $x$, $y$, and $z$ directions takes us to the position of the $B$ atoms.

Each $A$ atom is thus at the centre of a cube of $B$ atoms and each $B$ atom is at the centre of a cube of atoms $A$.

To find the bond angles, assume that each $A$ atom is bonded only to its $8$ nearest neighbours. As each unit is the same, consider the simplest unit, which is the $A$ atom at the origin and the $B$ atoms found at coordinates $(\pm 0.5, \pm 0.5, \pm 0.5)$.

To find angles, it is easiest to use the scalar product. The angles present in each of the cube are found by taking scalar products over all possible combinations of signs

$$\left(\begin{array}{c}\pm 0.5 \\ \pm 0.5 \\ \pm 0.5\end{array}\right)\cdot \left(\begin{array}{c}\pm 0.5 \\ \pm 0.5 \\ \pm 0.5\end{array}\right) = \pm 0.25 \pm 0.25 \pm 025$$

These scalar products are either $\pm 0.75$ or $\pm 0.25$.

Since $\bf{a}\dot {\bf b} = |{\bf a}||{\bf b}|\cos\theta$ and the squared distances of the $B$ atoms from the origin are $0.75$ we have that

$\pm 0.75 = 0.75 \cos\theta$ or $\pm 0.25 = 0.75 \cos\theta$

Thus $\cos\theta = \pm 1$ or $\cos\theta = \frac{1}{3}$, giving bond angles of $0, ^\circ, 70.5^\circ, 180^\circ$.

## Teachers' Resources

### Why do this problem?

This problem is a useful exercise in visualisation and 3D vectors. It provides a natural context for the mathematics and has many extension possibilities. It is an example of a problem where a clear geometric image really facilitates the work with vectors and where the use of the scalar product really facilitates the calculation of the angles.### Possible approach

Initially focus on trying to understand the atomic structure.
Encourage discussion and the drawing of diagrams? Share these.
Which ways of thinking about the atomic structure are the simplest
and clearest?

To understand how close the various atoms are to each other
requires clear thinking. It will be easiest to think in terms of
each atom $A$ surrounded by a 'box' of $B$ atoms, in which case it
will be easier to see which distances, and therefore angles, are
possible.

The extension concerning the other crystal configurations is
mathematically very interesting.

You could consider structures well known from chemistry or
encourage students to research the idea following the link from the
problem.

### Key questions

What sort of atom lies at the origin?

What is the configuration of all of the $A$ atoms or all of
the $B$ atoms?

What angle is formed between the atom at the origin and its
two closest neighbours?

### Possible extension

Consider creating a version of the problem with face-centred
cubic packing, where the first challenge is to determine an
algebraic form of the location of the different atoms.