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Atoms of an element 'A' occupy $ \large\frac{2}{3}$ tetrahedral voids in the hexagonal closed packed ( hcp) unit cell lattice formed by the element 'B' . The formula of the compound formed by 'A' and 'B' is

$\begin {array} {1 1} (1)\;A_2B & \quad (2)\;AB_2 \\ (3)\;A_4B_3 & \quad (4)\;A_2B_3 \end {array}$

1 Answer

(3) $A_4B_3$
answered Nov 7, 2013 by pady_1
 
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