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A compound formed by elements A and B has a cubic structure in which A atoms are at the corners of the cube and B atoms are at face centres. Derive the formula of the compound

$\begin{array}{1 1}(A)\;AB_3&(B)\;AB_2\\(C)\;AB&(D)\;\text{None of these}\end{array} $

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As ‘A’ atom are present at the 8 corners of the cube therefore no of atoms of A in the unit cell =$\large\frac{1}{8}$$\times 8=1$
As B atoms present at the face centres of the cube, therefore no of atoms of B in the unit cell =$\large\frac{1}{2}$$\times 6=3$
Hence the formula of compound is $AB_3$
Hence (A) is the correct answer.
answered May 28, 2014 by sreemathi.v

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