Worked examples — Cubic systems — SCC, BCC, FCC; packing fraction calculations

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The scenario matrix

Every exam question about cubic cells is really one of these cells. We will hit each at least once.

#	Case class	What makes it tricky	Example that covers it
C1	Find $a$ from $r$ (each of SCC/BCC/FCC)	must pick the right touch-direction	E1
C2	Find $r$ from $a$ (reverse direction)	rearranging, not memorising	E2
C3	Density forward ($\rho$ from $a,M,Z$)	pm→cm unit trap	E3
C4	Density backward (find $M$ or $Z$ from $\rho$)	solve for the unknown	E4
C5	Empty-space / packing as a percentage	$100\%-\varphi$ and comparisons	E5
C6	Degenerate / limiting input ($r\to 0$, "does it touch?")	sanity of the model itself	E6
C7	Real-world word problem (which structure is this metal?)	translate words → cell type	E7
C8	Exam twist (mixed: given $\rho$, find $r$)	chain two formulas together	E8

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The two relations we lean on

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Worked examples

E1 — Find $a$ from $r$, all three types (cell C1)

1. Pick the touch-direction for each. Why this step? The whole $a$–$r$ link changes depending on where spheres touch; using the wrong direction is the #1 error in the parent's [!mistake] box.

2. SCC: touch along the edge → $a=2r=2(150)=300$ pm.

3. BCC: touch along the body diagonal → $a=\frac{4r}{\sqrt{3}}=\frac{4(150)}{1.732}=346.4$ pm. Why divide by $\sqrt{3}$? The body diagonal is $a\sqrt{3}$ (Pythagoras twice) and equals $4r$; to isolate $a$ we divide both sides of $\sqrt{3}a=4r$ by $\sqrt{3}$, giving $a=4r/\sqrt{3}$.

4. FCC: touch along the face diagonal → $a=\frac{4r}{\sqrt{2}}=2\sqrt{2}r=2(1.414)(150)=424.3$ pm. Why the $\sqrt{2}$? Same isolation move on $\sqrt{2}a=4r$; note $4/\sqrt{2}=2\sqrt{2}$ after rationalising.

Verify: SCC $300<$ BCC $346<$ FCC $424$. Careful reading: for the same atom, the structure whose atoms touch across the longest line needs the biggest edge — here that is FCC (face diagonal) among our three, so FCC has the largest edge. Forecast: FCC largest.

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E2 — Find $r$ from $a$ (reverse) (cell C2)

1. Write the BCC link the "solve-for-$r$" way. From $\sqrt{3}a=4r$ we get $r=\frac{\sqrt{3}a}{4}$. Why this algebra? We're given $a$ and want $r$, so we divide both sides by $4$ to leave $r$ alone on one side — no new physics, just isolating the unknown.

2. Substitute: $r=\frac{1.732\times 316}{4}=\frac{547.3}{4}=136.8$ pm.

Verify: $a/2=158$ pm. Our $r=136.8$ pm is smaller than $a/2$ — correct, because in BCC corner atoms do not touch across the edge (only the centre bridges them), so each atom is smaller than the naive edge-touch guess would give. Forecast confirmed.

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E3 — Density forward, the unit trap (cell C3)

1. Get $Z$. FCC → $Z=8\times \frac{1}{8}+6\times \frac{1}{2}=4$. Why this step? Relation 2 needs the atom count first.

2. Convert $a$ to cm — two named sub-steps. Why cm? We want g/cm³, so every length must be in cm; skipping this is off by $10^{30}$.

   • pm → m: $1$ pm $=10^{-12}$ m, so $a=361\times 10^{-12}$ m $=3.61\times 10^{-10}$ m.
   • m → cm: $1$ m $=100$ cm $=10^2$ cm, so multiply by $10^2$: $a=3.61\times 10^{-10}\times 10^2=3.61\times 10^{-8}$ cm.

3. Cube it. $a^3=(3.61\times 10^{-8}{)}^3=4.70\times 10^{-23}{\text{ cm}}^3$. Why cube? Volume is edge${}^3$; we cube both the number ($3.61^3=47.0$) and the power ($10^{-8\times 3}=10^{-24}$), then combine: $47.0\times 10^{-24}=4.70\times 10^{-23}$.

4. Plug in. $\rho =\frac{4\times 63.5}{(4.70\times 10^{-23})(6.022\times 10^{23})}=\frac{254}{28.30}=8.97$ g/cm³.

Verify: Units: $\frac{\text{g/mol}}{{\text{cm}}^3\cdot {\text{mol}}^{-1}}={\text{g/cm}}^3$ ✓. Value $\approx 8.97$ — real copper is $8.96$ g/cm³. Forecast (near 9) confirmed.

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E4 — Density backward: find $Z$, then name the structure (cell C4)

1. Rearrange Relation 2 for $Z$. $\rho =\frac{ZM}{a^3{N}_A}\Rightarrow Z=\frac{\rho a^3{N}_A}{M}$. Why this algebra? $Z$ is the unknown; multiply both sides by $a^3{N}_A$ (undoing the division) then divide by $M$ (undoing the multiplication) to leave $Z$ alone.

2. Convert & cube $a$. $a=2.89\times 10^{-8}$ cm (pm→m→cm as in E3), so $a^3=2.414\times 10^{-23}$ cm³. Why cube here too? Relation 2 contains $a^3$, so we need the volume, not the edge.

3. Substitute — carry the numerator and denominator separately. $Z=\frac{(7.20)(2.414\times 10^{-23})(6.022\times 10^{23})}{52.0}$. Why group like this? The two powers $10^{-23}\times 10^{23}=10^0=1$ cancel, leaving $7.20\times 2.414\times 6.022=104.66$ on top; dividing by $52.0$ gives $Z=2.01\approx 2$.

4. Read off the structure. $Z=2\Rightarrow$ BCC (this is chromium, in fact).

Verify: $Z$ landed on $2.01$, essentially an integer — a good check that we didn't botch the units (a non-integer would scream "unit error"). $Z=2$ ⇒ BCC per the Coordination Number in Crystals family. Forecast confirmed.

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E5 — Empty space & the packing comparison (cell C5)

1. Recall the packing fractions $\varphi$ (fraction filled): SCC $=\pi /6$, BCC $=\pi \sqrt{3}/8$, FCC $=\pi \sqrt{2}/6$. Why this step? Empty space $=1-\varphi$; we need $\varphi$ first.

2. BCC empty: $1-\frac{\pi \sqrt{3}}{8}=1-0.680=0.320=32.0\%$. Why subtract from 1? The whole cube is $100\%$; whatever the spheres don't fill is air.

3. FCC empty: $1-0.740=0.260=26.0\%$. SCC empty: $1-0.524=0.476=47.6\%$. Why compute all three? The matrix cell C5 asks for the comparison, and this is a favourite exam ranking.

Verify: Air ranking (most → least): SCC $47.6\%>$ BCC $32\%>$ FCC $26\%$. More atoms per cell ($1<2<4$) ⇒ tighter packing ⇒ less air. Forecast confirmed; matches the mnemonic "52–68–74, more atoms means more floor."

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E6 — Degenerate / limiting input: "does it even touch?" (cell C6)

1. Write $\varphi$ without assuming touching. $\varphi =\frac{Z\cdot \frac{4}{3}\pi r^3}{a^3}$. Here $Z=1$, $a$ is fixed. So $\varphi =\frac{\frac{4}{3}\pi r^3}{a^3}$. Why this step? The famous $52.4\%$ only holds when the atoms touch ($a=2r$). If $a$ is fixed while $r$ shrinks, that condition breaks, so we must keep $r$ and $a$ independent.

2. Take the limit $r\to 0$. $\varphi =\frac{\frac{4}{3}\pi r^3}{a^3}\to 0$. Why this algebra? With $a$ held constant, $\varphi$ is a constant times $r^3$; as $r\to 0$, $r^3\to 0$, so the whole thing $\to 0$.

3. When is it $52.4\%$? Only at the touching size $r=a/2$: $\varphi =\frac{\frac{4}{3}\pi (a/2{)}^3}{a^3}=\frac{\pi }{6}=52.4\%$. Why substitute $r=a/2$? That is exactly the SCC touch condition; plugging it in must reproduce the known $\pi /6$, which it does.

Verify: The claim is false. $52.4\%$ is the packing at the maximum radius that fits (touching). Below that, $\varphi$ scales as $r^3$ and drops toward $0$. Sanity: at $r=a/2$ our formula reproduces exactly $\pi /6$. Forecast confirmed: smaller atoms ⇒ less filled.

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E7 — Real-world word problem: name the structure (cell C7)

1. Clue 1 — coordination number 12. From Coordination Number in Crystals, only FCC (=CCP) has $12$ nearest neighbours (SCC has $6$, BCC has $8$). Why this step? Coordination number is a fingerprint of the structure.

2. Clue 2 — $Z=4$. FCC is the only cubic type with $Z=4$. Consistent.

3. Clue 3 — touch across face diagonal. That is exactly the FCC contact line ($\sqrt{2}a=4r$). All three clues agree ⇒ FCC.

4. Its packing fraction — and why it is $74\%$. Pack $Z=4$ spheres of radius $r$ into volume $a^3$, and use the FCC link $a=2\sqrt{2}r$: ${\varphi }_{FCC}=\frac{4\cdot \frac{4}{3}\pi r^3}{(2\sqrt{2}r{)}^3}=\frac{\frac{16}{3}\pi r^3}{16\sqrt{2}{r}^3}=\frac{\pi }{3\sqrt{2}}=\frac{\pi \sqrt{2}}{6}=0.740=74\%.$ Why this reminder? It shows the number isn't memorised magic — it's (4 sphere volumes) ÷ (cube volume) with $r$ cancelling. See also Close Packing — HCP vs CCP: CCP is the cubic cousin of HCP, both $74\%$.

Verify: All three independent clues (CN$=12$, $Z=4$, face-diagonal touch) converge on FCC — no contradiction, so the identification is safe. $\varphi =74\%$. Forecast confirmed.

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E8 — Exam twist: chain density → find $r$ (cell C8)

1. Step A — get $a$ from density. Rearrange Relation 2 for $a^3$: $a^3=\frac{ZM}{\rho N_A}$, with $Z=4$ (FCC). Why this algebra? We can't get $r$ directly from $\rho$. Multiply both sides of $\rho =ZM/(a^3{N}_A)$ by $a^3$ and divide by $\rho$ to isolate $a^3$.

2. Compute $a^3$. $a^3=\frac{4\times 108}{(10.5)(6.022\times 10^{23})}=\frac{432}{6.323\times 10^{24}}=6.833\times 10^{-23}{\text{ cm}}^3.$

3. Cube-root for $a$. $a=(6.833\times 10^{-23}{)}^{1/3}=4.09\times 10^{-8}$ cm $=409$ pm. Why cube-root? $a^3$ is a volume; the edge $a$ is its cube-root — the inverse of the cubing we did in E3. (m→pm: multiply cm by $10^{10}$ to get pm.)

4. Step B — get $r$ from the FCC touch link. $\sqrt{2}a=4r\Rightarrow r=\frac{\sqrt{2}a}{4}=\frac{1.414\times 409}{4}=144.6$ pm. Why this algebra? Divide $\sqrt{2}a=4r$ by $4$ to isolate $r$ — the same "solve-for-$r$" move as E2, now for the face-diagonal link.

Verify: $r\approx 145$ pm — the accepted radius of silver is $\approx 144$ pm ✓. Units: $a^3$ came out in cm³, cube-root gives cm, converted to pm cleanly. Forecast (nearer 100 than 400) confirmed.

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Recall Quick self-test across the matrix

Given a metal's density and structure, which do you find first, $a$ or $r$? ::: $a$ (from Relation 2, the density formula), then $r$ from the touch-direction link. A computed $Z$ comes out $2.01$ — what does that tell you? ::: It's essentially $2$ ⇒ BCC; being near an integer confirms your units were right. As $r\to 0$ with $a$ fixed, what does $\varphi$ do? ::: It goes to $0$ (scales as $r^3$); $52.4\%$ only holds when atoms touch. Three clues say CN$=12$, $Z=4$, face-diagonal touch — which structure? ::: FCC (CCP), $74\%$ packing.

Connections

• Parent topic (Hinglish)
• Density of a Unit Cell
• Coordination Number in Crystals
• Avogadro Number and Molar Mass
• Pythagoras Theorem
• Close Packing — HCP vs CCP
• Voids — Tetrahedral and Octahedral