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

Before we start, one figure to keep the three touch-directions in your eye — refer back to it whenever a problem says "atoms touch."

Figure — Cubic systems — SCC, BCC, FCC; packing fraction calculations

What this figure shows: three cubes. In SC (orange line) the two corner atoms kiss along the edge. In BCC (teal line) the centre atom bridges two opposite corners along the body diagonal. In FCC (plum line) a corner atom kisses the face-centre atom along the face diagonal. The line you pick decides the $a$ – $r$ equation — nothing else.

Level 1 — Recognition

Can you read a structure and pull out the three basic numbers: $Z$ , touch-direction, and the $a$ – $r$ link?

L1.1

A cubic cell has atoms only at its 8 corners. State (a) $Z$ , (b) the direction atoms touch, (c) the relation between edge $a$ and radius $r$ .

Recall Solution

What kind of cell? Corners only = Simple Cubic (SC). (a) Each corner atom is shared by 8 cubes, so it contributes $1/8$ . There are 8 corners: $Z = 8 \times \frac{1}{8} = 1.$ (b) In SC the corner atoms along one edge are the ones that actually touch (look at the orange line in the figure). (c) Two corner-atom radii span the edge, so $a = 2 r$ .

L1.2

For a BCC metal, write $Z$ and the exact relation linking $a$ and $r$ . Explain in one sentence why it is the body diagonal and not the edge.

Recall Solution

$Z = 8 \times \frac{1}{8} + 1 = 2$ (8 corners each $1/8$ , plus one whole body-centre atom). Why body diagonal? The corner atoms are too far apart to touch each other; only the centre atom, sitting exactly in the middle, reaches out and touches two opposite corners. That line of contact runs corner → centre → opposite corner, i.e. the body diagonal. The body diagonal of a cube of edge $a$ is $a 3$ (see the diagonal shortcuts above), and it is spanned by $r + 2 r + r = 4 r$ : $3 a = 4 r .$

L1.3

State the packing fractions (as percentages) for SC, BCC and FCC, and the empty-space percentages.

Recall Solution

Type	Packing	Empty
SC	$\frac{π}{6} \approx 52.4%$	$47.6%$
BCC	$\frac{π 3}{8} \approx 68.0%$	$32.0%$
FCC	$\frac{π 2}{6} \approx 74.0%$	$26.0%$

Empty space $= 100% - packing %$ , since the box is either atom or air.

Level 2 — Application

Plug numbers into the touch-relation and the density formula.

L2.1

Chromium is BCC with atomic radius $r = 125$ pm. Find the edge length $a$ (in pm).

Recall Solution

Why start with the touch-relation? It is the only correct bridge between $r$ and $a$ for BCC. $3 a = 4 r \Rightarrow a = \frac{4 r}{3} = \frac{4 ( 125 )}{1.732} = \frac{500}{1.732} \approx 288.7 pm .$

L2.2

Gold is FCC with edge $a = 408$ pm. Find its atomic radius $r$ (in pm).

Recall Solution

FCC → touch along the face diagonal, $2 a = 4 r$ , so $r = \frac{2 a}{4} = \frac{1.414 \times 408}{4} = \frac{576.9}{4} \approx 144.2 pm .$

L2.3

Copper is FCC, $a = 361$ pm, molar mass $M = 63.5$ g/mol. Find the density $ρ$ in g/cm³. (Use $N_{A} = 6.022 \times 1 0^{23}$ .)

Recall Solution

Formula: $ρ = \frac{Z M}{a ^{3} N _{A}}$ — see Density of a Unit Cell and Avogadro Number and Molar Mass. Here $ρ$ is the mass density (mass per unit volume, in g/cm³): mass of one cell $= Z \times (M / N_{A})$ divided by cell volume $a^{3}$ . Step 1 — convert the edge to cm. Density is wanted in g/cm³, so $a$ must be in cm: $a = 361 pm = 361 \times 1 0^{- 10} cm = 3.61 \times 1 0^{- 8} cm .$ Step 2 — cube it. $a^{3} = (3.61 \times 1 0^{- 8})^{3} = 4.705 \times 1 0^{- 23} cm^{3} .$ Step 3 — assemble. FCC → $Z = 4$ : $ρ = \frac{4 \times 63.5}{4.705 \times 1 0 ^{- 23} \times 6.022 \times 1 0 ^{23}} = \frac{254}{28.33} \approx 8.97 g/cm^{3} .$

Level 3 — Analysis

Derive a result you weren't handed, or compare two structures.

L3.1

Derive the packing fraction of BCC from scratch, i.e. get $\frac{π 3}{8}$ .

Recall Solution

Idea of packing fraction: $ϕ = \frac{volume of the atoms the cell owns}{volume of the cell}$ . Numerator. $Z = 2$ atoms, each a sphere of volume $\frac{4}{3} π r^{3}$ : $V_{atoms} = 2 \cdot \frac{4}{3} π r^{3} = \frac{8}{3} π r^{3} .$ Denominator. We need $a^{3}$ in terms of $r$ . From the touch-relation $a = \frac{4 r}{3}$ : $a^{3} = (\frac{4 r}{3})^{3} = \frac{64 r ^{3}}{3 3} .$ Divide. $ϕ = \frac{\frac{8}{3} π r ^{3}}{\frac{64 r ^{3}}{3 3}} = \frac{8 π}{3} \cdot \frac{3 3}{64} = \frac{8 π 3}{64} = \frac{π 3}{8} \approx 0.680.$ The $r^{3}$ cancels — packing fraction never depends on how big the atoms are, only on the arrangement.

L3.2

Iron changes from BCC to FCC on heating, but the atomic radius $r$ stays essentially the same. Does the density go up or down? Explain physically, then confirm by comparing $Z / a^{3}$ .

Recall Solution

Physical read: FCC packs 74% vs BCC 68% — the atoms are squeezed tighter, so the same atoms occupy less space per atom, meaning higher density. Confirm with $ρ \propto Z / a^{3}$ (since $M$ , $N_{A}$ fixed and $r$ fixed):

BCC: $a = \frac{4 r}{3} = 2.309 r$ , so $a^{3} = 12.32 r^{3}$ ; $Z / a^{3} = 2/12.32 r^{3} = 0.1623/ r^{3}$ .
FCC: $a = 22 r = 2.828 r$ , so $a^{3} = 22.63 r^{3}$ ; $Z / a^{3} = 4/22.63 r^{3} = 0.1768/ r^{3}$ . $\frac{( Z / a ^{3} ) _{FCC}}{( Z / a ^{3} ) _{BCC}} = \frac{0.1768}{0.1623} \approx 1.089.$ So FCC iron is about 8.9% denser — density goes up, matching the packing-fraction intuition.

L3.3

A metal (FCC, $a = 350$ pm) and the same-radius metal rebuilt as SC — what fraction of the FCC edge is the SC edge? Comment on what this says about wasted space.

Recall Solution

Same $r$ in both.

FCC: $r = \frac{2 a _{FCC}}{4} = \frac{1.414 \times 350}{4} = 123.7$ pm.
SC uses $a_{SC} = 2 r = 2 (123.7) = 247.5$ pm. $\frac{a _{SC}}{a _{FCC}} = \frac{247.5}{350} \approx 0.707 = \frac{1}{2} .$ The SC box is smaller per atom's-worth but holds only $Z = 1$ vs $Z = 4$ , so per atom SC wastes far more room (47.6% empty vs 26%). The ratio $1/ 2$ is exactly the face-diagonal geometry showing its hand.

Level 4 — Synthesis

Chain several tools together; multi-step reasoning.

L4.1

A metal crystallises FCC with density $ρ = 10.5$ g/cm³ and molar mass $M = 108$ g/mol. Find (a) the edge $a$ in cm and pm, and (b) the atomic radius $r$ in pm. (This is silver.)

Recall Solution

Rearrange the density formula for $a^{3}$ . From $ρ = \frac{Z M}{a ^{3} N _{A}}$ (with $ρ$ the mass density in g/cm³): $a^{3} = \frac{Z M}{ρ N _{A}} = \frac{4 \times 108}{10.5 \times 6.022 \times 1 0 ^{23}} = \frac{432}{6.323 \times 1 0 ^{24}} = 6.833 \times 1 0^{- 23} cm^{3} .$ Cube-root for $a$ . $a = (6.833 \times 1 0^{- 23})^{1/3} = 4.087 \times 1 0^{- 8} cm = 408.7 pm .$ (b) Radius via face-diagonal relation ( $2 a = 4 r$ ): $r = \frac{2 a}{4} = \frac{1.414 \times 408.7}{4} \approx 144.5 pm .$

L4.2

An element has BCC structure, edge $a = 300$ pm and density $ρ = 6.15$ g/cm³. Estimate its molar mass $M$ , then guess the element.

Recall Solution

Solve the density formula for $M$ : $M = \frac{ρ a ^{3} N _{A}}{Z} .$ Convert & cube the edge: $a = 3.00 \times 1 0^{- 8}$ cm, $a^{3} = 2.70 \times 1 0^{- 23}$ cm³. BCC → $Z = 2$ : $M = \frac{6.15 \times 2.70 \times 1 0 ^{- 23} \times 6.022 \times 1 0 ^{23}}{2} = \frac{6.15 \times 2.70 \times 6.022}{2} = \frac{100.0}{2} \approx 50.0 g/mol .$ $M \approx 50$ g/mol, BCC → this is vanadium ( $M \approx 50.9$ ).

L4.3

In FCC, find the radius of the octahedral void (largest sphere that fits the gap) in terms of $r$ , using the void that sits at the centre of a cube edge. (Link: Voids — Tetrahedral and Octahedral.)

Recall Solution

Which void, and why the edge one is valid. FCC has octahedral holes in two apparently different-looking places: one at the body-centre of the cube, and one at the centre of each edge. They are the same kind of hole — in fact they are related by a translation of the lattice: shift the whole crystal by half an edge and a body-centre hole lands exactly on an edge-centre hole. Because the crystal repeats identically under that shift, the two holes have identical surroundings and therefore identical size. So we may compute $r_{v}$ at whichever one is easiest — and the edge is easiest, because the touching atoms lie in a straight line along a single edge. Geometry along the edge. Along one cube edge we meet: a corner atom (radius $r$ ), then the void sphere (radius $r_{v}$ ) sitting at the edge-centre, then the next corner atom (radius $r$ ), all touching in a line: $a = r + 2 r_{v} + r = 2 r + 2 r_{v} .$ Insert the FCC edge $a = 22 r$ (from $2 a = 4 r$ ): $22 r = 2 r + 2 r_{v} \Rightarrow r_{v} = \frac{2 2 r - 2 r}{2} = (2 - 1) r \approx 0.414 r .$ So the octahedral hole holds a sphere up to $0.414 r$ — the famous radius-ratio bound for octahedral coordination, and by the shift argument the body-centre hole gives the very same number.

Level 5 — Mastery

Non-routine: prove, generalise, or reconcile competing facts.

L5.1

Prove that the packing fraction of any close-packed arrangement of equal spheres where each sphere touches its neighbours only along a face diagonal (FCC) equals $\frac{π 2}{6}$ , and show it is strictly greater than the BCC value $\frac{π 3}{8}$ .

Recall Solution

FCC packing. $Z = 4$ , $a = 22 r$ : $ϕ_{FCC} = \frac{4 \cdot \frac{4}{3} π r ^{3}}{( 2 2 r ) ^{3}} = \frac{\frac{16}{3} π r ^{3}}{16 2 r ^{3}} = \frac{π}{3 2} = \frac{π 2}{6} \approx 0.7405.$ Compare to BCC $ϕ_{BCC} = \frac{π 3}{8} \approx 0.6802$ . Both share the factor $π$ , so compare $\frac{2}{6}$ vs $\frac{3}{8}$ : $\frac{2}{6} = 0.2357, \frac{3}{8} = 0.2165.$ Since $0.2357 > 0.2165$ , $ϕ_{FCC} > ϕ_{BCC}$ strictly. FCC is the tighter pack.

L5.2

A student claims: "FCC and HCP have different packing because their unit cells look different." Reconcile this with the known fact that both are 74% packed. (Link: Close Packing — HCP vs CCP and Coordination Number in Crystals.)

Recall Solution

Both are 74%. FCC (= CCP) and HCP are two ways of stacking identical close-packed layers:

CCP/FCC stacking order: ABCABC…
HCP stacking order: ABAB… In both, every sphere sits in the dimples of the layer below, each sphere touches 12 neighbours (coordination number 12), and each layer already achieves the densest possible 2D packing. Because the local contact geometry (a sphere nestled among 12 others) is identical, the fraction of space filled is identical: $74%$ . What differs is only the long-range repeat pattern (and hence the cell shape), not how tightly any sphere is surrounded. The unit cells look different; the packing efficiency does not. Same 74%, different bookkeeping.

L5.3

Show that for all three cubic types the packing fraction is independent of the atomic radius $r$ , and explain in one line why that must be true before doing any algebra.

Recall Solution

The one-line reason (before algebra): packing fraction is a ratio of volumes of the same kind of object (atoms vs box), and scaling every atom by a factor $k$ scales the box edge by the same $k$ (contacts are preserved), so numerator and denominator scale by $k^{3}$ and the ratio is unchanged. Packing is a shape property, not a size property. Algebra confirms it — the $r^{3}$ always cancels:

SC: $\frac{1 \cdot \frac{4}{3} π r ^{3}}{( 2 r ) ^{3}} = \frac{\frac{4}{3} π r ^{3}}{8 r ^{3}} = \frac{π}{6}$ .
BCC: $\frac{2 \cdot \frac{4}{3} π r ^{3}}{( 4 r / 3 ) ^{3}} = \frac{π 3}{8}$ .
FCC: $\frac{4 \cdot \frac{4}{3} π r ^{3}}{( 2 2 r ) ^{3}} = \frac{π 2}{6}$ . In each, every $r^{3}$ in the top meets an $r^{3}$ in the bottom and vanishes. The result is a pure number.

Recall One-glance answer key

L1.1 $Z = 1$ , edge, $a = 2 r$ ::: L1.2 $Z = 2$ , $3 a = 4 r$ ::: L1.3 52.4/68/74% L2.1 $a \approx 288.7$ pm ::: L2.2 $r \approx 144.2$ pm ::: L2.3 $ρ \approx 8.97$ g/cm³ L3.1 $π 3 /8 \approx 0.680$ ::: L3.2 FCC ~8.9% denser ::: L3.3 ratio $= 1/ 2 \approx 0.707$ L4.1 $a \approx 408.7$ pm, $r \approx 144.5$ pm ::: L4.2 $M \approx 50$ g/mol (V) ::: L4.3 $r_{v} = (2 - 1) r \approx 0.414 r$ L5.1 FCC > BCC ::: L5.2 same 74%, different stacking ::: L5.3 $r^{3}$ cancels

Connections

Parent topic (Hinglish)
Density of a Unit Cell — used in L2.3, L4.1, L4.2
Pythagoras Theorem — face & body diagonals
Voids — Tetrahedral and Octahedral — L4.3
Close Packing — HCP vs CCP — L5.2
Coordination Number in Crystals — L5.2
Avogadro Number and Molar Mass — density problems