Solid state — crystalline vs amorphous; unit cell, Bravais lattices: Exercises

Level 1 — Recognition

L1.1

Classify each as crystalline or amorphous, and state the one test that decides it: (a) common window glass, (b) table salt (NaCl), (c) rubber, (d) copper metal.

Recall Solution L1.1

The deciding test is melting behaviour (sharp point = crystalline; softens over a range = amorphous), NOT hardness.

(a) Glass — softens over a range → amorphous (a supercooled liquid).
(b) NaCl — sharp melting point, cleaves on flat faces → crystalline.
(c) Rubber — no long-range order, softens gradually → amorphous.
(d) Copper — metallic crystal, sharp melting point → crystalline. Why the test is melting, not hardness: hardness measures how tightly bonded particles are; order measures whether the pattern repeats. A hard thing can still be jumbled inside.

L1.2

Name the crystal system with $a = b = c$ and all angles $= 9 0^{\circ}$ , and list its Bravais lattices.

Recall Solution L1.2

$a = b = c$ , all $9 0^{\circ}$ → the cubic system. Its three Bravais lattices are P (primitive/simple), I (body-centred), and F (face-centred). So cubic contributes $3$ of the $14$ lattices.

Level 2 — Application

L2.1

Compute $Z$ for a simple cubic (SC), a body-centred cubic (BCC), and a face-centred cubic (FCC) cell using the sharing rule. Show every fraction.

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

Recall Solution L2.1

Look at the figure: corner atoms (blue-black dots) sit where 8 cells meet, the body-centre (red) sits alone inside, face atoms straddle 2 cells.

SC: $8$ corners $\times \frac{1}{8} = 1$ . So $Z = 1$ .
BCC: $8 \times \frac{1}{8} + 1 \times 1 = 1 + 1 = 2$ . So $Z = 2$ .
FCC: $8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 1 + 3 = 4$ . So $Z = 4$ . Why the body-centre counts as a whole 1: it lies fully inside — no neighbour shares it, so nothing is subtracted. See Packing efficiency and density of unit cell for where these $Z$ values feed next.

L2.2

A metal crystallises as FCC. In one unit cell, how many atoms come from corners alone, and how many from faces alone?

Recall Solution L2.2

Corners: $8 \times \frac{1}{8} === 1 ==$ atom. Faces: $6 \times \frac{1}{2} === 3 ==$ atoms. Together $1 + 3 = 4 = Z$ . The split ( $1$ from corners, $3$ from faces) is exactly why FCC feels "denser" than SC.

Level 3 — Analysis

L3.1

For BCC, the atoms touch along the body diagonal of the cube. If edge length is $a$ and atomic radius is $r$ , derive the relation between $r$ and $a$ .

Recall Solution L3.1

WHAT touches WHAT: in BCC the corner atom and the body-centre atom kiss along the body diagonal (the red line in the figure). Along that diagonal we cross: half a corner atom + one whole centre atom + half a corner atom $= 4$ radii, i.e. the diagonal length $= 4 r$ . WHY use the body diagonal and not a face diagonal: because that is the only line along which atoms actually make contact in BCC. Contact is the physical fact that ties $r$ to $a$ . The geometry: a cube of edge $a$ has body diagonal of length $a 3$ (Pythagoras twice: face diagonal $a 2$ , then combine with the third edge). $a 3 = 4 r ⟹ r = \frac{3}{4} a \approx 0.433 a$

L3.2

For FCC, atoms touch along the face diagonal. Derive $r$ in terms of $a$ .

Recall Solution L3.2

WHAT touches: on a cube face, the two corner atoms and the face-centre atom line up along the face diagonal. Crossing it: half-corner + full face-atom + half-corner $= 4 r$ . WHY the face diagonal here: in FCC contact happens on the face, not through the body — the body centre is empty. Face diagonal $= a 2$ , so $a 2 = 4 r ⟹ r = \frac{2}{4} a = \frac{a}{2 2} \approx 0.354 a$

Level 4 — Synthesis

L4.1

Silver crystallises FCC with edge length $a = 408 pm$ . Find (a) the atomic radius $r$ , and (b) the packing efficiency of FCC.

Recall Solution L4.1

(a) From L3.2, $r = \frac{a}{2 2} = \frac{408}{2 \times 1.4142} \approx== 144.2 pm ==$ . (b) Packing efficiency $= \frac{volume of atoms in cell}{volume of cell} \times 100%$ .

Atoms in cell: $Z = 4$ , each a sphere of volume $\frac{4}{3} π r^{3}$ .
Substitute $r = \frac{a}{2 2}$ , so $r^{3} = \frac{a ^{3}}{16 2}$ . $Efficiency = \frac{4 \cdot \frac{4}{3} π r ^{3}}{a ^{3}} \times 100% = \frac{\frac{16}{3} π \cdot \frac{a ^{3}}{16 2}}{a ^{3}} \times 100% = \frac{π}{3 2} \times 100% \approx== 74.05% ==$ This $74%$ is the maximum possible for equal spheres — see Close packing in solids — HCP, CCP, void fraction.

L4.2

Compute the density of silver from L4.1. Given: molar mass $M = 108 g mol^{- 1}$ , Avogadro number $N_{A} = 6.022 \times 1 0^{23} mol^{- 1}$ , $a = 408 pm = 4.08 \times 1 0^{- 8} cm$ .

Recall Solution L4.2

WHY this formula: density = mass of the cell ÷ volume of the cell. The cell holds $Z$ atoms, so its mass is $\frac{Z M}{N _{A}}$ . $ρ = \frac{Z M}{N _{A} a ^{3}}$

$a^{3} = (4.08 \times 1 0^{- 8})^{3} = 6.79 \times 1 0^{- 23} cm^{3}$ .
$ρ = \frac{4 \times 108}{6.022 \times 1 0 ^{23} \times 6.79 \times 1 0 ^{- 23}} = \frac{432}{40.89} \approx== 10.56 g cm^{- 3} ==$ . This matches silver's real density ( $\approx 10.5 g cm^{- 3}$ ) — a good sanity check. See Packing efficiency and density of unit cell.

Level 5 — Mastery

L5.1

A cubic solid has atoms of element X at all corners and atoms of element Y at the body centre. (a) Give the formula of the compound. (b) If instead Y atoms sit at the face centres (X still at corners), give the new formula.

Recall Solution L5.1

(a) X at corners: $8 \times \frac{1}{8} = 1$ . Y at body centre: $1 \times 1 = 1$ . Ratio $X : Y = 1 : 1 \Rightarrow$ formula $== X Y ==$ . (b) X at corners: $8 \times \frac{1}{8} = 1$ . Y at faces: $6 \times \frac{1}{2} = 3$ . Ratio $X : Y = 1 : 3 \Rightarrow$ formula $== X Y_{3} ==$ . Why the sharing rule still rules: the type of atom doesn't change how many cells share a site; position (corner/face/body) does. This is exactly the logic behind Ionic solids — NaCl, ZnS, CaF2 structures.

L5.2

A compound forms an FCC lattice of anions (A) with cations (B) occupying all the tetrahedral voids. There are $2$ tetrahedral voids per anion. Find the formula.

Recall Solution L5.2

Step 1 — count anions: FCC of A gives $Z_{A} = 4$ anions per cell (from L2.1). Step 2 — count voids: tetrahedral voids $= 2 \times (number of anions) = 2 \times 4 = 8$ per cell. Step 3 — fill: all $8$ voids hold a B cation → $Z_{B} = 8$ . Step 4 — ratio: $A : B = 4 : 8 = 1 : 2 \Rightarrow$ formula $== A B_{2} ==$ (this is the antifluorite-type ratio; compare $CaF_{2}$ in Ionic solids — NaCl, ZnS, CaF2 structures and void ideas in Radius ratio rule and coordination number).

Recall Quick self-check (reveal after all solutions)

SC gives Z equal to ::: $1$ BCC gives Z equal to ::: $2$ FCC gives Z equal to ::: $4$ BCC radius–edge relation ::: $r = \frac{3}{4} a$ FCC radius–edge relation ::: $r = \frac{a}{2 2}$ FCC packing efficiency ::: $\approx 74%$ Density formula for a unit cell ::: $ρ = \frac{Z M}{N _{A} a ^{3}}$