Question

Answer the following in brief.

Cesium chloride crystallizes in a cubic unit cell with Cl^– ions at the corners and a Cs⁺ ion in the center of the cube. How many CsCl molecules are there in the unit cell?

Answer 1

Given: Cl^– ions are present at the corners of the cube. Cs+ ion is at the center of the cube.

To find: Number of CsCl molecules in the unit cell

Calculation:
i. Cl^– ions are present at the 8 corners. The contribution of each corner particle to the unit cell is 1/8. Hence, the number of Cl^– ions that belong to the unit cell = 8 × (1/8) = 1

ii. Cs⁺ ion is at the center of the unit cell. The contribution of a particle at the center to the unit cell is 1. Hence, the number of Cs⁺ ions that belong to the unit cell = 1
There is one Cl^– ion and one Cs⁺ ion in the unit cell.
Hence, the number of CsCl molecules in the unit cell = 1.

Number of CsCl molecules in the unit cell = 1.