Homework 2 Solution

Homework 2 Solution - MEEN 222-502 Fall 2008 Homework #2...

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MEEN 222-502 Fall 2008 Homework #2 Due: 3 October 2008 at the beginning of class 1. Callister 12.5 (a) For CaO, using data from Table 12.3 714 . 0 nm 140 . 0 nm 100 . 0 2 2 O Ca = = + r r Now, from Table 12.2, the coordination number for each cation (Ca 2+ ) is six, and, using Table 12.4, the predicted crystal structure is sodium chloride. (b) For MnS, using data from Table 12.3 364 . 0 nm 184 . 0 nm 067 . 0 2 2 S Mn = = + r r The coordination number is four (Table 12.2), and the predicted crystal structure is zinc blende (Table 12.4). (c) For KBr, using data from Table 12.3 704 . 0 nm 196 . 0 nm 138 . 0 Br K = = + r r The coordination number is six (Table 12.2), and the predicted crystal structure is sodium chloride (Table 12.4). (d) For CsBr, using data from Table 12.3 867 . 0 nm 196 . 0 nm 170 . 0 Br Cs = = + r r The coordination number is eight (Table 12.2), and the predicted crystal structure is cesium chloride (Table 12.4).
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MEEN 222-502 Fall 2008 2. Callister 12.6 First of all, the possibilities would include only the monovalent cations Cs + , K + , and Na + . Furthermore, the coordination number for each cation must be 8, which means that 0.732 < r C / r A < 1.0 (Table 12.2). From Table 12.3 the r C / r A ratios for these three cations and the F - ion are as follows: 77 . 0 nm 133 . 0 nm 102 . 0 04 . 1 nm 133 . 0 nm 138 . 0 28 . 1 nm 133 . 0 nm 170 . 0 F Na F K F Cs = = = = = = + + + r r r r r r Thus, only sodium will form the CsCl crystal structure with fluorine. 3. Callister 12.14 (a) This part of the problem calls for us to determine the unit cell edge length for FeO. The density of FeO is 5.70 g/cm 3 and the crystal structure is rock salt. From Equation 12.1 A A 3 O Fe O Fe ) ( ) ( N a A A n
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Homework 2 Solution - MEEN 222-502 Fall 2008 Homework #2...

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