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CAF123
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Homework Statement
The CsCl structure is stable when the ratio of the smaller ion radius to larger ion radius is ##r > 0.73##. Why is this so? Show that the packing fraction, ##p##, is given by $$p = \frac{\sqrt{3} \pi}{2} \frac{1+r^3}{(1+r)^3}$$
volume of sphere, volume of structure
3. The Attempt at a Solution
I have obtained the correct answer for ##p## but I am not sure how to explain why ##r > 0.73## but I have some calculations. The CsCl structure has a Cl in the middle of a cube with 8 Cs around it (so like a BCC structure and the unit cell is BCC cubic). Suppose that along the edge of a cube, the two adjacent Cs ions are touching. Then ##a = 2R_1## where ##a## is the side of the cube and ##R_1## is the radius of a Cs ion. The face diagonal is then ##h = 2\sqrt{2}R_1##, from which we find the cube diagonal to be ##2\sqrt{3}R_1##. Now if we further suppose that the Cl ion is touching two Cs from either side (so close packing of spheres as much as possible) then we have ##2R_1 + 2R_2 = 2 \sqrt{3}R_1## where ##R_2## is the radius of Cl. Rearranging for ##R_2## gives ##R_2 = (\sqrt{3} - 1)R_1 \approx 0.73 R_1##. So I seemed to have extracted the correct number (0.73) but not the correct ratio. (i.e I wanted to derive ##R_1 = 0.73 R_2##), Cl is certainly bigger than Cs. So what went wrong here? Actually I have found the same calculation in this doc (bottom of first page), but nothing was said about it.
http://depts.washington.edu/chemcrs/bulkdisk/chem364A_spr05/notes_Lecture-Note-6-7.pdf 1.
Many thanks.
The CsCl structure is stable when the ratio of the smaller ion radius to larger ion radius is ##r > 0.73##. Why is this so? Show that the packing fraction, ##p##, is given by $$p = \frac{\sqrt{3} \pi}{2} \frac{1+r^3}{(1+r)^3}$$
Homework Equations
volume of sphere, volume of structure
3. The Attempt at a Solution
I have obtained the correct answer for ##p## but I am not sure how to explain why ##r > 0.73## but I have some calculations. The CsCl structure has a Cl in the middle of a cube with 8 Cs around it (so like a BCC structure and the unit cell is BCC cubic). Suppose that along the edge of a cube, the two adjacent Cs ions are touching. Then ##a = 2R_1## where ##a## is the side of the cube and ##R_1## is the radius of a Cs ion. The face diagonal is then ##h = 2\sqrt{2}R_1##, from which we find the cube diagonal to be ##2\sqrt{3}R_1##. Now if we further suppose that the Cl ion is touching two Cs from either side (so close packing of spheres as much as possible) then we have ##2R_1 + 2R_2 = 2 \sqrt{3}R_1## where ##R_2## is the radius of Cl. Rearranging for ##R_2## gives ##R_2 = (\sqrt{3} - 1)R_1 \approx 0.73 R_1##. So I seemed to have extracted the correct number (0.73) but not the correct ratio. (i.e I wanted to derive ##R_1 = 0.73 R_2##), Cl is certainly bigger than Cs. So what went wrong here? Actually I have found the same calculation in this doc (bottom of first page), but nothing was said about it.
http://depts.washington.edu/chemcrs/bulkdisk/chem364A_spr05/notes_Lecture-Note-6-7.pdf 1.
Many thanks.
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