Problem 13 - Entrance Test

Let's determine the effective number of each ion per unit cell: 1. **Calcium ions (Ca²⁺) at the corners:** There are 8 corners in a cube, and each corner ion is shared by 8 unit cells. Number of Ca²⁺ ions = 8 corners × (1/8 atom per corner) = 1 Ca²⁺ ion. 2. **Oxide ions (O²⁻) at the face centers:** There are 6 face centers in a cube, and each face-centered ion is shared by 2 unit cells. Number of O²⁻ ions = 6 faces × (1/2 atom per face) = 3 O²⁻ ions. 3. **Titanium ions (Ti⁴⁺) at the body center:** There is 1 body center in a cube, and the ion at the body center belongs entirely to that unit cell. Number of Ti⁴⁺ ions = 1 body center × (1 atom per body center) = 1 Ti⁴⁺ ion. So, the ratio of Ca²⁺ : Ti⁴⁺ : O²⁻ is 1 : 1 : 3. Therefore, the empirical formula of the perovskite compound is CaTiO₃.