Problem 14 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

In a face centered cubic lattice atoms A are at the corner points and atoms \(\mathrm{B}\) at the face centered points. If atom \(\mathrm{B}\) is missing from one of the face centered points, the formula of the ionic compound is: (a) \(\mathrm{AB}_{2}\) (b) \(\mathrm{A}_{5} \mathrm{~B}_{2}\) (c) \(\mathrm{A}_{2} \mathrm{~B}_{3}\) (d) \(\mathrm{A}_{2} \mathrm{~B}_{5}\)

Step-by-Step Solution

Answer

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

Step 1: Determine Total Atoms A

In a face centered cubic (FCC) lattice, there are atoms A located at each corner of the cube. Each corner atom is shared by 8 unit cells. Therefore, the contribution of each atom A to a single unit cell is \( \frac{1}{8} \times 8 = 1 \) atom A.

Step 2: Determine Total Atoms B

Atoms B are situated at each of the 6 face-centered positions, with each face atom being shared by 2 unit cells. This results in \( \frac{1}{2} \times 6 = 3 \) atoms B in a perfect unit cell.

Step 3: Calculate Effect of Missing B Atom

If one atom B is missing from the face-centered positions, the number of B atoms is reduced to \( 3 - 1 = 2 \) atoms B.

Step 4: Determine Formula of Compound

Now, considering the contributions from the unit cell, we have 1 atom A and 2 atoms B. Therefore, the empirical formula based on the composition is \( AB_2 \).

Step 5: Choose the Correct Answer

From the options provided, the correct answer that matches the derived formula is \( AB_2 \). It corresponds to option (a).

Key Concepts

Face Centered Cubic Lattice

Imagine stacking oranges in a box where each orange touches others along their surfaces. This is a bit like atoms in a face-centered cubic (FCC) lattice. In this arrangement, atoms are strategically placed at every corner and the centers of each of the six cube faces. The face-centered cubic structure is commonly seen in metals and ionic compounds.

Here's how it works in an FCC unit cell:

• Each corner atom is shared among eight different cubes, meaning its contribution to a single unit cell is only one-eighth.
• Each face-centered atom is shared between two cubes, so each one contributes one-half to a single unit cell.

Understanding this structure helps us figure out how atoms are distributed in the lattice, which is essential for calculating the formula of compounds.

Unit Cell Contributions

To determine how many atoms effectively belong to one unit cell, we sum their contributions.

In our given scenario:

• Atoms A are located at the corners. Since each corner atom is shared equally among eight cells, the contribution from atoms A is 1 full atom per unit cell \( \left(\frac{1}{8} \times 8\right)\).
• Atoms B are situated at each of the six faces. Each of these atoms is shared between two adjacent cells. Thus, the contribution from atoms B is three full atoms in an intact unit cell \( \left(\frac{1}{2} \times 6\right)\).

So, what's internal to the singular unit cell is vital in understanding the assembly and breakdown of the entire lattice.

Ionic Compound Formula

When dealing with ionic compounds, the arrangement and number of atoms dictate the resulting chemical formula. In face-centered cubic lattices, knowing the positions of atoms gives clarity on their contribution to the compound's formula.

In our example, atoms A take on the corner positions, giving us a total contribution of 1 atom per unit cell. Meanwhile, atoms B at face-centered positions would yield 3 per cell, assuming none are missing. However, when an atom is absent, this affects the total count and influences the overall balance and neutrality of the compound. This balance is crucial in determining the compound's empirical formula.

Empirical Formula Calculation

Empirical formulas emerge by assessing the smallest whole-number ratio of atoms within a compound. Given a face-centered cubic lattice:

• Start by noting the total count of atoms per unit cell. For instance, if we have 1 atom A and, adjusting for a missing B, 2 atoms B within the unit cell, the ratio gives us the empirical formula for the compound.
• Using our example, the final composition of the unit cell comprises 1 atom of A and 2 atoms of B.
• From these totals, the empirical formula is directly evident as \( AB_2 \).

By following the outlined steps precisely, one can efficiently convert the positional arrangements and contributions of atoms into a coherent and correct chemical formula.