Crystal structure of the oxide AB_(2)O_(4) is bassed on the cubic close packed (ccp) array of O^(2-) ions, with A^(2+) ions occupying some tetrahedral voids and B^(3+) ions occupying some octahedral voids. If a crystal of this oxide contains 1 mol AB_(2)O_(4) formula unit, then calculate (1) total number of voids in the crystal (2) total number of tetrahedral voids occupied by A^(2+) ions (3) total number of octahedral voids occupied by B^(3+) ions.

Crystal structure of the oxide AB_(2)O_(4) is bassed on the cubic clos

1.	Crystal structure of the oxide AB_(2)O_(4) is bassed on the cubic close packed (ccp) array of O^(2-) ions, with A^(2+) ions occupying some tetrahedral voids and B^(3+) ions occupying some octahedral voids. If a crystal of this oxide contains 1 mol AB_(2)O_(4) formula unit, then calculate (1) total number of voids in the crystal (2) total number of tetrahedral voids occupied by A^(2+) ions (3) total number of octahedral voids occupied by B^(3+) ions.
Answer» Solution :TOTAL NUMBER of `O^(2-)` ions in 1 mol of oxide `(AB_(2)O_(4))` crystal = `4xx6.022xx10^(23)`. The ccp structure is formed by `O^(2-)` ions. So, total number of TETRAHEDRAL voids = 2 `xx` no. of packed `O^(2-)` ions = `2xx4xx6.022xx10^(23)=8xx6.022xx10^(23)` and total number of octahedral voids = number of packed `O^(2-)` ions = `4xx6.023xx10^(23)`. (1) Total number of voids in 1 mol oxide crystal = `(8+4)xx6.022xx10^(23)=7.2264xx10^(24)`. (2) To maintain STOICHIOMETRY and electrical neutrality, one-eighth of the total tetrahedral voids will be occupied by `A^(2-)` ions. Hence, total number of tetrahedral voids occupied by `A^(2-)` ions = `1/8xx8xx6.022xx10^(23)` `=6.022xx10^(23)`. (3) For the same reasons, half of the total octahedral voids will be occupied by `B^(3+)` ions. Hence, total number of octahedral holes occupied by `B^(3+)` ions. `=1/2xx4xx6.022xx10^(23)=1.2044xx10^(24)`