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.

Explanation

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)`