If the anions $(A)$ form hexagonal closed packing and cations $(C)$ occupy $2 / 3^{r d}$ of all 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 $C_{2} A_{3}$
Number of given anions $A$ in the HCP = Number of atoms per unit cell = 6
Number of octahedral voids in HCP = 6
C atoms occupy $\frac{2}{3}$ of the octahedral voids
So, number of C atoms $6 \times \frac{2}{3}$ = 4
i.e $A_{6} C_{4}$ or $A_{3} C_{2}$

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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

\frac{1}{3} r d

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