Question

If the unit cell of a mineral has cubic close packed (ccp) array of oxygen atoms with $m$ fraction of octahedral holes occupied by aluminium ions and $n$ fraction of tetrahedral holes occupied by magnesium ions. $m$ and $n$ respectively are:

• A. $\frac{1}{2},\frac{1}{8}$
• B. $1,\frac{1}{4}$
• C. $\frac{1}{2},\frac{1}{2}$
• D. $\frac{1}{4},\frac{1}{8}$

Answer 1

The correct option is A

$\frac{1}{2},\frac{1}{8}$

For ccp, Z = 4 = no. of O-atoms
No. of octahedral voids $=4$
No. of tetrahedral voids $=2\times 4=8$
No. of $A{l}^{3+}$ ions $=m\times 4$
No. of $M{g}^{2+}$ ions $=n\times 8$
$\text{Thus, the formula of the mineral is}A{l}_{4m}$ $M{g}_{8n}$$O_4$.

$4m(+3)+8n(+2)+4(-2)=012m+16n-8=04(3m+4n-2)=03m+4n=2$

$\text{Possible values of}m\text{and}n\text{are}\frac{1}{2}\text{and}\frac{1}{8}\text{respectively}$.