Question

In a solid, oxide ions (${\mathrm{O}}^{2-}$) are arranged in ccp, cation (${\mathrm{A}}^{3+}$) occupies one-sixth of the tetrahedral void, and cations (${\mathrm{B}}^{3+}$) occupy one-third of the octahedral voids. What is the formula of the compound?

• A. ${\mathrm{ABO}}_2$
• B. ${\mathrm{ABO}}_3$
• C. ${\mathrm{A}}_2{\mathrm{BO}}_3$
• D. ${\mathrm{AB}}_2{\mathrm{O}}_3$

Answer 1

The correct option is B

${\mathrm{ABO}}_3$

Explanation:

Option(B);

Step 1: Given data:

In the given solid, oxide ions (${\mathrm{O}}^{2-}$) are arranged in CCP lattice

Hence, the effective number of oxide ions $\left({\mathrm{O}}^{2-}\right)$ in CCP lattice structure is given by $=4$

Here the cation (${\mathrm{A}}^{3+}$) occupies one-sixth of the tetrahedral void, and cations (${\mathrm{B}}^{3+}$) occupy one-third of the octahedral voids

Step 2: Calculation of voids in the given lattice:

We know in a given lattice structure,

The number of tetrahedral voids = effective number$\left(\mathrm{Z}\right)$ of the lattice

Whereas the number of octahedral voids =$2\times \mathrm{effective}\mathrm{number}\mathrm{of}\left(\mathrm{Z}\right)\mathrm{of}\mathrm{the}\mathrm{lattice}$

Hence as the number of ${\mathrm{O}}^{2-}$ ions in the given packing $=4$
Number of octahedral voids $=4$
Number of tetrahedral voids $=4\times 2=8$

Step 3: Calculation of the effective number of cations:

We can calculate the effective number of cations as ;

$$\begin{gathered} {\mathrm{A}}^{+3}=\frac{1}{6}\times \mathrm{No}.\mathrm{of}\mathrm{Tetrahedral}\mathrm{voids} \\ =\frac{1}{6}\times 8=\frac{4}{3} \end{gathered}$$​

And the effective number of cations can be calculated as;

$$\begin{gathered} {\mathrm{B}}^{+3}=\frac{1}{3}\times \mathrm{Octahedral}\mathrm{voids} \\ =\frac{1}{3}\times 4=\frac{4}{3} \end{gathered}$$

​Step 4: Calculation of the formula of the compound:

Therefore, the formula of the compound can be calculated as follows:

Thus, the formula of the compound is ${\mathrm{ABO}}_3$.

Hence, option (B) is the correct answer.