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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.

Two images are shown. The first image, labeled “Hexagonal closest packed,” shows seven green spheres arranged in a circular sheet lying atop another sheet that is the same except the spheres are purple. The second sheet is offset just a bit so that the spheres of the top sheet lie in the grooves of the second sheet. Two more alternating green and purple layers of spheres lie below the first pair. The second image shows seven blue spheres, labeled “Layer C,” arranged in a circular sheet laying atop another sheet, labeled “Layer B” that is the same except the spheres are green. The second sheet is offset just a bit so that the spheres of the top sheet lie in the grooves of the second sheet. Two more alternating purple and then blue layers of spheres lie below the first pair. The purple layer is labeled “Layer A” and the phrase written below this image reads “Cubic closest packed.”
In both types of closest packing, atoms are packed as compactly as possible. Hexagonal closest packing consists of two alternating layers (ABABAB…). Cubic closest packing consists of three alternating layers (ABCABCABC…).

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

An image of a cube is shown. A one-eighth section of a sphere is shown inside each corner of the cube and a full sphere is shown in the middle of the cube. A line stretched across the front face of the cube from the top left corner to the bottom right corner is labeled “4 r.” The left and bottom sides of the cube are labeled “a.”

(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 ( 558.8 pm ) 2 + ( 558.5 pm ) 2 = ( 4 r ) 2

Solving this gives r = ( 558.8 pm ) 2 + ( 558.5 pm ) 2 16 = 197.6 pmg for a Ca radius .

(b) Density is given by density = mass 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 ( 8 × 1 8 = 1 atom) and one-half of an atom on each of the six faces 6 × 1 2 = 3 atoms), for a total of four atoms in the unit cell.

The mass of the unit cell can be found by:

1 Ca unit cell × 4 Ca atoms 1 Ca unit cell × 1 mol Ca 6.022 × 10 23 Ca atoms × 40.078 g 1 mol Ca = 2.662 × 10 −22 g

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

V = a 3 = ( 558.8 × 10 −10 cm ) 3 = 1.745 × 10 −22 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 Po = 2.662 × 10 −22 g 1.745 × 10 −22 cm 3 = 1.53 g/cm 3

Check your learning

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.

Answer:

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

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Source:  OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9
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