Question

The ionic radii of $R{b}^{+}$ and $I^{-}$ are $1.46$ and $2.16\stackrel{o}{\text{A}}.$ The coordination number for the cation is :

• A. $2$
• B. $4$
• C. $6$
• D. $8$

Answer 1

The correct option is C $6$

$\text{Ionic radii of}R{b}^{+},r_{R{b}^{+}}=1.46\stackrel{o}{\text{A}}$
$\text{Ionic radii of}{I}^{-},r_{I^{-}}=2.16\stackrel{o}{\text{A}}$
$\Rightarrow \frac{r_{R{b}^{+}}}{r_{I^{-}}}=\frac{1.4\text{A}}{2.16}=0.67$
Radius ratio for $RbI$ is $0.67$
Since,
$0.414<0.67<0.77$
Octahedral void is occupied.
Thus, the coordination number is $6$.