JEE Advanced 2024 Paper 1 · Chemistry · Solutions, Solid State and Surface Chemistry · medium

In a normal spinel AB2O4, the oxide ions form a cubic close-packed lattice, the A^2+ ions occupy one-eighth of the tetrahedral voids and the B^3+…

Numerical answer

In a normal spinel

\mathrm{AB_2O_4}

, the oxide ions form a cubic close-packed lattice, the

\mathrm{A^{2+}}

ions occupy one-eighth of the tetrahedral voids and the

\mathrm{B^{3+}}

ions occupy one-half of the octahedral voids. The total number of voids (tetrahedral plus octahedral) that remain unoccupied per unit cell is

Show answer and solution

Answer: 9

A fcc unit cell of the oxide lattice contains

4

\mathrm{O^{2-}}

ions,

4

octahedral voids and

8

tetrahedral voids (a total of 12 voids). Occupied tetrahedral voids:

\tfrac18\times8=1

(

\mathrm{A^{2+}}

). Occupied octahedral voids:

\tfrac12\times4=2

(

\mathrm{B^{3+}}

). This corresponds to

4\ \mathrm{O^{2-}}

1\ \mathrm{A}

2\ \mathrm{B}

, i.e. the formula

\mathrm{AB_2O_4}

, as required. Unoccupied tetrahedral voids:

8-1=7

; unoccupied octahedral voids:

4-2=2

. Total unoccupied voids per unit cell:

7+2=9

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