Question

In a compound, oxide ions are arranged in CCP arrangement. Cations A occupy onesixth of the tetrahedral voids and cations B occupy one-third of the octahedral voids. The formula of the compound is (a) \(\mathrm{AB}_{2} \mathrm{O}_{4}\) (b) \(\mathrm{ABO}_{3}\) (c) \(\mathrm{ABO}_{2}\) (d) \(\mathrm{ABO}_{4}\)

Short answer

\(ABO_3\)

Step-by-step solution

Identify Number of Oxide Ions

In a cubic close-packed (CCP) arrangement, if we consider that there is one oxide ion for each lattice point, then for each unit cell, there would be 4 oxide ions as the CCP has 4 lattice points per unit cell.

Calculate Number of Tetrahedral Voids

In a CCP structure, the number of tetrahedral voids is twice the number of anions. Therefore, with 4 oxide ions, there will be 8 tetrahedral voids.

Determine Occupancy of Tetrahedral Voids by Cations A

Cations A occupy one-sixth of the tetrahedral voids. So the number of cations A would be 8 x (1/6) = 4/3 or approximately 1.33 cations A per unit cell.

Calculate Number of Octahedral Voids

In a CCP structure, the number of octahedral voids is equal to the number of anions. Hence, there would be 4 octahedral voids.

Determine Occupancy of Octahedral Voids by Cations B

Cations B occupy one-third of the octahedral voids. So the number of cations B would be 4 x (1/3) = 4/3 or approximately 1.33 cations B per unit cell.

Derive the Formula of the Compound

The ratio of cations A to oxide ions O is 1.33:4, which simplifies to 1:3 when divided by 1.33. Similarly, the ratio of cations B to oxide ions O is 1.33:4, also simplifying to 1:3. Combining these, we get a stoichiometry of \(A_1O_3\) and \(B_1O_3\), giving us the overall formula \(ABO_3\).

Key concepts

CCP arrangement

When studying crystal structures, the cubic close-packed (CCP) arrangement, also known as face-centered cubic (FCC) structure, is essential to understand. In this configuration, oxide ions are often considered as reference points, organizing themselves into a lattice where each ion is surrounded by others at each face of the cube. This arrangement offers high packing efficiency — about 74% of the space in the crystal is filled.

The CCP structure is characterized by layers of spheres (in this case, oxide ions) where each sphere is surrounded by six others in the same layer, three in the layer above, and three in the layer below. This sequence repeats in the pattern ABCABC... offering tight packing and leaving little unoccupied space. This pattern forms both tetrahedral and octahedral voids, which are spaces between the oxide ions that can be occupied by cations of a smaller size.

Tetrahedral voids occupancy

Tetrahedral voids are formed at the points where four oxide ions form a tetrahedron — a four-faced structure with triangular sides. Every oxide ion lies at one corner of such a tetrahedron. In a CCP arrangement, the number of tetrahedral voids is twice as many as the number of oxide ions. This is because each gap between the spheres can accommodate a small cation without disturbing the existing oxide framework.

When cations occupy these voids, it's important to consider the ratios. As shown in the exercise, cations A occupy only one-sixth of the available tetrahedral voids. Calculating the occupancy helps determine the ratio of cations to oxygens, ultimately leading to the formulation of the compound's chemical formula. It is crucial to precisely understand the concept of occupancy to predict the composition and properties of a crystal effectively.

Octahedral voids occupancy

Octahedral voids in a CCP structure are created at the points where six oxide ions are arranged in the shape of an octahedron — a polyhedron with eight triangular faces. These voids exist in the spaces between layers of the tightly packed ions, where each void is coordinated by six oxide ions. The total number of octahedral voids in a CCP arrangement equals the number of oxide ions.

In the given exercise, the octahedral voids are partially filled by cations B, with an occupancy rate of one-third. This partial occupancy highlights the versatility of crystal structures, where not every potential location for a cation is filled. Understanding how many voids are occupied by which type of cation helps in deducing the correct stoichiometry of the compound, a fundamental skill in crystal chemistry and material science.