In a mixed oxide of $A$ and $B, A$ occupies all the octahedral voids while $B$ occupies $(2 / 3)^{\text {rd…
In a mixed oxide of $A$ and $B, A$ occupies all the octahedral voids while $B$ occupies $(2 / 3)^{\text {rd }}$ the tetrahedral voids. The molecular formula of this oxide is
$\mathrm{A}_3 \mathrm{~B}_4 \mathrm{O}_3$
$\mathrm{A}_3 \mathrm{~B}_2 \mathrm{O}_3$
$\mathrm{A}_3 \mathrm{BO}_3$
$A B_2 \mathrm{O}_3$
Solution
According to mixed oxide, number of $\mathrm{O}^{2-}$ ions $=4$
So, number of tetrahedral voids $=8$
and number of octahedral voids $=4$
$B$ occupies 2/3rd of tetrahedral void $=8 \times \frac{2}{3}=16 / 3$
$A$ occupies all octahedral voids $=4$
Now,
$A: B: \mathrm{O}$
$4: \frac{16}{3}: 4$
$
\Rightarrow \quad 12: 16: 12
$
$\Rightarrow \quad 3: 4: 3 \quad$ (in simplest ratio)
Hence, molecular formula of this oxide is $A_3 B_4 \mathrm{O}_3$