Question

In a cubic closed packed structure of mixed oxides, the lattice is made up of oxide ions, \(20 \%\) of tetrahedral Voids are occupied by divalent \(\mathrm{A}^{2+}\) ions and \(50 \%\) of the octahedral voids by trivalent \(\mathrm{B}^{3+}\). The formula of the oxide is? (a) \(\mathrm{A}_{4} \mathrm{~B}_{5} \mathrm{O}_{10}\) (b) \(\mathrm{A}_{2} \mathrm{~B}_{5} \mathrm{O}_{5}\) (c) \(\mathrm{A}_{2} \mathrm{BO}\) (d) \(\mathrm{A}_{4} \mathrm{~B}_{5} \mathrm{O}_{8}\)

Short answer

Option (a) \(\mathrm{A}_{4} \mathrm{~B}_{5} \mathrm{O}_{10}\) is correct.

Step-by-step solution

- Determine Number of Oxide Ions

In a cubic closed packed (ccp) structure, oxide ions make up the lattice points. Let's assume there are 1 mole of oxide ions. Therefore, there are 1 mole of available sites.

- Calculate Tetrahedral Voids

For each oxide ion in a ccp structure, there are 2 tetrahedral voids, since there are 1 mole of oxide ions, there are 2 moles of tetrahedral voids.

- Calculate Occupied Tetrahedral Voids

Given that 20% of the tetrahedral voids are occupied by divalent \(\mathrm{A}^{2+}\) ions:\[\text{Occupied Tetrahedral Voids} = 0.2 \times 2 = 0.4 \text{ moles of } \mathrm{A}^{2+}\]Thus, 0.4 \( \mathrm{A}^{2+}\) ions are present.

- Calculate Octahedral Voids

For each oxide ion, there is one octahedral void, so for 1 mole of oxide ions, there is 1 mole of octahedral voids.

- Calculate Occupied Octahedral Voids

Given that 50% of the octahedral voids are occupied by trivalent \(\mathrm{B}^{3+}\) ions:\[\text{Occupied Octahedral Voids} = 0.5 \times 1 = 0.5 \text{ moles of } \mathrm{B}^{3+}\]Therefore, 0.5 \( \mathrm{B}^{3+}\) ions are present.

- Determine the Formula of the Compound

To determine the chemical formula, equate the moles of \(\mathrm{A}^{2+}\), \(\mathrm{B}^{3+}\), and the oxide ions \( \mathrm{O}^{2-}\). The empirical formula is structured based on these moles:\[\mathrm{A}_{0.4} \mathrm{B}_{0.5} \mathrm{O}_{1}\]By multiplying through by 2 to eliminate the decimals, we get:\[\mathrm{A}_{0.4 \times 2} \mathrm{B}_{0.5 \times 2} \mathrm{O}_{1 \times 2} = \mathrm{A}_{0.8} \mathrm{B}_1 \mathrm{O}_2\]

- Simplify the Ratio for the Final Formula

Upon simplifying this expression further to whole numbers for clarity and comparison with options provided, the empirical formula proportionally corresponds to \(\mathrm{A}_{4} \mathrm{~B}_{5} \mathrm{O}_{10}\)Thus, this is the formula.

Key concepts

Cubic Closed Packed Structure

The cubic closed packed (CCP) structure is a type of crystalline arrangement in solids where spheres (often ions or atoms) are packed as closely as possible. This structure is also known as face-centered cubic (FCC) due to the arrangement of atoms along the corners and faces of a cube.

In a CCP structure, each atom is surrounded by 12 neighbors, which is called the coordination number. This high coordination number ensures maximum packing efficiency, where approximately 74% of the available volume is occupied by atoms.

The CCP structure is common in metals and many ionic compounds due to its stability and low energy configuration. The efficient use of space and stable structure is crucial in forming both metallic and ionic compounds, helping to determine properties like density and hardness.

Tetrahedral Voids

In the context of close packing structures, voids or interstitial sites are small spaces between packed atoms or ions. These spaces can host additional atoms or ions and are categorized based on the geometry of the surrounding particles. Tetrahedral voids form when a triangular base, made of three atoms from one layer, is topped by a single atom from the next layer.

In a CCP structure, each atom is associated with two tetrahedral voids. Tetrahedral voids are typically smaller than octahedral voids, their counterparts. However, they are crucial in the formation of ionic compounds, where smaller ions usually occupy these positions.

Understanding tetrahedral voids is important for predicting the chemical formula of compounds, as these voids often determine how many and which type of ions an arrangement can accommodate.

Octahedral Voids

Octahedral voids occur in close-packed structures where six atoms form an octahedral geometry around an empty center. These voids are formed by aligning three atoms from one layer directly above or below three atoms in the next layer.

Within a CCP structure, each lattice point contributes to one octahedral void, hence there is one octahedral void per atom. These voids are larger than tetrahedral ones and can accommodate larger ions. Understanding which ions occupy these voids helps in determining the stoichiometry of a compound.

In many ionic compounds, the larger cations and anions occupy octahedral voids due to their size, playing a vital role in the stability and composition of the compound's crystal lattice.

Empirical Formula Calculation

The empirical formula represents the smallest whole number ratio of the elements in a compound, providing fundamental insights into a substance's composition. Calculating this involves determining the ratio of atoms or ions by their moles.

In the context of ionic compounds using voids in a CCP structure, knowing how atoms fill the tetrahedral and octahedral voids helps determine the moles of each ion present. This ratio is crucial for deriving the empirical formula.

For example, if 20% of tetrahedral voids are filled by divalent ions and 50% of octahedral voids by trivalent ions, we calculate their respective moles and find the simplest ratio compared to anions present in the structure. This results in an empirical formula that describes the proportionate presence of each element in the compound effectively.