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# Coordinated Crystals

##### Stage: 5 Challenge Level:

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 $0, ^\circ, 70.5^\circ, 180^\circ$.