Problem Set 9_Sol

Problem Set 9_Sol - Answer Key Side 1 of 8 Problem One...

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Answer Key Ch 1a, Problem Set Nine Side 1 of 8 Problem One – Oxidation Number (20 points, 5 each) Determine the oxidation number of the metal in each of the following complexes: a. [Pt(Cl) 6 ] 2- Pt(IV) b. [Pt(Cl) 4 ] 2- Pt(IV) c. Mn(CO) 5 Cl, Mn(I) d. [Rh(SO 4 )(NH 3 ) 5 ] 1+ , Rh(III) Problem One – Packing Efficiencies, 40 Points, 10 Each a. Case one – Rock salt structure (FCC) Here ions treated as hard spheres “touch” each other along the edges of the unit cell. That means that two radii of the corner atom (say, the anion) and one diameter of the cation make up this edge. Therefore the side of this cube has 2 r a + 2 r c and because it’s FCC, there are 4 of each ion in the cell. Therefore, the packing efficiency is thus: Packing FCC = 4 4 3 " r a 3 # $ % ( + 4 4 3 r c 3 # $ % ( 2 r a + 2 r c ( ) 3 = 2 3 r a 3 + r c 3 ( ) r a + r c ( ) 3 Case two – Cesium chloride structure (BCC) In a BCC structure, ions touch along the inner diagonal of the unit cell. From simple geometry, a cube of side x has an inner diagonal of 3 x , or if it has an inner diagonal of x , it has a side of length x / 3 . Therefore, the length of a side is 2 r a + 2 r c ( ) / 3 , and because it’s BCC, there is only 1 of each ion in the cell… Packing BCC = 4 3 r a 3 + 4 3 r c 3 2 r a + 2 r c ( ) / 3 [ ] 3 = 3 2 r a 3 + r c 3 ( ) r a + r c ( ) 3 Case three – Zinc blende structure (ZnS) This is trickier, but still possible. In zinc blende half of the tetrahedral holes are filled. Therefore it is easier to focus on one of the tetrahedral hole regions, or a single octant of the cube. In one octant, a cation occupies every other corner of the cube, and the anion is in the center. Therefore the inner diagonal of this octant is 2 r a + 2 r c and the side of this octant is 2 r a + 2 r c ( ) / 3 . Again, recall that we must consider the entire unit cell, whose sides are twice the length of the octant’s sides. Therefore, a side of the unit cell is 2 2 r a + 2 r c ( ) / 3 long, and there are 4 of each ions inside…
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Answer Key Ch 1a, Problem Set Nine Side 2 of 8 Packing ZincBlende = 4 4 3 " r a 3 + 4 3 r c 3 # $ % ( 2 2 r a + 2 r c ( ) / 3 [ ] 3 = 3 4 r a 3 + r c 3 ( ) r a + r c ( ) 3 As you can see, all three have a similar format, and only differ by a constant. Clearly this constant is important or there would not be a need for these different structures.
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Answer Key Ch 1a, Problem Set Nine Side 3 of 8 Problem one continued…
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This note was uploaded on 03/30/2010 for the course CH 1a taught by Professor Lewis during the Fall '08 term at Caltech.

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Problem Set 9_Sol - Answer Key Side 1 of 8 Problem One...

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