In a solid AB having the NaCl structure, A atom occupies the cornerns of the cubic unit cell. If all the face-centred atoms along one of the axes are removed, then the resultant stoichiometry of the solid is

A B_{2}

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A_{2} B

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A_{4} B_{3}

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A_{3} B_{4}

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Solution

The correct option is D $A_{3} B_{4}$
AB has NaCl-type structure (fee system).
$∴$ Number of A atoms
$= 8 (c o r n e r) \times \frac{1}{8} (p e r c o r n e r s h a r e) + 6 (f a c e s) \times (p e r f a c e c e n t e r s h a r e)$
= 1 + 3 = 4/unit cell

Number of B atoms
$= 12 (e d g e s) \times \frac{1}{4} (p e r e d g e c e n t r e s h a r e) + 1 (b o d y c e n t r e)$
= 1 + 3 = 4/unit cell

Number of A atoms removed (face centred atoms of one axis)
$= 2 (f a c e s) \times \frac{1}{2} (p e r f a c e c e n t r e s h a r e)$
= 1/unit cell

Number of A atoms left = 4 - 1 = 3/unit cell.

Formula = $A_{3} B_{4}$

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