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In a solid $$''AB''$$ having $$NaCl$$ type structure $$''A''$$ atoms occupy the FCC lattice points of the cubic unit cell. If all the face-centered atoms along one of the axis are removed, then the resultant stoichiometry of the solid is :


A
AB2
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B
A2B
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C
A4B3
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D
A3B4
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Solution

The correct option is C $$A_{3}B_{4}$$
Since $$A$$ atoms occupy the corners of the cubic unit cell, the number of atoms occupying the corners is $$1$$. 

If all the face-centred atoms along one of the axes are removed, then only $$4$$ faces remain, which counts in total to $$2$$ atoms of $$A$$. 

$$B$$ is present in octahedral voids, so the total number of atoms of $$B$$ is $$4$$.

Hence, the stoichiometry is $$A_3B_4$$.

Hence, option is D.

Chemistry

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