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प्रश्न
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?
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उत्तर
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.
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