Question

For the spinel structure $MgA{{l}_{2}}{{O}_{4}}$, the correct statement is / are
This question has multiple correct options
A) 50% of the octahedral voids are occupied by ions
B) $A{{l}^{3+}}$ ions are equally distributed in tetrahedral voids and octahedral voids.
C) Oxide ions occupy ccp lattice
D) 12.5% of the tetrahedral voids are occupied by ions.

Answer 1

Hint: Normal spinel structure has eight tetrahedral voids and four octahedral voids and are occupied by the ions.
- In $MgA{{l}_{2}}{{O}_{4}}$, the oxide ions are arranged in cubic close packing.

Complete step by step answer:
The spinel structure with $A{{B}_{2}}{{O}_{4}}$, is having the normal spinel structure. An example of a spinel structure with the general formulae of $A{{B}_{2}}{{O}_{4}}$ is $MgA{{l}_{2}}{{O}_{4}}$. In the $MgA{{l}_{2}}{{O}_{4}}$, there is one Mg atom, two Al atom and four O atoms, which resembles to $A{{B}_{2}}{{O}_{4}}$ structure. In this structure the A atoms occupy one-eighth of the tetrahedral voids and B atoms occupy half of the octahedral voids.
- Now let’s see the basic idea about cubic close packing, the unit cell associated, number of tetrahedral and octahedral voids.
Cubic close packing – It is the way of packing of atoms in which the atoms are so closely packed. ‘The unit cell in cubic close packing (ccp) is face centered cubic (fcc).
- FCC is an arrangement of four cubic close packed layers. The fcc unit cell contains four atoms.
- FCC contributes four atoms, one eighth of the atom from the atoms placed in the corner, one atom which is placed in the centered of the cube and is not shared by nearby lattice, and half of an atom from the six atoms present in the faces.
- So now let’s see the options given in the question and opt for the correct statements.
We know that in the spinel structure, the oxide ions are arranged in ccp and have an fcc unit cell.
And we know that the number of oxide ions, ${{O}^{2-}}$ = 4
Number of tetrahedral voids present in ccp =8
Number of octahedral voids present in ccp = 4
So, as explained above the $M{{g}^{2+}}$ ions will occupy half of the tetrahedral voids,
$No.\,of\,M{{g}^{2+}}\,ions=\dfrac{1}{8}\times {{T}_{d}}\,voids$
$No.\,of\,M{{g}^{2+}}\,ions=\dfrac{1}{8}\times 8=1$
 Number of $A{{l}^{3+}}$ ions =$\dfrac{1}{2}\times {{O}_{d}}\,voids$
 $No.\,of\,A{{l}^{3+}}\,ions=\dfrac{1}{2}\times 4=2$
$Percentage\,occupied\,by\,M{{g}^{2+}}=\dfrac{1}{8}\times 100=12.5%$
$Percentage\,occupied\,by\,A{{l}^{3+}}=\dfrac{1}{2}\times 100=50%$
Hence from the above option (A), (B) and (D) are the correct options.
The crystal lattice structure of $MgA{{l}_{2}}{{O}_{4}}$ is as follows.

So the correct answer is “A”:

Note: Here the value of Z in FCC is equal to 4.
As the FCC unit cell has 4 atoms (N). The number of octahedral voids in ccp will be N and tetrahedral voids will be 2N.