In a solid AB having the NaCl structure,A ato | vedclass

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(D) In an $NaCl$ structure,$A$ atoms (cations) occupy the corners and face centers,while $B$ atoms (anions) occupy the edge centers and body center.In

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$A_3B_4$

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(D) In an $NaCl$ structure,$A$ atoms (cations) occupy the corners and face centers,while $B$ atoms (anions) occupy the edge centers and body center.Initial number of $A$ atoms per unit cell $= 8 	imes \frac{1}{8} (	ext{corners}) + 6 	imes \frac{1}{2} (	ext{face-centers}) = 1 + 3 = 4$.Initial number of $B$ atoms per unit cell $= 12 	imes \frac{1}{4} (	ext{edge-centers}) + 1 (	ext{body-center}) = 3 + 1 = 4$.Removing face-centered atoms along one of the axes means removing $2$ face-centered $A$ atoms.New number of $A$ atoms $= 4 - 2 = 2$.Wait,the question states $A$ atoms occupy corners. In $NaCl$,$A$ is typically the cation. If $A$ is at corners and face centers,and $B$ is at edge centers and body center,removing $2$ face-centered $A$ atoms leaves $A = 8 	imes \frac{1}{8} + 4 	imes \frac{1}{2} = 1 + 2 = 3$.Number of $B$ atoms remains $4$.Therefore,the stoichiometry is $A_3B_4$.

In a solid $AB$ having the $NaCl$ structure,$A$ atoms occupy the corners of the cubic unit cell. If all the face-centered atoms along one of the axes are removed,then the resultant stoichiometry of the solid is

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