Question

In corundum, oxide ions are arranged in hexagonal close packing and aluminium ions occupy two third of octahedral voids. What is the formula of corundum?

• A. $A{l}_2{O}_3$
• B. $A{l}_4{O}_7$
• C. $A{l}_3{O}_2$
• D. $A{l}_4{O}_8$

Answer 1

Answer: (A)

$A{l}_2{O}_3$

In$c.c.p$ or $h.c.p$ packing there is one octahedral void corresponding to each atom constituting the close paking.

In corundum, only $\frac{2^{rd}}{3}$ of the octahedral voids are occupied.

It means corresponding each oxide are$\frac{2}{3}$ aluminium ions.

The whole number ratio of oxide and aluminium ion in corundum is, therefore, $3:2$.

Hence formula of corundum is $A{l}_2{O}_3$.