An ionic solid AB crystallizes in a NaCl-type | vedclass

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(D) In a $NaCl$ structure,$A$ ions (cations) are at the corners and face centers,while $B$ ions (anions) are at the body center and edge centers. Init

Vedclass · Accepted Answer

$A_3B_4$

Vedclass · Answer

(D) In a $NaCl$ structure,$A$ ions (cations) are at the corners and face centers,while $B$ ions (anions) are at the body center and edge centers. Initially,$A = 8 	imes (1/8) + 6 	imes (1/2) = 4$ and $B = 1 + 12 	imes (1/4) = 4$. If all face-centered ions along one axis are removed,two face-centered $A$ ions are removed. New number of $A$ ions = $4 - 2 	imes (1/2) = 3$. The number of $B$ ions remains $4$. Therefore,the stoichiometry is $A_3B_4$.

An ionic solid $AB$ crystallizes in a $NaCl$-type structure,where $A$ ions are at the corners of the cube. If all face-centered ions along one axis are removed,then the stoichiometry of the resulting solid will be .....

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