In an FCC arrangement of A and B atoms, where A atoms are at the corners of the unit cell, B atoms at the face centers, two atoms are missing from two corners in each unit cell, then the simplest formula of a compound is

$A B_{6}$

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$A_{2} B_{3}$

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$A_{7} B_{24}$

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$A B_{4}$

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Solution

The correct option is D
$A B_{4}$

Since there are six atoms (A) in the corner of the unit cell, the contribution of atoms in 1 unit cell is 6/8 = 3/4. Since 3 face-centered atoms (B) contribute to one unit cell, the formula is $A_{0.75} B_{3}$

= $A_{3} B_{12}$

= $A B_{4}$ .

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In an FCC arrangement of A and B atoms, where A atoms are at the corners of the unit cell, B atoms at the face centers, two atoms are missing from two corners in each unit cell, then the simplest formula of a compound is

The correct option is D AB4 Since there are six atoms (A) in the corner of the unit cell, the contribution of atoms in 1 unit cell is 6/8 = 3/4. Since 3 face-centered atoms (B) contribute to one unit cell, the formula is A0.75B3 = A3B12 = AB4.

The correct option is D
$A B_{4}$

Since there are six atoms (A) in the corner of the unit cell, the contribution of atoms in 1 unit cell is 6/8 = 3/4. Since 3 face-centered atoms (B) contribute to one unit cell, the formula is $A_{0.75} B_{3}$

= $A_{3} B_{12}$

= $A B_{4}$ .