In antifluorite structure, `50%` of `TVs` are occupied by anion.

Explanation

To determine whether the statement about the antifluorite structure is true or false, we need to analyze the structure and the occupancy of the tetrahedral voids.

### Step-by-Step Solution:

1. **Understanding Antifluorite Structure**:
- The antifluorite structure is a type of ionic crystal structure where the anions form a face-centered cubic (FCC) lattice, and the cations occupy the tetrahedral voids.

2. **Composition of Antifluorite Structure**:
- In the antifluorite structure, the general formula can be represented as \(A_2B\), where \(A\) is the cation and \(B\) is the anion. For example, in Na2O, sodium ions (Na\(^+\)) are the cations, and oxide ions (O\(^{2-}\)) are the anions.

3. **Lattice Arrangement**:
- The oxide ions (O\(^{2-}\)) form the FCC lattice. In an FCC lattice, there are 4 formula units per unit cell, which means there are 4 oxide ions.

4. **Tetrahedral and Octahedral Voids**:
- In an FCC lattice, there are 8 tetrahedral voids and 4 octahedral voids per unit cell.
- In the antifluorite structure, the tetrahedral voids are occupied by cations (Na\(^+\) in the case of Na2O).

5. **Occupancy of Tetrahedral Voids**:
- In the antifluorite structure, all tetrahedral voids are occupied by cations, not anions. Therefore, the statement that "50% of tetrahedral voids are occupied by anion" is incorrect.

6. **Conclusion**:
- Since the statement claims that 50% of tetrahedral voids are occupied by anions, which is false because they are actually occupied by cations, we conclude that the statement is **false**.

### Final Answer:
The statement is **false**.

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