Problem 144 - A Complete Resource Book in Chemistry for JEE Main

Question

In an face centred cubic lattice, atom A occupies the corner positions and atom B occupies the face centre positions. If one atom of \(\mathrm{B}\) is missing from one of the face centred points, the formula of the compound is [2011] (a) \(\mathrm{A}_{2} \mathrm{~B}_{5}\) (b) \(\mathrm{AB}_{2}\) (c) \(\mathrm{A}_{2} \mathrm{~B}\) (d) \(\mathrm{A}_{2} \mathrm{~B}_{3}\)

Step-by-Step Solution

Answer

The formula of the compound is \( \mathrm{AB}_{2} \).

Step 1: Calculating Atoms of A

In a face-centered cubic (FCC) lattice, there are 8 corners. Each corner atom is shared by 8 unit cells, so the contribution of each corner atom is \( \frac{1}{8} \). Therefore, the total contribution of atom A at the corners is \( 8 \times \frac{1}{8} = 1 \) atom of A.

Step 2: Calculating Atoms of B

In an FCC lattice, there are 6 face-centered positions, and each face atom is shared by 2 unit cells. So, the total contribution from face-centered atoms is \( 6 \times \frac{1}{2} = 3 \) atoms. But since one atom is missing from the face-centered points, the number of B atoms becomes \( 3 - 1 = 2 \).

Step 3: Determine the Formula

Given 1 atom of A and 2 atoms of B, the empirical formula for the compound based on their numbers is \( \mathrm{AB}_{2} \).

Key Concepts

Empirical Formula Determination

Determining the empirical formula of a compound involves finding the simplest whole number ratio of the moles of elements present. In a face-centered cubic (FCC) lattice, each type of atom's contribution to the structure helps determine this formula. For instance, if different atoms occupy specific positions within the lattice, their contributions can be calculated to find the empirical formula.

In our problem, atom A occupies the corner positions, while atom B is mainly at the face-centered positions. To find the empirical formula:

• Calculate the contribution of each atom within the lattice.
• Add up the contributions to find the total number of atoms of each type in the unit cell.
• Simplify the ratio of the atoms to deduce the empirical formula.

End result: Given 1 atom of A and 2 atoms from B, the empirical formula is determined to be \( \text{AB}_2 \).

Atom Contribution in Lattices

Atoms in a crystal structure often share positions in multiple unit cells, making it crucial to understand how much each atom contributes to the lattice. An FCC lattice, like the one in the exercise, provides a clear example.

In an FCC lattice:

• Corners: There are 8 corner atoms, each shared by 8 unit cells. This means each corner atom contributes \( \frac{1}{8} \) of an atom to the unit cell.
• Face Centers: There are 6 face-centered atoms, each shared by 2 unit cells, contributing \( \frac{1}{2} \) of an atom.

The total contribution from A at the corners is 1 (since \( 8 \times \frac{1}{8} = 1 \)). For B, we calculate \( 6 \times \frac{1}{2} = 3 \) atoms, but given one atom is missing, 2 remain. Understanding atom contribution is key to calculating the empirical formula.

Crystal Lattice Structure Analysis

Analyzing crystal lattice structures involves examining how atoms are arranged and how they interact to form the solid. Each structure has distinct geometric patterns and atom-sharing properties.

Face-centered cubic (FCC) lattices are a common type of crystal lattice where each unit cell comprises:

• Corner atoms that contribute very little to the cell as they are extensively shared.
• Face-centered atoms, contributing more significantly since they are less shared than corner atoms.

By analyzing the positions and contributions of each atom, we can determine how these atoms collectively stabilize and form the larger crystal structure. Such analysis is crucial in materials science as it influences the compound's properties, like density and strength, and directly impacts the chemical formula as seen in our example with \( \text{AB}_2 \).