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Concept 02 / 07Solid State

Packing Efficiency, Density of Unit Cells and Radius Relations

Relation between edge length and atomic radius for simple cubic, body-centred and face-centred cubic lattices, calculation of packing efficiency (52.4%, 68%, 74%), the density formula d = zM/(a^3 NA), and its use to find atomic mass, Avogadro's number, the type of cubic lattice or atomic radius.

Figure 1
Relation between edge aa and radius rr (see figure). Atoms touch along the line where they are closest. Simple cubic: along the edge, a=2ra = 2r. Body-centred cubic: along the body diagonal, 3a=4r\sqrt{3}a = 4r, so r=34ar = \frac{\sqrt{3}}{4}a. Face-centred cubic: along the face diagonal, 2a=4r\sqrt{2}a = 4r, so r=a22r = \frac{a}{2\sqrt{2}}. Nearest-neighbour distance =2r= 2r: aa (sc), 32a\frac{\sqrt{3}}{2}a (bcc), a2\frac{a}{\sqrt{2}} (fcc). Packing efficiency =z×43πr3a3×100%= \frac{z \times \frac{4}{3}\pi r^{3}}{a^{3}} \times 100\%. Simple cubic: z=1z = 1, a=2ra = 2r: 43πr38r3=π6=52.4%\frac{\frac{4}{3}\pi r^{3}}{8r^{3}} = \frac{\pi}{6} = 52.4\%. bcc: z=2z = 2, a=4r3a = \frac{4r}{\sqrt{3}}: 3π8=68%\frac{\sqrt{3}\pi}{8} = 68\%. fcc (ccp) and hcp: z=4z = 4, a=22ra = 2\sqrt{2}r: π32=74%\frac{\pi}{3\sqrt{2}} = 74\%. Coordination numbers: sc 66, bcc 88, fcc and hcp 1212. Density of a unit cell. Mass of the unit cell =z×MNA= z \times \frac{M}{N_{A}}, volume =a3= a^{3}: d=zMa3NAd = \frac{zM}{a^{3}N_{A}}. Take care with units: with aa in pm, a3a^{3} in cm3\mathrm{cm^{3}} is a3×10−30a^{3} \times 10^{-30}. This formula can be rearranged to find MM, NAN_{A}, aa or zz (and hence the type of lattice) when the others are known. The density of the unit cell equals the density of the substance (for a perfect crystal). Key Takeaways: - sc a=2ra = 2r; bcc 3a=4r\sqrt{3}a = 4r; fcc 2a=4r\sqrt{2}a = 4r. - Packing efficiency: sc 52.4%52.4\%, bcc 68%68\%, fcc and hcp 74%74\%. - d=zMa3NAd = \frac{zM}{a^{3}N_{A}} with z=1z = 1, 22, 44. - Pitfall: convert pm to cm (1 pm=10−10 cm1\,\mathrm{pm} = 10^{-10}\,\mathrm{cm}), so a3a^{3} carries a factor 10−3010^{-30}. Brain-Teaser 1: Copper crystallises in an fcc lattice with edge length 361 pm361\,\mathrm{pm}. Find its density (M=63.5 g mol−1M = 63.5\,\mathrm{g\,mol^{-1}}) and its atomic radius. Solution: a3=(3.61×10−8)3=4.70×10−23 cm3a^{3} = (3.61 \times 10^{-8})^{3} = 4.70 \times 10^{-23}\,\mathrm{cm^{3}}. d=4×63.54.70×10−23×6.022×1023=25428.3=8.97 g cm−3d = \frac{4 \times 63.5}{4.70 \times 10^{-23} \times 6.022 \times 10^{23}} = \frac{254}{28.3} = 8.97\,\mathrm{g\,cm^{-3}} (measured 8.968.96). r=36122=127.6 pmr = \frac{361}{2\sqrt{2}} = 127.6\,\mathrm{pm}. Brain-Teaser 2: An element with density 7.2 g cm−37.2\,\mathrm{g\,cm^{-3}} forms a bcc lattice with edge 288 pm288\,\mathrm{pm}. How many atoms are in 208 g208\,\mathrm{g} of it? Solution: Volume of 208 g208\,\mathrm{g} =2087.2=28.9 cm3= \frac{208}{7.2} = 28.9\,\mathrm{cm^{3}}. Volume of a unit cell =(2.88×10−8)3=2.39×10−23 cm3= (2.88 \times 10^{-8})^{3} = 2.39 \times 10^{-23}\,\mathrm{cm^{3}}. Number of cells =28.92.39×10−23=1.21×1024= \frac{28.9}{2.39 \times 10^{-23}} = 1.21 \times 10^{24}; atoms =2×1.21×1024=2.42×1024= 2 \times 1.21 \times 10^{24} = 2.42 \times 10^{24}. Brain-Teaser 3: A metal has atomic radius 143 pm143\,\mathrm{pm} and forms an fcc lattice. Find the edge length and the fraction of empty space. Solution: a=22r=2.828×143=404 pma = 2\sqrt{2}r = 2.828 \times 143 = 404\,\mathrm{pm} (aluminium). Empty space =100−74=26%= 100 - 74 = 26\%.

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