ts1201f05 - EMSE 201 Introduction to Ma teria ls Science &...

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EMSE 201 — Introduction to Materials Science & Engineering 28 September 2005 Name: SOLUTION Department of Materials Science and Engineering 1 of 6 Case Western Reserve University Test #1 — 75 minutes; 150 points; 6 questions; 6 pages; 15% of course grade No calculators or formula sheets are allowed. Where numerical answers are requested, full credit will be given for correctly setting up the calculation and for specifying the correct units . Use the backs of these sheets if necessary. 1) a) (12 points) Briefly state the rules for solid solution formation, as they would be applied to oxides. 3 pts each: For complete solid solution, … The two oxides must have the same crystal structure. The cations must have similar EN (within 0.6). The cationic radii must differ from each other by <15%. The cations must have at least one valence state in common. b) (8 points) Identify which (if any) pair(s) of the oxides listed below would be expected to form complete solid solution(s) with each other. Justify your answer. α -Al 2 O 3 Cr 2 O 3 Y 2 O 3 In 2 O 3 cation radius, nm 0.0535 0.0615 0.0900 0.0800 cation electronegativity 1.61 1.66 1.22 1.78 structure corundum corundum cubic M 2 O 3 cubic M 2 O 3 1-2 pts for each of the following statements ( 6 pts max): All rules must be satisfied to form a complete solid solution. Only α -Al 2 O 3 -Cr 2 O 3 and Y 2 O 3 -In 2 O 3 have the same structure only these two pairs could be complete solid solutions. All the cations are trivalent All of the pairs satisfy the valence rule. Only Al-Cr and Y-Cr satisfy the radius rule All pairs satisfy the EN rule. Predict α -Al 2 O 3 -Cr 2 O 3 and Y 2 O 3 -In 2 O 3 to be complete solid solutions. (They are, too.) ( 1 pt for each pair)
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EMSE 201 — Introduction to Materials Science & Engineering 28 September 2005 Department of Materials Science and Engineering 2 of 6 Case Western Reserve University 2) a) (5 points) In the corundum structure (mentioned in problem 1), oxygen ions are in a nearly-HCP arrangement, with cations in octahedral interstices. What fraction of the octahedral interstices is filled? Justify your answer. CP arrangement of oxygens one octahedral interstice per oxygen ( 3 pts ) three octahedral interstices per formula unit Cation:anion ratio of 2:3 and Al in octahedral coordination two octahedral interstices are filled per formula unit, i.e. two-thirds of the octahedral interstices are filled ( 2 pts ). b) (5 points) In the cubic M 2 O 3 structure (mentioned in problem 1), some of the cations are in 6- fold coordination and some are in 7-fold coordination. What characteristic of Y 2 O 3 and In 2 O 3 accounts for this larger coordination number of the cation, compared to that of the cations in the corundum structure? Larger cations permit larger coordination numbers (
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ts1201f05 - EMSE 201 Introduction to Ma teria ls Science &...

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