Q7(c)(i). Packing fractions of SC, BCC and FCC

Method

• The atoms are taken as equal hard spheres of radius r. The packing fraction f is the fraction of the volume of the unit cell filled by the spheres: f=(volume of the spheres in one cell)/(volume of the cell)=Z×4/3π r³a³
• Z, atoms per cell: an atom at a corner is shared by 8 cells (counts 18), one at a face centre by 2 cells (12), one at the body centre by none (1).
• a in terms of r: find the direction in which neighbouring spheres touch. Along that line the length in the cell equals a whole number of radii. Pythagoras gives the face diagonal a²+a²=√2 a and the body diagonal a²+2a²=√3 a.

1. Simple cubic (SC)

1. Atoms only at the 8 corners: Z=8×18=1
2. Neighbouring corner atoms touch along the cube edge. The edge holds one radius of each: a=r+r=2r
3. Cell volume a³=8r³; volume of the spheres 1×43π r³: 3pt] f_(SC)=4/3π r³8r³=(π)/(6)=0.524 (52.4 %)
4. Each atom touches 6 others (CN 6). The cube centre is a large empty hole; 47.6 % of the space is empty.

2. Body-centred cubic (BCC)

1. 8 corner atoms and one at the body centre: Z=8×18+1=2
2. The corner atoms do not touch one another. Each touches the atom at the body centre, so the contact line is the body diagonal. It holds one radius of each of two corner atoms and the diameter of the centre atom: 3pt] √3 a=r+2r+r=4r, a=(4r)/(√3)=2.309 r
3. Cell volume a³=64r³3√3; volume of the spheres 2×43π r³=83π r³: 3pt] f_(BCC)=8/3π r³64r³/3√3=(√3 π)/(8)=0.680 (68.0 %)
4. Each atom touches 8 others (CN 8). Six more lie at a=1.15×2r, only 15 % farther. Empty space 32.0 %.

3. Face-centred cubic (FCC)

1. 8 corner atoms and 6 face-centre atoms: Z=8×18+6×12=1+3=4
2. Each corner atom touches the atom at the centre of the face, so the contact line is the face diagonal. It holds one radius of each of two corner atoms and the diameter of the face atom: 3pt] √2 a=4r, a=(4r)/(√2)=2√2 r=2.828 r
3. Cell volume a³=16√2 r³; volume of the spheres 4×43π r³=16/3π r³: 3pt] f_(FCC)=16/3π r³16√2 r³=(π)/(3√2)=0.740 (74.0 %)
4. Each atom touches 12 others (CN 12): 4 in its own layer and 4 above and 4 below. Empty space 26.0 %.

Comparison

Conclusion: FCC is the most economically packed (74.0 %) and SC the least (52.4 %). BCC lies between them (68.0 %).

ORDERFCC 74.0 % > BCC 68.0 % > SC 52.4 %FOLLOWScoordination number 12 > 8 > 6

• r cancels in every ratio, so the result holds for spheres of any size, that is for any metal.
• The packing fraction rises with the coordination number. In SC the spheres touch only along the edges and leave a large hole at the centre of the cube. BCC places one sphere in that hole. FCC is a stacking of close-packed layers (ABC…), in which every sphere sits in a hollow of the layer below.
• 74.0 % is the largest fraction possible for equal spheres. Hexagonal close packing (ABAB…) has the same CN 12 and the same 74.0 %. SC wastes almost half the space, which is why polonium is the only element with this structure.

Application. the packing fraction decides the volume per atom, V=43π r³/f. When iron changes from b.c.c. to f.c.c. on heating, with the same radius the volume would fall in the ratio f_(BCC)/f_(FCC)=0.680/0.740=0.919, a contraction of 8.1 %. The densities of metals follow from ρ=ZM/N_Aa³.