Worked examples — Solid state — crystalline vs amorphous; unit cell, Bravais lattices

This page is the practice arena for Solid state — crystalline vs amorphous; unit cell, Bravais lattices (index 2.4.12). The parent taught you the rules: the sharing fractions (corner $\frac{1}{8}$, edge $\frac{1}{4}$, face $\frac{1}{2}$, body $1$), how to count $Z$ (atoms per cell), and the 7 systems / 14 Bravais lattices. Here we use those rules on every kind of case an exam can hand you — including the weird degenerate ones.

Before we touch numbers, let's agree on every symbol so nothing appears unexplained.

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The scenario matrix

Every problem this topic throws at you is one of these cells. The worked examples below are tagged with the cell they cover, so together they hit all of them.

Cell	Case class	What makes it tricky
A	Simple / primitive cubic ($Z=1$)	Only corners — the "baseline"
B	Body-centred cubic ($Z=2$)	The un-shared centre atom
C	Face-centred cubic ($Z=4$)	Faces shared by 2
D	Edge + body-centre mix (rock-salt style)	Edge atoms contribute $\frac{1}{4}$
E	Degenerate: missing / removed atom	A corner or face atom deleted
F	Real-world: formula from a picture	Translate a drawing → chemical formula
G	Limiting geometry: edge along a diagonal	Relate $a$ and $r$ per lattice type
H	Exam twist: 2 different atoms, find ratio	Non-cubic contributions, hexagonal

Read the statement, forecast your answer before scrolling, then check your reasoning against the steps.

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1. List what's present. 8 corner atoms, nothing else. Why this step? You can only count what the picture actually contains.
2. Apply the corner fraction. Each corner is shared by 8 cells → contributes $\frac{1}{8}$. Why this step? The sharing rule prevents double-counting the same physical atom across neighbouring cells.
3. Add up: $Z=8\times \frac{1}{8}=1$. Why this step? Summing fractions gives the whole atoms you actually own.

Verify: 8 corners each giving one-eighth is literally "eight one-eighths make one whole." $Z=1$, not 8. ✓

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1. Corners: $8\times \frac{1}{8}=1$. Why this step? Same corner rule as Cell A.
2. Body centre: it lies entirely inside the cube, touching no wall → shared by 1 cell → contributes $1$. Why this step? Fraction $=\frac{1}{\text{number of cells sharing}}=\frac{1}{1}=1$. The centre is yours alone.
3. Add: $Z=1+1=2$.

Verify: BCC always gives $Z=2$; sodium's known structure confirms 2 atoms per cell. ✓

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1. Corners: $8\times \frac{1}{8}=1$.
2. Faces: each face is a wall between 2 cells → each face atom contributes $\frac{1}{2}$. Six faces: $6\times \frac{1}{2}=3$. Why this step? A face atom sits on the boundary plane, so exactly two cubes claim half of it each.
3. Add: $Z=1+3=4$.

Verify: FCC $Z=4$; copper's density calculations rely on exactly this. ✓

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1. Count ${\text{Cl}}^{-}$ (corners + faces): $8\times \frac{1}{8}+6\times \frac{1}{2}=1+3=4$. Why this step? This is exactly an FCC arrangement of chloride, so $Z_{\text{Cl}}=4$.
2. Count ${\text{Na}}^{+}$ edges: each edge is shared by 4 cells → contributes $\frac{1}{4}$. Twelve edges: $12\times \frac{1}{4}=3$. Why this step? An edge line is the meeting seam of 4 cubes stacked around it — hence quarter each.
3. Count ${\text{Na}}^{+}$ body centre: $1\times 1=1$.
4. Total ${\text{Na}}^{+}$: $3+1=4$.

Verify: 4 Na${}^{+}$ and 4 Cl${}^{-}$ → formula ratio $1:1$ → NaCl, matching Ionic solids — NaCl, ZnS, CaF2 structures. ✓

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1. Original count: $Z=4$ (from Cell C).
2. What we remove: one face atom, which contributed only $\frac{1}{2}$ (not a whole atom). Why this step? Removing a shared atom removes only the fraction the cell owned, never a full atom.
3. New count: $Z=4-\frac{1}{2}=\frac{7}{2}=3.5$.

Verify: $4-0.5=3.5$ atoms per cell. Non-integer $Z$ is normal for defective / mixed lattices — see Defects in solids — Schottky and Frenkel. ✓

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1. Count X: corners $=8\times \frac{1}{8}=1$.
2. Count Y: body centre $=1\times 1=1$.
3. Ratio: $X:Y=1:1$. Why this step? The chemical formula is just the smallest whole-number ratio of owned atoms.

Verify: $1:1\Rightarrow$ formula XY (like CsCl). ✓

Twist check: if instead X were at faces (6 faces $\times \frac{1}{2}=3$) and Y at corners (1), the ratio would be $X_{3}Y$ — showing how the positions alone decide the formula.

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1. Length of the body diagonal. For a cube of edge $a$, the body diagonal has length $\sqrt{a^2+a^2+a^2}=a\sqrt{3}$. Why this step? Pythagoras in 3-D: the diagonal spans all three perpendicular edges.
2. Atoms along that diagonal. Corner atom, then the centre atom, then the far corner atom — they touch, so the diagonal covers $r+2r+r=4r$. Why this step? Along the diagonal you cross one radius of the near corner, a full diameter ($2r$) of the centre atom, one radius of the far corner.
3. Set equal: $a\sqrt{3}=4r\Rightarrow a=\frac{4r}{\sqrt{3}}$.

Verify: With $r=\sqrt{3}$ (test value), $a=4$. Check: body diagonal $=4\sqrt{3}=a\sqrt{3}$ ✓, and $4r=4\sqrt{3}$ ✓ — they match. This feeds directly into Packing efficiency and density of unit cell. ✓

Case contrast: in FCC atoms touch along the face diagonal ($a\sqrt{2}=4r$); in simple cubic along the edge ($a=2r$). Different lattice → different touching line.

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1. Corner P atoms: each shared by 6 cells → $\frac{1}{6}$. Twelve corners: $12\times \frac{1}{6}=2$. Why this step? In a hexagonal prism, 6 prisms meet at each vertical edge line, so the fraction is $\frac{1}{6}$, not $\frac{1}{8}$ — the cubic rule does not apply here.
2. Face-centre P atoms: 2 atoms $\times \frac{1}{2}=1$. Why this step? Top and bottom hexagon faces are each shared between 2 stacked cells.
3. Total P: $2+1=3$.
4. Interior Q atoms: $3\times 1=3$ (fully inside → whole).
5. Ratio: $P:Q=3:3=1:1$.

Verify: $P:Q=1:1$. This matches the HCP motif discussed in Close packing in solids — HCP, CCP, void fraction, where the standard hexagonal cell famously owns 6 atoms — here we split them across two atom types 3 and 3. ✓

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Recall Quick self-test

Corner atom in a cube contributes what fraction, and why? ::: $\frac{1}{8}$, because 8 cubes meet at that corner. Edge atom contributes what, and why? ::: $\frac{1}{4}$, because 4 cells share that edge line. Removing one face atom from FCC changes $Z$ by how much? ::: By $\frac{1}{2}$, giving $Z=3.5$ — you only lose the owned fraction. For BCC, atoms touch along which line, giving what relation? ::: The body diagonal, giving $a\sqrt{3}=4r$. Corner atom in a hexagonal prism contributes what? ::: $\frac{1}{6}$, because 6 prisms meet there.