Question

In an oxide, oxide ions are packed in an FCC unit cell in which $B^{3+}$ ions occupy x% of octahedral voids and $A^{2+}$ ions occupy y% of tetrahedral voids, then which of the following is/are correct about the unit cell?

• A. Formula is $A_{2}B{O}_8$ in which x= 12.5% and y= 12.5%
• B. Formula is $A_2{B}_4{O}_6$ in which x= 50% and y= 50%
• C. Formula is $A{B}_2{O}_4$ in which x= 50% and y= 12.5%
• D. Formula is $A_{2}B{O}_2$ in which x= 12.5% and y= 50%

Answer 1

Answer: (A), (C)

Formula is $A{B}_2{O}_4$ in which x= 50% and y= 12.5%

Let's backtrack this question.
$O^{2-}$ ions are packed in a FCC unit cell.
Number of $O^{2-}$ ions = 4
This means, there are 8 T-voids and 4 O-voids present.
$B^{3+}$ occupy x% of O-voids, $A^{2+}$ ions occupy tetrahedral voids and y% are occupied.
The options talk about x and y being either 50% or 12.5 %.
So, if x is 50%, we have 2 ions, if it is 12.5, we have 0.5 ions per unit cell. Similarly y can be either 4 or 1 ions per unit cell.
The possible formulae then can be:
x= 50%, y = 50% $A_4{B}_2{O}_4\equiv A_{2}B{O}_2$
x= 50%, y = 12.5% $A{B}_2{O}_4$
x= 12.5%, y = 50% $A_4{B}_{0.5}{O}_4\equiv A_{8}B{O}_8$
x= 12.5%, y = 12.5% $A{B}_{0.5}{O}_4\equiv A_{2}B{O}_8$