Ciencia dos materiais

12.5 On the basis of ionic charge and ionic radii given in Table 12.3, predict crystal structures for the following materials: (a) CsI, (b) NiO, (c) KI, and (d) NiS. Justify your selections. Solution This problem calls for us to predict crystal structures for several ceramic materials on the basis of ionic charge and ionic radii. (a) For CsI, using data from Table 12.3

rCs + rI−

=

0.170 nm = 0.773 0.220 nm

Now, from Table 12.2, the coordination number for each cation (Cs+) is eight, and, using Table 12.4, the predicted crystal structure is cesium chloride. (b) For NiO, using data from Table 12.3

rNi2+ rO 2−

=

0.069 nm = 0.493 0.140 nm

The coordination number is six (Table 12.2), and the predicted crystal structure is sodium chloride (Table 12.4). (c) For KI, using data from Table 12.3

rK + rI−

=

0.138 nm = 0.627 0.220 nm

The coordination number is six (Table 12.2), and the predicted crystal structure is sodium chloride (Table 12.4). (d) For NiS, using data from Table 12.3

rNi2+ rS 2−

=

0.069 nm = 0.375 0.184 nm

The coordination number is four (Table 12.2), and the predicted crystal structure is zinc blende (Table 12.4).
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12.13 Calculate the density of FeO, given that it has the rock salt crystal structure. Solution We are asked to calculate the theoretical density of FeO. This density may be computed using Equation (12.1) as

ρ =

n′( AFe + AO ) VC N A

Since the crystal structure is rock salt, n' = 4 formula units per unit cell. Using the ionic radii for Fe2+ and O2- from Table 12.3, the unit cell volume