If the anions (A) form hexagonal closed packing and cations (C) occupy only two-third of octahedral voids in it, then the general formula of the compound is :

C A

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C A_{2}

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C_{2} A_{3}

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C_{3} A_{2}

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Solution

The correct option is C $C_{2} A_{3}$
A for hcp and effective number of atoms $= 6$

Number of octahedral voids $= 6$

Number of cations (C) $= 6 \times \frac{2}{3} = 4$

The formula is $C_{4} A_{6}$ i.e.., $C_{2} A_{3}$

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