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: Calculate Atoms of A

In a face-centered cubic (FCC) lattice, the atom A occupies the corner positions. A cube has 8 corners and each corner atom is shared by 8 unit cells. Therefore, the total contribution of atom A in one unit cell is \( 8 \times \frac{1}{8} = 1 \).

Step 2: Calculate Atoms of B in Perfect FCC Lattice

Atom B is positioned at the face centers in the FCC lattice. There are 6 faces, and each face-centered atom is shared between 2 unit cells. Thus, the total contribution of atom B in one unit cell without any atoms missing is \( 6 \times \frac{1}{2} = 3 \).

Step 3: Adjust for Missing Atom B

The problem states that one atom of B is missing from the face-centered points. Thus, the total number of B atoms becomes \( 3 - 1 = 2 \).

Step 4: Determine the Empirical Formula

Based on the calculations, for every atom of A there are 2 atoms of B when one B atom is missing. Therefore, the empirical formula of the compound is \( \mathrm{AB}_{2} \).

Key Concepts

Atomic Positions in a Face-Centered Cubic Lattice

In a face-centered cubic (FCC) lattice, understanding atomic positions is crucial for calculating the overall composition of a unit cell.
Let's break it down into simpler terms:
- **Corners**: Each corner of the cubic lattice houses an atom. In the case of our example, atom A is located at these corner positions.- There are 8 corners in a cubic unit cell, and each of these corner atoms is shared among 8 neighboring unit cells. This means that the total contribution of corner atoms (atom A) in one individual unit cell is calculated as follows:
\[8 \text{ corners} \times \left(\frac{1}{8}\right) = 1 \text{ atom A per unit cell}\] - **Face Centers**: In an FCC lattice, one atom is located at the center of each face. Atom B takes these positions on our lattice.- A cube has 6 faces, and each face-centered atom is shared by 2 unit cells. The contribution of face-centered atoms (atom B) in one unit cell is:
\[6 \text{ faces} \times \left(\frac{1}{2}\right) = 3 \text{ atoms B per unit cell}\]

Unit Cell Calculation

The unit cell calculation is foundational to understanding the stoichiometry of a crystalline structure. Let's explore how to assess and adjust the unit cell based on given conditions.
Firstly, the perfect face-centered cubic (FCC) lattice, before any adjustment for missing atoms, contains a specific number of each type of atom that sums to represent the unit cell. Using our example with atoms A and B:
- **Atom A Contribution**: As calculated, the net contribution of atom A from the corners is 1.
- **Atom B Contribution**: Initially, before accounting for missing atoms, the contribution of B at the face centers totals to 3.
When one atom of B is removed from the face-centered positions, we need to adjust the count accordingly based on the scenario's conditions.
After removing one atom of B, the remaining contribution of B in one unit cell is \[3 - 1 = 2 \text{ atoms of B}\] Now we understand the atomic positioning and contributions within the unit cell, we're ready to determine its empirical formula.

Empirical Formula Determination

Determining the empirical formula of a substance involves comparing the relative quantities of each type of atom present within the unit cell.
From our calculations, here's what we've observed:
- **Atom A Contribution**: Since atom A has a total contribution of 1 in each unit cell, A is our baseline for comparison.
- **Atom B Contribution**: Adjusting for the missing B atom, we find 2 atoms of B in the unit cell.
To find the empirical formula, consider the ratio between the numbers of atoms A and B. Here, it has been reduced to the smallest, whole-number ratio:
- For every 1 atom of A, there are 2 atoms of B
Thus, the empirical formula for the compound becomes: \[ \text{AB}_2 \] The formula reflects the simplified ratio of atoms present in the face-centered cubic lattice under the conditions given, offering a clear view of the compound's composition.