The crystal structure of Zinc blende (ZnS) consists of cubic close-packed (ccp) array of `S^(2-)` ions. If the radii of `Zn^(2+)andS^(2-)` ions are 0.74 and 1.84Å respectively, then which type of voids (tetrahedral or octahedral) formed in ccp array of `S^(2-)` ions are occupied by `Zn^(2+)` ions? What fraction of total number of this void remains unoccupied?

Explanation

The radius of `Zn^(2+)` ion = 0.74Å and that of `S^(2-)` ion = 1.84Å.
Therefore, `r_(+)/r_(-)=0.74/1.84=0.402`
This value lies between 0.225 and 0.414. Therefore, `Zn^(2+)` ions occupy the tetrahedral voids in the crystal structure of zinc blende.
In the number of `S^(2-)` ions in the ccp array of `S^(2-)` ions is 'n', then the number of tetrahedral voids in the close-packed structure = 2n.
To maintain 1 : 1 stoichiometry and electrical neutrality, half of the tetrahedral voids are occupied by `Zn^(2+)` ions and the remaining half is left unoccupied.