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For the spinel structure `(MgAl_(2)O_(4))`, the correct statement is/areA. 50% Ovs are occupied by ions.B. `Al^(3+)` is equally distributed in TVs and Ovs.C. Oxide ions occupy ccp lattice.D. 12.5% TVs are occupied by ions. |
Answer» For spinel structure `O^(2-)` ions form fcc arrangement. `therefore ` Number of `O^(2-)` ions =4. Number of TV=8, Number of OV=4 Number of `Mg^(2+)` ions `=(1)/(8)xxTV =(1)/(8)xx8=1` Number of `Al^(3+)` ions `=(1)/(2)xxOV=(1)/(2)xx4=2` So, 50%OVs are occuped and 1/8th , i.e., 12.5% TVs are occupied. |
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