The anions $(A)$ form hexagonal closest packing and atoms $(M)$ occupy only two-thirds of octahedral voids in it; then the general formula of the compound is:

M A

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M A_{2}

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M_{2} A_{3}

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M_{3} A_{2}

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Solution

The correct option is D $M_{2} A_{3}$
Effective numbers of $A, Z_{A} = 6$ (Hexagonal closest packing)
Effective number of $M, Z_{M} = \frac{2}{3} \times o c t a h e d r a l v o i d s = \frac{2}{3} \times 6 = 4$
Molecular formula $M_{4} A_{6} = M_{2} A_{3}$

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The anions (A) form hexagonal closest packing and cation (C) occupy only 2/3 of octahedral voids in it, then the general formula of the compound is

\frac{2}{3}

\frac{2}{3} r d

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