Exercises — Coordination number, voids (tetrahedral, octahedral)

Tool A — the void counts. For $N$ close-packed atoms: $\text{Octahedral voids}=N,\text{Tetrahedral voids}=2N$ Tool B — the fcc atom count. In a face-centred-cubic cell: $N=8\times \frac{1}{8}+6\times \frac{1}{2}=1+3=4$ Tool C — the radius-ratio map. $r$ = small (cation) radius, $R$ = big (anion) radius.

WHAT the question wants: the single number 12 and where it comes from. A sphere in a close-packed layer touches:

• $6$ spheres in its own flat layer (a ring of six around it),
• $3$ spheres nestled in the dimples above,
• $3$ spheres nestled in the dimples below. $6+3+3=12$ Answer: coordination number $=12$. This is the densest any equal spheres can pack (Kepler's limit) — you physically cannot make a 13th identical sphere touch.

• Tetrahedral void: surrounded by $4$ spheres → coordination number $=4$.
• Octahedral void: surrounded by $6$ spheres → coordination number $=6$.
• Larger void: the octahedral one. Its limiting radius ratio is $\approx 0.414$, bigger than the tetrahedral $\approx 0.225$. More spheres around the pocket means more space in the middle, not less.

WHY these rules: Tool A says voids scale directly with the number of packed atoms $N$. $N=2.0\times 10^{22}$

• Octahedral voids $=N=2.0\times 10^{22}$.
• Tetrahedral voids $=2N=4.0\times 10^{22}$. Answer: $2.0\times 10^{22}$ octahedral and $4.0\times 10^{22}$ tetrahedral voids.

WHAT we count: every position equidistant from exactly 6 atoms.

• Body centre: $1$ void, fully inside the cell → contributes $1$.
• Edge centres: there are $12$ edges; each edge-centre void is shared between $4$ neighbouring cells → contributes $12\times \frac{1}{4}=3$. $\text{Octahedral voids}=1+3=4=N✓$ This matches Tool A ($\text{Oct}=N$) and the fcc count $N=4$.

Step 1 — pick $N$. Let there be $N$ anions forming the ccp skeleton. Step 2 — count filled cation sites. Octahedral voids $=N$ (Tool A). "All filled" → cations $=N$. Step 3 — take the ratio. A (cation) : B (anion) $=N:N=1:1$. Answer: formula $\mathbf{AB}$ — the rock-salt (NaCl) type. See NaCl, ZnS, CaF2 crystal structures.

Step 1: $N$ anions. Step 2: tetrahedral voids $=2N$; all filled → cations $=2N$. Step 3: cation : anion $=2N:N=2:1$. Answer: ${\mathbf{A}}_2\mathbf{B}$ (antifluorite, e.g. ${\text{Na}}_2\text{O}$). Twice as many tetrahedral voids as packed atoms is what doubles the cation count.

Step 1: $N$ anions. Step 2: octahedral voids $=N$; half filled → cations $=\frac{N}{2}$. Step 3: cation : anion $=\frac{N}{2}:N=1:2$. Answer: $\mathbf{A}{\mathbf{B}}_2$ — this is exactly how a ${\text{CdCl}}_2$-type layer arrangement gets its 1 : 2 stoichiometry.

Step 1 — anions. Let $N=$ number of ${\text{O}}^{2-}$ ions (ccp) $=N$. Step 2 — cations. Octahedral voids $=N$; two-thirds filled → ${\text{Al}}^{3+}=\frac{2}{3}N$. Step 3 — ratio. $\text{Al}:\text{O}=\frac{2}{3}N:N=2:3$. Formula: $\mathbf{A}{\mathbf{l}}_2{\mathbf{O}}_3$. Charge check: $2\times (+3)+3\times (-2)=+6-6=0$ ✓ neutral. This is corundum.

Step 1: $N$ sulphide ions. Step 2: tetrahedral voids $=2N$; half filled → ${\text{Zn}}^{2+}=\frac{1}{2}\times 2N=N$. Step 3: $\text{Zn}:\text{S}=N:N=1:1\Rightarrow$ ZnS ✓. Coordination number of ${\text{Zn}}^{2+}$: it sits in a tetrahedral void → CN $=4$. (And by symmetry each ${\text{S}}^{2-}$ is also surrounded by 4 zincs — a $4:4$ structure.)

(a) Read Tool C: $0.40$ lies in the band $0.225-0.414$ → tetrahedral void, CN $=4$. It is just below the octahedral threshold $0.414$, so it would rattle in an octahedral hole but fits snugly in a tetrahedral one. (b) $r=0.40\times R=0.40\times 1.40=0.56\text{A˚}$. (c) The octahedral limiting ratio is $\sqrt{2}-1\approx 0.414$, so $r=(\sqrt{2}-1)\times 1.40=0.414\times 1.40\approx 0.580\text{A˚}.$ Just $0.02\text{A˚}$ more turns a snug-tetrahedral cation into a snug-octahedral one — this is the razor-thin boundary that decides real crystal structures.

WHAT the picture shows: four big spheres form a square; big spheres touch along each edge, and the small sphere sits dead centre touching two opposite big spheres along the diagonal. Step 1 — the edge. Two big spheres touch along a side, so the side length $=R+R=2R$. Step 2 — the diagonal. A square of side $2R$ has diagonal $2R\sqrt{2}$ (Pythagoras: $\sqrt{(2R{)}^2+(2R{)}^2}$). Step 3 — what lies along the diagonal. From one big-sphere centre across to the opposite one, we pass: one big radius $R$, the whole small sphere $2r$, then another big radius $R$. So the diagonal also equals $R+2r+R$. Step 4 — equate and solve. $2R+2r=2R\sqrt{2}\Rightarrow R+r=R\sqrt{2}\Rightarrow r=R(\sqrt{2}-1)$ $\boxed{\frac{r}{R}=\sqrt{2}-1\approx 0.414}$

Step 1 — face diagonal (big–big touch). Two big spheres touch across a face diagonal of length $a\sqrt{2}$, so $a\sqrt{2}=2R\Rightarrow R=\frac{a\sqrt{2}}{2}$. Step 2 — body diagonal (big–small–big touch). Along the body diagonal, length $a\sqrt{3}$, we cross $R+2r+R=2R+2r$. So $a\sqrt{3}=2R+2r$. Step 3 — divide by $2R$. $\frac{a\sqrt{3}}{2R}=1+\frac{r}{R}$ Step 4 — substitute $2R=a\sqrt{2}$. $\frac{a\sqrt{3}}{a\sqrt{2}}=1+\frac{r}{R}\Rightarrow \sqrt{\frac{3}{2}}=1+\frac{r}{R}$ $\boxed{\frac{r}{R}=\sqrt{\frac{3}{2}}-1\approx 0.225}$

(a) fcc → $N=4$ atoms per cell (Tool B). (b) Density formula (see Density of a unit cell): $\rho =\frac{N\cdot M}{N_A\cdot a^3}$

• $a^3=(4.00\times 10^{-8}{)}^3=6.40\times 10^{-23}{\text{cm}}^3$.
• Numerator $N\cdot M=4\times 200=800{\text{g mol}}^{-1}$.
• Denominator $N_A\cdot a^3=6.022\times 10^{23}\times 6.40\times 10^{-23}=38.54{\text{cm}}^3{\text{mol}}^{-1}$. $\rho =\frac{800}{38.54}\approx 20.8{\text{g cm}}^{-3}$ Answer: $\approx 20.8{\text{g cm}}^{-3}$.

"Oct = One, Tet = Two; fill-fraction times the crew." Octahedral voids $=N$, tetrahedral $=2N$; multiply by the fraction actually filled, then read the ratio. Sizes: oct 0.414, tet 0.225 ("4-1-4 is the wide door, 2-2-5 is the tight one").