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Concept 05 / 07Solid State

Structures of Ionic Solids: Radius Ratio Rule, NaCl, CsCl, ZnS, CaF2 and Na2O

Radius ratio rule and coordination number in ionic crystals, and the structures of rock salt (NaCl), caesium chloride, zinc blende and wurtzite (ZnS), fluorite (CaF2) and antifluorite (Na2O) in terms of close packing and void occupancy, with coordination numbers and formula units per unit cell.

Figure 1
In ionic solids, the larger ions (usually anions) form the lattice and the smaller ions occupy voids. For stability, each cation should touch as many anions as possible without the anions touching each other too closely. This depends on the radius ratio r+r−\frac{r_{+}}{r_{-}} (see figure): 0.1550.155 to 0.2250.225: CN 33 (trigonal planar); 0.2250.225 to 0.4140.414: CN 44 (tetrahedral), as in ZnS; 0.4140.414 to 0.7320.732: CN 66 (octahedral), as in NaCl; 0.7320.732 to 1.01.0: CN 88 (cubic), as in CsCl. The rule is a useful guide, though it fails for significantly covalent compounds. Rock salt, NaCl (type AB). Cl−\mathrm{Cl^{-}} ions form an fcc (ccp) lattice; Na+\mathrm{Na^{+}} ions occupy all the octahedral voids (edge centres and body centre). CN 6:66 : 6. Formula units per unit cell =4= 4. The edge a=2(r++r−)a = 2(r_{+} + r_{-}). Examples: KCl, NaI, AgCl, AgBr, MgO, CaO, FeO, NiO. Caesium chloride, CsCl (AB). Cl−\mathrm{Cl^{-}} at the corners of a cube and Cs+\mathrm{Cs^{+}} at the body centre (or vice versa); it is not bcc, but two interpenetrating simple cubic lattices. CN 8:88 : 8. One formula unit per cell. Body diagonal: 3a=2(r++r−)\sqrt{3}a = 2(r_{+} + r_{-}). Examples: CsBr, CsI, TlCl, TlBr. On heating to about 760 K760\,\mathrm{K}, CsCl converts to the NaCl structure; high pressure converts NaCl to the CsCl structure (higher CN). Zinc blende, ZnS (AB). S2−\mathrm{S^{2-}} ions form ccp; Zn2+\mathrm{Zn^{2+}} ions occupy half of the tetrahedral voids (alternate ones). CN 4:44 : 4. Four formula units per cell. Examples: CuCl, CuBr, CuI, AgI, BeS. Wurtzite is the hexagonal form of ZnS (hcp S2−\mathrm{S^{2-}}, half the tetrahedral voids filled). Fluorite, CaF2\mathrm{CaF_{2}} (AB2\mathrm{AB_{2}}). Ca2+\mathrm{Ca^{2+}} ions form ccp; F−\mathrm{F^{-}} ions occupy all the tetrahedral voids. CN 8:48 : 4 (each Ca2+\mathrm{Ca^{2+}} has 88 F−\mathrm{F^{-}}; each F−\mathrm{F^{-}} has 44 Ca2+\mathrm{Ca^{2+}}). Four formula units per cell. Examples: SrF2\mathrm{SrF_{2}}, BaCl2\mathrm{BaCl_{2}}, CdF2\mathrm{CdF_{2}}. Antifluorite, Na2O\mathrm{Na_{2}O} (A2B\mathrm{A_{2}B}). The reverse: O2−\mathrm{O^{2-}} ions form ccp and Na+\mathrm{Na^{+}} ions occupy all the tetrahedral voids. CN 4:84 : 8. Examples: Li2O\mathrm{Li_{2}O}, K2O\mathrm{K_{2}O}, Na2S\mathrm{Na_{2}S}. Key Takeaways: - Radius ratio: 0.2250.225 (CN 44), 0.4140.414 (CN 66), 0.7320.732 (CN 88) are the thresholds. - NaCl: fcc anions, cations in all octahedral voids, 6:66 : 6, z=4z = 4. - CsCl: simple cubic interpenetrating, 8:88 : 8, z=1z = 1. - ZnS: ccp S2−\mathrm{S^{2-}}, half the tetrahedral voids, 4:44 : 4; CaF2\mathrm{CaF_{2}}: 8:48 : 4; Na2O\mathrm{Na_{2}O}: 4:84 : 8. - Pitfall: CsCl is not a bcc lattice, because the body-centre ion differs from the corner ions. Brain-Teaser 1: The radii of Na+\mathrm{Na^{+}} and Cl−\mathrm{Cl^{-}} are 9595 and 181 pm181\,\mathrm{pm}. Predict the coordination number of Na+\mathrm{Na^{+}} and estimate the edge length of the NaCl unit cell. Solution: 95181=0.525\frac{95}{181} = 0.525, in the range 0.4140.414 to 0.7320.732: CN 66 (octahedral). Edge a=2(95+181)=552 pma = 2(95 + 181) = 552\,\mathrm{pm} (the measured value is 564 pm564\,\mathrm{pm}; ionic radii are approximate). Brain-Teaser 2: KCl has the NaCl structure with a=629 pma = 629\,\mathrm{pm}. Find its density (M=74.5 g mol−1M = 74.5\,\mathrm{g\,mol^{-1}}). Solution: z=4z = 4. a3=(6.29×10−8)3=2.49×10−22 cm3a^{3} = (6.29 \times 10^{-8})^{3} = 2.49 \times 10^{-22}\,\mathrm{cm^{3}}. d=4×74.52.49×10−22×6.022×1023=298149.9=1.99 g cm−3d = \frac{4 \times 74.5}{2.49 \times 10^{-22} \times 6.022 \times 10^{23}} = \frac{298}{149.9} = 1.99\,\mathrm{g\,cm^{-3}}. Brain-Teaser 3: In the CsCl structure, a=412 pma = 412\,\mathrm{pm}. If the radius of Cl−\mathrm{Cl^{-}} is 181 pm181\,\mathrm{pm}, find the radius of Cs+\mathrm{Cs^{+}}. Solution: Ions touch along the body diagonal: 3a=2(r++r−)\sqrt{3}a = 2(r_{+} + r_{-}), so r++r−=1.732×4122=356.8 pmr_{+} + r_{-} = \frac{1.732 \times 412}{2} = 356.8\,\mathrm{pm}, giving r(Cs+)=356.8−181=176 pmr(\mathrm{Cs^{+}}) = 356.8 - 181 = 176\,\mathrm{pm} (radius ratio 0.970.97, consistent with CN 88).

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Structures of Ionic Solids: Radius Ratio Rule, NaCl, CsCl, ZnS, CaF2 and Na2O — DPP

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