Question

The arrangement of $X^{-}$ ions around $A^{+}$ ions in solid $AX$ is given in the figure (not drawn to scale).
If the radius of $X^{-}$ is $250\text{pm}$, the radius of $A^{+}$ is:

• A. $103.5\text{pm}$
• B. $125\text{pm}$
• C. $183.5\text{pm}$
• D. $56.75\text{pm}$

Answer 1

The correct option is A $103.5\text{pm}$

From the figure it can be inferred,
void is octahedral.
We know that, $\frac{r_{+}}{r_{-}}=0.414$
$r_{-}=250\text{pm}$
$\therefore r_{+}=0.414\times 250\text{pm}=103.5\text{pm}$