Question

If the radius of an anion $\left(r^{-}\right)$ is $0.20$ nm, the maximum radius of cation $\left(r^{+}\right)$ which can be filled in respective voids is correctly matched in :

• A. $r^{+}=0.0828$ nm for tetrahedral void
• B. $r^{+}=0.045$ nm for triangular void
• C. $r^{+}=0.1464$ nm for octahedral void
• D. none of the above

Answer 1

Answer: (A), (B), (C)

The correct options are
A $r^{+}=0.0828$ nm for tetrahedral void
B $r^{+}=0.045$ nm for triangular void
C $r^{+}=0.1464$ nm for octahedral void

The radius ratio, $\frac{r^{+}}{r^{-}}$ is $0.732,0.414$ and $0.225$ for octahedral void, tetrahedral void and triangular void respectively.
The radius of anion is $0.20$ nm.

The maximum radius of cations which can be filled in respective voids are given below.
$r^{+}=0.0828$ nm for tetrahedral void.
Radius ratio $=\frac{\text{radius of the cation}}{\text{radius of the anion}}$ $⟹0.414=\frac{\text{radius of the cation}}{0.20}$
Therefore, radius of the cation is $0.0828$ nm.

$r^{+}=0.045$ nm for triangular void.
Radius ratio $=\frac{\text{radius of the cation}}{\text{radius of the anion}}$ $⟹0.225=\frac{\text{radius of the cation}}{0.20}$
Therefore, radius of the cation is $0.045$ nm.

$r^{+}=0.1464$ nm for octahedral void.
Radius ratio $=\frac{\text{radius of the cation}}{\text{radius of the anion}}$ $⟹0.732=\frac{\text{radius of the cation}}{0.20}$
Therefore, radius of the cation is $0.1464$ nm.