Problem 141 - A Complete Resource Book in Chemistry for JEE Main

Question

Copper crystallizes in fcc with a unit cell length of \(361 \mathrm{pm}\). What is the radius of copper atom? [2009] (a) \(127 \mathrm{pm}\) (b) \(157 \mathrm{pm}\) (c) \(181 \mathrm{pm}\) (d) \(108 \mathrm{pm}\)

Step-by-Step Solution

Answer

The radius of the copper atom is approximately 127 pm, so option (a) is correct.

Step 1: Understand the FCC Structure

Copper crystallizes in a face-centered cubic (FCC) structure. In an FCC unit cell, atoms are located at each corner and the centers of all the faces. The key feature of the FCC cell is that along the face diagonal, four atomic radii fit perfectly.

Step 2: Identify the Relationship

In an FCC structure, the face diagonal is equal to four times the atomic radius, i.e., the relation could be expressed as \( ext{face diagonal} = 4r \), where \( r \) is the radius of the atom.

Step 3: Calculate the Face Diagonal

For an FCC unit cell with edge length \( a \), the face diagonal is calculated using the Pythagorean theorem: \[ ext{face diagonal} = \ \sqrt{a^2 + a^2} = \sqrt{2a^2} = a\sqrt{2} \] Substitute \( a = 361 \mathrm{pm} \): \( 361 \sqrt{2} \mathrm{pm} \).

Step 4: Substitute into the Relationship

Using the relation \( a\sqrt{2} = 4r \), substitute the value of \( a = 361 \mathrm{pm} \): \[ 361\sqrt{2} = 4r \] Solve for \( r \).

Step 5: Solve for Atomic Radius

Re-arrange the equation to find \( r \): \[ r = \frac{361\sqrt{2}}{4} \] Calculate \( \sqrt{2} \approx 1.414 \), then \( r = \frac{361 \times 1.414}{4} \).

Step 6: Final Calculation

Calculate the value: \[ r \approx \frac{510.754}{4} \approx 127.69 \mathrm{pm} \] Hence, the radius of a copper atom is approximately \(127 \mathrm{pm}\).

Key Concepts

Atomic Radius Calculation

The atomic radius is a crucial parameter in understanding the structure and properties of a crystalline solid. For face-centered cubic (FCC) structures, calculating the atomic radius involves geometry and specific relationships based on the cell structure itself.
In the FCC crystal system, the atoms are tightly packed in such a way that within each unit cell, the face diagonal accommodates four atomic radii. From this understanding, we can outline the formula:

• Face Diagonal = 4r
• Where "r" is the atomic radius

Knowing the unit cell edge length allows us to use geometry to determine the face diagonal and, subsequently, the atomic radius. This requires an application of the Pythagorean theorem in the context of the crystallography of the FCC structure.

Face-Centered Cubic Structure

An FCC structure is one of the most efficient ways atoms can pack together. Imagine a cube where each corner and the center of each cube face contains one atom. This configuration allows for the maximum packing efficiency and is commonly found in metals like copper.
Each unit cell in an FCC structure has:

• 8 atoms at the corners (shared with adjacent cells)
• 6 atoms at the face centers

The coordination number, or the number of nearest-neighbor atoms, is 12, indicating a densely packed structure. The edge length of the unit cell is related to the atomic radius through the face diagonal, enabling the calculation of atomic dimensions from macroscopic crystal properties.

Pythagorean Theorem in Crystallography

The Pythagorean theorem is not just confined to classroom geometry; it is a pivotal mathematical tool in crystallography. When applied to the FCC structure, it helps calculate distances within the crystal lattice.
The Pythagorean theorem states: With a unit cell edge "a," the face diagonal can be represented as:

• Face diagonal = \(\sqrt{a^2 + a^2}\)
• This simplifies to \(a\sqrt{2}\)

The formula is derived because a face diagonal is the hypotenuse of a right triangle formed by two edge lengths in the plane. This mathematical relationship is crucial for translating microscopic atomic radii into measurable macroscopic unit cell parameters, crucial for understanding material science at a more profound level.