The Foundations of Chemistry

In certain types of solids, including ionic crystals, particles differentfrom those at the corners
of the unit cell occupy extra positions within the unit cell. For example, the face-centered cubic
unit cell of sodium chloride can be visualized as having chloride ions at the corners and middles
of the faces; sodium ions are on the edges between the chloride ions and in the center (see
Figure 13-28). Thus, a unit cell of NaCl contains the following.

eight at six in middles

Cl: 
 1 
 3 4 Clions/unit cell

corners of faces

8( 8 ^1 ) 
 6(^12 )

twelve on

Na
: 
one in center 3 
 1 4 Na
ions/unit cell

edges

12(^14 ) 
 1

The unit cell contains equal numbers of Na
and Clions, as required by its chemical
formula. Alternatively, we could translate the unit cell by half its length in any axial direc-
tion within the lattice, and visualize the unit cell in which sodium and chloride ions
have exchanged positions. Such an exchange is not always possible. You should confirm
that this alternative description also gives four chloride ions and four sodium ions per
unit cell.
Ionic radii such as those in Figure 6-1 and Table 14-1 are obtained from X-ray crys-
tallographic determinations of unit cell dimensions, assuming that adjacent ions are in
contact with each other.

EXAMPLE 13-11 Ionic Radii from Crystal Data

Lithium bromide, LiBr, crystallizes in the NaCl face-centered cubic structure with a unit cell
edge length of abc5.501 Å. Assume that the Brions at the corners of the unit cell
are in contact with those at the centers of the faces. Determine the ionic radius of the Brion.
One face of the unit cell is depicted in Figure 13-30.

Plan

We may visualize the face as consisting of two right isosceles triangles sharing a common
hypotenuse, h, and having sides of length a5.501 Å. The hypotenuse is equal to four times
the radius of the bromide ion, h 4 rBr.

Solution

The hypotenuse can be calculated from the Pythagorean theorem, h^2 a^2
a^2. The length
of the hypotenuse equals the square root of the sum of the squares of the sides.

ha^2 
a^2  2 a^2  2 (5.5 0  1 Å)^2 7.780 Å

The radius of the bromide ion is one fourth of h, so

rBr1.945 Å

EXAMPLE 13-12 Ionic Radii from Crystal Data

Refer to Example 13-11. Calculate the ionic radius of Li
in LiBr, assuming anion–cation
contact along an edge of the unit cell.

7.780 Å



4

Figure 13-30 One face of the face-

centered cubic unit cell of lithium

bromide (Example 13-11).

13-16 Bonding in Solids 525

5.501 Å

Br

Li


5.501 Å

h

a

a