There are four formula units in fluorite and antifluorite structure.

Explanation

To solve the problem regarding the formula units in fluorite and antifluorite structures, we will analyze both structures step by step.

### Step 1: Understanding Fluorite Structure
1. **Identify the Compound**: The fluorite structure is represented by the formula CaF₂.
2. **Cation Arrangement**: In this structure, the Ca²⁺ ions occupy the face-centered cubic (FCC) lattice.
3. **Anion Arrangement**: The F⁻ ions occupy the tetrahedral voids in the FCC lattice.

### Step 2: Calculate the Number of Cations (Ca²⁺)
1. **FCC Lattice Contribution**: In an FCC lattice:
- There are 8 corner atoms, each contributing 1/8.
- There are 6 face-centered atoms, each contributing 1/2.
2. **Total Contribution**:
\[ \text{Total Ca²⁺} = 8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 1 + 3 = 4 \]
Thus, there are 4 Ca²⁺ ions in the unit cell.

### Step 3: Calculate the Number of Anions (F⁻)
1. **Tetrahedral Voids**: In an FCC lattice, there are 8 tetrahedral voids.
2. **F⁻ Ion Occupation**: Each tetrahedral void is occupied by one F⁻ ion.
3. **Total Contribution**:
\[ \text{Total F⁻} = 8 \]
Thus, there are 8 F⁻ ions in the unit cell.

### Step 4: Determine the Effective Number of Formula Units (Z)
1. **Formula Units**: The effective number of formula units (Z) is determined by the lower of the two counts (cations and anions).
2. **Calculation**:
\[ Z = \text{min}(4, 8) = 4 \]

### Step 5: Understanding Antifluorite Structure
1. **Identify the Compound**: The antifluorite structure is represented by compounds like Na₂O.
2. **Cation Arrangement**: In this structure, the Na⁺ ions occupy the tetrahedral voids.
3. **Anion Arrangement**: The O²⁻ ions form the FCC lattice.

### Step 6: Calculate the Number of Cations (Na⁺)
1. **Tetrahedral Voids**: As previously mentioned, there are 8 tetrahedral voids in the FCC lattice.
2. **Total Contribution**:
\[ \text{Total Na⁺} = 8 \]
Thus, there are 8 Na⁺ ions in the unit cell.

### Step 7: Calculate the Number of Anions (O²⁻)
1. **FCC Lattice Contribution**: The calculation remains the same as in the fluorite structure.
2. **Total Contribution**:
\[ \text{Total O²⁻} = 4 \]
Thus, there are 4 O²⁻ ions in the unit cell.

### Step 8: Determine the Effective Number of Formula Units (Z)
1. **Formula Units**: The effective number of formula units (Z) is determined by the lower of the two counts (cations and anions).
2. **Calculation**:
\[ Z = \text{min}(8, 4) = 4 \]

### Final Answer:
- For both the fluorite and antifluorite structures, the effective number of formula units (Z) is 4.