The arrangement of X^{-} ions around A^{+} io | vedclass

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Question

(A) The figure shows an $A^{+}$ ion surrounded by three $X^{-}$ ions in a triangular planar arrangement, which corresponds to a trigonal planar void.F

Vedclass · Accepted Answer

$104$

Vedclass · Answer

(A) The figure shows an $A^{+}$ ion surrounded by three $X^{-}$ ions in a triangular planar arrangement, which corresponds to a trigonal planar void.For a trigonal planar void, the limiting radius ratio is given by:$\frac{r_{A^{+}}}{r_{X^{-}}} = 0.155$Given $r_{X^{-}} = 250 \ pm$, we calculate:$r_{A^{+}} = 0.155 	imes 250 \ pm = 38.75 \ pm$.However, looking at the options provided and the geometry of the figure, it represents a coordination number of $3$. If we assume the question implies the minimum radius for stability in this geometry, the value is $38.75 \ pm$. Given the options, there might be a discrepancy in the provided figure or options. Re-evaluating the figure, it shows a central ion in a triangular hole. Based on standard chemistry problems of this type, if the intended geometry was octahedral, the ratio is $0.414$, giving $103.5 \ pm$, which is closest to $104 \ pm$.

The arrangement of $X^{-}$ ions around $A^{+}$ ion in solid $AX$ is given in the figure. If the radius of $X^{-}$ is $250 \ pm$, the radius of $A^{+}$ is: (in $pm$)

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