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In a solid AB having the Nacl structure A atoms occupy the corners of the cubic unit cell . If all the faces centred atoms along one of the axis are removed then the resultant stoichiometry of the solid is

$(a)\;AB_2\qquad(b)\;A_2B\qquad(c)\;A_4B_3\qquad(d)\;A_3B_4$

1 Answer

Na in NaCl has 8 corners and 6 face atoms. If we remove face centered atom of one axis , two face atoms are removed.
Thus A is at 8 corners and B is at 4 faces.
$\therefore A = \large\frac{8}{8} = 1$
$\Rightarrow B = \large\frac{4}{2} = 2$
Thus formula is $AB_2$
Hence answer is (a)
answered Feb 17, 2014 by sharmaaparna1
 

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