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Solid Structures

>> Parts of this equation/concept include:

A -	Unit Cells >> Types >> Stoichiometry >> Size and Density
B -	Close Packing >> Types >> Holes

Define the unit cell by choosing the largest type of atom. (Normally, this is the anion.) Search in one dimension until you find that atom again. Check the same distance in the same direction to see that all the atoms are the same along the second segment of the line as the first. If so, you have found one edge of the unit cell. If not, check both segments together as the potential edge of that unit cell. Return to the original atom and repeat for the other two dimensions.

>> Types

Three types of unit cells are discussed in this chapter. Each unit cell is defined by one type of atom. Consequently, whichever atom you choose when defining that unit cell is the only atom used in defining the unit cell. (Ignore all other types.)

The simple cubic will have that atom only at the corners of the unit cell. The other types will have that atom at the corners and in additional locations.

The body-centered cubic will have that atom in the center of the unit cell as well as at the corners.

The face-centered cubic will have that atom in the center of each face as well as at the corners.

There are other types of unit cells, so it is possible that your answer will be "none of the above."

>> Example 1

Describe each of the following unit cells.

Solution:

The unit cell on the left is a simple cubic. Although there is an atom in the center of the unit cell, it is not of the same type as the atoms at the corners. Therefore it does not count as part of the unit cell description. If the atom in the center was of the same type, it would be a body-centered cubic.

The unit cell on the left is a face-centered cubic. It has only one type of atom. It too has an atom in the center, but more important, there is an atom on each face. Face-centered cubics have an atom in the center as well as on the faces.

>> Stoichiometry

For ionic and network solids, the stoichiometry is the simplest ratio of particles in an atom. Since the unit cell represents the pattern of the atom, the ratio in the unit cell is also the ratio for the entire compound. The way we have drawn unit cells is with the center of the atom at the corners. Consequently, not all of the atom is in the unit cell. The part of the atom in the unit cell is counted according to the following (Table 10.1):

corner = 1/8

edge = 1/4

face = 1/2

body = 1

Once the number of atoms of each type within a unit cell is determined, the simplest ratio of those atoms can be determined.

>> Example 2

What is the stoichiometry of the following compound if black dots = X and unfilled circles = M? (M goes first in the formula.)

Solution:

In this cell there are eight black atoms, one at each corner, but no more. So the total number of X atoms in the cell = 81/8 = 1.

There is one circle in the center (body), so the number of M atoms = 11 = 1 So the ratio is 1:1 and the formula = MX.

>> Example 3

What is the stoichiometry of the following compound if black dots = X and unfilled circles = M? (M goes first in the formula.)

Solution:

This is a body-centered cubic with X on the corners and in the body, and M on each face.

X:	8 corners (1/8)	= 1
	1 body (1)	= 1
	total	= 2

M:

6 faces (1/2)

= 3

So the stoichiometry is M₃X₂ (There is no simpler ratio.)

>> Size and Density

The volume (size) of a unit cell depends on the ionic radii (size) of the ions and their arrangement (lattice structure). Figure 9.3 lists ionic radii. The length of an edge can be determined by x-ray diffraction.

To determine the edge length from ionic radii, the exact type of lattice is required. For a lattice where ions touch along the edge of a unit cell, the edge length can be determined by adding the relevant ionic radii.

To determine edge length for unit cell where the ions touch along a face, it is normally assumed that the atoms along the diagonal across the face are touching. The length of this diagonal is determined from the sum of the ions along the diagonal (c). The Pythagorean theorem

a² + b² = c²

is used to determine edge length. If this is a cubic structure, both edges are the same length (a = b) and the formula simplifies to

2a² = c²

To determine edge length for a unit cell where ions touch through the body of the cell, the Pythagorean theorem must be used twice. The sum of the ionic radii corresponds to a diagonal through the center of the unit cell (c). A right triangle can be made from a diagonal through the face (b) and an edge (a).

a² + b² = c²

The value for the diagonal across the face (b) can be determined by a right triangle along each edge

2a² = b²

Putting the two equations together, the edge length can be determined from

a² + 2a² = 3a² = c²

>> Example 4

What is the edge length of a cubic unit cell if iodide ions touch along the face of the unit cell?

Solution:

The radius of an iodide ion is 220 pm. The corner of the unit cell starts in the center of an iodide ion, so from the corner of the unit cell to the next ion is one radius. The diagonal goes completely through the ion on the face, so that is two radii. The diagonal continues to the center of the ion on the other corner, one more radius, for a total of four.

Therefore the length of the diagonal (c) = 4(220 pm) = 880 pm

Using the Pythagorean theorem,

a² + b² = c²

and it is a cubic cell, a = b, so

2a² = c²

Then

2a² = (880)²

2a² = 774,400

a² = 387,200

a = 622 pm

>> Example 5

What is the edge length of a cubic unit cell where a corner chloride ion (r = 181 pm) touches a body-centered cesium ion (r = 174 pm) that touches the opposite corner chloride ion?

Solution:

The ions are touching along a body diagonal. The body diagonal consists of two radii of chloride ions and two radii of cesium ions. Therefore its length (c) is

c = 2(181) + 2(174) = 710 pm

In the introduction, the relationship between edge length and body diagonal was derived to be

3a² = c²,

so

3a² = (710)²

3a² = 504,100

a² = 16,803.3333

a = 410 pm

To determine the volume from edge length, recall that for cubic lattices, each edge length is the same. Consequently, for cubic lattice structures, the volume is simply the edge length cubed.

Density is mass/volume. The density of the unit cell is the same as the density of the compound as a whole. Volume of the unit cell is determined from edge length. The number of particles in the unit cell is determined in the same way as the stoichiometry of the cell. (Do not simplify to the smallest ratio.) The mass can be determined from the atomic weight. (Remember, this time you are working in atoms, not moles!)

>> Example 6

The cesium chloride unit cell has a chloride ion at each corner and a cesium ion in the center. The ions touch through a body diagonal. What is the density of this solid in g/cm³? (Hint: See the previous example.)

Solution:

The edge length of this unit cell was determined in Example 2 to be 410 pm. To get the volume in cubic centimeters, it is probably more convenient to convert picometers to centimeters rather than cubic picometers to cubic centimeters, so

410 pm

10–12 m 1 pm

102 m 1 m

=

4.10 x 10–8 cm

The volume is the edge length cubed

(4.10 x 10^–8 cm)³ = 6.8921 x 10^–23 cm³ = 6.89 x 10^–21 cm³
(with correct significant figures)

In the unit cell, there are eight chlorides, one on each corner, so that is a total of 1/8, or 1 chloride ion. The cesium ion is in the center, so it counts as one atom. The mass, in grams, of one atom of chloride is

1 atom Cl

1 mol 6.022 x 1023 atoms

35.453 g 1 mol

=

5.887 x 10–23 g Cl

The four significant figures are due to the value used for Avogadro's number. 1 atom is a counted number, therefore it has infinite significant figures. The mass in grams of cesium is

1 atom Cs

1 mol 6.022 x 1023 atoms

132.9 g 1 mol

=

2.200 x 10–22 g Cs

The total mass of the unit cell is the sum of these two values

2.200 x 10^–22 g + 5.887 x 10^–23 g = 2.789 x 10^–22 g

So the density is

density

=

mass volume

=

2.789 x 10–22 g 6.8921 x 10–23 cm3

=

4.05 g/cm3

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Close packing descriptions are an alternative to unit cell descriptions. It is not obvious, but a face-centered cubic has the same structure as a cubic close-packed unit cell. A simple cubic might be the same as simple cubic packing. However, a body-centered cubic is not the same as hexagonal close packing.

The comparisons are not obvious because the layers are not oriented the same way as they are for unit cells. First, you must determine the close-packed layer, which is a layer where each row is slightly offset. Each particle ends up surrounded by six other particles.

>> Types

The type of close packing is then determined by how these layers are stacked. If one layer (which is not necessarily close-packed itself) is stacked directly on top of the other, it is simple cubic packing (aaaa...). If the layers alternate every other layer (abababab...), it is hexagonal close packing. If the layers repeat on the third row (abcabcabc...), it is cubic close packing.

>> Holes

The easiest way to determine the type of hole is by looking at how the atoms surround the ion in the hole. Hexagonal and cubic close packing have octahedral and tetrahedral holes. Ions in tetrahedral holes are surrounded by ions in a tetrahedral orientation. Ions in octahedral holes are surrounded by ions oriented octahedrally. For simple cubic packing, the holes are cubic holes. The ion is surrounded by eight other ions arranged in a cube.

>> Example 7

What kind of hole is the white ion in if it is surrounded as follows?

Solution:

The center ion is surrounded by six other ions. The ions are oriented top, bottom, left, right, front, back, which is an octahedral geometry. Therefore the ion is in an octahedral hole.

The size of the hole depends on the type and size of the atoms making the hole. Larger close-packed atoms leave larger holes. Cubic holes are larger than octahedral holes, which are larger than tetrahedral holes. However, there are more tetrahedral holes than octahedral holes. The ratio of anion and cation size can be used to predict what type of hole the ion will occupy. Table 10.2 lists the ratios appropriate for each type of hole.

>> Example 8

For the following cations and anion, predict the type of hole the smaller ion will occupy.

1. Fe²⁺ (r = 55 pm) and S^2– (r = 184 pm)
2. K⁺ (r = 151 pm) and Cl^– (r = 181 pm)
3. Cs⁺ (r = 174 pm) and F^– (r = 133 pm)

Solution:

1. Sulfide is the larger ion, so it forms the packing structure and iron(II) is in the holes. The ratio of smaller/larger = 55/184 = 0.30. That makes iron(II) small enough to fit in tetrahedral holes.
2. Chloride is the larger ion, so potassium will fit in the holes. The ratio of smaller/larger = 151/181 = 0.83. So the potassium will only fit in cubic holes.
3. Cesium is the larger ion, so the ratio of smaller/larger = 133/174 = 0.76. Fluorides are also in cubic holes.

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