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If the anions (A) from hexagonal closest packing and cations (C) occupy only 2/3 octahedral voids in it, then the general formula of the compound is:

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CA₂

C₂A₃

C₃A₂

answer is C.

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Detailed Solution

A form HCP (Z=6) -‘A6’, T.V = 12, O.V = 6
‘C’ occupies 2/3 O.V = 6 x 2 / 3 = 4
i.e., C4, C₄A₆ = C₂A₃

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