Problem 138 - A Complete Resource Book in Chemistry for JEE Main

Question

An ionic compound has a unit cell consisting of A ions at the corners of a cube and \(\mathrm{B}\) ions on the centres of the faces of the cube. The empirical formula for this compound would be [2005] (a) \(\mathrm{AB}\) (b) \(\mathrm{A}_{2} \mathrm{~B}\) (c) \(\mathrm{AB}_{3}\) (d) \(\mathrm{A}_{3} \mathrm{~B}\)

Step-by-Step Solution

Answer

The empirical formula for the compound is \( \mathrm{AB}_3 \).

Step 1: Identify A ion position

The A ions are located at the corners of the cubic unit cell. A cube has 8 corners.

Step 2: Calculate contribution of A ions

Each corner ion is shared by 8 adjacent unit cells. Therefore, the contribution of one A ion at the corner is \( \frac{1}{8} \). Thus, the total contribution of A ions is \( 8 \times \frac{1}{8} = 1 \).

Step 3: Identify B ion position

The B ions are located at the centers of the 6 faces of the cubic unit cell.

Step 4: Calculate contribution of B ions

Each face-centered ion is shared by 2 adjacent unit cells. Therefore, the contribution of one B ion at the face is \( \frac{1}{2} \). Thus, the total contribution of B ions is \( 6 \times \frac{1}{2} = 3 \).

Step 5: Determine empirical formula

From the contributions, we have 1 A ion and 3 B ions per unit cell. Therefore, the empirical formula is \( \mathrm{AB}_3 \).

Key Concepts

Ionic Compounds

Understanding ionic compounds is a crucial part of studying crystal structures. An ionic compound consists of positively and negatively charged ions. These ions are bound together by strong electrostatic forces known as ionic bonds. In these compounds, metals usually donate electrons to become positively charged cations, while non-metals accept electrons to become negatively charged anions. This transfer ensures both types of ions achieve a stable electronic configuration. Some characteristics of ionic compounds include:

• High melting and boiling points because of the strong forces of attraction between the ions.
• They are hard and brittle in solid form.
• When dissolved in water, they conduct electricity because the ions are free to move.

Ionic compounds commonly form a crystalline structure. This structure is composed of a repeated arrangement of ions, known as a lattice. The specific positions of ions within the lattice ensure that each cation is surrounded by anions and vice versa, maintaining overall electrical neutrality. Understanding how these structures form and interact aids in predicting the properties of ionic compounds.

Unit Cell

The unit cell is a fundamental concept in crystallography, as it serves as the building block of a crystalline structure. It defines the smallest repeatable unit that, when stacked together in three-dimensional space, creates the entire lattice of the crystal. Think of it like a single brick that, when repeated, constructs a complete wall. In our specific example, the ions inside a unit cell were arranged as follows:

• A ions are located at the corners of the cubic unit cell.
• B ions are positioned at the centers of the cube's faces.

The precise arrangement of these ions within the unit cell significantly influences the properties of the ionic compound, such as density, symmetry, and overall geometry. Calculating the contribution of each ion involved looking at their shared positioning across adjacent unit cells. Recognizing the unit cell’s configuration helps in deriving essential aspects like the empirical formula and understanding the compound's macroscopic properties.

Empirical Formula

The empirical formula of a compound represents the simplest whole number ratio of the ions present in a compound. Unlike the molecular formula, which tells you the exact number of each type of atom in a compound, the empirical formula provides a simplified version that is useful for identifying the proportionate amounts of each element.In the case of the ionic compound presented in the exercise, deriving the empirical formula involved:

• Recognizing that each A ion at the corners contributes \( \frac{1}{8} \) to the unit cell because it was shared by 8 cubes.
• Understanding that each B ion in the face of the cube had a contribution of \( \frac{1}{2} \) because it was shared with an adjacent unit cell.
• Summarizing these contributions resulted in 1 A ion and 3 B ions per unit cell.

Thus, the empirical formula was determined as \( \mathrm{AB}_3 \) based on the calculated contributions. This formula provides valuable insights into the composition of the crystal, helping in the comprehension of its stoichiometric properties and material characteristics.