Problem 81 - A Complete Resource Book in Chemistry for JEE Main

Question

In a cubic unit cell, seven of the eight corners are occupied by atom A and having of faces are occupied of \(\mathrm{B}\). The general formula of the substance having this type structure would be (a) \(\mathrm{A}_{7} \mathrm{~B}_{14}\) (b) \(\mathrm{A}_{14} \mathrm{~B}_{7}\) (c) \(\mathrm{A}_{7} \mathrm{~B}_{24}\) (d) \(\mathrm{A}_{9} \mathrm{~B}_{24}\)

Step-by-Step Solution

Answer

The general formula is \( \mathrm{A}_{7} \mathrm{~B}_{24} \).

Step 1: Determine Atom A Contribution

In a cubic unit cell, there are 8 corners. Each corner atom contributes \( \frac{1}{8} \) of an atom to the unit cell. For 7 corners occupied by atom A, the contribution is \( 7 \times \frac{1}{8} = \frac{7}{8} \) atom A per unit cell.

Step 2: Determine Atom B Contribution

Each face on a cube shares an atom. There are 6 faces of a cube, and if all faces are occupied by atom B, the contribution is \( 6 \times \frac{1}{2} = 3 \) atoms B per unit cell.

Step 3: Write the Formula

Calculating the contributions, we find that for each unit cell there are \( \frac{7}{8} \) atom A and 3 atoms B. To get a whole number ratio, multiply both numbers by 8 to eliminate the fraction: \( A_{7}B_{24} \).

Step 4: Match With Options

Given options are (a) \( \mathrm{A}_{7} \mathrm{~B}_{14} \), (b) \( \mathrm{A}_{14} \mathrm{~B}_{7} \), (c) \( \mathrm{A}_{7} \mathrm{~B}_{24} \), and (d) \( \mathrm{A}_{9} \mathrm{~B}_{24} \). The correct match is (c) \( \mathrm{A}_{7} \mathrm{~B}_{24} \).

Key Concepts

Atom Contribution

In a cubic unit cell, understanding how atoms contribute to the entire structure is crucial. Cubic unit cells are commonly formed in crystalline solids, where the positioning and contribution of atoms determine the overall symmetry and chemical properties.
Each of the eight corners of a cubic cell can host an atom. However, these corner atoms are not solely part of one cell; instead, they've shared across eight adjacent unit cells. Hence, each corner atom contributes only one-eighth (\( \frac{1}{8} \) of an atom to a single unit cell.
As specified in the problem, seven out of these eight corners are occupied by atom A, leading to an atomic contribution of:

• Atom A: Seven corners each contribute \( \frac{1}{8} \) which becomes \( 7 \times \frac{1}{8} = \frac{7}{8} \),

which simplifies to less than a whole atom A per unit cell. This information is vital in determining the final unit cell formula.

Unit Cell Formula

The unit cell formula summarizes the chemical composition of the unit cell based on contributions from each type of atom.
Given our analysis of how many atoms of each element definitively occupy a unit cell, we need to express these contributions using the smallest whole numbers.
Initially, from the atomic contributions calculated:

• Atom A, present at the corners, contributes \( \frac{7}{8} \) of an atom per unit cell.
• Atom B, occupying each face and contributing \( \frac{1}{2} \) per face, results in \( 6 \times \frac{1}{2} = 3 \) atoms B per unit cell

To eliminate fractions, multiply each term by 8, giving:

• Atom A: \( \frac{7}{8} \times 8 = 7 \)
• Atom B: \( 3 \times 8 = 24 \)

Which results in a final formula of \( A_7B_{24} \). Matching this with the options provided shows that option (c) \( A_7B_{24} \) is the correct representation.

Face-Centered Cubic

The face-centered cubic (FCC) structure is one of the most efficient ways to pack atoms in a solid structure.
When dealing with FCC arrangements, atoms are positioned at each of the cube's faces, which contributes significantly to the unit cell's density. Each face-centered atom is shared between two adjacent cubes, thus contributing half (\( \frac{1}{2} \)) per unit cell. In a full cube, there are six faces, and when fully occupied, they together contribute \( 6 \times \frac{1}{2} = 3 \) atoms to the unit cell.
Understanding the FCC structure is crucial in many applications, especially in determining material properties like strength and conductivity. The densely packed arrangement results in minimal empty space, leading to highly compact structures akin to the one described in the exercise.