A solid AB crystallises in NaCl structure. Th | vedclass

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(A) In an $NaCl$ structure,there are $4$ $A$ atoms and $4$ $B$ atoms per unit cell. $B$ atoms are at corners and face centers,while $A$ atoms are in o

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$AB$

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(A) In an $NaCl$ structure,there are $4$ $A$ atoms and $4$ $B$ atoms per unit cell. $B$ atoms are at corners and face centers,while $A$ atoms are in octahedral voids (at edge centers and body center).The diagonal plane passing through the cube contains $4$ corners and $2$ edge centers.$1.$ Removal of $B$ atoms: The plane passes through $4$ corners. Each corner atom contributes $1/8$ to the unit cell. Total $B$ atoms removed = $4 	imes (1/8) = 0.5$. Remaining $B$ atoms = $4 - 0.5 = 3.5$.$2.$ Removal of $A$ atoms: The plane passes through $2$ edge centers. Each edge center atom contributes $1/4$ to the unit cell. Total $A$ atoms removed = $2 	imes (1/4) = 0.5$. Remaining $A$ atoms = $4 - 0.5 = 3.5$.The ratio of $A:B$ is $3.5:3.5$,which simplifies to $1:1$. Thus,the formula is $AB$.

$A$ solid $AB$ crystallises in $NaCl$ structure. The anion $B$ forms $fcc$ lattice,while the cation $A^{+}$ occupies all the octahedral holes. If all the particles along a diagonal plane (as shown in the figure) are removed,then the formula of the solid becomes

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