Face-Centered Cubic Packing

Introduction

Face-centered cubic packing places equivalent particles at eight corners and six face centres of a conventional cube. It has four effective atoms per cell and twelve nearest neighbors around each atom. Equal hard spheres touch along face diagonals, giving the familiar close-packed fraction of about 74%.

Core explanation

The eight corner positions contribute 8×1/8=1 atom. Six face-centred positions contribute 6×1/2=3 atoms. Thus the conventional FCC cell contains Z=4 equivalent lattice points for a monatomic structure. Counting all fourteen visible spheres as whole atoms would overstate cell mass and density dramatically.

Each atom has twelve nearest neighbors in the ideal FCC lattice. One way to see this is to examine layers of close-packed spheres: six neighbors lie in the same layer, three in the layer above and three below. A corner-focused cube picture can hide some of these because neighbors extend into adjacent cells. Coordination number counts the infinite periodic environment, not only the particles inside one drawn cube.

Spheres touch along a face diagonal from one corner through the face centre to the opposite corner. A face diagonal has length √2a. The two corner-to-face-centre contact distances total 4r, so √2a=4r, or a=2√2r. There is no contact along the cube edge in this hard-sphere model, so a=2r would be wrong.

Packing fraction is four sphere volumes divided by cube volume: 4(4πr³/3)/(2√2r)³ = π/(3√2)≈0.7405. This is the same ideal fraction as hexagonal close packing, though their layer stacking differs. Close packing here is about equal hard spheres; real atoms have variable electron densities, and different bonding can favor another lattice despite a lower geometric fraction.

Several common metals, including copper and aluminium under ordinary conditions, have FCC structures. Their ductility is related partly to available slip systems, but packing fraction alone does not fully determine mechanical behavior. Grain size, impurities and temperature matter. FCC can also describe a sublattice in a compound, such as chloride sites in rock-salt NaCl; the full compound structure requires a basis with both ion types.

For a monatomic FCC metal, cell density is ρ=4M/(N Aa³). If r is known, use a=2√2r first. Proper length conversion to cm is required for g cm⁻³. This connects geometry, cell occupancy and a measurable bulk property.

Step-by-step reasoning

1. Count 8 corner eighths and 6 face halves to get Z=4. 2. Count neighbors in close-packed layers to get CN=12. 3. Identify contact along face diagonal √2a=4r. 4. Derive packing fraction π/(3√2). 5. Use four atoms per cell for monatomic density calculations.

Visual explanation

Draw a cube with corner and face-centred spheres. Highlight one face diagonal connecting corner, face centre and opposite corner, labeled √2a=4r. Add a layer sketch with six same-layer neighbors, three above and three below.

Real-world analogy

Stacked rows of oranges fit into gaps between oranges below, giving each interior orange many close neighbors. FCC is one repeating way to stack such dense layers, though atoms are not rigid fruit.

Real-world example

Aluminium metal's FCC structure is one part of its material behavior. Its low density follows atomic mass and cell dimensions as well, while alloying and defects affect strength and processing.

Why?

Why do FCC and HCP share packing fraction but remain different structures? Both pack equal spheres densely in layers, but their stacking sequences differ: ABC for FCC and AB for HCP.

Common misconception

“The six face atoms each lie wholly inside the cell.” A face is shared by two cells, so each contributes one half to the conventional cell count.

Worked example

For FCC r=0.140 nm, a=2√2r≈2.828×0.140=0.396 nm. The face diagonal √2a≈0.560 nm, matching 4r=0.560 nm. Effective atom count is 4, and ideal packing fraction is about 74.0%. These values follow one consistent contact model.

Quick check

1. What is the effective particle count in a monatomic FCC conventional cell? Answer: Four.

Exam focus

State Z=4, CN=12, face relation √2a=4r and packing fraction ≈74%. Keep close-packing geometry distinct from actual material strength or bonding.

Advanced insight

FCC has multiple close-packed {111} planes and slip directions, contributing to deformation behavior in many metals. Actual yield strength still depends strongly on dislocations and obstacles to their motion.

Summary

FCC packs equivalent spheres at corners and face centres, giving four effective atoms and twelve nearest neighbors. Face-diagonal contact yields a=2√2r and ideal packing fraction about 74%.

Practice questions

1. Count FCC atoms from boundaries. Answer: 8(1/8)+6(1/2)=4. 2. If a=0.400 nm, what is FCC radius? Answer: r=a/(2√2)≈0.141 nm. 3. Does FCC's 74% fraction prove every FCC metal is stronger than every BCC metal? Answer: No. Bonding, defects, alloying and microstructure strongly affect strength.