Solid State - Result Question 16
16. The arrangement of $X^{-}$ions around $A^{+}$ion in solid $A X$ is given in the figure (not drawn to scale). If the radius of $X^{-}$is $250 $ $pm$, the radius of $A^{+}$is
(2013 Adv.)
(a) $104 $ $pm$
(b) $125 $ $pm$
(c) $183 $ $pm$
(d) $57$ $ pm$
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Answer:
Correct Answer: 16. (a)
Solution:
- PLAN: Given arrangement represents octahedral void and for this
$ \begin{aligned} & \frac{r _{+} \text {(cation) }}{r _{-} \text {(anion) }}=0.414 \\ \frac{r\left(A^{+}\right)}{r\left(X^{-}\right)} & =0.414 \\ r\left(A^{+}\right) & =0.414 \times r\left(X^{-}\right)=0.414 \times 250 \hspace{2mm} pm \\ & =103.5 \hspace{2mm} pm \approx 104 \hspace{2mm} pm \end{aligned} $
BETA
