Cubic systems — SCC, BCC, FCC; packing fraction calculations: Question bank

Before any trap, we lay down every symbol and abbreviation so nothing on this page is assumed.

Definition The words and symbols used everywhere below

A unit cell is the smallest cube that, stacked in 3D, rebuilds the whole crystal. All its edges are equal.
$a$ = the edge length of that cube — the length of one straight side of the box (look at the red side in the figure below).
$r$ = the atomic radius — half the width of one atom, treated as a hard sphere (the green arrow from an atom's centre to its surface).
SCC = Simple Cubic — atoms only at the 8 corners.
BCC = Body-Centred Cubic — 8 corners + 1 atom at the very centre of the cube.
FCC = Face-Centred Cubic — 8 corners + 1 atom at the centre of each of the 6 faces.
CCP = Cubic Close Packing — the same lattice as FCC, drawn a different way (identical $74%$ packing).
HCP = Hexagonal Close Packing — a non-cubic close packing, also $74%$ , differing only in stacking order.
$Z$ = number of whole atoms owned by one cell.
$M$ = molar mass (grams per mole); $N_{A}$ = Avogadro's number (atoms per mole). See Avogadro Number and Molar Mass.

The three geometric facts this whole page rests on

Every trap below is a disguised test of which atoms touch and what length that contact line spans. Here is each one derived, with a picture.

The one idea behind every question below: you must know which atoms actually touch before you can relate radius $r$ to edge $a$ . SCC touches along the edge, BCC along the body diagonal, FCC along the face diagonal. Everything traps you the moment you forget this.

True or false — justify

In BCC the eight corner atoms touch each other along the edge.

False. In BCC only the body-centre atom bridges the corners; corner atoms are held apart, so contact runs along the body diagonal (

3 a = 4 r

), never the edge.

Packing fraction depends on the atomic radius

r

False. Every

r^{3}

in the numerator cancels an

r^{3}

a^{3}

, leaving a pure number (

π /6

π 3 /8

π 2 /6

). Packing fraction is a property of the arrangement, not the atom size.

FCC and CCP describe the same packing.

True. The face-centred cubic cell is just one way to draw the cubic close-packed lattice; they have identical packing (

74%

) and coordination number 12. See Close Packing — HCP vs CCP.

HCP and FCC have different packing fractions because they look different.

False. Both are close-packed with

74%

; they differ only in stacking order (ABAB vs ABCABC), not in how tightly space is filled.

A body-centred cubic cell contains 2 whole atoms.

True.

8 \times \frac{1}{8}

(corners)

+ 1

(body centre)

= 2

. The centre atom lies entirely inside, so it counts as a full 1.

Doubling the edge

a

of an SCC cell doubles its packing fraction.

False. Packing fraction is dimensionless and fixed at

52.4%

for SCC regardless of

a

; scaling the box scales the atoms too, so the ratio is unchanged.

The empty space in FCC is larger than in SCC.

False. FCC empty

= 26%

, SCC empty

= 47.6%

. FCC is the tightest of the three, so it has the least empty space.

Coordination number and packing fraction always rise together across SCC→BCC→FCC.

True here. SCC (6, 52.4%), BCC (8, 68%), FCC (12, 74%): denser packing means each atom touches more neighbours. See Coordination Number in Crystals.

Spot the error

"BCC copper: I used

a = 2 r

a = 248

pm."

Error:

a = 2 r

is the SCC relation. BCC needs

3 a = 4 r

, i.e.

a = 4 r / 3

. Using the edge relation gives a wrong (too small) edge and wrong density.

"FCC has

Z = 1

because I counted the 8 corners."

Error: the 6 face-centre atoms were forgotten.

Z = 8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 1 + 3 = 4

. Corner-only counting is the SCC habit leaking in.

"Density:

a = 361

pm, so

a^{3} = (361)^{3}

pm³, done."

Error: units never converted. Density in g/cm³ needs

a

in cm;

1 pm = 1 0^{- 10}

cm, so cubing multiplies the error by

1 0^{30}

. Convert before cubing. See Density of a Unit Cell.

"Face diagonal of a cube of edge

a

a 3

Error:

a 3

is the body diagonal (spans three dimensions). The face diagonal lies in one square face and is

a^{2} + a^{2} = a 2

. See Pythagoras Theorem.

"In BCC the body diagonal equals

2 r

Error: three atoms lie along the body diagonal (corner–centre–corner), touching over two radii on each side, so the diagonal

= r + 2 r + r = 4 r

, giving

3 a = 4 r

"Packing fraction of FCC is

π /6

Error:

π /6 \approx 52.4%

is SCC. FCC is

π 2 /6 \approx 74%

. The mix-up comes from reusing the SCC edge relation instead of the face-diagonal one.

"Mass of one atom is

M \times N_{A}

Error: it is

M / N_{A}

M

is grams per mole and

N_{A}

atoms make a mole, so one atom weighs

M

divided by

N_{A}

. See Avogadro Number and Molar Mass.

Why questions

Why do we count a corner atom as

1/8

and not a whole atom?

A corner is the meeting point of

2 \times 2 \times 2 = 8

neighbouring cubes, so each cube owns only one-eighth of that shared sphere.

Why does the atom radius cancel out of every packing fraction?

Atom volume goes as

r^{3}

and the cube volume

a^{3}

is proportional to

r^{3}

(since

a

is a fixed multiple of

r

), so their ratio is a pure geometric constant.

Why must we identify the touch direction before writing the

a

–

r

relation?

Only along the touch direction do sphere surfaces actually meet, so only there does the sum of radii equal a known length of the cube — that is the single equation linking

r

a

Why is FCC called "closest packing"?

74%

it fills more of space than any other cubic arrangement; spheres nestle into the dimples of the layer below, leaving the least possible gaps.

Why does BCC pack tighter than SCC even though both have corner atoms?

The extra body-centre atom slips into the large central void, adding filled volume for the same box, raising packing from

52.4%

68%

Why do we need

a

in centimetres for density in g/cm³?

Density is mass over volume; if mass is in grams the volume must be in cm³, so the edge must be in cm before it is cubed.

Why can two different unit-cell drawings (FCC vs CCP) describe one crystal?

A unit cell is just a repeating tile choice; different tiles can regenerate the same infinite lattice, so the same close-packed solid can be framed as face-centred cubic. See Close Packing — HCP vs CCP.

Edge cases

If an SCC cell had zero atoms at corners (empty box), what is its packing fraction?

0%

— with no spheres the numerator

Z \cdot \frac{4}{3} π r^{3}

vanishes; packing fraction is only meaningful once atoms occupy positions.

Could a hypothetical structure ever reach

100%

packing?

No for hard spheres — round objects always leave gaps; the densest possible sphere packing tops out near

74%

, the FCC/HCP value.

If atoms stay touching but shrink together with the box (both

r

and

a

halved), does the packing fraction change?

No — since

a = 2 r

still holds,

ϕ

stays exactly

52.4%

; the ratio of atom volume to box volume is scale-free, so shrinking everything together changes nothing.

If instead the box

a

is held fixed while atoms shrink toward a point (

r \to 0

, atoms no longer touching), what happens to the packing fraction?

It falls toward

0%

— the numerator

\frac{4}{3} π r^{3} \to 0

while

a^{3}

stays fixed, so the filled fraction vanishes. (This is a different scenario from the touching case above: here

a \neq = 2 r

Does an atom sitting on an edge (not corner) contribute

1/8

No — an edge is shared by only

4

cubes, so an edge atom contributes

1/4

; only corners (shared by 8) give

1/8

If we picked a bigger, non-primitive box containing 8 SCC cells, does the packing fraction change?

No — it stays

52.4%

; enlarging the tile multiplies both atom volume and box volume by 8, and the ratio is unchanged.

In BCC, if you wrongly assume corners touch, is your predicted density too high or too low?

Too high — assuming

a = 2 r

gives a smaller edge than the true

4 r / 3

, shrinking

a^{3}

in the denominator and inflating

ρ = Z M / (a^{3} N_{A})

Between two octahedral-void-rich structures, does having more voids mean looser packing?

Not necessarily — FCC has many voids yet is the tightest packing; voids are just leftover gaps between touching spheres, catalogued in Voids — Tetrahedral and Octahedral.

Recall One-line summary of every trap

Almost every mistake on this page is the same mistake wearing a costume: using the wrong touch direction (edge vs face vs body diagonal), or the wrong atom count $Z$ . Fix the touch direction first, count $Z$ second, and no trap survives.