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If oxide ions are arranged in hcp and the aluminium ions occupy two thirds of octahedral voids. What will be the formula of the compound?
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The correct answer is:
$\mathrm{Al}_2 \mathrm{O}_3$
In hcp arrangement of oxide ion.
Number of octabedral (oh) voids for each oxide ion $=1$
Number of tetrahedral $(\mathrm{Td}$ ) voids for each oxide ion $=2$
$\mathrm{Al}^{3+}$ occupies only $2 / 3$ of octahedral voids.
Thus, number of $\mathrm{Al}^{3+}$ for each oxide ion $=2 / 3$
$\Rightarrow$ Ratio of oxide ion : Aluminium ion
$\Rightarrow 3: 2$
Thus, formula is $\mathrm{Al}_2 \mathrm{O}_3$.
Number of octabedral (oh) voids for each oxide ion $=1$
Number of tetrahedral $(\mathrm{Td}$ ) voids for each oxide ion $=2$
$\mathrm{Al}^{3+}$ occupies only $2 / 3$ of octahedral voids.
Thus, number of $\mathrm{Al}^{3+}$ for each oxide ion $=2 / 3$
$\Rightarrow$ Ratio of oxide ion : Aluminium ion
$\Rightarrow 3: 2$
Thus, formula is $\mathrm{Al}_2 \mathrm{O}_3$.
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