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Concept 04 / 07Solid State

Close Packing in Solids: hcp, ccp, Tetrahedral and Octahedral Voids

Close packing of identical spheres in one, two and three dimensions, square and hexagonal layers, ABAB (hexagonal close packing) and ABCABC (cubic close packing) arrangements, coordination numbers, tetrahedral and octahedral voids with their numbers and limiting radius ratios, and deducing formulas of compounds from void occupancy.

Figure 1
Particles in crystals tend to pack as closely as possible. Treat them as identical hard spheres. One and two dimensions. In a row each sphere touches 22 neighbours. Rows can be stacked into a square layer (each sphere has 44 neighbours, AAA arrangement, leading to simple cubic packing) or a hexagonal layer (each sphere has 66 neighbours, with rows offset, ABAB in 2D), which is more efficient. Three dimensions from hexagonal layers. A second layer B placed in the hollows of layer A covers half of the hollows. Where a B sphere sits over an A hollow, the hollow is surrounded by 44 spheres at the corners of a tetrahedron: a tetrahedral void. Where a hollow of B lies over a hollow of A, the gap is surrounded by 66 spheres at the corners of an octahedron: an octahedral void (see figure). Third layer: over the tetrahedral voids, directly above layer A, giving ABAB...: hexagonal close packing (hcp) (Mg, Zn, Be, Co); over the octahedral voids, in a new position C, giving ABCABC...: cubic close packing (ccp), which is identical to the face-centred cubic lattice (Cu, Ag, Au, Al, Ni). In both, each sphere touches 1212 others (coordination number 1212) and 74%74\% of space is filled. Number of voids. If a close-packed structure contains NN spheres, there are NN octahedral voids and 2N2N tetrahedral voids. In an fcc unit cell (44 atoms), there are 44 octahedral voids (at the body centre and the 1212 edge centres: 1+12×14=41 + 12 \times \frac{1}{4} = 4) and 88 tetrahedral voids (one in each of the eight small cubes, on the body diagonals at one quarter of their length from each corner). Size of voids. The largest sphere of radius rr that fits in a void between spheres of radius RR: tetrahedral, rR=0.225\frac{r}{R} = 0.225; octahedral, rR=0.414\frac{r}{R} = 0.414; cubic (at the centre of a simple cube), rR=0.732\frac{r}{R} = 0.732. An octahedral void is larger than a tetrahedral one. Formulas from void occupancy. In ionic compounds, the larger ions (usually anions) form the close-packed array and the smaller ions occupy voids. Count: close-packed ions NN, octahedral NN, tetrahedral 2N2N, then apply the fraction occupied. Key Takeaways: - hcp: ABAB; ccp = fcc: ABCABC; both CN 1212, 74%74\% filled. - NN spheres give NN octahedral and 2N2N tetrahedral voids. - Void radius ratio: tetrahedral 0.2250.225, octahedral 0.4140.414, cubic 0.7320.732. - In fcc: octahedral voids at the body and edge centres; tetrahedral voids on body diagonals. - Pitfall: there are twice as many tetrahedral voids as atoms, but they are smaller than the octahedral voids. Brain-Teaser 1: Atoms of element B form ccp, and atoms of A occupy two-thirds of the tetrahedral voids. What is the formula? Solution: If B =N= N, tetrahedral voids =2N= 2N, and A =23×2N=4N3= \frac{2}{3} \times 2N = \frac{4N}{3}. A : B =43:1=4:3= \frac{4}{3} : 1 = 4 : 3: A4B3\mathrm{A_{4}B_{3}}. Brain-Teaser 2: In a compound, oxide ions form ccp and aluminium ions occupy two-thirds of the octahedral voids. Find the formula. Solution: O2−=N\mathrm{O^{2-}} = N, octahedral voids NN, Al3+=2N3\mathrm{Al^{3+}} = \frac{2N}{3}. Al : O =23:1=2:3= \frac{2}{3} : 1 = 2 : 3: Al2O3\mathrm{Al_{2}O_{3}} (corundum); the charges also balance: 2(+3)+3(−2)=02(+3) + 3(-2) = 0. Brain-Teaser 3: Derive the limiting radius ratio for an octahedral void. Solution: In an octahedral void, four spheres of radius RR lie in a square, touching along the sides (side 2R2R), with the void sphere of radius rr at the centre touching all four. The diagonal of the square is 2R2=2R+2r2R\sqrt{2} = 2R + 2r, so r=R(2−1)r = R(\sqrt{2} - 1) and rR=0.414\frac{r}{R} = 0.414.