# 10.6 Lattice structures in crystalline solids  (Page 4/29)

 Page 4 / 29

Because closer packing maximizes the overall attractions between atoms and minimizes the total intermolecular energy, the atoms in most metals pack in this manner. We find two types of closest packing in simple metallic crystalline structures: CCP, which we have already encountered, and hexagonal closest packing (HCP)    shown in [link] . Both consist of repeating layers of hexagonally arranged atoms. In both types, a second layer (B) is placed on the first layer (A) so that each atom in the second layer is in contact with three atoms in the first layer. The third layer is positioned in one of two ways. In HCP, atoms in the third layer are directly above atoms in the first layer (i.e., the third layer is also type A), and the stacking consists of alternating type A and type B close-packed layers (i.e., ABABAB⋯). In CCP, atoms in the third layer are not above atoms in either of the first two layers (i.e., the third layer is type C), and the stacking consists of alternating type A, type B, and type C close-packed layers (i.e., ABCABCABC⋯). About two–thirds of all metals crystallize in closest-packed arrays with coordination numbers of 12. Metals that crystallize in an HCP structure include Cd, Co, Li, Mg, Na, and Zn, and metals that crystallize in a CCP structure include Ag, Al, Ca, Cu, Ni, Pb, and Pt.

## Calculation of atomic radius and density for metals, part 2

Calcium crystallizes in a face-centered cubic structure. The edge length of its unit cell is 558.8 pm.

(a) What is the atomic radius of Ca in this structure?

(b) Calculate the density of Ca.

## Solution

(a) In an FCC structure, Ca atoms contact each other across the diagonal of the face, so the length of the diagonal is equal to four Ca atomic radii (d = 4 r ). Two adjacent edges and the diagonal of the face form a right triangle, with the length of each side equal to 558.8 pm and the length of the hypotenuse equal to four Ca atomic radii:

${a}^{2}+{a}^{2}={d}^{2}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\left(558.8\phantom{\rule{0.2em}{0ex}}\text{pm}\right)}^{2}+{\left(558.5\phantom{\rule{0.2em}{0ex}}\text{pm}\right)}^{2}={\left(4r\right)}^{2}$

Solving this gives $r\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\sqrt{\frac{{\left(558.8\phantom{\rule{0.2em}{0ex}}\text{pm}\right)}^{2}+{\left(558.5\phantom{\rule{0.2em}{0ex}}\text{pm}\right)}^{2}}{16}}\phantom{\rule{0.2em}{0ex}}=\text{197.6 pmg for a Ca radius}.$

(b) Density is given by $\text{density}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{\text{mass}}{\text{volume}}.$ The density of calcium can be found by determining the density of its unit cell: for example, the mass contained within a unit cell divided by the volume of the unit cell. A face-centered Ca unit cell has one-eighth of an atom at each of the eight corners $\left(8\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{1}{8}\phantom{\rule{0.2em}{0ex}}=1$ atom) and one-half of an atom on each of the six faces $6\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{1}{2}\phantom{\rule{0.2em}{0ex}}=3$ atoms), for a total of four atoms in the unit cell.

The mass of the unit cell can be found by:

$\text{1 Ca unit cell}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{\text{4 Ca atoms}}{\text{1 Ca unit cell}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{\text{1 mol Ca}}{6.022\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{23}\phantom{\rule{0.2em}{0ex}}\text{Ca atoms}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{40.078\phantom{\rule{0.2em}{0ex}}\text{g}}{\text{1 mol Ca}}\phantom{\rule{0.2em}{0ex}}=2.662\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-22}\phantom{\rule{0.2em}{0ex}}\text{g}$

The volume of a Ca unit cell can be found by:

$V={a}^{3}={\left(558.8\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-10}\phantom{\rule{0.2em}{0ex}}\text{cm}\right)}^{3}=1.745\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-22}\phantom{\rule{0.2em}{0ex}}{\text{cm}}^{3}$

(Note that the edge length was converted from pm to cm to get the usual volume units for density.)

Then, the density of $\text{Po}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{2.662\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-22}\phantom{\rule{0.2em}{0ex}}\text{g}}{1.745\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-22}\phantom{\rule{0.2em}{0ex}}{\text{cm}}^{3}}\phantom{\rule{0.2em}{0ex}}={\text{1.53 g/cm}}^{3}$

Silver crystallizes in an FCC structure. The edge length of its unit cell is 409 pm.

(a) What is the atomic radius of Ag in this structure?

(b) Calculate the density of Ag.

(a) 144 pm; (b) 10.5 g/cm 3

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