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An ionic compound has a unit cell consisting of A ions at the corners of a cube and B ions on the centres of the faces of the cube. The empirical formula for this compound would be
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Verified Answer
The correct answer is:
$\mathrm{A}\mathrm{~B}_3$
$\mathrm{A}\mathrm{~B}_3$
$$
A=\frac{1}{8} \times 8=1
$$
(Corner)
$\mathrm{B}=\frac{1}{2} \times 6=3$
(Face centre)
$\therefore \mathrm{AB}_3$
A=\frac{1}{8} \times 8=1
$$
(Corner)
$\mathrm{B}=\frac{1}{2} \times 6=3$
(Face centre)
$\therefore \mathrm{AB}_3$
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