Question

In chromium (III) chloride, $\left(CrC{l}_3\right)$, chloride ions have cubic close packed arrangement and $Cr\left(III\right)$ ions are present in octahedral voids.
Choose the correct option(s):

• A. $\text{Fraction of total voids occupied}=\frac{1}{6}$
• B. $\text{Fraction of total voids occupied}=\frac{1}{9}$
• C. $\text{Fraction of octahedral voids occupied}=\frac{1}{3}$
• D. $\text{Fraction of octahedral voids occupied}=\frac{1}{2}$

Answer 1

Answer: (B), (C)

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

According to the formula $CrC{l}_3$ if there are $3C{l}^{-}$ ions , then there is $1C{r}^{3+}$ ion.
$C{l}^{-}$ ions have a ccp arrangement.

So total number of octahedral voids is equal to the number of $C{l}^{-}$ ions in ccp, i.e., 3.
Number of tetrahedral voids is twice the no of $C{l}^{-}$ ions in ccp, i.e., 6

$1C{r}^{3+}$ will occupy one octahedral void.
Total number of octahedral voids are 3
$\text{Fraction of octahedral voids occupied}=\frac{1}{3}$
$\text{Total number of voids}=\text{Tetrahedral voids +Octahedral voids}=6+3=9$
$\text{Fraction of total voids occupied}=\frac{1}{9}$
Option (c) and (d) are correct