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In a solid ' $A B$ ' having the $\mathrm{NaCl}$ 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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$A_3 B_4$
There were $6 A$ atoms on the face-ceptres removing face-ceptred atoms along one of the axes means removal of $2 A$ atoms. Now, number of $A$ atoms per unit cell
$=\underset{\text { (corners) }}{8 \times \frac{1}{8}}+\underset{\text { (face-centred) }}{4 \times \frac{1}{2}}=3$
Number of $B$ atoms per unit cell
$\begin{aligned} & =12 \times \frac{1}{4} \quad 1=4 \\ & \text { (edgecentred) }+ \text { (body centred) } \\ & \end{aligned}$
Hence the resultant stoichiometry is $A_3 B_4$
$=\underset{\text { (corners) }}{8 \times \frac{1}{8}}+\underset{\text { (face-centred) }}{4 \times \frac{1}{2}}=3$
Number of $B$ atoms per unit cell
$\begin{aligned} & =12 \times \frac{1}{4} \quad 1=4 \\ & \text { (edgecentred) }+ \text { (body centred) } \\ & \end{aligned}$
Hence the resultant stoichiometry is $A_3 B_4$
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