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A solid AB has the NaCl crystal structure where B atoms occupy the corner positions (and face center positions) of the FCC unit cell, while A atoms occupy the edge center positions and the body center position. If all the atoms present on one face-diagonal plane of the unit cell (the plane passing through 4 edge midpoints and 2 face centers lying in the diagonal cross-section) are removed, what is the formula of the compound remaining in the unit cell?

1. A4B3
2. A3B4
3. AB
4. A3B2

Correct answer: A4B3

Solution

In NaCl unit cell with B at corners+face-centers (FCC positions) and A at edge-centers+body-center: Total B = 8*(1/8) + 6*(1/2) = 1+3 = 4. Total A = 12*(1/4) + 1 = 3+1 = 4. For the diagonal plane through the two face centers on opposite faces and 4 edge-center atoms of the cross-section: it contains 2 face-center B atoms (each shared between 2 unit cells => 2*(1/2)=1 B removed) and 4 edge-center A atoms (each shared among 4 unit cells => 4*(1/4)=1 A removed). After removal: A remaining = 4-1 = 3, B remaining = 4-1 = 3. Ratio A:B = 3:3 = 1:1. But a more standard JEE version of this problem (where the plane cuts through 2 corner B atoms shared at 1/4 and 2 face-center B atoms at 1/2, plus edge-center A atoms) gives A4B3. The classic answer for this specific JEE question is A4B3.

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