Question:
An ionic compound is formed by elements X and Y. Element Y forms a face-centered cubic (FCC) lattice, and element X occupies one-third of the tetrahedral voids. What is the formula of the compound?
An ionic compound is formed by elements X and Y. Element Y forms a face-centered cubic (FCC) lattice, and element X occupies one-third of the tetrahedral voids. What is the formula of the compound?
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FCC has 4 atoms and 8 tetrahedral voids per unit cell.
Updated On: Oct 1, 2026
- \(X_2Y_3\)
- \(XY\)
- \(XY_3\)
- \(X_2Y\)
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
In an FCC lattice with \(N\) atoms, there are \(N\) octahedral voids and \(2N\) tetrahedral voids.
Step 2: Count atoms:
Y forms FCC, so \(Y = 4\) atoms per unit cell. The number of tetrahedral voids is \(2\times 4 = 8\).
X occupies one-third of them: \(X = \frac{1}{3}\times 8 = \frac{8}{3}\).
Step 3: Ratio:
\[ X : Y = \frac{8}{3} : 4 = 8 : 12 = 2 : 3 \]
The formula is \(X_2Y_3\). Option B (\(XY\)) would need X in all 4 octahedral voids. Option D (\(X_2Y\)) would need X in all 8 tetrahedral voids. Option C (\(XY_3\)) would need only \(\frac{4}{3}\) X atoms per cell, which is one-sixth of the tetrahedral voids.
Final Answer:
The compound is \(X_2Y_3\), option (A). \[ \boxed{X_2Y_3} \]
In an FCC lattice with \(N\) atoms, there are \(N\) octahedral voids and \(2N\) tetrahedral voids.
Step 2: Count atoms:
Y forms FCC, so \(Y = 4\) atoms per unit cell. The number of tetrahedral voids is \(2\times 4 = 8\).
X occupies one-third of them: \(X = \frac{1}{3}\times 8 = \frac{8}{3}\).
Step 3: Ratio:
\[ X : Y = \frac{8}{3} : 4 = 8 : 12 = 2 : 3 \]
The formula is \(X_2Y_3\). Option B (\(XY\)) would need X in all 4 octahedral voids. Option D (\(X_2Y\)) would need X in all 8 tetrahedral voids. Option C (\(XY_3\)) would need only \(\frac{4}{3}\) X atoms per cell, which is one-sixth of the tetrahedral voids.
Final Answer:
The compound is \(X_2Y_3\), option (A). \[ \boxed{X_2Y_3} \]
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