Chemistry - A Molecular Science

Chapter 8 Solid Materials

170

Figure 8.14 NaCl structure

(a)

(b)

Figure 8.15 CsCl unit cells (a) CsCl in the NaCl structure.

(b) CsCl in the CsCl structure.

rrClCl 2r2rLiLi rrClCl

fd

Example 8.7 The void space between chloride ions is highlighted in red. fd is the face diagonal of the LiCl unit cell. The Li

1+ ion touches the Cl

1-^

along the edge, which forces the Cl

1- ions apart.

chloride anions are moved even farther apar

t to accommodate the larger sodium ions.

While this expansion of the packing in the un

it cell lowers the packing efficiency, there is

little energy cost because each ion is still in di

rect contact with six ions of opposite charge.

The major change in this expanded structur

e is that the like charges (anion-anion and

cation-cation) are further separated,

which is energetically favorable.

The radius of a cesium ion is slightly greater than that of a chloride anion, so the fcc

lattice is expanded considerably to introduce

a great deal of void space. The resulting low

packing efficiency is essentially the same as in

a simple cube. Indeed, the similarity of the

ionic radii results in a unit cell that has the a

ppearance of a simple cube in which four of

the corners are occupied by cations and four

are occupied by anions, as shown in Figure

8.15a. Under special conditions, CsCl can be made

to crystallize with this sodium chloride

structure type, but under normal

conditions, the packing efficiency of these nearly equally

sized ions is optimized in the arrangement,

as shown in Figure 8.15b. This structure,

known as the

cesium chloride structure

, can be described as having chloride anions

forming a simple cube with a cesium cati

on in the body center. The unit cell has the

packing efficiency of a body-centered cubic unit cell, and it can be described as a simple cube of cesium cations with a chloride anion in the body center. Example 8.7

Use ionic radii to determine the void space between chloride ions in LiCl. The ionic radii from Table 8.3 are r

= 0.90 Å and rLi

= 1.67 Å. LiCl

1+ and Cl

1- ions touch

along the edge of the unit cell, so the

length of the unit cell edge (a) is two Li

1+ ionic radii

plus two Cl

1- ionic radii (see figure in margin):

a = 2r

+ 2rLi

= 2(0.90) + 2(1.67) = 5.14 Å Cl

The length of the face diagonal of LiCl

is obtained from the edge length (a) and the

Pythagorean theorem to be

222 2fd = a + a = 2a

fd = 2a = 2 (5.14) = 7.27 Å

⇒

Four chloride ionic radii is 4r

= 4(1.67) = 6.68 Å, so the acCl

tual face diagonal is 7.27 -

6.68 = 0.59 Å longer than the length of the chloride ions that lie on the diagonal. This difference is twice the void space between Cl

1- ions (red bars in figure). The void space

between chloride ions in LiCl is ~0.59/2 or ~0.3 Å.

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