• coordination # increases with issue: how many anions can you arrange around a cation?

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Coordination # increases with Issue: How many anions can you arrange around a cation? r cation r anion r cation r anion Coord # < .155 .155-.225 .225-.414 .414-.732 .732-1.0 ZnS (zincblende) NaCl (sodium chloride) Cs Cl (cesium chloride) 2 3 4 6 8 Adapted from Table 12.2, Callister 6e. Adapted from Fig. 12.2, Callister 6e. Adapted from Fig. 12.3, Callister 6e. Adapted from Fig. 12.4, Callister 6e. COORDINATION # AND IONIC RADII

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COORDINATION # AND IONIC RADII. • Coordination # increases with Issue: How many anions can you arrange around a cation?. Adapted from Fig. 12.4, Callister 6e. Adapted from Fig. 12.2, Callister 6e. Adapted from Fig. 12.3, Callister 6e. Adapted from Table 12.2, Callister 6e. - PowerPoint PPT Presentation

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Page 1: •  Coordination # increases with Issue:  How many anions can you        arrange around a cation?

• Coordination # increases with Issue: How many anions can you arrange around a cation?

rcationranion

rcationranion

Coord #

< .155 .155-.225 .225-.414 .414-.732 .732-1.0

ZnS (zincblende)

NaCl (sodium chloride)

CsCl (cesium chloride)

2 3 4 6 8

Adapted from Table 12.2, Callister 6e.

Adapted from Fig. 12.2, Callister 6e.

Adapted from Fig. 12.3, Callister 6e.

Adapted from Fig. 12.4, Callister 6e.

COORDINATION # AND IONIC RADII

Page 2: •  Coordination # increases with Issue:  How many anions can you        arrange around a cation?

• On the basis of ionic radii, what crystal structure would you predict for FeO?

• What are the lattice parameter, the density and the packing factor? Cation

Al3+

Fe2+

Fe3+

Ca2+ Anion

O2-

Cl-

F-

Ionic radius (nm)

0.053

0.077

0.069

0.100

0.140

0.181

0.133

• Answer:

rcationranion

0.0770.140

0.550

based on this ratio,--coord # = 6--structure = NaCl

Data from Table 12.3, Callister 6e.

EXAMPLE: PREDICTING THE STRUCTURE OF FeO

a = 2(0.077) + 2(0.14) = 0.434 nm

= {(4x55.85) + (4x16)}/{(0.434x10-7 cm)3 (6.02x1023 atoms/mol)} = 5.84 g/cm3

P.F. = {(4/3)(4 ions/cell)(0.0773 + 0.143)}/(0.434)3 = 0.656