In chromium (III) chloride `CeCl_(3)` chloride ions have cubic close packed arrangement and Cr (III) ions present in the octahedral voids. What fraction of the octahedral void is occupied ? What fraction of the total number of voids is occupied?

Answer

1/3, 1/9

Explanation

In ccp arangement, each chloride ions is associated with one octahedral and two tetrahedral voids.
Number of octahedral voids with three `Cl^(-)` ions = 3.
Number of tetrahedral voids with three `Cl^(-)` ions = 6
Total number of voids with three `Cl^(-)` ions = 3 + 6 = 9
Number of octahedral voids occupied by `Cr^(3+)` ions = 1
Fraction of octahedal voids occupied by `Cr^(3+)` ion `=(1)/(3)`
Fraction of total number of voids occupied `=(1)/(9)`.