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

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Question

(A) According to the given figure,$A^{+}$ is present in the octahedral void of $X^{-}$ ions.The limiting radius ratio for an octahedral void is given

Vedclass · Accepted Answer

$104$

Vedclass · Answer

(A) According to the given figure,$A^{+}$ is present in the octahedral void of $X^{-}$ ions.The limiting radius ratio for an octahedral void is given by the formula: $r_{void} = 0.414 	imes r_{sphere}$.Here,$r_{A^{+}} = 0.414 	imes r_{X^{-}}$.Substituting the given value: $r_{A^{+}} = 0.414 	imes 250 \ pm = 103.5 \ pm$.Rounding to the nearest integer,we get $104 \ pm$.Therefore,option $A$ is correct.

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 ............. $pm$.

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