The crystal AB has a structure similar to NaC | vedclass

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(B) In an $NaCl$ type structure,$A$ atoms are at the corners and face centers,while $B$ atoms are at the edge centers and the body center.Initial numb

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$A_3B_4$

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(B) In an $NaCl$ type structure,$A$ atoms are at the corners and face centers,while $B$ atoms are at the edge centers and the body center.Initial number of $A$ atoms: Corners = $8 	imes \frac{1}{8} = 1$,Face centers = $6 	imes \frac{1}{2} = 3$. Total $A = 4$.Initial number of $B$ atoms: Edge centers = $12 	imes \frac{1}{4} = 3$,Body center = $1 	imes 1 = 1$. Total $B = 4$.When atoms along one axis passing through face centers are removed,the two face-centered $A$ atoms on that axis are removed.Remaining $A$ atoms = $4 - 1 = 3$ (Note: The calculation accounts for the contribution of face-centered atoms,removing one axis removes two face centers,each contributing $1/2$,so $2 	imes 1/2 = 1$ atom is removed).$B$ atoms remain unchanged as they are at edge centers and the body center.Thus,the resulting stoichiometry is $A_3B_4$.

The crystal $AB$ has a structure similar to $NaCl$. Atoms of $A$ are present at each corner of the cubic unit cell. If all atoms along one of the axes passing through the face centers are removed,what will be the resulting stoichiometry?

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