Switch the cell and the view, and drag the canvas to rotate it. Change the edge a and the molar mass M to see how density responds.
Controls
Choose a crystal structure
View
Readings
- Atoms per cell (Z)
- 8 × ⅛ + 6 × ½ = 4
- Coordination number
- 12
- Edge a versus radius r
- a = 2√2 r ≈ 2.83 r
- Atomic radius r (pm)
- 127.3
- Packing efficiency
- 74.0 %
- Empty space
- 26.0 %
- Density ρ (g/cm³)
- 9.11
- Examples
- Copper, aluminium, gold, silver
How to use this simulation
- FCC is selected to begin with. Press play and the cell turns slowly by itself. Drag the canvas to rotate it your own way; let go and it drifts again.
- Blue atoms sit at corners, green on faces, orange at the centre. The readout counts atoms per cell: each corner contributes 1/8, each face atom 1/2 and a centre atom all of it.
- Switch to "Touching spheres". The atoms swell to their true size and touch exactly along the red contact line. That line is what fixes the link between a and r.
- Pick SC, BCC and FCC in turn and compare coordination number, the a-versus-r relation and packing efficiency. Every one is computed from atom positions.
- Move the edge a and molar mass M sliders and watch the density: it falls as a grows and rises with M.
A pile of oranges and the inside of a metal
How does a fruit seller stack oranges? One layer at the bottom, the next layer sitting in the gaps. That way the most oranges fit in the least space. Metal atoms behave like spheres, and they are answering the same question: how densely can you pack them?
In a crystal the pattern repeats. The smallest piece that, copied in every direction, builds the whole crystal is the unit cell. Here we look at three cube-shaped ones.
A great deal about copper, gold, iron or sodium, such as hardness, density and whether it can be drawn into wire, depends on how its atoms are arranged. That is why unit cells matter so much in the solid-state chapter.
The unit cell, edge a and radius r
A cube has three equal edges of length a. Treat an atom as a sphere of radius r. Along the direction in which atoms touch, the distance between centres is exactly 2r. That touching direction is what ties a to r through geometry.
In the animation the touching line is red. In SC it runs along an edge, in BCC along the cube's body diagonal (corner through the centre to the opposite corner), in FCC along a face diagonal.
How to read the animation
The twelve grey lines are the cube's edges and the balls are atoms. In "Ball and stick" the balls are small so you can see the framework. In "Touching spheres" they are drawn at true radius, so you see how much space is filled and how much is empty.
Nearer atoms are drawn larger and farther ones a little smaller, which gives depth as you rotate. The relation between a and r is written at the bottom, with its numeric coefficient.
Every number in the readout is computed from the atom positions. The coordination number is found by counting nearest neighbours of one atom; the packing efficiency is the ratio of sphere volume to cube volume.
Atoms per unit cell
A corner atom is shared among eight cells, so only 1/8 of it lies inside one cell. A face atom sits between two cells, so 1/2 is inside. A body-centre atom is entirely inside.
SC: 8 corners × 1/8 = 1. BCC: 8 × 1/8 + 1 = 2. FCC: 8 × 1/8 + 6 × 1/2 = 4. That is why the animation colours corner, face and body atoms differently.
| Cell | Corners | Faces | Centre | Atoms per cell |
|---|---|---|---|---|
| SC | 8 × 1/8 | – | – | 1 |
| BCC | 8 × 1/8 | – | 1 × 1 | 2 |
| FCC | 8 × 1/8 | 6 × 1/2 | – | 4 |
Coordination number: how many neighbours
The number of atoms touching one atom is its coordination number. In SC it is 6 (front, back, left, right, up, down), in BCC 8, in FCC 12. More neighbours means tighter packing.
An atom belongs to the neighbouring cells as well as this one, so count in the whole crystal and not only the drawn cell.
Edge a versus radius r
In SC atoms touch along an edge, so a = 2r (coefficient 2.00). In BCC they touch along the body diagonal, which is √3 a long and holds four radii, so 4r = √3 a and a = 4r/√3 (coefficient 2.31). In FCC they touch along the face diagonal, √2 a long, again four radii, so a = 2√2 r (coefficient 2.83).
The larger the coefficient, the emptier the cell compared with the atoms in it. That is why SC is the loosest.
SC: a = 2rtouch along an edge
BCC: √3 a = 4ralong the body diagonal
FCC: √2 a = 4ralong the face diagonal
Packing efficiency: how much space is filled
Packing efficiency = volume of the atoms in the cell ÷ volume of the cell. Each atom is a sphere of volume (4/3)πr³ and the cell volume is a³. So efficiency = Z × (4/3)πr³ ÷ a³, and writing r in terms of a gives a pure number.
Working it out gives SC 52.4%, BCC 68.0% and FCC 74.0%. The empty space is 47.6%, 32.0% and 26.0%. Equal spheres cannot be packed more densely than FCC.
FCC is one form of close packing, which is why so many soft, easily shaped metals such as copper, silver and gold use it.
| Cell | Z | Coordination number | a versus r | Packing efficiency |
|---|---|---|---|---|
| SC | 1 | 6 | a = 2.00 r | 52.4% |
| BCC | 2 | 8 | a = 2.31 r | 68.0% |
| FCC | 4 | 12 | a = 2.83 r | 74.0% |
From cell to density
Mass of a cell = Z × (mass of one atom) = Z × M ÷ N_A, with N_A = 6.02 × 10²³ per mole. Volume = a³. Density = mass ÷ volume, so ρ = ZM ÷ (N_A a³). If a is in picometres, convert to centimetres first: 1 pm = 10⁻¹⁰ cm.
It also runs backwards: given ρ, a³ = ZM ÷ (ρ N_A). Measuring density can therefore reveal the structure.
ρ = Z M ÷ (N_A a³)ρ in g/cm³, M in g/mol, a in cm
a³ = Z M ÷ (ρ N_A)edge from a known density
One complete example for each cell
Take one hypothetical metal with atomic radius 125 pm and molar mass 64 g/mol, and see what changes across the three arrangements. Every figure below is calculated by the model.
The SC cell
Z = 8 × 1/8 = 1. Coordination number 6. Atoms touch along an edge, so a = 2r = 2.00 × 125 = 250.0 pm. Packing efficiency 52.4%, so 47.6% is empty. Density = 1 × 64 ÷ (6.02 × 10²³ × a³) = 6.80 g/cm³.
The BCC cell
Z = 8 × 1/8 + 1 = 2. Coordination number 8. They touch along the body diagonal, so a = 4r ÷ √3 = 2.31 × 125 = 288.7 pm. Packing efficiency 68.0%, empty 32.0%. Density = 8.84 g/cm³.
The FCC cell
Z = 8 × 1/8 + 6 × 1/2 = 4. Coordination number 12. They touch along the face diagonal, so a = 2√2 r = 2.83 × 125 = 353.6 pm. Packing efficiency 74.0%, empty 26.0%. Density = 9.62 g/cm³.
The three side by side
Same atom, same mass, only the arrangement differs, and density goes 6.80 then 8.84 then 9.62. More atoms fit per cell, and the cell volume does not grow in proportion. One atom has volume 8.18 × 10⁶ pm³ and the FCC cell has volume 44.19 × 10⁶ pm³. Add the volumes of four atoms, divide by the cell volume, and you get exactly 74.0%.
Empty space: the holes between the spheres
However tightly you stack spheres, gaps remain. In FCC the empty space is 26.0%. The gaps are not random: they are regular holes of two kinds. An octahedral hole is surrounded by six spheres, a tetrahedral hole by four.
An FCC cell has 4 octahedral holes and 8 tetrahedral holes, which is one and two per atom. The largest sphere that fits in an octahedral hole has radius 0.414 × r, and in a tetrahedral hole 0.225 × r. With r = 125 pm that is 51.8 pm and 28.1 pm.
Smaller atoms or ions can sit in these holes. Carbon dissolves in the holes of iron's lattice, which is part of why steel is harder than pure iron. Ionic crystals are built the same way.
Octahedral hole: r(hole) = (√2 − 1) rsix neighbours
Tetrahedral hole: r(hole) = (√(3/2) − 1) rfour neighbours
In the real world: metals and table salt
How metals behave matches how they are stacked. In FCC metals (copper, aluminium, gold) the layers of atoms slide over one another easily, so the metals are soft and can be drawn into wire. BCC metals (sodium, potassium) are less dense, 68.0% against 74.0%, so their densities tend to be lower.
In a table-salt crystal the chloride ions sit on an FCC pattern, which gives 4 per cell. The sodium ions fill the octahedral holes, and there are 4 of those, so 4 sodium ions. The ratio 4 : 4 is 1 : 1, which is the formula NaCl. Each ion is surrounded by six of the opposite charge, because the hole is octahedral.
This also explains why salt grains are cube-shaped: the compact pattern inside shows up as flat faces outside, and a grain cracks along flat planes when you crush it.
Try it yourself in the simulation
Write your prediction first, then test it.
- Experiment 1: choose FCC and switch the view. Which way does the red line run? Is there another atom between the two corner atoms it joins?
- Experiment 2: rotate and count the face atoms. Six faces, one atom each, each half inside. What is their total?
- Experiment 3: write the packing efficiency of SC, BCC and FCC side by side and order them.
- Experiment 4: in FCC move a from 250 to 600 pm. How does the radius readout change? Does a ÷ r stay fixed?
- Experiment 5: double M. How does density change? Now double a instead. By what factor does density change this time?
Worked problems
Every number comes from the model's own calculation, and the metals in the problems are hypothetical.
Problem 1: How many atoms per FCC cell?
8 corners at 1/8 each and 6 faces at 1/2 each. Z = 8 × 1/8 + 6 × 1/2 = 4.
Problem 2: A metal is FCC with atomic radius 125 pm. What is the edge?
a = 2√2 r = 2.83 × 125 = 353.6 pm.
Problem 3: Metal X, FCC, M = 64 g/mol, a = 360 pm. Find the density.
a = 360 × 10⁻¹⁰ cm, so a³ = 4.67 × 10⁻²³ cm³. Z = 4, so the cell mass = 4 × 64 ÷ 6.02 × 10²³ = 4.25 × 10⁻²² g. ρ = 9.11 g/cm³.
Problem 4: Metal Y, BCC, M = 50, a = 300 pm. Density?
Z = 2. ρ = 2 × 50 ÷ (6.02 × 10²³ × (300 × 10⁻¹⁰)³) = 6.15 g/cm³.
Problem 5: An FCC metal has density 9 g/cm³ and M = 64. Find the edge.
a³ = 4 × 64 ÷ (9 × 6.02 × 10²³); taking the cube root, a ≈ 361 pm.
Mistakes everyone makes
These cost marks every year.
- Counting a corner atom as a whole one. It is shared by eight cells, so 1/8.
- Taking Z = 1 for BCC. The centre atom makes it 2.
- Using a = 2r for every cell. That holds for SC only. BCC is 4r = √3 a and FCC is 4r = √2 a.
- Putting a in picometres straight into the density formula. Convert with 10⁻¹⁰ to get centimetres.
- Mixing up packing efficiency and coordination number. One is a count, the other a percentage of volume.
- Swapping BCC and FCC in the ranking. FCC is denser: 74% against 68%.
Exam corner (Class 12, JEE, NEET, A Level)
Answer without looking, then check.
- Give atoms per cell and coordination number for SC, BCC and FCC.
- What links a and r in FCC? If r = 125 pm, find a. Answer: 353.6 pm.
- What is the packing efficiency of FCC? Answer: 74.0%.
- A BCC metal has M = 50 and a = 300 pm. Density? Answer: 6.15 g/cm³.
- Rank SC, BCC and FCC by packing efficiency.
Revision: the whole page in a minute
Read this just before the exam.
- Z: SC 1, BCC 2, FCC 4. Corner 1/8, face 1/2, centre 1.
- Coordination number: 6, 8, 12.
- a and r: SC a = 2r; BCC 4r = √3 a; FCC 4r = √2 a.
- Packing efficiency: 52.4%, 68.0%, 74.0%.
- Density ρ = ZM ÷ (N_A a³), with a in cm.
Frequently asked questions
What is a unit cell?
The smallest repeating piece of a crystal. Copied in every direction, it builds the whole crystal.
How many atoms are there per cell in SC, BCC and FCC?
1, 2 and 4. A corner atom contributes 1/8, a face atom 1/2 and a centre atom 1.
What are the packing efficiencies?
SC 52.4%, BCC 68.0% and FCC 74.0%. FCC is the densest.
How do I find density from a unit cell?
Use ρ = ZM ÷ (N_A a³). Remember to convert a to centimetres first: 1 pm = 10⁻¹⁰ cm.
What is coordination number?
The number of nearest neighbours of one atom: 6 in SC, 8 in BCC and 12 in FCC.
What is the relation between a and r in BCC?
Atoms touch along the body diagonal, so 4r = √3 a and a = 2.31 r.
Keep studying this topic
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