A solid $A B$ crystallizes in $N a C l$ structure in which $B$ atoms are at the corners of the cube. If all the atoms present in the plane $a b c d$ are removed (as shown in the figure), then the formula of the compound will be :

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A_{3} B_{4}

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A_{4} B_{3}

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A_{3} B_{2}

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Solution

The correct option is D AB
Since the structure is of NaCl , there are 4 A atoms and 4 B atoms in the unit cell. B atoms are present at the corners and face centers and A atoms are present in octahedral voids.
If all the atoms present in the plane abcd are removed (as shown in the figure), 4 corner B atoms and 2 A atoms at edge centers are removed. Each B atom at corner contributes one eight to the unit cell
4 corner B atoms had contribution $4 \times \frac{1}{8} = \frac{1}{2}$ to the unit cell. Thus 0.5 B atoms are removed and $4 - 0.5 = 3.5$ B atoms remain in the unit cell.
2 A atoms are removed from the corner and each has contribution of one fourth to the unit cell.
They had contribution of $2 \times \frac{1}{4} = 0.5$ to the unit cell.
Thus out of 4 A atoms in the unit cell, 0.5 atoms are removed and 3.5 atoms remain.
Hence, the ratio of the number of A atoms to B atoms that remain in the unit cell is $3.5 : 3.5$ or $1 : 1$
Hence, the formula of the compound will be $A B$

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A solid AB crystallizes in NaCl structure in which B atoms are at the corners of the cube. If all the atoms present in the plane abcd are removed (as shown in the figure), then the formula of the compound will be :

A solid $A B$ crystallizes in $N a C l$ structure in which $B$ atoms are at the corners of the cube. If all the atoms present in the plane $a b c d$ are removed (as shown in the figure), then the formula of the compound will be :