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A compound $AB_x$ is formed such that A atoms are at corners and B atoms at face centre. Find the value of x .

$(a)\;2\qquad(b)\;3\qquad(c)\;4\qquad(d)\;5$

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A atoms are at the corners.
$\therefore$ No of A atoms per unit cell = $8\times\large\frac{1}{8}$
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = 1$
B atoms are face centre
$\therefore$ No of B atoms per unit cell = $6\times\large\frac{1}{2}$
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=3$
$\therefore AB_x = AB_3$
$\Rightarrow x = 3$
Hence answer is (b)
answered Feb 24, 2014 by sharmaaparna1
 

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