In chromium (III) chloride. CrCl₃. Chloride ions have cubic close packed arrangement and Cr(III) ions are present in the octahedral holes. The fraction of the octahedral holes occupied and also the fraction of total number of holes occupied are -- & -- respectively.

According to the formula ${\mathrm{CrCl}}_3$ if there are 3 $3{\mathrm{Cl}}^{-}$ ions, then there is $1{\mathrm{Cr}}^{3+}$ ion.

${\mathrm{Cl}}^{-}$ ions have a ccp arrangement.

So total number of octahedral voids are equal to the number of ions in ccp i.e. 3

Number of tetrahedral voids are twice the no of ions in ccp i.e. 6

$1{\mathrm{Cr}}^{3+}$ ion will occupy octahedral void.

Total number of octahedral voids are 3

Fraction of octahedral voids occupied = $\frac{1}{3}$

Total number of voids = Tetrahedral voids + Octahedral voids = 6 + 3 = 9

Fraction of total voids occupied = $\frac{1}{9}$