Question

If the unit cell of a material 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) $\dfrac{1}{2},\dfrac{1}{8}$
(b) $1,\dfrac{1}{4}$
(c) $\dfrac{1}{2},\dfrac{1}{2}$
(d) $\dfrac{1}{4},\dfrac{1}{8}$

Answer 1

Hint: As we know that in ccp or cubic centred packing there are two types of common voids are formed. In this number of octahedral voids present in the lattice is equal to the number of lattice points and the number of tetrahedral voids is twice the number of lattice points.

Complete step by step solution:
As we know that each unit cell is adjacent to another and most of the atoms are shared by neighbouring unit cells due to which only some portion of each atom belongs to a particular unit cell.

Let us recall that in given solids generally the bigger ions that are anions form the cubic centred packing and smaller ions that are cations occupy the voids. The position of these voids can be determined as we know in a unit cell is divided into eight small cubes in which each cube has atoms at alternate corners. So in tetrahedron, there are $8$ tetrahedral voids and we also know that a ccp or fcc structure of a unit cell possesses $8$ lattice point at $8$ corners and $6$ lattice point at the face-centres. On joining the face centres atoms we get an octahedron which will have $4$ octahedral voids per unit cell.

Now, the number of voids formed depends on the number of closed packed spheres. So that in ccp array,
The number of oxygen atom $ = 4$
The numbers of octahedral voids formed be $N = $$4$
The number of tetrahedral voids formed be $2N$$ = 8$
Now we are given that aluminium ions occupied octahedral voids or holes. So, we can write:
Number of $A{l^{3 + }}$ ions $ = 4 \times m$
And magnesium ions occupied tetrahedral voids. So, we can write:
Number of $M{g^{2 + }}$ ions $ = 8 \times n$
Lastly we know that molecules of material will be neutral, hence we all add all of the calculated terms and we will get:
$\Rightarrow 4 \times ( - 2) + 4 \times m \times ( + 3) + 8 \times n \times ( + 2) = 0$
$\Rightarrow 12m + 16n = 8$
$\Rightarrow 3m + 4n = 2$
On solving for m and n we will get:
$m = \dfrac{1}{2}$ and $n = \dfrac{1}{8}$.

Therefore, the correct answer is option(A).

Note: By knowing the types of voids, locations of voids and the fractions occupied by these voids we can also determine the formula of the compound as we know that octahedral voids per unit cell is $4$ and tetrahedral voids per unit cell is double the number of lattice points that is $8$.