Exercises — Ionic crystals — NaCl, CsCl, ZnS, fluorite, antifluorite structures

WHAT we look at: the table of cutoffs. WHY: the CN is decided purely by which band $\rho$ lands in. $0.414\le 0.60<0.732$ → the octahedral band → CN 6 → NaCl (rock-salt) structure.

A corner ion is shared between 8 neighbouring cubes, so it contributes $\frac{1}{8}$ each.

• Cl⁻: $8\times \frac{1}{8}=1$
• Cs⁺: $1\times 1=1$ (body centre is wholly inside)

Ratio $1:1$, so one formula unit CsCl per cell → $Z=1$.

NaCl, CsCl, ZnS are all 1:1 (MX). CaF₂ is 1:2 (MX₂) — Ca²⁺ FCC with F⁻ in all tetrahedral holes → 4 Ca : 8 F. So CaF₂ is the odd one out.

WHAT: divide radii. WHY: $\rho$ selects the band. $\rho =\frac{148}{181}=0.818$ $0.732\le 0.818<1.000$ → cubic band → CN 8 → CsCl-type. (Real RbCl is actually rock-salt at ordinary pressure — a reminder the rule is a guide, not a law. But the ratio prediction is CsCl-type.)

WHAT LOOKS LIKE: look at Figure 1 — Na⁺ (edge midpoint) and Cl⁻ (corner) touch along the cube edge, so the whole edge is anion–cation–anion.

$a=2(r_{+}+r_{-})=2(95+181)=2(276)=552\text{pm}$

WHY the diagonal, not the edge? In CsCl the cation sits at the body centre, so it touches corner anions along the body diagonal, whose length is $\sqrt{3}a$ (Figure 2).

$\sqrt{3}a=2(r_{+}+r_{-})=2(169+181)=700\text{pm}$ $a=\frac{700}{\sqrt{3}}=404.1\text{pm}$

WHY $Z=1$: CsCl has one formula unit per cell (L1.2). Convert edge: $a=404.1\text{pm}=4.041\times 10^{-8}$ cm, so $a^3=(4.041\times 10^{-8}{)}^3=6.599\times 10^{-23}{\text{cm}}^3.$ ${\rho }_{\text{density}}=\frac{ZM}{N_A{a}^3}=\frac{1\times 168.4}{6.022\times 10^{23}\times 6.599\times 10^{-23}}=\frac{168.4}{39.74}\approx 4.24{\text{g/cm}}^3.$ Experimental CsCl density $\approx 3.99$ g/cm³ — close, confirming $Z=1$. See Density of crystals formula.

WHAT: rearrange the density formula for $N_A$. WHY: the same equation runs backwards. $N_A=\frac{ZM}{{\rho }_{\text{density}}{a}^3},a^3=(5.64\times 10^{-8}{)}^3=1.794\times 10^{-22}{\text{cm}}^3.$ $N_A=\frac{4\times 58.5}{2.16\times 1.794\times 10^{-22}}=\frac{234}{3.875\times 10^{-22}}\approx 6.04\times 10^{23}{\text{mol}}^{-1}.$ Within rounding of the true $6.022\times 10^{23}$. ✓

FCC has $N=4$ ions per cell → $2N=8$ tetrahedral holes.

• ZnS: Zn²⁺ count $=4$, so filled fraction $=4/8=\frac{1}{2}$ → half (alternate holes). CN 4:4. ✓
• CaF₂: the FCC framework is Ca²⁺ (4 per cell), giving $8$ tetrahedral holes, and F⁻ count $=8$ → filled fraction $=8/8=1$ → all holes. Stoichiometry $4\text{Ca}:8\text{F}=1:2$. ✓

Step 1 — ratio. $\rho =118/133=0.887$. This is in the cubic band ($\ge 0.732$), consistent with the 8-coordinate cation of fluorite (CaF₂-type) — an MX₂ solid with big cations. ✓

Step 2 — edge. Cation–anion contact along a quarter body diagonal: $\sqrt{3}a=4(r_{+}+r_{-})=4(118+133)=4(251)=1004\text{pm}$ $a=\frac{1004}{\sqrt{3}}=579.7\text{pm}=5.797\times 10^{-8}\text{cm}.$

Step 3 — density. Fluorite has $Z=4$ formula units (4 Ca, 8 F). $a^3=(5.797\times 10^{-8}{)}^3=1.948\times 10^{-22}{\text{cm}}^3$ ${\rho }_{\text{density}}=\frac{4\times 125.6}{6.022\times 10^{23}\times 1.948\times 10^{-22}}=\frac{502.4}{117.3}\approx 4.28{\text{g/cm}}^3.$ (Experimental SrF₂ $\approx 4.24$ g/cm³ — excellent.) ✓

(a) Antifluorite is fluorite with roles swapped. In fluorite cation CN 8, anion CN 4. Swap: here the anion (O²⁻) has CN 8, the cation (Li⁺) has CN 4. So CN is 4:8 (Li:O). (b) O²⁻ FCC $=4$ per cell; all $8$ tetrahedral holes filled by Li⁺ $=8$. Ratio $8\text{Li}:4\text{O}=2:1$ → Li₂O. ✓ (c) O²⁻–O²⁻ close-packing contact is the FCC framework (along face diagonals); Li⁺–O²⁻ contact runs along a quarter of the body diagonal, exactly as in fluorite. See Close packing FCC HCP and voids.

The picture (Figure 3): put the anions at 4 alternate corners of a cube of edge $b$; the tetrahedral hole (cation) sits at the cube's centre. Two anions on the same face diagonal are the touching pair; a cation–anion contact runs along half a body diagonal.

• Anion–anion contact = face diagonal $=\sqrt{2}b=2r_{-}$.
• Cation–anion contact = half body diagonal $=\frac{\sqrt{3}}{2}b=r_{+}+r_{-}$.

WHY these two: the limiting case is when both touchings happen at once. Take the ratio to kill $b$: $\frac{r_{+}+r_{-}}{r_{-}}=\frac{\frac{\sqrt{3}}{2}b}{\frac{\sqrt{2}}{2}b}=\frac{\sqrt{3}}{\sqrt{2}}=\sqrt{\frac{3}{2}}.$ $1+\rho =\sqrt{\frac{3}{2}}=1.2247\Rightarrow \rho =\sqrt{\frac{3}{2}}-1=\mathrm{0.225.}✓$

The octahedral band needs $\rho \ge 0.414$. Here $0.35<0.414$, so the cation is too small for the octahedral hole — it cannot touch all 6 anions at once and would rattle, letting anions crush together (repulsion). Energy is lowered by dropping to the tetrahedral band ($0.225\le 0.35<0.414$) → CN 4. This is the physical meaning of the lower cutoff. See Coordination number.

Step 1 — molar mass from density rearranged: $M=\frac{{\rho }_{\text{density}}{N}_A{a}^3}{Z},a^3=(5.0\times 10^{-8}{)}^3=1.25\times 10^{-22}{\text{cm}}^3.$ $M=\frac{3.20\times 6.022\times 10^{23}\times 1.25\times 10^{-22}}{4}=\frac{3.20\times 75.28}{4}=\frac{240.9}{4}\approx 60.2\text{g/mol}.$

Step 2 — cation radius from the rock-salt edge relation $a=2(r_{+}+r_{-})$: $r_{+}=\frac{a}{2}-r_{-}=\frac{500}{2}-167=250-167=83\text{pm}.$

Step 3 — consistency: $\rho =83/167=0.497$, which lies in $0.414$–$0.732$ → CN 6, exactly the rock-salt band we assumed. Self-consistent. ✓

Predict CN from $\rho$ ::: $0.225$–$0.414$→4, $0.414$–$0.732$→6, $0.732$–$1$→8. Rock-salt edge–radius relation ::: $a=2(r_{+}+r_{-})$. CsCl edge–radius relation ::: $\sqrt{3}a=2(r_{+}+r_{-})$. Fluorite cation–anion contact ::: quarter body diagonal, $\sqrt{3}a=4(r_{+}+r_{-})$. Density formula ::: ${\rho }_{\text{density}}=ZM/(N_A{a}^3)$. Tetrahedral limiting ratio derivation ::: $\sqrt{3}/\sqrt{2}=1+\rho \Rightarrow \rho =\sqrt{3/2}-1=0.225$.