Problem

An ionic compound has X atoms in FCC and Y atoms filling only the ALTERNATE tetrahedral voids. Find the formula.

Solution

Step 1: Identify Z_X X in FCC → Z_X = 4 atoms per unit cell.

Step 2: Count tetrahedral voids For n = 4 atoms → tetrahedral voids = 2n = 2 × 4 = 8.

Step 3: Identify which voids are filled "Alternate" tetrahedral voids = half = 8/2 = 4 voids filled by Y.

$Z_Y = 4$

Step 4: Write formula X : Y = 4 : 4 = 1 : 1 → formula = XY

Interpretation: This is the ZnS (zinc blende) structure.

• $S^{2-}$ (X) in FCC; $Zn^{2+}$ (Y) in alternate tetrahedral voids
• Formula = ZnS (1:1) ✓

Contrast with NaCl: If Y filled ALL octahedral voids (n = 4), Z_Y = 4 → still XY. But in NaCl, the void type is octahedral (not tetrahedral).

Additional example: If Y fills ALL tetrahedral voids: Z_Y = 8; formula = $X_{4}Y_{8}$ = $XY_{2}$ (fluorite type, e.g., $CaF_{2}$)