MSE 331 Midterm Examination –I(2018) Key .pdf

10 points each x 2 20 points total direction ca 132 1

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(10 points each x 2 = 20 points total) Direction CA = (1/3)[2 -1 -1 0]+[0 0 0 -1/2] = [2/3 -1/3 -1/3 -1/2] The Miller index must be a set of integers, so multiplying by 6, you get [ 4 -2 -2 -3] Check sum of the first three digit must be zero: 4-2 -2 = 0 Alternative approach: C=(0 0 1/2), A=(1 0 0). (-1 (010) (100) (-100) (001) (01 (011) (-101) (101) (11 (111) (-111) (0-11) (110) (1-11) (-1-11) (-110) a 1 a 2 a 3 c 1/2 2 A B C
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Therefore, direction CA = [1 0 0] – [0 0 ½] = [1 0 -1/2] and convert it to 4 coordinates, you get [4 -2 -2 -3] [4]. Ionic radii of Ni +2 and O -2 are 0.069nm and 0.140nm, respectively. What crystal structure do you predict for NiO? (10 points total) NiO is an A-X type compound. Next determine the cation-anion radius ratio. r(Ni +2 )/r(O -2 ) = 0.069/0.140=0.493 This value lies between 0.414 and 0.732, and, therefore, from the Table, the coordination number for the both ions is 6. The predicted crystal structure is “rock-salt” or “NaCl” type. 4 r cation r anion Coord # < .155 .155-.225 .225-.414 .414-.732 .732-1.0 ZnS (zincblende) NaCl (sodium chloride) CsCl (cesium chloride) 2 3 4 6 8 Coordination # increases with r cation r anion How many anions can you arrange around a cation?
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