Body-Centered Cubic (BCC) Packing

What Is BCC Packing?

In a body-centered cubic (BCC) unit cell, atoms sit at each of the 8 corners of a cube, plus one additional atom at the exact centre (body centre) of the cube. The body-centre atom touches all 8 corner atoms.

BCC is adopted by many important metals at room temperature, including iron (α-Fe), chromium, tungsten, sodium, and potassium.

Learning Goals: By the end of this guide, you should be able to:

Describe the BCC unit cell and count atoms per cell.

Derive the relationship $a = \frac{4r}{\sqrt{3}}$ .

Calculate packing efficiency (68%).

Compare BCC with SC and FCC structures.

Anatomy of the BCC Unit Cell

Atom Count

8 corner atoms: each shared by 8 unit cells → $8 \times \frac{1}{8} = 1$
1 body-centre atom: entirely within the cell → $1$

Z = 1 + 1 = 2 \text{ atoms per unit cell}

Edge Length and Atomic Radius

In BCC, atoms touch along the body diagonal of the cube.

The body diagonal has length $a\sqrt{3}$ and spans 4 atomic radii (corner atom radius + body-centre diameter + corner atom radius):

a\sqrt{3} = 4r \quad \Rightarrow \quad a = \frac{4r}{\sqrt{3}}

Coordination Number

Each atom is surrounded by 8 nearest neighbours. The body-centre atom contacts all 8 corners; each corner atom contacts 8 body-centre atoms from adjacent cells.

\text{Coordination number} = 8

Packing Efficiency

\text{Packing efficiency} = \frac{Z \times \frac{4}{3}\pi r^3}{a^3} \times 100\%

Substituting $a = \frac{4r}{\sqrt{3}}$ :

a^3 = \frac{64r^3}{3\sqrt{3}}

\text{Efficiency} = \frac{2 \times \frac{4}{3}\pi r^3}{\frac{64r^3}{3\sqrt{3}}} \times 100\% = \frac{\pi\sqrt{3}}{8} \times 100\% \approx 68.0\%

Structure	Z	Coordination	Packing Efficiency	Example
SC	1	6	52.4%	Po
BCC	2	8	68.0%	Fe, Cr, W, Na
FCC (CCP)	4	12	74.0%	Cu, Al, Au

Worked Examples

Example 1: Density of Iron (α-Fe)

Given: $M = 55.85\ g/mol$ , $r = 124\ pm$ , BCC.

a = \frac{4 \times 124}{\sqrt{3}} = \frac{496}{1.732} = 286.4\ pm = 2.864 \times 10^{-8}\ cm

\rho = \frac{2 \times 55.85}{6.022 \times 10^{23} \times (2.864 \times 10^{-8})^3} = \frac{111.7}{1.416 \times 10^{-23} \times 6.022 \times 10^{23}} = 7.87\ g/cm^3

(Experimental: 7.87 g/cm³ ✅)

Example 2: Is This Metal BCC or FCC?

Given: Metal X, $\rho = 19.3\ g/cm^3$ , $M = 197\ g/mol$ , $r = 144\ pm$ .

Test BCC ( $Z = 2$ ): $\rho_{calc} = \frac{2 \times 197}{6.022 \times 10^{23} \times (332\ pm)^3} = 17.9\ g/cm^3$ ❌

Test FCC ( $Z = 4$ , $a = 2\sqrt{2}r = 407\ pm$ ): $\rho_{calc} = \frac{4 \times 197}{6.022 \times 10^{23} \times (407\ pm)^3} = 19.3\ g/cm^3$ ✅

This is gold — FCC structure.

Common Mistakes

Using the wrong diagonal — BCC atoms touch along the body diagonal ( $a\sqrt{3} = 4r$ ), NOT the face diagonal or edge.
Confusing Z = 2 with Z = 9 — The body-centre atom is ONE complete atom; don't count it as another set of shared atoms.
Applying FCC formulas to BCC — The relationship is different: FCC uses $a\sqrt{2} = 4r$ (face diagonal); BCC uses $a\sqrt{3} = 4r$ (body diagonal).
Forgetting that BCC is NOT close-packed — BCC has no close-packed layers (unlike FCC/HCP). It's efficient but not a close-packing arrangement.

Exam Tips (A-Level / AP / IB)

Draw the body diagonal clearly in your diagram — show the line from one corner through the body centre to the opposite corner.
In density problems, state $Z = 2$ explicitly and show your unit conversion.
Remember that iron transitions: BCC (α) → FCC (γ) → BCC (δ). This is commonly examined.
BCC metals are generally harder and more brittle than FCC metals because BCC has fewer slip systems.

Frequently Asked Questions

Why do some metals prefer BCC over FCC?

The stability of a crystal structure depends on electronic band structure and interatomic potential, not just packing efficiency. Many transition metals with partially filled d-bands prefer BCC because the electronic energy is minimised in that geometry.

Is BCC a close-packed structure?

No. Close-packed structures (FCC and HCP) have a packing efficiency of 74%. BCC (68%) is not close-packed — it lacks the hexagonal close-packed layers found in FCC and HCP.

What voids exist in a BCC structure?

BCC has both octahedral and tetrahedral voids, but they are asymmetric and smaller than those in FCC. Tetrahedral voids in BCC are at $(0, \frac{1}{2}, \frac{1}{4})$ positions.

Simple Cubic Packing — The simplest unit cell for comparison.
Cubic Close Packing (CCP) — FCC structure with maximum packing efficiency.
Tetrahedral Voids — How interstitial sites differ between crystal types.