In a cubic close packed structure(ccp) of mixed oxide, it is found that lattice has O2- ions and one half of the octahedral voids are occupied by trivalent cations (X3+) and one eighth of the tetrahedral voids are occupied by divalent cations (Y2+). What is the formula of the mixed oxide?
We have:
For ccp, we know:
Number of Oxide ions = 4
We know that in any closed packed structure consisting of n atoms, the number of Octahedral voids is n while the number of Tetrahedral voids is equal to 2n
Number of trivalent ions in octahedral voids
Similarly, number of divalent ions in tetrahedral voids
Thus the formula is
Therefore, ?Option(4) is correct