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陶瓷材料 (1011) Homework 2 - Solution 1. (a) Show that the minimum cation/anion radius ratio for a coordination number of 6 is 0.414. (b) Repeat part (a) for coordination number 3. (c) Which interstitial site is larger, the tetrahedral or the octahedral? Calculate the ratio of the sizes for the tetrahedral and octahedral sites. (c) The octahedral site with an ionic radius ratio of 0.414 is larger than that of the tetrahedral site with an ionic radius ratio of 0.225. The volumetric size ratio is then 3 . 6 225 . 0 414 . 0 3 3 tetra oct r r

Homework 2 - Solution

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  • (1011) Homework 2 - Solution

    1. (a) Show that the minimum cation/anion radius ratio for a coordination number of 6 is 0.414. (b) Repeat part (a) for coordination number 3. (c) Which interstitial site is larger, the tetrahedral or the octahedral? Calculate the ratio of the sizes for the tetrahedral and octahedral sites.

    (c) The octahedral site with an ionic radius ratio of 0.414 is larger than that of the tetrahedral site with an ionic radius ratio of 0.225. The volumetric size ratio is

    then 3.6225.0414.0 3

    3

    tetra

    oct

    rr

  • 2. Find the ionic radii of Ti4+ and O2- from your textbook, and apply the Paulings 1st

    and 2nd rules in predicting the coordination number for the cation and the anion, respectively. Sol.

    The ionic radii of Ti4+ and O2- are 0.60 and 1.40 , respectively. Therefore,

    43.040.160.0

    2

    4

    O

    Ti

    a

    c

    rr

    rr

    There should exist a polyhedron with 6 anions (O2-) surrounding 1 cation (Ti4+) according to the 1st Paulings rule. For maintaining the charge neutrality in the polyhedron (i.e., the 2nd Paulings rule), we obtain

    3

    )4()1)(12(

    ))(1()1)()(6( 42

    CNCN

    TiCN

    O

    Therefore, the coordination numbers for the cations (Ti4+) and anions (O2-) are 6 and 3, respectively. Note that the coordination number for the anions (O2-) can also be solved alternatively by recognizing that the stoichiometric form of TiO2 is 1:2 in molar ratio. Since 6 O2- ions are required to form a polyhedron for surrounding 1 Ti4+ ion, the number of Ti4+ ions for 1 O2- ion must therefore be one half of 6 in order to maintain the stoichiometry.

    3. Starting with the cubic close packing of oxygen ions:

    1. How many tetrahedral and how many octahedral sites are there per unit cell? 2. What is the ratio of octahedral sites to oxygen ions? What is the ratio of

    tetrahedral sites to oxygen ions? 3. What oxide would you get if one-half of the octahedral sites are filled?

    Two-thirds? All? 4. Locate all the tetrahedral sites and fill them up with cations. What structure

    do you obtain? If the anions are oxygen, what must be the charge on the cation for charge neutrality to be maintained?

    Sol. (a) 8 tetrahedral and 4 octahedral (b) Octahedral sites to oxygen ions = 1:1; tetrahedral sites to oxygen ions = 2:1

  • (c) 1/2 filled = MO2; 2/3 filled = M2O3; and all filled = MO (d) M2O (anti-fluorite); charge = 1+

    4. We have mentioned in class about ceramic materials with crystalline structures of rocksalt, fluorite, anti-fluorite, zincblende, wurtzite, etc. Try to use any search engines available on the web, and find more details of ONE ceramic material that interests you. Give me what kind of structure it is, specific properties and applications about this particular ceramics. Suggestions There are plenty of examples that can be found from the web for almost all existing ceramics. My favorite website (or recommendation) is Wikipedia (), which is an on-line encyclopedia written by experts from all over the world in particular area of interests.

    (Any queries about the solution or if you have any difficulties in understanding my lecture, feel free to come in and talk with me

    during the office hours.)