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

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AB₂

A₂B

A₄B₃

A₃B₄

answer is C.

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Detailed Solution

According to the structure of NaCl

′A′ will occupy 8 corners of cubic unit cell which will contribute 1/8 each to the cube

8×1/8=1

'6' face centered atoms will contribute1/2 to the cube
Hence 6×1/2=3

Therefore total contribution of atom 'A' is 3+1=4
Now B occupies octahedral voids which contributes 1/4
therefore 1+12×1/4=4
So the formula is A is- A₄B₄or AB
Now we are removing all face centered atoms along one axis-
so the number of atoms removed from cube are 2
so contribution from 'A' remains 8×1/8 (corner atoms)+4×1/2 (face centered atoms)=3
so the resultant becomes A₃B₄

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