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A certain oxide of metal M crystallises in such a way that O^2– ions occupy hcp arrangement following AB AB....pattern. The metal ions, however, occupy 2/3rd of the octahedral voids. The formula of the compound is

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MO₂

M₃O

M_8/3O₃

M₂O₃

answer is A.

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Detailed Solution

$AB - AB - HCPZ = 6, T . V = 12, OV = 6$
$O^{- 2}$ in HCP array So, $' O_{6}'; 'M' occupies 2 / 3$
of O.V= 2 / 3 x 6 = 12 / 3 ;
$M 12 / 3 O_{6} \Rightarrow M_{12} O_{18} \Rightarrow M_{2} O_{3}$

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