Question

A mineral consists of a cubic close-packed structure formed by $O^{2-}$ ions where half the octahedral voids are occupied by $A{l}^{3+}$ and one-eighth of the tetrahedral voids are occupied by $M{n}^{2+}$. The chemical formula of the mineral is:

• A. $M{n}_{3}A{l}_2{O}_6$
• B. $MnA{l}_2{O}_4$
• C. $MnA{l}_4{O}_7$
• D. $M{n}_{2}A{l}_2{O}_5$

Answer 1

The correct option is B $MnA{l}_2{O}_4$

As oxide ions are present in ccp so number of $O^{2-}$ ions=4.

Number of octahedral voids=4

Number of tetrahedral voids=8

As $1/8$ of the tetrahedral voids are occupied and $1/2$ of the octahedral voids are occupied.

Number of $M{n}^{2+}$ ions in tetrahedral voids=1

Number of $A{l}^{3+}$ ions in octahedral voids=2

So the formula of the mineral is $MnA{l}_2{O}_4$.