Question

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

• A. $104$ pm
• B. $125$ pm
• C. $183$ pm
• D. $57$ pm

Answer 1

The correct option is A $104$ pm

According to the given data, ${\text{A}}^{+}$ is present in the octahedral voids of $X^{-}$. The limiting radius in the octahedral void is related to the radius of the sphere as:

${\text{r}}_{void}=0.414{\text{r}}_{sphere}$

${\text{r}}_{{\text{A}}^{+}}=0.414{\text{r}}_{X^{-}}=0.414\times 250\text{pm}$

$=103.5\approx 104\text{pm}$

Hence, the correct option is $\text{A}$