Crystalline solids are composed of regular repeating patterns over relatively long distances ( >1000 Å). This is known as having a **long range order**.

ONLY crystalline solids produce macroscopic crystals (visible crystals).

Crystals have **flat faces** which can vary in size from crystal to crystal. The **angle between similar faces is constant**, and they break (**cleave**) in preferred directions (a feature exploited by diamond cutters).

**Crystal Structure Studies tell us:**

- Chemical Characteristics
- How are atoms connected?
- What are the bond lengths and angles between them?
- What does this say about the bonding?

- Inorganic Solids
- Structural features which may lead to an important property.
- How changing a metal coordination sphere may alter the property (eg. in superconductors and electrical materials).

- Organic Compounds
- Which points of stereochemistry are important.
- How solubility is affected by the way the molecules pack together.
- How many ways the molecules can pack together.

- Solution vs Solid State Structure
- Structures may differ between different phases.

**A couple of things about Molecule Packing:**

- Atoms, ions or molecules always try to pack into the lowest energy configuration.
- This configuration can then be repeated for a large number of units.
- As the configuration is repeated, a regular pattern forms and a
**lattice**emerges through the crystalline as a whole. - This pattern may interact with certain wavelengths of radiation and lead to diffraction (constructive interference) which provides a means of studying the pattern.

**3D Lattices and Translations – Unit Cells:
**

We can use three translations (a, b and c) to illustrate distance between atoms, ions or molecules; and three angles between each of these translations (α, β and γ). These parameters define the size and shape of the unit cell.

*Note that angle α corresponds to the translation a, angle β to translation b…etc.*

The unit cell is:

- A small volume defined by 6 faces, consisting of 3 identical pairs.
- A small unit of the larger structure which is then repeated.

“The smallest repeating unit that shows the full symmetry of the crystal structure.”

There are 7 Crystal Systems

**Triclinic**– a≠b≠c – α≠β≠γ (ALL Different!)**Monoclinic**– a≠b≠c – α=γ=90°, β**Orthohombic**– a≠b≠c – α=β=γ=90°**Tetragonal**– a=b≠c – α=β=γ=90°**Hexagonal**– a=b≠c – α=β=90°, γ=120° (Need a and c)**Rhombohedral**– a=b=c – α=β=γ≠90°**Cubic**– a=b=c – α=β=γ=90° (ALL Equal, Only need a)

Let’s look at some unit cells. Try and work out the number of atoms in each 2D cell:

You should have:

i) 1

ii) 1

iii) 2

Why? This is something we’ll explore a little more later, but for now consider this. i) has 4 dots (atoms, ions or molecules) in it’s shape. Each of these four dots can participate in a total of four unit cells.

Look at the red dot. It is shared between four unit cells and so only contributes 1/4 to the contents of that cell. The same can be said for each other dot at the corner of a cell, as we assume that it is only attributing 1/4.

So we know that i) has 4 corners, each worth 1/4. That totals 1!

ii) is just the same, but iii) is different. It has 4 corners each worth 1/4 BUT it also has a single dot in the middle that is not shared with any other cell…meaning it is worth a whole 1. So (4×1/4)+1 = 2.

**– Bravais Lattices**

Although there may be an infinite number of chemical structures, there are only 14 3D lattice types, known as Bravais lattices.

We’re just going to look at the cubic system to start with. Remember, a=b=c – α=β=γ=90°. In the cubic system there are 3 lattice types, P (primitive), I (body centered), and F (face centered).

- Primitive cubic cells contain only corner dots, which can be shared between 8 different unit cells, meaning each dot is worth 1/8. 8*(1/8) is 1 lattice point.
- Body Centered (I) cubic cells have the primitive structure but feature a central dot, worth 1 (as it is exclusive to that cell. This gives 8*(1/8) + 1 = 2 lattice points.
- Face Centered (F) cubic cells once again follow the primitive structure but this time feature a dot on each of the 6 faces. Each of these dots can be shared between two cells and so each is worth 1/2. This gives 8*(1/8) + 6*(1/2) = 1+3 = 4 lattice points.

So, when counting particles in a unit cell:

- An atom at a corner counts for 1/8
- An atom on an edge counts for 1/4
- An atom on a Face counts for 1/2
- An atom within the cell counts for 1

Technically this is only true for cells with 90 degree angles, but it does actually work for all cells.

**– Projection Drawings:**

Although we could draw the 3D shape of a cell every time, it is easier to draw a projection – a top down sketch of the cell.

Each atom’s height within the cell is indicated as a fraction of the cell height. Note the cell heights in the drawing below.

Here are the projection drawings for each cubic lattice:

**– Packing Densities:**

To find the structure offering the maximum density of atoms, we can calculate the atomic volume and packing density per unit cell. This involves some fairly basic maths which I have recapped here. We’ll start with Primitive cubic cell.

Volume of cubes = a³ (a*a*a)

Volume of a Sphere/Atom = 4/3 x π x r³

(r = atomic radius)

Packing Density = n x p / a³

(n = number of lattice points per cell, p = Volume of Atom)

**Primitive:**

There is 1 lattice point per cell – and the atomic radius is half the distance between the closest atom points. In this case, r = 1/2a.

Therefore:

- Volume of atom = 4/3 x 3.142 x (1/2a)³ = 0.524a³

The Packing Density of these atoms is then simply the volume of one atom (which we have found to be 0.524 a³) divided by the volume of the cube (which is a³):

- Packing Density = (0.524a³) / a³ = 0.524

**Body Centered:**

There are 2 lattice points per cell, and in this case the radius is not simply 1/2a as the central atom is the closest point. More trigonometry!

The distance between the 2 corner points is √3a, as shown in the first image. Thus the distance between the two closest atoms is half of this…√3a/2. This means the atomic radii must be √3a/4.

Therefore:

- Volume of Atom = 4/3 x 3.142 x (√3a/4)³ = 0.34a³
- Packing Density = (2 x 0.34a³) / a³ = 0.68

(Note the doubling of the atomic volume in the packing density calculation – this is because Body Centered cubes contain 2 lattice points.)

**Face Centered:**

4 lattice points per cell, with more trigonometry.

The distance between a corner atom and the corner atom diagonally across the same face is √2a. This means the distance between a corner atom and a face atom is half of this, √2a/2. This means the atomic radii must be √2a/4.

Therefore:

- Volume of Atom = 4/3 x 3.142 x (√2a/4)³ = 0.185a³
- Packing Density = (4 x 0.185a³) / a³ = 0.74

**Summary:**

Primitive Density (P) = 0.524 = 52.4%

Body Density (I) = 0.68 = 68%

Face Density (F) = 0.741 = 74.1%

You can see the density increasing as you move down the list.

**– Using this knowledge**

- Finding the metallic radius of α-tungsten. We know the unit cell is cubic (so we can use what’s above), a = 3.15Å, Atomic Mass = 183.85g mol^-1 and it’s measured density is 19.25g cm^-3 near room temperature and pressure. Step by step:
- We need to calculate the number of atoms in the unit cell (the number of lattice points). We know:

Density = (Mass of 1 atom x Number of Atoms in Cell) / Volume of Unit Cell

Therefore:

19.2 = ((183.85/6.023×10^23)* x N) / (3.15×10^-8)*^3

and Number of Atoms = 2 (rounded to leave an integer).We now know that there are 2 lattice points, or atoms per unit cell. Looking above we know that this unit cell is Body Centered.

(* Finding the mass of 1 atom is done by taking the molar mass of an element and dividing it by Avogadro’s number, AKA the number of atoms per mole.)

(** This value has been converted into Metres from Angstroms. 1Å = 1×10^8 Metres.) - Calculate the metallic radius – we have learnt that the unit cell is body centered, so we already know the distances involved.

The atomic radius in this lattice is √3a/4, and we know that a for α-tungsten = 3.15Å. Therefore the metallic radius is (√3 x 3.15)/4 = 1.36Å.

- We need to calculate the number of atoms in the unit cell (the number of lattice points). We know:

Audie SibellLong time viewer / first time poster. Really enjoy reading the blog, keep up the excellent work. Will definitely start posting more oftenin the near future.