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A crystal is made of particles A and B. A forms FCC packing and B occupies all the octahedral voids. If all the particles along the plane as shown in figure are removed, then, the formula of the crystal would be:

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A₂B

$A_{5} B_{7}$

$A_{7} B_{5}$

$A B$

answer is A.

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Detailed Solution

Atoms "A" forming FCC packing

Effective atoms of A = $8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 4$

Octahedral voids = 4 ; Tetrahedral voids = 8

Atoms along diagonal plane is removed, octahedral voids remains same.Only tetrahedral voids are present along diagonal plane

Number of "B" atoms = 4

Formula A₄B₄ = AB

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