Chemistry - Solid State Question with Solution | TestHub

1. A atoms: Present at $6$ corners, which is $\frac{3}{4}$ of the $8$ corners. This gives $6\times \frac{1}{8}=\frac{3}{4}$.

2. B atoms: Present at $2$ pairs of diagonally opposite edge centers, totaling $4$ edge centers. This gives $4\times \frac{1}{4}=1$.

3. C atoms: Vacant corners are $2$ corners, which contributes $2\times \frac{1}{8}=\frac{1}{4}$. The remaining edges are $12-4=8$ edge centers, which contribute $8\times \frac{1}{4}=2$. Total C atoms = $\frac{1}{4}+2=\frac{9}{4}$.

4. Ratio A:B:C = $\frac{3}{4}:1:\frac{9}{4}=3:4:9$.

Simplest formula = A₃B₄C₉