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Problem set 4 Crib 2011

Problem set 4 Crib 2011 - CHM 241 Spring 2011 Problem Set 4...

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Unformatted text preview: CHM 241 Spring 2011 Problem Set 4 Due February 28 Answer Key Exercises 3.4) The unit cell in Figure 3.20a contains 2 atoms of A and four tetrahedral holes. If X occupies of the latter, the formula is XA2. 3.7) (2 Na atoms)(1 mol/6.022 x 1023 atoms)(22.99 g/mol)(1 m3/970 kg)(kg/1000 g)= 7.871 x 10-29 m3 but that's ao3, so... ao=(7.871 x 10-29)1/3 =428 pm 3.11) CsCl was done in class so NaCl is done here. There are 12 second nearest neighbor Na+. Looking at Fig. 3.30a we see that there are 4 second neighbors in the same layer of any Na+, 4 in the layer below, and 4 in the layer above. (3)(4)=12 second nearest neighbors. 3.13c) BeO Be and O CN=4 ratio Be/O = 27/138=0.195 and adopts a wurtzite structure. 3.18) In general a greater size difference favors solubility. With this in mind the following are more soluble: (a)MgSO4 (b)NaBF4. 3.19) The simplified Born-Mayer proportion on pg. 89 can be used. The lattice energy is directly proportional to the absolute product of the charges and inversely proportional to the distance between them. So the answer in increasing lattice energy is RbCl<LiF<CaO<NiO Problems 3.5) Re in blue and O in red. A) Re CN=6, O=2 B) Perovskite structure. 3.7) In order to do this and get a unit cell that repeats when projected in all directions you need the CaF2 unit cell. A diagram of what this looks like is Figure 3.38a in the book. 3.8a) Employing the same strategy as before in a CCP array with all Oh holes filled you have MX as in NaCl. Filling only of these with cations you get M1/2X or making them integers MX2. The lattice is the CaF2 lattice. 3.14a) We must make some assumptions here. We use the Madelung constant for fluorite=2.519. For the K2+ radius we use 8 coordinate Ca2+ because it is the closest to it=112 pm. Putting these into equation 3.2 we get a lattice energy of 2471 kJ/mol. Knowing this value and looking up the ionization and dissociation energies from the reference sections in the back of the book we can treat this exactly as example 3.11 is treated. The final equation is as follows: Hf (kJ/mol)= (89 + 418.75 + 3050.86) + (157.98 655.901) + (-2471)= +589.7 kJ/mol. This is endothermic and would need more energy put in so not readily made due to the high 2nd ionization energy of K. Show in block form too. See figure. Questions I) CuBr crystallizes in the zinc-blend lattice so Cu+ are blue and Br- are yellow. Counting the number of spheres in the cell. Cu+=(8 x 1/8)+(6 x )=4 Br-=4 Both Cu+ and Br- are tetrahedral. So CN=4 for Cu+ and Br-. Shortest Cu+-Cu+ bond distance is calculated by using the density provided, 4.98 g/cm3. First, we need to find the edge length, ao. (4 CuBr atoms)(1 mol/6.022 x 1023 atoms)(143.45 g/mol)(1 cm3/4.98 g)= 1.9 x 10-22 cm3 but that's ao3, so... ao=(1.9 x 10-22)1/3 = 576 pm Cu+-Cu+ = Br--Br- = ao/sqrt(2) = 407 pm Cu+-Br- = (ao x sqrt(3))/4 = 249 pm II) III) A. CH4 (g) + H2O (g) -> 3H2 (g) + CO (g) B. Li (s) + Al (s) + 2H2 (g) -> LiAlH4 (s) C. 3NaBH4 (s) + BF3 (s) -> 2B2H6 (g) + 3NaF (s) D. As shown in class: HF is significantly different because of hydrogen bonding. Van der waal interactions explain the increase in boiling point as you go down the group, less electronegative. ...
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