A solid contains A and B ions. The structure of solid is FCC for B ions and A ions are present in on — Solid State Chemistry Question
Question
A solid contains A$^{n+}$ and B$^{m-}$ ions. The structure of solid is FCC for B$^{m-}$ ions and A$^{n+}$ ions are present in one-fourth of tetrahedral voids as well as in one fourth of octahedral voids. What is the simplest formula of solid?
Answer: A
💡 Solution & Explanation
Let the number of B$^{m-}$ ions in FCC = 4. Number of tetrahedral voids = 8, so A in TV = $\frac{1}{4} \times 8 = 2$. Number of octahedral voids = 4, so A in OV = $\frac{1}{4} \times 4 = 1$. Total A ions = $2 + 1 = 3$. The ratio of A:B is 3:4. The formula is A$_3$B$_4$. Therefore, correct answer is A.
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