Problem 7 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Copper crystallises in fcc with a unit length of \(361 \mathrm{pm}\). What is the radius of copper atom? (a) \(157 \mathrm{pm}\) (b) \(128 \mathrm{pm}\) (c) \(108 \mathrm{pm}\) (d) \(181 \mathrm{pm}\)

Step-by-Step Solution

Answer

The radius of the copper atom is approximately 128 pm, option (b).

Step 1: Understand the FCC Structure

Copper crystallizes in a face-centered cubic (fcc) structure where atoms are located at each of the corners and the centers of all the faces of the cube. Knowing this helps us understand how to find the radius of an atom based on the unit cell parameters.

Step 2: Determine Relationship of Radius and Edge Length

In a face-centered cubic structure, the face diagonal is equal to four times the radius of the atoms since the face diagonal passes through the center of two corner atoms and one face-centered atom. The relationship is given by: \[ 4r = \sqrt{2}a \, where \, r \, is \, the \, radius \, and \,a \, is \, the \, edge \, length. \]

Step 3: Calculate Using the Given Data

Given the unit cell edge length \(a = 361\, \mathrm{pm}\), we substitute into the formula: \ \[ 4r = \sqrt{2} \times 361 \, \mathrm{pm} \] \ \[ r = \frac{\sqrt{2} \times 361}{4}\, \mathrm{pm} \]

Step 4: Simplify and Calculate

Simplify the expression: \ \[ r = \frac{1.414 \times 361}{4} \approx \frac{510.554}{4} \approx 127.6385 \, \mathrm{pm} \]

Step 5: Select the Closest Answer

Among the given options, the closest to our calculated radius of approximately \(128\, \mathrm{pm}\) is option (b) \(128\, \mathrm{pm}\).

Key Concepts

Copper Crystallization

Copper crystallizes in a face-centered cubic (fcc) crystal structure, which is one of the most common types of atomic arrangements. The fcc structure is highly efficient in packing because atoms in the structure are tightly packed.
A cube represents the fcc structure where atoms are placed at each corner and at the center of each face. This means there are 8 corner atoms and 6 face-centered atoms in the fcc structure.
However, it's important to remember that these corner atoms are shared among eight adjacent unit cells, so only 1/8 of each corner atom belongs to one unit cell. Similarly, the face-centered atoms are shared between two adjacent cells, contributing half of each atom to the unit cell. This results in a total of four effective atoms per fcc unit cell.
Understanding the copper crystallization process through the fcc structure is pivotal for calculating atomic properties, such as atomic radius.

Atomic Radius Calculation

The atomic radius is a crucial measurement in understanding the dimensions of an atom in a solid state system. In an fcc structure, to find the atomic radius, one must understand how the atoms are positioned. The atoms in fcc are so arranged that the face diagonal of the cube equals four times the atomic radius.
This relationship can be mathematically defined as:

• Face diagonal = 4r (where r is the radius of the atom)

The face diagonal can also be expressed in terms of the edge length of the cube unit cell using the Pythagorean theorem, which gives the equation:

• \(4r = \sqrt{2}a\)

By rearranging this equation, the atomic radius \(r\) can be found:

• \(r = \frac{\sqrt{2}a}{4}\)

Understanding this relationship allows one to calculate the atomic radius when the unit cell edge length is known.

Unit Cell Parameters

Unit cell parameters provide essential information about the dimensions and structure of a crystal. They include the edge length and angles between the edges. In cubic systems like the face-centered cubic (fcc) structure, angles are all 90 degrees, making calculations straightforward.
The primary parameter for an fcc unit cell is the edge length, denoted as \(a\), which directly influences atomic calculations. A given edge length enables the calculation of the face diagonal as \(\sqrt{2}a\), essential in determining atomic radius.
These parameters are central to understanding material properties such as density, packing efficiency, and even electrical conductivity. For instance, knowing the edge length allows the calculation of how densely atoms are packed within the structure, affecting the physical properties of the material such as hardness and melting point.
The unit cell parameters thus serve as building blocks for more complex calculations and provide insight into material characteristics.