An $A C$ solid has $N a C l$ type of unit cell. If atoms from the edge centres along an axis connecting the opposite edge centres on a face are removed, the empirical formula of the remaining solid would be :

A C

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A_{2} C_{4}

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A_{3} C_{4}

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A_{3} C_{2}

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Solution

The correct option is B $A_{3} C_{4}$
According to $N a C l$ structure, A atom will occupy all the corners and face centers whereas C atoms will occupy all the octahedral voids.
Total number of A atoms in NaCl structure $= 4$
Total number of C atoms in NaCl structure $= 4$
Contribution from atoms at corners $= \frac{1}{8} \times 8 = 1$
If atoms along one axis are removed, 4 atoms will remain at the face centers.
Contribution from atoms at face centers $= \frac{1}{2} \times 4 = 2$
Total number of A atoms $= 1 + 2 = 3$
Total number of C atoms $= 4$
Therefore, formula of the compound is $A_{3} C_{4}$

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